SCHOLAR Study Guide SQA Advanced Higher Chemistry Unit 3 ... · SCHOLAR Study Guide Unit 3:...

SCHOLAR Study Guide

SQA Advanced Higher ChemistryUnit 3Organic Chemistry

Peter JohnsonHeriot-Watt University

Brian T McKercharBalerno High School

Arthur A SandisonSt Thomas of Aquin’s High School

Heriot-Watt University

Edinburgh EH14 4AS, United Kingdom.

First published 2001 by Heriot-Watt University.

This edition published in 2009 by Heriot-Watt University SCHOLAR.

Copyright © 2009 Heriot-Watt University.

Members of the SCHOLAR Forum may reproduce this publication in whole or in part foreducational purposes within their establishment providing that no profit accrues at any stage,Any other use of the materials is governed by the general copyright statement that follows.

All rights reserved. No part of this publication may be reproduced, stored in a retrieval systemor transmitted in any form or by any means, without written permission from the publisher.

Heriot-Watt University accepts no responsibility or liability whatsoever with regard to theinformation contained in this study guide.

Distributed by Heriot-Watt University.

SCHOLAR Study Guide Unit 3: Advanced Higher Chemistry

1. Advanced Higher Chemistry

ISBN 978-1-906686-02-4

Printed and bound in Great Britain by Graphic and Printing Services, Heriot-Watt University,Edinburgh.

AcknowledgementsThanks are due to the members of Heriot-Watt University’s SCHOLAR team who planned andcreated these materials, and to the many colleagues who reviewed the content.

We would like to acknowledge the assistance of the education authorities, colleges, teachersand students who contributed to the SCHOLAR programme and who evaluated these materials.

Grateful acknowledgement is made for permission to use the following material in theSCHOLAR programme:

The Scottish Qualifications Authority for permission to use Past Papers assessments.

The Scottish Government for financial support.

All brand names, product names, logos and related devices are used for identification purposesonly and are trademarks, registered trademarks or service marks of their respective holders.

i

Contents

1 Introduction to Organic Chemistry 11.1 Revision of naming of hydrocarbons . . . . . . . . . . . . . . . . . . . . 21.2 The importance of organic chemistry . . . . . . . . . . . . . . . . . . . . 41.3 Key concepts throughout the Unit . . . . . . . . . . . . . . . . . . . . . . 61.4 Resources . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

2 Hydrocarbons and Halogenoalkanes 112.1 Hydrocarbons . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 132.2 Halogenoalkanes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 302.3 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 412.4 Resources . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 412.5 End of Topic test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41

3 Alcohols and Ethers 433.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 453.2 Classification and nomenclature . . . . . . . . . . . . . . . . . . . . . . 463.3 Physical properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 493.4 Preparation of alcohols . . . . . . . . . . . . . . . . . . . . . . . . . . . 533.5 Reactions of alcohols . . . . . . . . . . . . . . . . . . . . . . . . . . . . 573.6 Preparation of ethers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 603.7 Reactions of ethers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 613.8 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 623.9 Resources . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 623.10 End of Topic test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63

4 Aldehydes, Ketones and Carboxylic Acids 654.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 674.2 Physical properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 684.3 Reactions of aldehydes and ketones . . . . . . . . . . . . . . . . . . . . 744.4 Carboxylic acids . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 814.5 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 874.6 Resources . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 884.7 End of Topic test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 88

5 Amines 895.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 905.2 Naming and classification . . . . . . . . . . . . . . . . . . . . . . . . . . 905.3 Physical properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 955.4 Chemical properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 975.5 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100

ii CONTENTS

5.6 Functional groups: summary . . . . . . . . . . . . . . . . . . . . . . . . 1015.7 Resources . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1035.8 End of Topic test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103

6 Aromatics 1056.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1076.2 Benzene structure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1076.3 Benzene reactions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1106.4 Acidity of phenol and basicity of phenylamine (aniline) . . . . . . . . . . 1176.5 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1216.6 Resources . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1216.7 End of Topic test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 122

7 Stereoisomers 1237.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1247.2 Stereoisomerism . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1257.3 Resources . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1447.4 End of Topic test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 145

8 Elemental analysis and Mass spectrometry 1478.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1488.2 Elemental analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1488.3 Mass spectrometry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1508.4 High resolution mass spectrometry . . . . . . . . . . . . . . . . . . . . . 1608.5 Example analysis of an unknown compound . . . . . . . . . . . . . . . . 1618.6 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1628.7 Resources . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1638.8 End of Topic test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 163

9 Infrared and Nuclear magnetic resonance spectroscopy and X-ray . . . . 1659.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1679.2 Infrared spectroscopy . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1679.3 Nuclear magnetic resonance spectroscopy . . . . . . . . . . . . . . . . 1789.4 X-ray crystallography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1859.5 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1889.6 Resources . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1899.7 End of Topic test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 189

10 Medicines 19110.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19210.2 Aspirin development . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19210.3 How a medicine functions . . . . . . . . . . . . . . . . . . . . . . . . . . 19410.4 Case studies . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19910.5 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20710.6 Resources . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20810.7 End of Topic test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 208

11 End of Unit 3 Test (NAB) 209

Glossary 211

© HERIOT-WATT UNIVERSITY

CONTENTS iii

Further questions 216

Hints for activities 220

Answers to questions and activities 2211 Introduction to Organic Chemistry . . . . . . . . . . . . . . . . . . . . . 2212 Hydrocarbons and Halogenoalkanes . . . . . . . . . . . . . . . . . . . 2223 Alcohols and Ethers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2274 Aldehydes, Ketones and Carboxylic Acids . . . . . . . . . . . . . . . . . 2345 Amines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2386 Aromatics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2407 Stereoisomers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2428 Elemental analysis and Mass spectrometry . . . . . . . . . . . . . . . . 2449 Infrared and Nuclear magnetic resonance spectroscopy and X-ray . . . 24710 Medicines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 251


1

Topic 1

Introduction to Organic Chemistry

Contents

1.1 Revision of naming of hydrocarbons . . . . . . . . . . . . . . . . . . . . . . . . 2

1.2 The importance of organic chemistry . . . . . . . . . . . . . . . . . . . . . . . . 4

1.3 Key concepts throughout the Unit . . . . . . . . . . . . . . . . . . . . . . . . . . 6

1.4 Resources . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

Prerequisite knowledge

Before you begin this Unit, you should be able to understand the meanings of thefollowing terms:

homologous series; functional group; saturated; unsaturated; isomers;addition reaction; condensation reaction; hydrolysis; physical properties;chemical properties.

Look up the glossary near the end of this booklet for the definition of any term aboutwhich you are unsure.

Learning Objectives

After studying this Unit, you should be able to:

• recognise and write equations for different types of reaction;

• describe, when appropriate, reaction mechanisms in terms of electron shifts;

• explain physical properties, such as melting point, boiling point and miscibility withwater in terms of intermolecular forces.

2 TOPIC 1. INTRODUCTION TO ORGANIC CHEMISTRY

1.1 Revision of naming of hydrocarbons�

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Learning Objective

To revise the rules for the systematic naming of hydrocarbons

You should also be familiar with the rules for the nomenclature of simple alkanes andbranched alkanes.

Nomenclature of alkanes

20 min

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Learning Objective

To revise the rules for naming alkanes and practise the application of these rules

The website shows an animation which runs through the rules to name an isomer ofheptane, followed by a multiple choice test to give practice. This activity starts with adescription of the rules for naming branched alkanes using an isomer of heptane as anexample, followed by a multiple choice test to give practice.

Unbranched alkanes are named by counting the number of carbon atoms in the chain,choosing the correct prefix (Table 1.1) and adding the name ending ’-ane’.

Table 1.1: Number of carbon atoms

No. of C atoms Prefix No. of C atoms Prefix1 meth- 6 hex-2 eth- 7 hept-

3 prop- 8 oct-4 but- 9 non-

5 pent- 10 dec-

The naming of branched alkanes follows a set of rulesso that each individual alkane has a unique name.

This can be illustrated by using an isomer of heptane.

Step 1

Identify the longest continuouscarbon chain in the molecule (notnecessarily a straight chain). Thisgives the name for the alkane asdescribed above.

In this case, the longest chain is 5carbon atoms. So the basic name ispentane.


1.1. REVISION OF NAMING OF HYDROCARBONS 3

Step 2

Number the chain from the endnearest the first branch point. Thiskeeps the numbers as low aspossible.

Step 3 Identify the branch points on thechain.

Step 4

Identify the substituents. These willbe alkyl groups - an alkane minus ahydrogen atom. In this case thereare two methyl groups.

Step 5

Now assemble the name. The substituents are written in front of the basicname, in alphabetical order if there is more than one type. The position ofeach substituent is shown by writing the number of the appropriate branchpoint in front of the name of the substituent. This gives the correct name:

2,3-dimethylpentane

Note the ’comma’ between the numbers and the use of ’di-’ to show thatthere are two methyl groups attached to the chain.

Now try the test.

Name the following compounds from their structural formulae.

Q1:

a) 2,4-ethylpentaneb) 2,4-methylpentanec) 2,4-dimethylpentaned) 2,4-dimethylheptane

Q2:

a) 2-ethyl-4-methylpentane



b) 3,5-dimethylhexanec) 4-ethyl-2-methylpentaned) 2,4-dimethylhexane

Q3:

a) 3-ethyl-4-methylpentaneb) 3-ethyl-2-methylpentanec) 3-methyl-2-ethylpentaned) 2-ethyl-2-methylpentane

Q4:

a) 2-ethyl-2,3-dimethylbutaneb) 2,3-dimethyl-3-ethylbutanec) 3,3,4-trimethylpentaned) 2,3,3-trimethylpentane

See further questions on page 216.

1.2 The importance of organic chemistry�

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Learning Objective

After studying this Topic, you should be able to:

identify the type of bond fission occurring in a given reaction;

classify molecules or ions as nucleophiles or electrophiles;

classify appropriate ions as carbanions or carbocations.

Introduction

The term ’organic’ was first used to describe those substances produced by livingorganisms. It was believed that such compounds could only be created in the presenceof a special ’vital force’ found in living things and that they could not be made fromsubstances obtained from non-living sources (inorganic compounds). However, theproduction of urea, which had previously been found in urine, by heating the inorganiccompound, ammonium cyanate, was the beginning of the end for the ’vital force’ theory.

NH4OCN NH2CONH2ureaammonium cyanate


1.2. THE IMPORTANCE OF ORGANIC CHEMISTRY 5

Despite this, the classification of compounds as organic or inorganic continues to thisday. Since then it has become clear that the survival of living organisms depends onthe individual properties of the huge variety of compounds of carbon and not on some’vital force’. Carbon makes up only 0.2% of the earth’s crust and yet at least five millioncompounds have been identified, of which more than 95% are compounds of carbon.

Why is the element carbon so special?

The importance of carbon

5 min

Visit the web version of this Topic to find an interactive diagram that shows some of theunique properties of carbon.

On paper, try to write down as many reasons as you can to say why carbon is so special(i.e. what carbon can do that other elements cannot).

The ability of carbon to form such a huge variety of compounds is due to importantproperties of the carbon atom itself.

Organic chemistry is the chemistry of life itself and so is fundamental to biology andmedicine.



The importance of organic chemistry

5 min

On the website you can see another interactive diagram that can be used to revealimportant areas where organic chemistry has had an impact.

Organic chemistry has made, and continues to make, a huge contribution to modernday society.

Potential solutions to many of the world’s major problems may lie within the field oforganic chemistry.

1.3 Key concepts throughout the Unit

The study of such a huge variety of organic compounds is simplified by theirclassification into smaller groups. Any study of organic chemistry must involveconsideration and explanation of the physical and chemical properties of thesehomologous series.

Physical properties such as melting and boiling point and miscibility with water will beexplained in terms of the intermolecular forces involved.

Throughout the unit, you will encounter and be expected to identify a variety of differenttypes of reaction, some of which will be new to you. These will include:

Addition Condensation Hydrolysis Oxidation

Reduction Substitution Elimination Acid / Base

Possible mechanisms for some of these reactions will be discussed in terms of electronshifts.

Bond breaking

All chemical reactions involve the breaking and making of bonds. The products and themechanism of a reaction will be strongly influenced by the way in which the bonds break.In covalent bonding, electrons are generally shared in pairs between two atoms, e.g. inthe hydrogen bromide molecule (see Figure 1.1).

or a general bond

Figure 1.1: Covalent bond

During bond breaking (also called bond fission), the electrons are redistributed betweenthe two atoms. There are two ways in which this can occur.

In homolytic fission , the two shared electrons separate equally, one going to eachatom , as shown in Figure 1.2.

Figure 1.2: Homolytic fission


1.3. KEY CONCEPTS THROUGHOUT THE UNIT 7

The curly half arrow () is used to represent the movement of a single electron. Thesingle dot (�) beside each atom represents the unpaired electron that has been retainedby each atom from the shared pair. However, the atoms are highly reactive becauseunpaired electrons tend to attack other species. Highly reactive atoms or groups ofatoms containing unpaired electrons are called free radicals.

Free radicals are most likely to be formed when the bond being broken is essentiallynon-polar, i.e. the electrons are more or less equally shared.

In heterolytic fission, both of the shared electrons go to only one of the two atomsproducing ions, as shown in Figure 1.3.

Figure 1.3: Heterolytic fission

The full curly arrow (�) is used to represent the movement of a pair of electrons.Heterolytic fission is more likely when a bond is already polar. For example, the carbonto bromine bond in bromomethane is polar and can break heterolytically, the pair ofelectrons going to the more electronegative bromine atom (see Figure 1.4).

Figure 1.4: C-Br bond fission

Note that the H3C+ ion (sometimes written as CH3+) contains a carbon carrying a

positive charge. The H3C+ ion is an example of a carbocation (also called a carboniumion).

If the carbon atom is the more electronegative, heterolytic fission can lead to theformation of ions in which a carbon atom carries a negative charge (see Figure 1.5).

Figure 1.5: Carbanion formation

These negative ions are called carbanions. In general, both carbocations andcarbanions are highly reactive and so very short-lived.

Bond fission and carbocations

20 minNow try the online tests. These are designed to promote familiarity with the differentmodes of bond breaking.

If you score less than 70% in the first test, you should revise this section and try thesecond test.

Nucleophiles and electrophiles

Nucleophile and electrophile are two other important terms which are essential to an



understanding of the mechanisms of organic reactions.

Nucleophile means literally ’nucleus seeker’. Since the nucleus is positively charged,nucleophiles are negatively charged ions, e.g. hydroxide ion, OH-, carbanions.

Similarly, electrophile means literally ’electron seeker’. Since electrons are negativelycharged, electrophiles are positively charged ions, e.g. the hydrogen ion, H+,carbocations.

With polar molecules, the situation is more complicated. A partially negative atom willact as a nucleophile because it is electron-rich, as in Figure 1.6.

The nitrogen atom in an ammoniamolecule with its lone pair of electrons iselectron rich.

Figure 1.6: Ammonia

A partially positive atom will act as an electrophile because it is electron-deficient, as inFigure 1.7.

The hydrogen end of a water moleculewill be electron-deficient.

Figure 1.7: Water

So in a polar molecule, there will be both an electrophilic centre and a nucleophiliccentre, as shown in Figure 1.8.

Figure 1.8: Polar molecule

Nucleophiles and electrophiles

20 min

Now try the two online tests. These are designed to promote familiarity with the terms -nucleophile and electrophile.

If you score less than 70% in the first test, you should revise this section and try thesecond test.


1.4. RESOURCES 9

Summary of Key Points

10 min

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Learning Objective

To summarise the key points in the topic

The summary is presented as a cloze test and takes the place of an end of Topic test.

Copy and complete the summary using the words in the word bank.

Summary - Key Points

Organic Chemistry is the study of the compounds of .................... .

The study of Organic Chemistry is essential to the understanding of the chemistry of................. .

Bond ................. can occur in two different ways - heterolytic and homolytic fission.

Homolytic fission produces .......... .................... .

Heterolytic fission produces .................. .

......................... contain a carbon atom carrying a full positive charge.

........................ contain a carbon atom carrying a full negative charge.

Nucleophiles are ...................... ions or .......................... centres in molecules, whichare attracted to ................. centres.

Electrophiles are ..................... ions or ........................... centres in molecules, whichare attracted to ................... centres.

Word Bank

life electron-deficient free radicals negative positive

carbon electron-rich carbocations negative positive

ions breaking carbanions

1.4 Resources• Higher Still Support, Chemistry, Unit 3: Organic Chemistry, Learning and

Teaching Scotland, ISBN 1 85955 873 9

• Chemistry in Context: Hill and Holman, Nelson, ISBN 0-17-438401-7

• Organic Chemistry: J. McMurry, Brooks/Cole Publishing, ISBN 0-534-16218-5

• Chemistry: A. and P. Fullick, Heinemann, ISBN 0-435-57080-3


11

Topic 2

Hydrocarbons andHalogenoalkanes

Contents

2.1 Hydrocarbons . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13

2.1.1 Bonding in alkanes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

2.1.2 Reactions of alkanes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

2.1.3 Bonding in alkenes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20

2.1.4 Synthesis of alkenes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22

2.1.5 Reactions of alkenes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25

2.2 Halogenoalkanes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30

2.2.1 Naming halogenoalkanes . . . . . . . . . . . . . . . . . . . . . . . . . . 31

2.2.2 Reactions of monohalogenoalkanes . . . . . . . . . . . . . . . . . . . . 34

2.3 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41

2.4 Resources . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41

2.5 End of Topic test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41


Before you begin this Topic, you should be able to:

• use different ways to represent organic formulae (full structural formula, shortenedstructural formula, molecular formula);

• describe covalent and polar covalent bonding (Higher);

• describe types of bond breaking (Unit 3, Topic 1);

• recognise nucleophiles and electrophiles (Unit 3, Topic 1);

• recognise carbocations and carbanions (Unit 3, Topic 1).

Learning Objectives

After studying this Topic on Hydrocarbons, you should be able to:

• to describe the bonding in alkanes and alkenes in terms of hybridisation, � bondsand � bonds;

12 TOPIC 2. HYDROCARBONS AND HALOGENOALKANES

• to describe the mechanism of the free radical substitution of alkanes by chlorineand bromine;

• to state two methods used for the preparation of alkenes in the laboratory;

• to predict the products of the addition reactions of alkenes with hydrogen,hydrogen halides, halogens and water;

• to explain the mechanisms involved for each of the above reactions.

After studying this Topic on Halogenoalkanes, you should be able to:

• to name halogenoalkanes given the structural formula;

• to draw structural formulae given the correct name;

• to classify halogenoalkanes as primary, secondary or tertiary;

• to explain the mechanism of the SN1 reaction and the SN2 reaction;

• to predict the products of reaction of a monohalogenoalkane given the otherreactant and conditions.


2.1. HYDROCARBONS 13

2.1 Hydrocarbons

Hydrocarbons are compounds containing the elements carbon and hydrogen only.Hydrocarbons can be subdivided into smaller subsets including alkanes, alkenes,alkynes and cycloalkanes amongst others (Figure 2.1).

Figure 2.1: Hydrocarbon subsets

Alkanes are saturated hydrocarbons which means that their molecules contain onlysingle bonds. The simplest alkane is methane, CH4. The structure of methane can beshown in various ways.

Figure 2.2: Figure 2.3:

Figure 2.4: Methane 3D

’live’ molecule - is availableonly on the web

Figure 2.2 and Figure 2.3 show that there are four single covalent bonds but do not showthe correct shape. Figure 2.4 is a two dimensional representation to show that the fourbonds are tetrahedrally arranged around the carbon atom.

For more complicated molecules, drawings like that in Figure 2.3 will be used but itmust always be remembered that these do not show the true shape. Where shape isimportant, drawings like that in Figure 2.4 will be used, as well as rotating molecules on



the web. A structure for a more complicated alkane, 2,2-dimethylbutane, is shown inFigure 2.5.

Figure 2.5:

’live’ molecule - is available only on the web

Sometimes even the carbon and hydrogen symbolscan be missed out. Figure 2.6 also represents 2,2-dimethylbutane. The end of a bond or an anglerepresents a carbon atom and its hydrogen atoms.

Figure 2.6: wire framedimethylbutane

2.1.1 Bonding in alkanes�

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Learning Objective

To be able to describe the bonding in alkanes in terms of sp3 hybridisation and �

bonds

Covalent bonds are formed when atoms share electrons in pairs, usually one electronfrom each atom. This can be achieved by the overlap of half-filled atomic orbitals oneach atom. A carbon atom has the electron arrangement shown in Figure 2.7.

Figure 2.7

This shows that carbon atoms have two unpaired electrons and so might be expected toform only two covalent bonds. In fact, carbon atoms usually form four bonds. How canthis be?

A simple explanation might involve promotion of an electron from the 2s orbital to theempty 2p orbital, producing four unpaired electrons (Figure 2.8) which could then formfour bonds with hydrogen.

Figure 2.8

The energy required to promote the electron would be more than offset by the formationof two extra covalent bonds. However, the bonds might not be identical since one bondwould involve the 2s orbital whereas the others would involve 2p orbitals. Spectroscopic



measurements show that all four bonds in methane are identical.

A more satisfactory explanation involves the valence bond theory and hybridisation.The theory assumes that the 2s and three 2p orbitals combine during bonding to formfour new identical hybrid orbitals (Figure 2.9).

Figure 2.9

The shapes of atomic orbitals are derived from solutions to mathematical equations. Bymathematically combining the four atomic orbitals, it is possible to predict the shape andorientation of the hybrid orbitals. The hybrid orbitals are known as sp3 orbitals becausethey are formed by combining one s and three p orbitals.

Figure 2.10: a sorbital

Figure 2.11: a p orbitalFigure 2.12: a sp3 orbital

The four sp3 hybrid orbitals are arranged tetrahedrally, as would be predicted by simpleelectron pair repulsion theory (Unit 1, Topic 3). Compare the shape of the sp3 orbital forcarbon (Figure 2.12) with that of the p orbital (Figure 2.11). The hybrid orbital is morestrongly directed in one direction. This provides better overlap when forming bonds andso produces stronger bonds. The sp3 orbital consists of two lobes, one much smallerthan the other. For simplicity, the smaller lobe is often omitted from diagrams.

Overlap of a half-filled sp3 hybrid orbital with a half-filled 1s orbital of a hydrogen atomforms a new molecular orbital within which the shared pair of electrons now moveunder the influence of both nuclei.

��

�

��

� ��

��

� ��

Figure 2.13

The new molecular orbital (Figure 2.13 above) lies along the axis joining the two nucleiand is known as a sigma bond. A sigma bond (� bond) is formed by end-on overlap ofatomic orbitals.



Figure 2.14: Molecular orbitals in methane

In methane, all four hybrid orbitals are used in forming � bonds to hydrogen atoms(Figure 2.14).

The same type of hybridisation occurs in all other alkanes. Each carbon atom uses the2s and all three 2p orbitals in forming � bonds to other carbon atoms and hydrogenatoms, e.g. in ethane.

In Figure 2.15, three sigma bonds from carbon to hydrogen are already formed, leavingone half-filled hybrid orbital on each carbon atom.

Figure 2.15: Ethane forming

Overlap of the two hybrid orbitals forms a carbon to carbon � bond.

Figure 2.16: Ethane - molecular orbitals

This model (Figure 2.16) is consistent with the fact that all the bond angles in alkanesare 109.5Æ (the tetrahedral bond angle).



2.1.2 Reactions of alkanes�

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Learning Objective

To describe the mechanism of the free radical substitution of alkanes by chlorine andbromine

Alkanes are fairly unreactive. The only reactions of importance at normal temperaturesare those with oxygen and halogens. Reaction with oxygen (combustion) is highlyexothermic and so alkanes are widely used as fuels :

The reaction of methane with chlorine does not occur in the dark but occurs explosivelyon exposure to sunlight to produce chloromethane and hydrogen chloride. A hydrogenatom has been replaced by chlorine on each methane molecule. A substitutionreaction has taken place.

Use the data booklet to find the bond enthalpies of the C-H bond and the Cl-Cl bond.

Q1: C-H bond enthalpy (in kJ mol-1)

Q2: Cl-Cl bond enthalpy (in kJ mol-1)

Q3: Which bond will be easier to break ?

a) Cl-Clb) C-H

Q4: What type of bond fission is likely to occur ?

The following activity explains the mechanism of the reaction between methane andchlorine.

Reaction between methane and chlorine

15 minThis is a simulation (only available on the website) that illustrates the mechanism of thereaction between methane and chlorine.

The chain reaction mechanism includes the following steps:-

i) Initiation

Light of the appropriate wavelength causes homolytic fission of the Cl-Cl bond,producing some chlorine atoms. (Figure 2.17)

Figure 2.17: Initiation



ii) Propagation

The chlorine atoms are free radicals and so are highly reactive. Each atom will combinewith a hydrogen atom from a methane molecule, producing a new free radical.(Figure 2.18)

Figure 2.18: Propagation step 1

The methyl radical is also highly reactive and, on colliding with a chlorine molecule,produces another chlorine radical. (Figure 2.19)

Figure 2.19: Propagation step 2

The chlorine radical can then react with more methane as in Figure 2.18 and aself-sustaining cycle is set up. This is a free radical chain reaction.

iii) Termination

The chain reaction ends when two radicals collide and combine (Figure 2.20).

Figure 2.20: Termination

The concentration of radicals at any time is very small so these termination steps occurvery infrequently.



Learning point

Methane (and other alkanes) undergo substitution reactions with chlorine and bromineby a free radical chain reaction mechanism.

All the processes are very rapid, hence the explosive nature of the reaction. A similarbut slower reaction occurs when bromine is used instead of chlorine.

With an excess of halogen, multiple substitution can occur, producing mixtures whichmust be further separated.

With more complicated alkanes, containing many more hydrogen atoms, substitutioncan occur at a variety of different places in the molecule, giving complex mixtures, e.g:

Figure 2.21

In Figure 2.21, replacing any one of the lighter shaded hydrogen atoms by ahalogen atom would give rise to a different compound, i.e. there are nine differentmonohalogenoalkanes possible and many more dihalogenoalkanes.

Free radical chain reaction

15 minAnswer the following questions.

Hydrogen and chlorine react together to form hydrogen chloride. The following fourstatements describe aspects of the reaction mechanism.

1. No reaction occurs in darkness.

2. If light of an appropriate wavelength is supplied, there is a violent reaction.

3. Thousands of HCl molecules are produced for each photon absorbed.

4. The presence of substances with unpaired electrons slows down the reaction.

Answer the following questions:

Q5: Why is light needed?

Q6: What type of step is this?

Q7: Write an equation for this step.

Q8: Write the number of the statement which gives the best evidence that a chainreaction is involved.



Q9: Explain your answer to the previous question.

Q10: Write equations for the propagation steps.

Q11: Why should substances with unpaired electrons slow down the reaction?

A Free Radical Chain Reaction is often initiated by light and involves

• an initiation step

• followed by propagation steps

• and finished by termination steps.

2.1.3 Bonding in alkenes�

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Learning Objective

To be able to describe the bonding in alkenes in terms of sp2 hybridisation and �

bonds and � bonds

Alkenes are unsaturated hydrocarbons whose molecules contain at least one carbon tocarbon double bond. How is a double bond formed?

Consider the bonding in an ethene molecule. Whatever model is used to describe theformation of the bond, it must be able to explain the observed facts.

1. The ethene molecule is flat and all bond angles are 120Æ.

2. The carbon to carbon double bond is shorter than the single bond (Table 2.1).

3. The C=C bond is intermediate in strength between a single and a triple bond butnot twice as strong as a single bond (Table 2.1).

4. There is restricted rotation around the C=C bond but not around the C-C bond.There are two isomers of but-2-ene which differ only in the position of themethyl groups. In one, they are on the same side of the double bond, whilein the other they are on opposite sides. (This is covered more fully in Topic 7,Stereoisomerism).

Table 2.1: Carbon to carbon bonds

C-C C=C C�C

Bond enthalpy / kJ mol-1 346 602 835

Bond length / nm 0.154 0.134 0.121

These facts can again be explained by a model involving hybridisation. In this case,each carbon atom uses its 2s orbital and two of its 2p orbitals.



On mathematically mixing these, three identical hybrid orbitals are obtained. These lie inthe same plane at 120Æ to each other and at 90Æ to the remaining p orbital (Figure 2.22).This is called sp2 hybridisation and the new orbitals are called sp2 hybrid orbitals.

Figure 2.22: sp2 hybridisation

Two of the hybrid orbitals are used to form � bonds to hydrogen atoms as with ethane(Figure 2.15). The remaining hybrid orbitals can overlap to form a carbon to carbon �

bond (Figure 2.23).

Figure 2.23: C to C sigma bond forming

As this bond forms, the unhybridised p orbitals are in the right position to overlapsideways on. This produces a new molecular orbital with lobes above and below theplane of the molecule.



A bond formed by sideways overlap of two parallel atomic orbitals is called api bond ( � bond).

The carbon to carbon double bond involves the sharing of four electrons, one pair in a� bond and the other pair in a � bond. The two atoms are held more tightly togetherand so the bond is shorter and stronger than that in ethane. Sideways overlap is lesseffective than end-on overlap and so the � bond is weaker than the � bond (i.e. the C=Cbond strength is less than twice the C–C bond strength).

Bonding in hydrocarbons

15 min

This activity contains extension questions on bonding and hybridisation.

Answer the following questions.

It may help to draw Lewis dot structures for the molecule to identify the bonding andnon-bonding pairs of electrons before answering the question.

For the next three questions, consider the hydrocarbon ethyne, HC�CH.

Q12: What is the hybridisation of the carbon atoms required to form a C�C bond?

a) sp3

b) sp2

c) sp

Q13: Which of the following statements is true about an ethyne molecule?

a) There are three � bonds and two � bonds.b) There are two � bonds and two � bonds.c) There are two � bonds and three � bonds.d) There are three � bonds and three � bonds.

Q14: What shape will the ethyne molecule be?

2.1.4 Synthesis of alkenes�

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Learning Objective

To state two methods used for the preparation of alkenes in the laboratory

Alkenes, particularly ethene and propene, are extremely important as feedstocks in thesynthesis of a huge number of important chemicals. In industry, they are generally



obtained by the cracking of natural gas products or naphtha.

They can be prepared in the laboratory by two main methods, both of which involveelimination reations.

In an elimination reaction, a single reactant splits up to form two products, one of whichis a small molecule like water. The other product will contain a multiple bond.

Method 1 - Dehydration of an alcohol

a) Using aluminium oxide as the catalyst.

The alcohol vapour is passed over the hot aluminium oxide and the product gas iscollected. Water is lost and an alkene is formed (Figure 2.24).

Figure 2.24

This is only useful for volatile alcohols, e.g:

C2H5OH � C2H4

ethanol ethene

b) Using a strong, non-volatile acid like sulphuric or phosphoric acid.

Consider the dehydration of hexan-1-ol.

Substance hexan-1-ol hex-1-ene H2SO4 H3PO4

b.pt. ( ÆC) 157 63 330 213

The hexan-1-ol is warmed with the acid to about 80ÆC.

Q15: Which of the substances will boil at this temperature ?

Q16: From the information in the table, suggest how you could isolate the product fromthe reactant and catalyst.

Method 2 - Elimination of hydrogen halide from monohalogenoalkanes

A similar elimination reaction can occur with halogenoalkanes (often called haloalkanesor alkyl halides). In this case, potassium hydroxide in ethanol solution is used as areactant. A molecule of the appropriate hydrogen halide is eliminated and an alkene isformed.



Figure 2.25

In all the examples so far, there has been only one possible product. However, in mostcases, more than one alkene product is possible. Consequently, most reactions producemixtures of products.

Synthesis of alkenes

15 min


Example

Dehydration of butan-2-ol can give rise to two isomeric alkenes by loss of the hydroxylgroup and a hydrogen atom from one of the adjacent carbon atoms.

1. Loss of a hydrogen atom from carbon atom number 1.

2. Loss of a hydrogen atom from carbon atom number 3.

Q17:

How many isomeric alkenes couldbe obtained by the dehydration ofpropan-2-ol?

Q18:

How many isomeric alkenescould be obtained from thiscompound by elimination ofHCl?



Q19:

How many isomeric alkenescould be obtained by thedehydration of this alcohol?

Q20:

How many isomeric alkenes couldbe obtained by the dehydration ofthis alcohol?


Mixtures of alkenes are normally obtained unless the starting material is a simple orvery symmetrical molecule.

At this point it would be a good idea to try the Prescribed Practical Activity on’Preparation of Cyclohexene’.

PPA - Preparation of cyclohexene (Unit 3 PPA 1)

120 min

Consult with your tutor to find out whether the PPA on preparing cyclohexene is to becompleted at this stage.

2.1.5 Reactions of alkenes�

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Learning Objective

To be able to predict the products of the addition reactions of alkenes with hydrogen,hydrogen halides, halogens and water

To be able to explain the mechanisms involved for each of the above reactions

Alkene molecules contain a C=C bond. Two pairs of electrons are shared, making theregion between the two carbon atoms electron-rich. Consequently, alkenes can behaveas nucleophiles. The carbon to carbon double bond is able to donate a pair of electronsto an electrophile.



Figure 2.26: Forming carbocation intermediate

As the electrophile approaches the double bond, the pair of � electrons move to form abond between one of the carbon atoms and the electrophile, leaving the other carbonatom electron-deficient. This produces a carbocation intermediate (Figure 2.26).

Figure 2.27: Attack of nucleophile

This carbocation intermediate is then attacked by a nucleophile to form a saturatedmolecule (Figure 2.27). The overall reaction is addition. Since the initial attack is by anelectrophile, the reaction is called Electrophilic Addition.

This is the typical reaction of alkenes. A wide range of substances can be added acrossthe double bond by a similar mechanism. The various products can then be used tosynthesise other important organic chemicals.

1. Hydrogen halides

Hydrogen chloride, hydrogen bromide and hydrogen iodide can all be added across thedouble bond to give the appropriate monohalogenoalkane. The mechanism involved inthe reaction is just like the one discussed above ( Figure 2.26 and Figure 2.27).

Consider the reaction of ethene with hydrogen chloride. The HCl molecule is polar.



Figure 2.28

The partially positive hydrogen atom is anelectrophile. As the bond forms betweenthe H atom and the C atom, the H-Cl bondbreaks heterolytically.

Figure 2.29

The negative chloride ion formed is anucleophile which then attacks the positivecarbon of the carbocation to form theproduct, chloroethane.

Only one product is formed since it does not matter which carbon atom is attacked bythe electrophile.

What about the reaction of propene?

This molecule is not symmetrical. The twocarbon atoms of the double bond are notidentical.

So two different products are possible depending on which carbon is the subject of theelectrophilic attack.

Attack by HCl at the end carbon atom (carbonnumber 1) produces 2-chloropropane.

Attack by HCl at the middle carbon atom(carbon number 2) produces 1-chloropropane.

In fact, nearly all the product is 2-chloropropane.

As a result of studying many similar reactions, a general rule, known asMarkovnikov’s Rule, was produced enabling us to predict the major product of thistype of reaction. It can be stated:

’In the addition of HX to an alkene, the H atom attaches to the carbon atom with feweralkyl groups attached to it and the X atom attaches to the carbon atom with more alkyl



groups attached to it ’.

Markovnikov’s Rule can be explained in terms of the relative stabilities of thecarbocations involved (Figure 2.30).

Figure 2.30: Relative stabilities of carbocations

With propene, the two possible carbocations are:

The H atom has been attached tocarbon number 1.

The H atom has been attached tocarbon number 2.

The major product is produced from the more stable carbocation intermediate, i.e. theone formed by attaching the H atom to the carbon with fewer alkyl groups attached to it.

2. Acid-catalysed addition of water

Water can also be added to a C=C bond by a very similar mechanism to that describedin the previous section. As before, using ethene as an example, the first step involvestransfer of an H+ ion from the acid catalyst to one of the carbon atoms of the doublebond, forming the carbocation intermediate (in a similar way to that shown inFigure 2.28).

The second step (Figure 2.31) involves nucleophilic attack by water on the electron-deficient carbon to form intemediate 2 shown in Figure 2.31.

Figure 2.31: Ethanol formation - stage 2

Intermediate 2 (Figure 2.31) then loses an H+ ion to form the product, ethanol.



This is a general method for the conversion of alkenes into alcohols. Addition of water(hydration) to asymmetric alkenes can again give rise to mixtures of isomeric alcohols.Once more, Markovnikov’s Rule can be used to predict the major product because thisreaction involves a similar carbocation intermediate.

3. Addition reaction with halogens

Electrophilic addition of halogens, like bromine and chlorine, also readily takes place,although there is strong evidence to suggest that a slightly different mechanism occurs.Consider the reaction of ethene with bromine.

The bromine molecule is non-polar but, as the molecules approach, the electrons in theC=C bond repel those in the Br2 molecule causing it to become polarised.

Figure 2.32

As the molecules get closer, the Br-Br bond gradually breaks at the same time as the� electrons of the double bond form a bond to the electrophilic bromine atom. Thereis strong evidence that the intermediate is not a carbocation but a cyclic bromoniumion (Figure 2.32) . This ion is more stable because the positive charge is spread(delocalised) over three atoms.

Figure 2.33

The bromide ion formed in step 1 (Figure 2.32 ), then attacks one of the carbon atomsin the ring that opens to form the product, 1,2-dibromoethane, as shown in Figure 2.33.The addition of chlorine to alkenes is believed to occur by a similar mechanism.



4. Catalytic addition of hydrogen

A heterogeneous catalyst is used, usually a transition metal such as platinum, palladiumor nickel.

Adsorption of hydrogen onto thesurface of the catalyst breaks up themolecules into atoms. The alkenealso bonds to the surface using theelectrons in the � bond.

Any hydrogen atoms which are close enough can then bond to the adsorbed alkene.

The alkane which is formed has no� electrons and cannot bond to thesurface. As it departs, it leaves thesurface free to accept another alkenemolecule.

Addition reactions

15 minVisit the website to complete the drag and drop exercise designed to give practice atpredicting products and understanding mechanisms involved.

2.2 Halogenoalkanes�

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Learning Objective

To name halogenoalkanes given the structural formula.

To draw structural formulae given the correct name.

To classify halogenoalkanes as primary, secondary or tertiary.

To explain the mechanism of the SN1 reaction and the SN2 reaction.

To predict the products of reaction of a monohalogenoalkane given the other reactantand conditions.

Halogenoalkanes are saturated organic compounds derived from alkanes bysubstituting one or more hydrogen atoms by halogen atoms. They are frequentlyreferred to as alkyl halides or haloalkanes.


2.2. HALOGENOALKANES 31

Organic compounds containing halogens have a wide importance. Their uses includeindustrial solvents, anaesthetics, refrigerants and pesticides amongst many others.They also play an equally important role in the synthesis of other important organicchemicals.

2.2.1 Naming halogenoalkanes

Halogenoalkanes are named according to the IUPAC rules (International Union of Pureand Applied Chemistry) which is a worldwide agreed system for naming chemicals.In Topic 1 of this Unit, you were reminded of the rules as they applied to alkanes.These rules are easily extended to cover halogenoalkanes. The presence of halogensubstituents is shown by the appropriate prefix:

Halogen prefix

fluorine fluoro-chlorine chloro-bromine bromo-iodine iodo-

Examples

1. Using the rules to name a compound

Name the following compound:

An animation on the web site shows the steps involved in naming this compound.

Following these steps gives 2,3-dichloro-3-methylpentane. Note that the substituentsare listed in alphabetical order.

2. Working out the structure from the name

Draw the full structural formula for the following compound:

2-bromo-1-chloro-2-methylbutane



1.Identify the parentchain and number it. Inthis case - butane

2.

Add the substituentsto the correct carbonatoms - Br to carbon 2,Cl to carbon 1 and themethyl to carbon 2.

3.Finally, add hydrogenatoms to all theremaining bonds

Answer the following questions. In the first, you will be given a structural formula toname. In the second, you will be given the name and asked to draw the full structuralformula. Further sets of questions can be used online for practice.

Q21: Name this compound.

Q22: Draw the full structural formula for 2-chloro-2,3-dimethylbutane.

Monohalogenoalkanes can be classified as primary, secondary or tertiary depending onthe environment of the halogen (Table 2.2, where X represents any halogen).

Table 2.2

The halogen is attached to a carbon at the end of a chain.(This carbon has one alkyl group attached to it)

The halogen is attached to a carbon in the middle of achain. (This carbon has two alkyl groups attached to it)

The halogen is attached to a carbon at a branch point in achain. (This carbon has three alkyl groups attached to it)



In each of the following questions, classify the halogenoalkane as primary, secondaryor tertiary.

Q23:

What class of halogenoalkane is this?

a) Primaryb) Secondaryc) Tertiary

Q24:



Q25:



Q26:





2.2.2 Reactions of monohalogenoalkanes�

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Learning Objective

• be able to explain the mechanism of the SN1 reaction and the SN2 reaction

• be able to predict the products of reaction of a monohalogenoalkane given theother reactant and conditions.

Halogenoalkanes undergo two main types of reaction:

1. Elimination reactions in which a hydrogen halide molecule is eliminated byreaction with a strong base (this reaction was discussed earlier as a method forsynthesising alkenes, Figure 2.25).

2. Substitution reactions in which the halogen atom is replaced by anucleophile.

A nucleophile by definition is attracted to a positive centre and will be capable ofaccepting an H+ ion. This means that good nucleophiles tend to be good bases.Eliminations reactions are favoured by the presence of a strong base. Consequently,elimination reactions and substitution reactions can occur simultaneously and competewith one another, frequently providing mixtures of products.

2.2.2.1 Elimination reactions

In elimination reactions, a strong base, such as potassium hydroxide, is used in a solventof relatively low polarity such as ethanol.

An OH- ion attacks one of the hydrogen atoms of a carbon atom adjacent to the carbonatom carrying the halogen substituent. As the new O-H bond forms, the C-X bondbreaks and a � bond forms between the two carbon atoms. In effect, this eliminationreaction is the opposite of the addition reaction in which HX is added to an alkene. Asstated previously, if the halogenoalkane is not symmetrical, a mixture of products forms.

Products of elimination reactions

10 minThis is a problem solving activity that includes questions to give practice at predictingthe products of elimination reactions of halogenoalkanes.

For each of the following compounds, predict the product(s) that could be obtained bythe elimination of a hydrogen halide molecule. You will need to draw the structural



formula and name the products on paper before displaying the answer.

Q27:

What elimination product(s) will beproduced from this compound ?

Q28:


Q29:


Q30:



2.2.2.2 Mechanisms of substitution reactions

The other main type of reaction of monohalogenoalkanes involves substitution of thehalogen atom by another atom or group.

Consider this diagram of a bromoalkane:

Q31: What type of bond is the C-Br bond?

a) pure covalent



b) polar covalent with the carbon partially positivec) polar covalent with the carbon partially negative

Q32: From your answer to the previous question, how will the carbon atom behave?

a) as an electrophileb) as a nucleophilec) as a free radical

Q33: What type of reactive species is liable to attack the carbon atom

a) an electrophileb) a nucleophilec) a free radical

Q34: How is the C-Br bond liable to break?

a) homolyticallyb) heterolytically

Substitution reactions can occur by either of two different mechanisms.

A. SN1 reaction

In this mechanism, there are two steps. In the first, the C-Br bond breaks heterolyticallyto form a Br- ion and a carbocation intermediate. This step is slow.

In the second step, a nucleophile (Nu-) rapidly attacks the positive carbon atom to formthe product.

The overall rate of the reaction depends on the slow step (the rate determining step)which involves only one molecule. The reaction is said to have 1st order kinetics (see



Advanced Higher, Unit 2, Topic 9). Consequently, this reaction mechanism is describedas SN1.

B. SN2 reaction

In this mechanism, the nucleophile approaches the carbon atom from the opposite sideto the halogen and a bond begins to form between the two atoms. At the same time, thebond between the carbon atom and the halogen begins to break.

So there is only one step in this reaction and no carbocation intermediate. Instead, atransition state forms in which the new bond is half-formed and the old bond is half-broken. The transition state can break down either to reform the reactants or to form theproducts.

The overall rate of the reaction depends on the concentrations of both reactants. Thereaction is said to show 2nd order kinetics (see Advanced Higher, Unit 2, Topic 9). Sothis reaction is described as SN2.



2.2.2.3 Products of substitution reactions

Using a variety of nucleophiles, a wide range of useful products can be obtained.

Alcohols

The reaction is carried out using an aqueous solution of an alkali such as sodium orpotassium hydroxide.

This provides a convenient route to specific alcohols (see Topic 3).

Note that while the use of KOH in ethanol favours elimination ("E" for Ethanol - "E" forElimination), the use of KOH in aqueous solution favours substitution.

Ethers

Alkoxides (containing the R-O- ion) are formed by the reaction of reactive metals withthe appropriate dry alcohol (this reaction will be considered in Topic 3), e.g. sodium andethanol.

The nucleophilic ethoxide ion can react with a halogenoalkane to form an ether (also



considered in more detail in Topic 3)

Nitriles

The use of potassium cyanide in ethanol provides a very useful route to compoundscontaining the nitrile functional group, -C�N.

This introduces another carbon atom and so increases the chain length by one. Note thechange in name from chloroethane to propanenitrile. Nitriles can be readily hydrolysedto form carboxylic acids (see Topic 4).

Amines

In the first step of this reaction, the ammonia molecule acts as a nucleophile, using thelone pair on the nitrogen atom, and forms the positive methylammonium ion.

Q35: What type of covalent bond is formed between the nitrogen atom and the carbon



atom?

In the second step, an ammonia molecule (usually present in excess) acts as a base byremoving an H+ ion to form the product, methylamine - an example of an amine. Amineswill be considered in more detail in Topic 5.

Products of substitution reactions

15 minQuestions to give practice in predicting the products of reactions of halogenoalkanes.

All the questions in this activity refer to Figure 2.34.

Figure 2.34: Reaction Scheme

Q36: What type of compound is compound 1?

Q37: Write the name of reagent X.

Q38: Draw the structural formula for the compound in box 2.




2.3. SUMMARY 41

2.3 Summary• The concept of hybridisation of atomic orbitals provides a useful model for bonding

in organic compounds - sp3 hybridisation for the tetrahedral arrangement of thebonds around a saturated carbon atom and sp2 hybridisation for the formation ofa C=C bond.

• The reaction between alkanes and bromine takes place by a free radical chainreaction, involving initiation, propagation and termination steps.

• Alkenes can be prepared either by dehydration of suitable alcohols ordehydrohalogenation of suitable monohalogenoalkanes using KOH in ethanol.

• Addition reactions of alkenes provide an important route to various types oforganic molecule, including alkanes, mono- and dihalogenoalkanes, alcohols. Themechanisms of these reactions can be explained.

• Halogenoalkanes can be named systematically using IUPAC rules and can beclassfied as primary, secondary or tertiary.

• Halogenoalkanes react by either elimination reactions or substitution reactions.

• Substitution reactions occur by eitherSN1 or SN2 mechanisms and provide asynthetic route to many other types of organic compound, including alcohols,ethers, nitriles and amines.

2.4 Resources• Higher Still Support: Advanced Higher Chemistry - Unit 3:

Organic Chemistry, Learning and Teaching Scotland, ISBN 0-333-18153-0




Website for representing molecules in different ways, for hybridisation and reactionmechanisms:

• http://www.colby.edu/chemistry/OChem/demoindex.html

2.5 End of Topic test

An online assessment is provided to help you review this topic.


http://www.colby.edu/chemistry/OChem/demoindex.html

43

Topic 3

Alcohols and Ethers

Contents

3.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45

3.2 Classification and nomenclature . . . . . . . . . . . . . . . . . . . . . . . . . . 46

3.3 Physical properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49

3.4 Preparation of alcohols . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53

3.5 Reactions of alcohols . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57

3.5.1 Reaction with reactive metals . . . . . . . . . . . . . . . . . . . . . . . . 57

3.5.2 Dehydration to form alkenes . . . . . . . . . . . . . . . . . . . . . . . . . 58

3.5.3 Forming esters . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59

3.6 Preparation of ethers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60

3.7 Reactions of ethers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61

3.8 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62

3.9 Resources . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62

3.10 End of Topic test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63



• use systematic nomenclature to name alkanols (Higher, Unit 2);

• classify alcohols as primary, secondary or tertiary (Higher, Unit 2);

• describe the nature and relative strengths of intermolecular forces(Higher, Unit 1);

• identify the different types of bond breaking (Topics 3.1 and 3.2);

• identify carbanions, carbocations, electrophiles and nucleophiles in reactionmechanisms (Topics 3.1 and 3.2);

• describe how alcohols and carboxylic acids can react reversibly to form esters.

Learning Objectives


• use systematic nomenclature to name ethers;

44 TOPIC 3. ALCOHOLS AND ETHERS

• explain some physical properties of alcohols and ethers (such as solubility in waterand boiling point) in terms of intermolecular forces;

• describe methods used to prepare alcohols and ethers;

• describe some reactions of alcohols and ethers.


3.1. INTRODUCTION 45

3.1 Introduction

The intoxicating effect of ethanol (an alcohol) has been known since ancient times.The anaesthetic properties of diethyl ether were discovered much more recently (inthe 1840s). Both ethanol and diethyl ether are simple molecules containing carbon,hydrogen and oxygen atoms but arranged in different ways.

Answer the following questions by drawing structural formulae on paper before revealingthe answers.

Q1: Draw full structural formulae for the two isomers of formula C2H6O and for thethree isomers of formula C3H8O.

Q2: Study the structures carefully and divide the five compounds into two groupsaccording to similarities in structure.

Q3: Look at the group of three compounds. What structural feature do they have incommon?

These three substances belong to the homologous series known as the alkanols whichare a subset of the larger family of alcohols. A few examples of alcohols and their usesare shown in Figure 3.1.

Figure 3.1: Some important alcohols

Q4: Now look at the structure of the remaining two compounds. What structuralfeature do they have in common?



Such compounds are known as ethers. Some examples of ethers and their uses areshown in Figure 3.2.

Figure 3.2: Some important ethers

These photographs show models of the most common alcohol and a simple ether.

ethanol dimethyl ether

3.2 Classification and nomenclature�

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Learning Objective

To classify and name alcohols and ethers

Alcohols

At Higher level, the systematic naming of alkanols and their classification as primary,secondary or tertiary was introduced.

The web version of this Topic has two animations which you can use for revision. Thenanswer the following questions.


3.2. CLASSIFICATION AND NOMENCLATURE 47

Q5:

What class of alcohol is this?

a) primaryb) secondaryc) tertiary

Q6:

What class of alcohol is this?


Q7: What class of alcohol is cholesterol (Figure 3.1)?


Q8: Draw the structural formulae for the two primary alcohols of formula, C4H10O.Give the correct name for both your structures.


Ethers

Ethers are compounds containing the group R’-O-R". ( R1 and R2 are often used insteadof R’-O-R".)

Ethers can be classified as

• symmetrical - if R’ and R" are identical

• unsymmetrical - if R’ and R" are different.

Ethers are named in a different way. In fact, there are two possible ways:

• The alkyl groups R’ and R" are named according to the number of carbon atomsand these names are written in front of the word ’ether’. If the ether is symmetrical,the name is of the form - dialkyl ether. This method of naming is only useful forsimple ethers.

• More complex ethers are considered as substituted alkanes.



Consider the ether opposite:

An animation on the website shows in detail how to name this compound.

The smaller alkyl group and the oxygen atom areconsidered as a substituent.

The other alkyl group is treated as a hydrocarbon and named according to the rulescovered in Higher and summarised in Topic 1.

Following this procedure gives:

2-methoxy-2-methylpropane.

Note that the substituents are given in alphabetical order.

Now, try the following questions.

Q9: What is the alternative name for diethyl ether (shown in Figure 3.2)?

Q10:

What is the correct name for theether shown opposite?

Q11: What is the alternative name of ethyl methyl ether?

Q12:

What is the correct name for theether shown opposite?


3.3. PHYSICAL PROPERTIES 49

3.3 Physical properties�

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Learning Objective

To explain some physical properties of alcohols and ethers (such as solubility in waterand boiling point) in terms of intermolecular forces

The physical properties of organic compounds such as melting point, boiling point andsolubility in water are dependent on the intermolecular forces involved. These forces areelectrostatic in nature and involve attractions between molecules which contain dipoles.A dipole arises when there is an uneven distribution of charge in a molecule, suchthat one part has a partial positive charge and another part has an equal but oppositenegative charge.

Copy and complete the following table using the word bank.

Force Origin Example

van der WaalsPolar-polar attractions

Hydrogen bonding

Word bank:

between watermolecules

between H-Clmolecules

attraction betweentemporary dipoles

attraction betweenpermanent dipoles

attractions involvingHF, HO and HN

bonds

between CH4

molecules

Q13: Which of the following shows the intermolecular forces in the correct order ofstrength?

a) hydrogen bonding � polar-polar attraction � van der Waals forcesb) van der Waals forces � polar-polar attraction � hydrogen bondingc) polar-polar attraction � van der Waals forces � hydrogen bondingd) hydrogen bonding � van der Waals forces � polar-polar attraction

Boiling points of alcohols

15 min

Here is a problem-solving exercise based on a graph of the boiling points of someorganic compounds.

The following graph (Figure 3.3) shows the boiling points for the first four members of thealkane (R-H), chloroalkane (R-Cl) and alkanol (R-OH) series. Study the graph carefullyand answer the questions that follow on paper before revealing the answers.



Figure 3.3

First consider the alkanes.

Q14: What type of intermolecular force is involved when alkanes boil?

a) van der Waals forcesb) polar-polar attractionsc) hydrogen bonding

Q15: What happens to the boiling point as the number of carbon atoms increases?

a) it decreasesb) it increases

Q16: Explain this trend.

Now consider the chloroalkanes. Compare chloromethane (bp 249 K) with propane(bp 231 K).

Q17: Why compare chloromethane with propane rather than methane?

Q18: Explain why chloromethane has a higher boiling point than propane.

Now consider the alkanols.

Figure 3.4:

Q19: For fairness, which alkane should be compared with ethanol (Figure 3.4)?

Q20: For fairness, which chloroalkane should be compared with butanol?

Q21: Using Figure 3.3, write a general statement about the boiling points of alkanolscompared with alkanes and chloroalkanes.



Q22: Which is the most important type of intermolecular force being broken whenalkanols are boiled (see Figure 3.4)?

a) van der Waals forcesb) polar-polar attractionsc) hydrogen bonding


Alcohols exhibit hydrogen bonding and as a result have higher boiling points than otherorganic compounds of comparable relative formula mass.

Boiling points of ethers

10 min

Visit the web version of this Topic to find a drag and drop exercise as well as the followingquestions to investigate the boiling points of ethers.

Figure 3.5 Figure 3.6

By considering the structures of the compounds in the table below and the probableintermolecular forces, answer the questions which follow.

Ether bp / ÆC Alcohol bp / ÆC Alkane bp / ÆC

CH3OCH3 -25 C2H5OH 78 C3H8 -45CH3OC2H5 11 C3H7OH 97 C4H10 -1C2H5OC2H5 35 C4H9OH 117 C5H12 36

Q23: Which family of compounds has the highest boiling points?

a) ethersb) alcoholsc) alkanes

Q24: In which family will the intermolecular forces be strongest?

a) ethersb) alcoholsc) alkanes

Q25: Consider the bonding in the ether functional group (Figure 3.6) and compare withthe structure of an alkane. Which of the following statements is false?

a) Ethers are more polarb) Intermolecular forces will be stronger between ether molecules



c) Alkanes will have higher boiling pointsd) Intermolecular forces will be weaker between alkane molecules.

Q26: Consider the bonding in the ether functional group (Figure 3.6) and the alcoholfunctional group (Figure 3.5). Which of the following statements is true?

a) Hydrogen bonding occurs between ether moleculesb) Ether molecules will be more polar than alcohol moleculesc) Alcohols will tend to have higher boiling pointsd) Intermolecular forces will be stronger between ether molecules.


Due to the lack of hydrogen bonding, ethers have lower boiling points than thecorresponding isomeric alkanols.

Solubility in water

Table 3.1

bp = -25ÆC bp = 78ÆC bp = 100ÆC

From the information in Table 3.1, answer the following questions.

Q27: Is hydrogen bonding possible between one water molecule and another?

a) yesb) no

Q28: Is hydrogen bonding possible between one ethanol molecule and another?

a) yesb) no

Q29: Is hydrogen bonding possible between one methoxymethane molecule andanother?

a) yesb) no

Q30: Why is methoxymethane a gas at room temperature while the other two areliquids?


Table 3.2 shows solubility data in grams per 100 ml of water (g / 100 ml) for some ethersand alcohols.


3.4. PREPARATION OF ALCOHOLS 53

Table 3.2: Solubility in water of ethers and alcohols

EtherSolubility (g

/ 100 ml) AlcoholSolubility (g

/ 100 ml)

methoxymethane �10 ethanol �10ethoxyethane 6.9 butan-1-ol 6.32

1-ethoxybutane 0.21 hexan-1-ol 0.591-butoxybutane � 0.1 octanol �0.1

Q31: Compare the solubilities of ethers in water with alcohols in water. Which of thefollowing statements is true about the solubility of isomeric ethers and alcohols?

a) The ether is more soluble.b) The alcohol is more soluble.c) The solubilities are roughly the same.

Q32: Compare the solubilities of ethers in water with alcohols in water. What is thetrend in solubility in both families as the chain length increases?

In summary, ethers and alcohols of low molecular mass are soluble in water (misciblewith water) due to their ability to form hydrogen bonds with water molecules. As thechain length increases, the solubility decreases.

3.4 Preparation of alcohols�

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Learning Objective

To describe methods used to prepare alcohols from alkenes or from halogenoalkanes

For thousands of years, ethanol has been produced by fermentation. Nowadays, thisprocess is still used on a much larger scale to produce alcoholic drinks, see Figure 3.7,and in some places such as Brazil, to produce fuel.



Figure 3.7: Copper stills in the Benromach distillery and a ’wee dram’

There are two general methods used to prepare alcohols, both of which were describedin Topic 3.2:

• Hydration of alkenes

• Nucleophilic substitution using halogenoalkanes

Hydration of alkenes

10 min

On the website, you can see a simulation of the mechanism for the addition of water toa double bond.

Study the diagrams of the mechanism for addition of water to an alkene and then answerthe questions which follow.


3.4. PREPARATION OF ALCOHOLS 55

Table 3.3: Hydration mechanism

Q33: What role does the hydrogen atom of the acid catalyst play in the first stage of thereaction?

a) free radicalb) nucleophilec) electrophile

Q34: Conversely, what role does the alkene play in the reaction?

Q35: What type of carbocation intermediate is shown in Table 3.3?

Q36: If the attack had taken place on the other carbon atom of the double bond, whattype of carbocation would have formed?


Alcohols can be prepared from alkenes by the acid-catalysed addition of water. Ifthe alkene is unsymmetrical, a mixture of products is obtained which requires furtherseparation.

Preparation from halogenoalkanes

10 min

A series of questions to revise and reinforce the nucleophilic substitution of halogenatoms by a hydroxyl group.

Study the diagrams and answer the questions which follow.



Figure 3.8: Reaction mechanism 1

Figure 3.9: Reaction mechanism 2

Q37: How is the mechanism in Figure 3.8 normally described?

Q38: How is the mechanism in Figure 3.9 normally described?

Q39: Explain why halogenoalkanes are subject to nucleophilic attack.

Q40: Write a structural formula and name the product formed when 2-bromopropane isreacted with aqueous KOH.


Alcohols can be prepared from halogenoalkanes by nucleophilic substitution usingaqueous alkali.

Both the alkene hydration method and the halogenoalkane substitution method can be


3.5. REACTIONS OF ALCOHOLS 57

used in the laboratory. Preparation from halogenoalkanes has the advantage in thatgenerally only one product is formed, although care must be taken to choose conditionssuch that elimination does not occur (see Topic 3.2).

In industry, alcohols (except for methanol) tend to be manufactured on a large scale bythe acid-catalysed hydration of alkenes.

Q41: Explain why methanol cannot be prepared by this method.

3.5 Reactions of alcohols�

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Learning Objective

To describe some reactions of alcohols

Alcohols are amongst the most useful of organic compounds. They can be preparedfrom, and converted into, a wide variety of other compounds, some of which are shownin Figure 3.10.

Figure 3.10: Reactions of alcohols

3.5.1 Reaction with reactive metals�

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Learning Objective

To describe the reaction of alcohols with sodium.

Alcohols react with metals like sodium or potassium to produce hydrogen gas.Water also reacts with these metals to produce hydrogen gas.

ethanol water

There is an obvious similarity in structure between ethanol and water and so thesimilarity in reaction is not surprising.



Q42: Write a balanced equation for the reaction of sodium with water.

Q43: Apart from hydrogen, what other product is formed?

Q44: Look closely at the equation for the reaction of sodium with water. Predict theother product formed when sodium reacts with ethanol. Draw its structure and suggesta name.

In fact, this is a general reaction for alcohols.

Figure 3.11: Formation of alkoxides

The metal alkoxides are powerfulbases as well as nucleophiles andare frequently used in organicsynthesis.

3.5.2 Dehydration to form alkenes�

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Learning Objective

To describe the dehydration of alcohols to give alkenes

This reaction was discussed in Topic 3.2 under the heading ’Synthesis of alkenes’.

For simple alcohols, the alcohol vapour can be passed over hot aluminium oxide, forexample:

Figure 3.12: Dehydration of propan-2-ol

A more general method involves the reaction of an alcohol with a strong non-volatileacid such as concentrated sulphuric acid or phosphoric acid.

Q45: What class of reaction is this?

a) substitution


3.5. REACTIONS OF ALCOHOLS 59

b) eliminationc) neutralisationd) condensation

Q46: If butan-2-ol is warmed with phosphoric acid, what product (or products) will beformed?

You will not be asked to describe the mechanism of this type of reaction in detail but itis effectively the reverse of the mechanism described in Table 3.3 for the hydration of analkene.

3.5.3 Forming esters�

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Learning Objective

To describe the reaction of alcohols to form esters

The reaction between alcohols and carboxylic acids to form esters was introduced atHigher level (Unit 2, World of Carbon).

Figure 3.13: Esterification

Q47: What class of reaction is shown in Figure 3.13?

a) neutralisationb) eliminationc) neutralisationd) condensation

Q48: The reaction is frequently carried out in the presence of a small amount ofconcentrated sulphuric acid. Give two reasons for this.

Q49: Name the ester formed in Figure 3.13.

Even in the presence of a catalytic amount of concentrated sulphuric acid, the reactionbetween alcohols and carboxylic acids is slow.

Esters are normally prepared by an alternative two-stage process in which the carboxylicacid is first converted into an acid chloride by reaction with phosphorus pentachloride(PCl5) or thionyl chloride (SOCl2), see Figure 3.14.



Figure 3.14: Making ester (stage 1)

Q50: What type of reaction has taken place?

a) neutralisationb) eliminationc) substitutiond) addition

In stage 2, the acid chloride reacts with an alcohol. Acid chlorides are much morereactive than the parent acids and readily react with alcohols to form esters.

Figure 3.15: Acid chloride to ester

Q51: What type of reaction occurs in the second stage (Figure 3.15)?

a) neutralisationb) additionc) eliminationd) condensation

3.6 Preparation of ethers�

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Learning Objective

To describe the preparation of ethers from halogenoalkanes

The standard method used to prepare ethers involves the nucleophilic substitution of amonohalogenoalkane using a metal alkoxide (formed as described in Figure 3.11). Thisreaction was briefly described in Topic 3.2.


3.7. REACTIONS OF ETHERS 61

Figure 3.16: Forming an ether

This reaction (Figure 3.16) is a general one and can be used to prepare bothsymmetrical and unsymmetrical ethers.

3.7 Reactions of ethers

Ether molecules are slightly polar (Table 3.1) but sufficiently non-polar to make themresistant to attack by most nucleophiles and electrophiles. They are also relativelyresistant to reduction and, with some exceptions, oxidation.

However they are highly flammable,particularly those with low molecular mass.

In addition, on contact with air even for only a few days, ethers can form hydroperoxidesand peroxides.

Q52: The O-O bond in peroxides is fairly weak. What type of bond fission is likely tooccur?

As a result of this reactivity, these peroxidesare extremely unstable and potentiallyexplosive.

Ethers are widely used as solvents because they can dissolve a large number of organiccompounds. Because the ether molecule (Figure 3.6) is slightly polar, ethers readilydissolve non-polar molecules as well as molecules which are slightly polar.

Ethoxyethane is widely used in solvent extractions since the extracted compounds areeasily isolated by evaporation of the solvent. However, extreme care must be taken in



using such ethers, particularly if the solvent is removed by distillation. No naked flamesshould be used anywhere near the solvent and the procedure should be carried out ina fume cupboard. To make sure that the solvent is free from peroxides, it can first betreated with a solution of an iron(II) salt.

3.8 Summary• Ethers (general formula - R’-O-R") can be named according to IUPAC rules.

• Hydrogen bonding between alcohol molecules and lack of hydrogen bondingbetween ether molecules explains why alcohols (general formula - R-OH) havehigher boiling points than isomeric ethers and other organic molecules of similarrelative molecular mass.

• Hydrogen bonding between alcohol molecules and water, and between ethermolecules and water, explains why ethers and alcohols of similar relative molecularmass have similar solubilities in water.

• Alcohols can be prepared from alkenes by hydration or from halogenoalkanesby nucleophilic substitution. In industry, alcohols (except methanol) aremanufactured by the acid-catalysed hydration of alkenes.

• Alcohols react with some reactive metals to form alkoxides.

• Alcohols can be dehydrated to alkenes.

• Alcohols undergo condensation reactions slowly with carboxylic acids and morevigorously with acid chlorides to form esters.

• Ethers can be prepared by the reaction of halogenoalkanes with alkoxides.

• Ethers are highly flammable and on exposure to air may form potentially explosiveperoxides.

• Ethers are mainly used as solvents because they are relatively inert chemicallyand will dissolve many organic compounds.






Website




3.10. END OF TOPIC TEST 63




65

Topic 4

Aldehydes, Ketones and CarboxylicAcids

Contents

4.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 674.2 Physical properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 68

4.2.1 Solubility in water . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 704.3 Reactions of aldehydes and ketones . . . . . . . . . . . . . . . . . . . . . . . . 74

4.3.1 Oxidation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 744.3.2 Reduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 754.3.3 Nucleophilic addition reactions . . . . . . . . . . . . . . . . . . . . . . . 76

4.4 Carboxylic acids . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 814.4.1 Preparation of carboxylic acids . . . . . . . . . . . . . . . . . . . . . . . 814.4.2 Reactions of carboxylic acids . . . . . . . . . . . . . . . . . . . . . . . . 82

4.5 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 874.6 Resources . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 884.7 End of Topic test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 88



• recognise aldehydes, ketones, carboxylic acids and esters from their functionalgroups and give systematic names for simple examples;

• describe the oxidation of primary alcohols and secondary alcohols to aldehydesand ketones respectively and the further oxidation of aldehydes to carboxylic acids;

• describe oxidation as an increase in the oxygen to hydrogen ratio and reductionas a decrease in the oxygen to hydrogen ratio in an organic compound;

• describe how alcohols and acids can react reversibly to form esters.

Learning Objectives


• explain the physical properties, such as boiling points and solubility in water, ofaldehydes, ketones and carboxylic acids in terms of intermolecular forces;

66 TOPIC 4. ALDEHYDES, KETONES AND CARBOXYLIC ACIDS

• distinguish between aldehydes and ketones using either Tollens’ reagent orFehling’s solution;

• describe some reactions of aldehydes, ketones and carboxylic acids;

• describe some methods of preparation of carboxylic acids.



4.1 Introduction�

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Learning Objective

To recognise functional groups and describe the bonding in the carbonyl group

Molecules of aldehydes, ketones and carboxylic acids all contain a common feature, thecarbonyl group- C=O.

This is arguably the most important of functional groups. It is an essential part ofmany pharmaceuticals and most important biological molecules. The ability to identifycorrectly the various functional groups is crucial to a full understanding of organicchemistry.

Carbonyl group

The carbonyl group, C=O, appears as part of other functional groups. The formation ofthe carbon to oxygen double bond can be explained in a similar way to that of the carbonto carbon double bond (see Topic 3.2, section 1.3).

Hybridisation of the same orbitals on the oxygen atom also occurs and overlap of theorbitals on both atoms gives the structure shown in Figure 4.1.

Figure 4.1: Carbonyl bonding

There is one crucial difference between the formation of the C=O bond and the formationof the C=C bond. Oxygen being more electronegative than carbon attracts the bondedelectrons more strongly, particularly the � electrons, making the C=O bond polar(Figure 4.2).



Figure 4.2

The physical properties and chemical properties of aldehydes, ketones and carboxylicacids are largely determined by the polarity of the C=O bond.

Q1: Which orbitals in the carbon atom will be involved in hybridisation?

a) 2s onlyb) 2s and one 2p orbitalc) 2s and two 2p orbitalsd) 2s and three 2p orbitals

Q2: Will the carbon atom be an electrophilic centre or a nucleophilic centre?

a) electrophilicb) nucleophilic

Q3: Will the oxygen atom be an electrophilic centre or a nucleophilic centre?

a) electrophilicb) nucleophilic

Q4: What type of species will attack the carbon atom?

Q5: What type of species will attack the oxygen atom?

Q6: The oxygen atom is able to accept H+ ions. In doing this, it is behaving like a .. ?


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Learning Objective

To explain the physical properties, such as boiling points and solubility in water, ofaldehydes, ketones and carboxylic acids in terms of intermolecular forces

In the last Topic, the boiling points and solubility in water of alcohols and ethers wereexplained in some detail in terms of the intermolecular forces involved. The physicalproperties of aldehydes, ketones and carboxylic acids can be explained in a similar way.



Boiling points

15 min

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Learning Objective

To be able to explain the trends in boiling points of aldehydes, ketones and carboxylicacids.

A data handling exercise based on graphs of the boiling points of compounds fromdifferent homologous series.

The graph shows the boiling points for members of five different homologous series.Study it carefully and then answer the questions underneath.

Q7: Which are the most important intermolecular forces being overcome whenaldehydes and ketones boil?

a) covalent bondsb) hydrogen bondsc) polar-polar attractionsd) van der Waals forces

Q8: Explain why aldehydes and ketones have higher boiling points than the alkanes ofsimilar molecular mass.

Q9: Which are the most important intermolecular forces being overcome whenalcohols boil?

a) covalent bondsb) hydrogen bondsc) polar-polar attractionsd) van der Waals forces

Q10: Explain why aldehydes and ketones have lower boiling points than the alcohols ofsimilar molecular mass.


Figure 4.3



Figure 4.4

In pure carboxylic acids, the molecules pair up to form dimers ( Figure 4.4), in whicheach molecule forms two hydrogen bonds. This stronger bonding explains the higherboiling points of carboxylic acids.

Aldehydes and ketones have higher boiling points than the corresponding alkanesbecause polar-polar attractions are stronger than van der Waals forces. They havelower boiling points than the corresponding alcohols because polar-polar attractionsare weaker than hydrogen bonds. Both carboxylic acids and alcohols exhibit hydrogenbonding but this must be stronger in the acids since they have higher boiling points.

4.2.1 Solubility in water�

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Learning Objective

To explain the trends in solubility in water of aldehydes, ketones and carboxylic acidsin terms of intermolecular forces and explain the acidity of carboxylic acids.

The solubility of an organic compound in water depends on the ability of the substanceto form hydrogen bonds with water.



Study the structures in the diagram above.

Q11: Which of the structures shown will be able to form hydrogen bonds with water?

a) all of them except propanoneb) ethanoic acid onlyc) ethanoic acid and ethanol onlyd) all of them

Q12: Which substances shown are likely to be soluble in water?

a) all of them except propanoneb) ethanoic acid onlyc) ethanoic acid and ethanol onlyd) all of them

Q13: Use your knowledge to predict the effect of chain length on the solubility ofaldehydes and ketones in water.

Q14: Which of the following statements is false?

a) Carboxylic acids are likely to be more soluble than the corresponding aldehydes andketones.

b) As the number of carbon atoms increases, the solubility of carboxylic acidsdecreases.

c) Hexanoic acid is more soluble than ethanoic acid.d) Carboxylic acids of low molecular mass are very soluble in water.

Acidity

Although pure carboxylic acids exist as dimers ( Figure 4.4), such dimerisation does notoccur in aqueous solution. Instead there is tendency for carboxylic acid molecules todissociate according to the equation:



Figure 4.5

As a result, carboxylic acids are weak acids. Values for pKa can be used as a measureof the strength of an acid (see Unit 2, Topic 4).

Substance pKa

ethanoic acid 4.8ethanol 18water 15.7

Q15: Which is the strongest acid?

a) ethanoic acidb) ethanolc) water

Q16: Which is the weakest acid?

a) ethanoic acidb) ethanolc) water

With all three substances, the H+(aq) ion is produced by the heterolytic fission of an OHbond in a hydroxyl group. Why is the carboxyl group so acidic?

The dissociation (Figure 4.5) is reversible. Any factor which pushes the equilibrium tothe right will strengthen the acid.

The ethanoate ion can be represented by two equivalent resonance structures:

In fact, the electrons are spread (delocalised) over all three atoms of the group.Consideration of the bonding in the ethanoate ion helps to explain this (Figure 4.6).The three atoms in the carboxylate group are sp2 hybridised.



Figure 4.6: Delocalisation in the ethanoate ion

The unhybridised p orbitals are close enough to overlap to form a delocalised �-system(Figure 4.6). Delocalisation of the charge makes the ion more stable and makes it lesslikely to accept a H+(aq) ion.

Figure 4.7: Ethanol dissociation

Figure 4.7 shows the similar dissociation of ethanol.

Q17: Look at diagram (Figure 4.7). Can the alkoxide ion be stabilised in the same way?

a) yesb) no

Q18: In which group is the OH bond more likely to break?

a) hydroxylb) carboxyl

Q19: Which will be more acidic?

a) alkanolsb) alkanoic acids

Summary of solubility in water

Aldehydes and ketones of relatively low molecular mass are miscible with water.Carboxylic acids of similar molecular mass are even more soluble.

In both cases, as the chain length increases, the solubility decreases.



4.3 Reactions of aldehydes and ketones

4.3.1 Oxidation�

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Learning Objective

To distinguish between aldehydes and ketones using Tollens’ or Fehling’s reagents

The oxidation of aldehydes to carboxylic acids was introduced at Higher level (Unit 2,World of Carbon). This reaction can be used to distinguish between aldehydes andketones.

Figure 4.8

Aldehydes have a hydrogen atom attached to the carbonyl carbon atom (Figure 4.8).This can be readily lost during oxidation. Ketones do not have such a hydrogen atomand are much more resistant to oxidation.

Tollens’ reagent (named after its discoverer, Bernhard Tollens) contains complexedsilver(I) ions, [Ag(NH3)2]+, which on reduction produce metallic silver.

Figure 4.9: Reaction of an aldehyde with Tollens’ reagent

The formation of a silver mirror (Figure 4.9) is often used as a test for the presence ofaldehydes as opposed to ketones.

Fehling’s solution contains complexed copper(II) ions and will also oxidise aldehydes butnot ketones.


4.3. REACTIONS OF ALDEHYDES AND KETONES 75

Figure 4.10: Reaction of an aldehyde with Fehling’s solution

When reacted with an aldehyde, the copper(II) ions are reduced to copper(I) and areddish-brown precipitate of copper(I) oxide forms (Figure 4.10).

Figure 4.11

The aldehyde is oxidised to an alkanoic acid (Figure 4.11). The symbol, [O], is oftenused in organic equations to show that a compound has been oxidised.

4.3.2 Reduction�

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Learning Objective

To describe the reduction of aldehydes and ketones to alcohols

Both aldehydes and ketones can be reduced to alcohols using reducing agents likelithium aluminium hydride, LiAlH4, in ether.

Figure 4.12: Reduction mechanism

The AlH4- ion is able to provide a hydride ion, H-, which attacks the partially positive

carbon atom of the carbonyl group to form the negative ion shown (Figure 4.12, stage1).



Q20: What role does the hydride ion play in this step?

a) acidb) basec) nucleophiled) electrophile

When the ether solution is shaken with water, the negatively charged oxygen atomaccepts a proton from water to form the alcohol product(Figure 4.12, stage 2).

Q21: Which of the following describes the change in the arrangement of the bondsaround the carbonyl carbon atom?

a) from tetrahedral to trigonal planarb) from trigonal bipyramidal to tetrahedralc) from tetrahedral to tetrahedrald) from trigonal planar to tetrahedral

Q22: If the starting material was an aldehyde, which class of alcohol would beproduced?

Q23: If the starting material was a ketone, which class of alcohol would be produced?

Q24:

Give the full systematic name of theproduct when the compound opposite isreduced using lithium aluminium hydride inether:

Q25:

Give the full systematic name of theproduct when the compound opposite isreduced using lithium aluminium hydride inether.

4.3.3 Nucleophilic addition reactions�

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Learning Objective

To describe the addition of reagents to the carbonyl group in aldehydes and ketonesand the subsequent elimination of water where appropriate.

Aldehydes and ketones react with many reagents by nucleophilic addition reactions. It iseasier to work out the various products if you first understand what goes on in the basicmechanism.



In nucleophilic addition reactions, the mechanism is very similar to that described for thereduction using LiAlH4. Overall, the reduction of an aldehyde or ketone corresponds tothe addition of hydrogen across the carbonyl group (Figure 4.12).

Figure 4.13: Stage 1

In the first stage (Figure 4.13), nucleophilic attack on the carbon atom of the carbonylgroup is followed by electrophilic attack by a H+ ion (protonation) of the oxygen atom toproduce an alcohol. Note that the trigonal planar carbonyl compound is converted intoa tetrahedral alcohol. In effect, H-Nu has added across the carbonyl group (where Nu isthe nucleophile).

If the nucleophilic atom has a hydrogen atom attached to it, the reaction can proceedfurther. This happens with some compounds related to ammonia, where one of thehydrogen atoms of the ammonia molecule has been replaced by another group(i.e. R-NH2 ). Stage one of the reaction is as before (Figure 4.13).

Figure 4.14: Stage 2

In the second stage (Figure 4.14), an acid-catalysed elimination of water occursresulting in the formation of a carbon to nitrogen double bond.



Q26: Overall, what type of reaction has occurred? (i.e. Addition + elimination = ?)

Examples of nucleophiles that can react with aldehydes and ketones include thoseshown in Figure 4.15.

Figure 4.15: Hydrogen cyanide and hydrazines

In the presence of cyanide ions, nucleophilic addition of hydrogen cyanide to analdehyde or ketone produces a cyanohydrin (Figure 4.16).

Figure 4.16: Cyanohydrin

The cyanohydrin can be isolated because it does not react further by elimination of wateras described in Figure 4.14. The production of cyanohydrins is useful since hydrolysiscan convert the cyanohydrin into a carboxylic acid with one more carbon atom than thestarting material, e.g.

On the other hand, hydrazine, NH2-NH2 (Figure 4.15), is capable of nucleophilic additionfollowed by elimination of water. The product is known as a hydrazone(Figure 4.17).



Figure 4.17: Hydrazone formation

Compounds based on hydrazine, in which one of the hydrogen atoms has been replacedby another group, react easily with aldehydes and ketones in exactly the same wayto form crystalline products that are easily purified. These products are known asderivatives. The pure derivatives have sharp, distinctive melting points. A commonlyused example is 2,4-dinitrophenylhydrazine (Figure 4.15). When this is used, theproducts are known as 2,4-dinitrophenylhydrazones.

By comparing the melting point of a derivative with tables of the melting points of knownderivatives, it is possible to identify the original aldehyde or ketone.

Learning point

Hydrazine and 2,4-dinitrophenylhydrazine react with aldehydes and ketones bynuceophilic addition to form hydrazones and 2,4-dinitrophenylhydrazones withdistinctive melting points which can be used to identify unknown aldehydes andketones.

At this point it would be a good idea to try the Prescribed Practical Activity on’Identification by Derivative Formation’.

PPA - Identification by Derivative Formation (Unit 3 PPA 2)

120 min

Consult with your tutor to find out whether the PPA on ’Identification by DerivativeFormation’ is to be completed at this stage.



Although aldehydes and ketones differ in their ease of oxidation, the rest of theirchemistry is very similar, although in general aldehydes are more reactive than ketones.

If both the alkyl groups in a ketone are large and bulky, they may hinder the nucleophilefrom attacking the carbon atom of the C=O group and so reduce the reactivity of theketone.

However, this does not explain why simple ketones, which allow relatively unhinderedaccess to nucleophiles, are less reactive than the corresponding aldehydes.

In Topic 3.2 (section 2.1.3), the relative stabilities of carbocations were discussed(Figure 4.18). Alkyl groups are able to push electrons towards a positively chargedcarbon atom and so stabilise the positive charge. The more alkyl groups, the greaterthe stabilisation.

Figure 4.18: Relative stabilities of carbocations

Q27: Which carbocation is most susceptible to attack by a nucleophile?

Now look at the structures of aldehydes and ketones.

Q28: Which structure is better able to stabilise the positive charge on the carbon atom?

a) aldehydeb) ketone

Q29: In which is the carbonyl atom more susceptible to attack by nucleophiles?

a) aldehydeb) ketone


4.4. CARBOXYLIC ACIDS 81

4.4 Carboxylic acids

4.4.1 Preparation of carboxylic acids�

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Learning Objective

To describe some methods of preparation of carboxylic acids.

Most of the normal methods used to prepare carboxylic acids have been discussedalready and are summarised in Figure 4.19.

Figure 4.19: Preparation of acids

Amides contain the functional group shown opposite and theirpreparation from carboxylic acids is considered in the nextsection.

Q30: Name the class of compound X.

Q31: What does the symbol, [O], represent?

Q32: What general type of reaction occurs when esters, nitriles and amides areconverted into carboxylic acids?

Q33:



Which of the reactions in the grid above could not be used to prepare2-methylpropanoic acid?

a) Ab) Bc) Cd) D


PPA - Preparation of benzoic acid by hydrolysis of ethyl benzoate (Unit 3PPA3)

120 min

Consult with your tutor to find out whether the PPA on ’Preparation of benzoic acid byhydrolysis of ethyl benzoate’ is to be completed at this stage.

4.4.2 Reactions of carboxylic acids�

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Learning Objective

To describe some reactions of carboxylic acids.

Some of the reactions of carboxylic acids are shown in Figure 4.20.

Figure 4.20: Reactions of carboxylic acids



4.4.2.1 Salt formation

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Learning Objective

To describe the reactions of carboxylic acids that result in the formation of salts.

Carboxylic acids behave as typical acids by forming salts with alkalis, basic metal oxides,carbonates and some metals.

On paper, write balanced equations, including states, for the reactions of aqueousethanoic acid with the following substances:

Q34: sodium hydroxide solution

Q35: solid copper(II) carbonate

Q36: magnesium metal

Q37: solid copper(II) oxide

4.4.2.2 Amide formation

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Learning Objective

To describe the reaction of carboxylic acids with ammonia and amines to form amides.

Ammonia and the related compounds, amines, are examples of bases that can reactwith carboxylic acids to form salts (see Topic 3.5), e.g:

If these salts are heated strongly, further reaction occurs, producing amides, e.g:



Amides contain a functional group in which the carbon atom of a carbonyl group isbonded to nitrogen. There are three classes of amide.

Figure 4.21: Classification of amides

The functional group shown in Figure 4.21(b) was introduced at Higher level(Unit 2, World of Carbon, Polymers and Natural Products). It is formed when certainmonomers join to form polymers, some natural and some synthetic.

Q38: What important class of naturally occurring polymers contains this functionalgroup?

a) fats and oilsb) carbohydratesc) proteinsd) vitamins

Q39: What alternative name is given to this linking group?

Q40: Synthetic polymers containing this group are called polyamides. Give twoexamples of such polymers.

4.4.2.3 Ester formation

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Learning Objective

To describe the reaction of carboxylic acids with alcohols to form esters.

Carboxylic acids react reversibly with alcohols to form esters. This reaction was firstcovered at Higher level (Unit 2, World of Carbon). It was briefly discussed in Topic 3.3in the section on Reactions of alcohols. Since the reaction is reversible, it provides a



convenient method for preparing carboxylic acids from esters(this Topic, section 4.1, Figure 4.19).

The reaction is catalysed by acid. Although detailed knowledge of the reactionmechanism is not required, it is worth pointing out the similarity with other carbonylreactions.

Figure 4.22: Ester formation - mechanism

The first stage involves nucleophilic addition to the carbonyl group (Figure 4.22).Protonation of the oxygen atom of the carbonyl group makes the carbon atom moresusceptible to nucleophilic attack by the oxygen atom of the alcohol. In the secondstage, water is eliminated to form the ester product,

Q41: What other term can be used to describe such addition-elimination reactions?

Q42: Name and draw the structural formula of the ester formed when propanoic acidand ethanol react.

Q43: Name the ester formed when benzoic acid and methanol react.

Q44: Name the ester formed when octanoic acid and propanol react.

Q45:

Which of the structures in the above grid does not represent methyl propanoate?



4.4.2.4 Reduction

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Learning Objective

To describe the reduction of carboxylic acids to primary alcohols with lithiumaluminium hydride.

Earlier in this Topic (section 3.2), the reduction of aldehydes and ketones to alcoholsusing lithium aluminium hydride in ether was described (Figure 4.12). Carboxylic acidsare more resistant to reduction than aldehydes and ketones and LiAlH4 is one of the fewreagents powerful enough to reduce carboxylic acids to primary alcohols.

The mechanism for the reduction of acids is more complicated but the initial step is againlikely to be nucleophilic attack by hydride ion, H-, on the partially positive carbon atomof the carbonyl group.

Reactions of aldehydes, ketones and carboxylic acids

15 min

This is a problem solving activity to test your knowledge of reactions of aldehydes,ketones and carboxylic acids.

A hydrocarbon of formula, C4H8, was reacted to form two isomeric compounds, P andQ, which were separated and further reacted as shown in the reaction scheme below(Figure 4.23). Using knowledge gained in this and the previous Topic, identify theunknown compounds in the scheme. Write your answers on paper and then confirmthem by answering the questions which follow.

If you are really stuck, there is a hint for this activity at the back of the book.


4.5. SUMMARY 87

Figure 4.23

Q46: Which is the original hydrocarbon?

a) cyclobutaneb) but-1-enec) but-2-ened) 2-methylpropene

Q47: Compound P is


Q48: Which compound is a ketone?

Q49: Name compound S.

Q50: What type of compound is T?

Q51: What is the formula for reagent Z?

Q52: What is the formula for reagent Z?

4.5 Summary• Aldehydes and ketones have higher boiling points than the corresponding alkanes

and lower boiling points than the corresponding alcohols. These differences can



be explained in terms of the relative strengths of the intermolecular forces

• Pure carboxylic acids have relatively high boiling points due to the formation ofdimers held together by hydrogen bonding.

• Aldehydes, ketones and carboxylic acids of low relative formula mass are misciblewith water due to their ability to form hydrogen bonds with water molecules. Thesolubility in water decreases as chain length increases.

• Carboxylic acids are weak acids because they dissociate slightly in water. Thisis can be explained by the stability of the carboxylate anion caused by electrondelocalisation.

• Aldehydes and ketones can be reduced to primary and secondary alcoholsrespectively using lithium aluminium hydride in ether, and undergo nucleophilicaddition reactions with a variety of reagents.

• Nucleophilic addition with 2,4-dinitrophenylhydrazine followed by elimination ofwater (also described as condensation) produces 2,4-dinitrophenylhydrazonesthat have characteristic melting points useful in the identification of unknownaldehydes and ketones.

• Carboxylic acids can be prepared by the oxidation of primary alcohols or aldehydesand by the hydrolysis of esters, nitriles or amides.

• Carboxylic acids can react to form salts, esters and amides and can be reducedwith lithium aluminium hydride to form primary alcohols.






Website






89

Topic 5

Amines

Contents

5.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 90

5.2 Naming and classification . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 90

5.3 Physical properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 95

5.4 Chemical properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 97

5.5 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100

5.6 Functional groups: summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101

5.7 Resources . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103

5.8 End of Topic test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103


Before you begin this topic, you should be able to:

• identify the functional group present in amines and relate this to ammonia and itsproperties;

• describe the formation of amides and polyamides from amines and carboxylicacids. (Higher Unit 2);

• relate the strength of nylon polymers to the hydrogen bonding between chains.(Higher Unit 2).

Learning Objectives


• classify amines as primary, secondary or tertiary and name them according toIUPAC rules;

• explain why the boiling points of primary and secondary amines are higher thanalkanes of comparable relative formula mass and explain why the lower aminesare soluble in water;

• explain how amines can act as proton acceptors and describe some reactionsresulting from their basic nature.

90 TOPIC 5. AMINES

5.1 Introduction

The amines are a family of nitrogen-containing organic compounds derived fromammonia. Proteins are condensation polymers made up of many amino acids moleculescontain the amine functional group and the polymers called polyamides are made frommonomers containing amine groups. Nylon is one versatile polyamide that has importantengineering uses related to its strength. This strength comes from the hydrogenbonding between the polymer chains. The monomer unit used to make nylon is 1,6-diaminohexane.

The amines with low molecular mass like methylamine have a fishy smell.

Amines are released by all decaying animal flesh and this macabre aspect of aminechemistry is reflected in the names of two diamines:

Cadaverine (from cadaver-a dead body) NH2(CH2)5NH2.

Putrescine (from putrefy - to decay with an odour) NH2(CH2)4NH2.

The physical and chemical properties of the simpler members of the amines resemblethat of ammonia, but are modified by the presence of alkyl groups. Most of the propertiesare due to the lone pair of electrons on the nitrogen atom. The illustration showsammonia and a typical amine (trimethylamine).

5.2 Naming and classification�

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Learning Objective

To be able to classify amines as primary, secondary or tertiary and name themaccording to IUPAC rules.

Amines are named according to IUPAC rules. The system is quite complex however,and amines can be named in several ways depending on their structure. Either the ’e’from the parent alkane is removed and the suffix ’amine’ added, or the ’amine’ is addedto the name of the substituent group (Figure 5.1).


5.2. NAMING AND CLASSIFICATION 91

Figure 5.1

The prefix di- or tri- is sometimes necessary if there are substituent groups with thesame name, remembering to list the substituents alphabetically (Figure 5.2 (a)).

If the amine group is a branch from the main chain, a number is used to describe itsposition and the prefix amino- is used. Although IUPAC rules suggest that2-aminobutane (Figure 5.2(b)) should be named 1-methylpropylamine, texts like SaltersAdvanced Chemistry and A-Level Chemistry (Ramsden), use 2-aminobutane.To avoidconfusion with textbooks, the name 2-aminobutane is preferred. The website listed inthe resources section provides further detail.

Amino groups rank low in seniority in compounds with multiple groups, so in these cases,the prefix amino is used, with the compound being classed as belonging to the familywith the senior functional group. See Figure 5.2(c); in this case, an alcohol.

Figure 5.2


Q1: Name this compound:

a) methylamineb) dimethylaminec) diethylamined) 2-butylamine

Q2: Name this compound:


92 TOPIC 5. AMINES

a) methylethylamineb) propylaminec) propan-2-amined) 2-aminopropane

Amines can be classified as primary (1o), secondary (2o) or tertiary (3o), depending onthe number of carbon atoms connected to the nitrogen (Figure 5.3).

Figure 5.3: Primary, secondary and tertiary amines

Amines are regarded as organic compounds based on ammonia. When one hydrogenis replaced by an alkyl or aryl (benzene related) group, the amine is primary. Two alkylor aryl groups make the amine secondary, and three groups make it tertiary. The alkylor aryl groups may be the same or different. The simplest aryl amine has one hydrogenreplaced by benzene and is commonly named aniline. In this Topic, as in common withmost text books, the name phenylamine (Figure 5.4) will also be used for this compound.

Figure 5.4: Phenylamine

In summary:


5.2. NAMING AND CLASSIFICATION 93

Class General Formula Functional group

primary 1o RNH2

secondary 2o RR’NH

tertiary 3o RR’R”N

Classification and naming of amines

15 min

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Learning Objective

To be able to classify amines as primary, secondary or tertiary and name themaccording to IUPAC rules

A drag and drop version of the following exercise is available on the web sitesummarising the naming and classification of amines

Match up the structures (shown as (a), (b) and (c) in Figure 5.5), with the availablespaces and then match names to all the structures. Lastly, classify the amines asprimary (1Æ), secondary (2Æ), or tertiary (3Æ).


94 TOPIC 5. AMINES

Figure 5.5: Amine exercise

These questions will guide you through part of the exercise.

Q3: Which of these structures is tertiary: a, b, or c?

Q4: What name would you give to the tertiary structure you selected?

a) ethylamineb) diethylmethylaminec) 2-aminopropane



d) ethyldimethylamine

Q5: Classify diethylamine.


Q6: What name and class would you give to the structure in the lower left box?

a) ethyldimethylamine, tertiaryb) methyldiethylamine, tertiaryc) ethyldimethylamine, secondaryd) methydiethylamine, secondary

Amines are named according to IUPAC rules and can be classified as primary,secondary or tertiary, according to the number of carbon atoms connected to thenitrogen.


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Learning Objective

To explain why the boiling points of primary and secondary amines are higher thanalkanes of comparable relative formula mass.

To explain why the lower amines are soluble in water.

The physical properties of the amines such as melting point, boiling point and solubilityare largely influenced by the presence of polar N-H bonds and their capability of forminghydrogen bonds with other amines or with water. Primary and secondary aminesform hydrogen bonds readily (Figure 5.6) but tertiary amines have no N-H bonds andtherefore cannot associate with each other. The only intermolecular forces present in apure tertiary amine are van der Waals attractions.

Figure 5.6


96 TOPIC 5. AMINES

Examine the boiling point data (Table 5.1), comparing amines with alkanes ofcomparable relative formula mass.

Table 5.1: Amines and alkanes

AmineBoiling point

/CoAlkane of equivalent

massBoiling point

/Co

(a) CH3NH2 -7.5 (e) CH3CH3 -89(b) (CH3)2NH 7.5 (f) CH3CH2CH3 -42(c) (CH3)3N 3 (g) CH3CH(CH3)CH3 -12(d) CH3CH2CH2NH2 49 (h) CH3CH2CH2CH3 -1

Compare the isomeric amines (c) and (d). These have identical relative formula masses.

Q7: Which amine has the higher boiling point?

a) (c)b) (d)

Q8: Which amine has the stronger intermolecular bonds?

a) (c)b) (d)

Q9: Which amine is tertiary?

a) (c)b) (d)

Q10: Which class of amine tends to have lower boiling points because the moleculescannot associate by hydrogen bonding?

Q11: Compare amine (a) with alkane (e) which has approximately the same formulamass (Table 5.1).

Which has the higher boiling point?

a) (a)b) (e)

Q12: Which has the stronger intermolecular bonds?

a) (a)b) (e)

Q13: Which associate with hydrogen bonds as well as van der Waals bonds?

a) (a)b) (e)

Learning Point

Primary and secondary amines, but not tertiary amines, associate by hydrogenbonding. As a result, they have higher boiling points than isomeric tertiary amines andalkanes with comparable relative formula masses.


5.4. CHEMICAL PROPERTIES 97

Water is able to form hydrogen bonds with all amines, including tertiary amines.Because of this, amines with small alkyl groups are soluble. The larger the alkyl groups,the less soluble the amine, because large groups disrupt the hydrogen bonding withwater (Figure 5.7).

Figure 5.7: Hydrogen bonding in aqueous amines

Learning Point

Amine molecules can form hydrogen bonds with water molecules. This explains theappreciable solubility of the lower amines in water.

5.4 Chemical properties�

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Learning Objective

To explain how amines can act as proton acceptors and describe some reactionsresulting from their basic nature.

Basicity

The chemical properties of the amines are similar to the properties of ammonia and arestrongly influenced by the lone pair of electrons on the nitrogen atom.

The lone pair of electrons gives amines the ability to behave as proton acceptors, i.e. asbases. The lone pair forms a dative covalent bond in which both electrons come fromthe same atom (the nitrogen).

When an amine dissolves in water, an alkaline solution is formed as the reactionproduces an ammonium ion and a hydroxide ion.

In the example shown, methylamine dissolves to form the methylammonium ion andhydroxide (Figure 5.8). The equilibrium lies well to the left so methylamine is a weakbase and the solution is alkaline.


98 TOPIC 5. AMINES

Figure 5.8: Methylamine and water

One of the most convenient ways of measuring the basicity of an amine is to look at thepKa value of the conjugate acid (Figure 5.9). The higher the pKa of the conjugate acid,the stronger the base. Look back to Topic 4.4 of Unit 2 if revision is needed.

Figure 5.9

Table 5.2

Base Conjugate acid (ion) pKa

ammonia ammonium 9.3ethylamine ethylammonium 10.7

diethylamine diethylammonium 11.1phenylamine phenylammonium 4.6

Answer the following questions:

Q14: In general, are the alkylamines shown in the table (Table 5.2) weaker or strongerbases than ammonia?

Q15: Is phenylamine a weaker or stronger base than ammonia?

Q16: Which of these is the most likely pH of a phenylamine solution?

a) 2b) 5c) 8d) 14

Salt formation

When a bottle of concentrated hydrochloric acid is opened near a bottle of concentrated


5.4. CHEMICAL PROPERTIES 99

methylamine, dense white fumes of methylammonium chloride are produced. Themethylamine acts as a base, the hydrogen chloride as an acid and methylammoniumchloride is the salt produced (Figure 5.10).

Opening a bottle of concentrated methylamineand concentrated hydrochloric acid close to eachother allows the fumes from each to mix in the airaround the bottles. The reaction produces awhite cloud of tiny crystals of methylammoniumchloride. A neutralisation reaction has takenplace.

Figure 5.10: Methylamine and acid equation

Answer these questions:

Q17: The product of the reaction is a salt. What type of bonding is present in all salts?

Q18: If the amine used was ethylamine, what name would the salt have?

Amines react with other aqueous mineral acids and also with carboxylic acids to formsalts (Figure 5.11). The reaction with a carboxylic acid is often the first stage in theproduction of amides. The ammonium salt of the carboxylic acid is heated to producethe amide.

In this case (Figure 5.11), the methylammonium ethanoate will decompose on heatingto form a secondary amide. ( see Topic 4.2 of Unit 3 on Amide formation).


100 TOPIC 5. AMINES

Figure 5.11: Salt formation

5.5 Summary• Amines can be considered to be derivatives of ammonia (NH3), whose properties

are closely associated with the lone pair of electrons on the nitrogen and thepossibilities for hydrogen bonding.

• Amines can be classified as primary if there is one carbon containing groupattached to the nitrogen, secondary if there are two groups and tertiary if thereare three. The names of animes are governed by IUPAC rules, although in manycases the names used do not conform to these rules.

• Hydrogen bonding is possible between amines which have an N-H bond, i.e.primary and secondary amines. This leads to higher boiling points than thoseof isomeric tertiary amines (which have no N-H bonds) and higher boiling pointsthan alkanes with comparable relative formula masses.

• The presence of the nitrogen lone pair on amines of all classes allows hydrogenbonding with water molecules. This is responsible for the solubility of the shorterchain amines.

• The lone pair of electrons can act as a proton acceptor (a base) and, in water,produces hydroxide ions that lead to amines behaving as weak alkalis.

• One of the reactions that amines undergo when acting as a weak base is to formsalts with a range of mineral or carboxylic acids.


5.6. FUNCTIONAL GROUPS: SUMMARY 101

5.6 Functional groups: summary�

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Learning Objective

To recognise the characteristic chemical sections of an organic molecule that classifyits chemical family

Recognising functional groups

5 min

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Learning Objective

To give practice at recognising functional groups

A version of this drag and drop exercise is available on the web site.

The diagram contains all the classes of organic compound which have been introducedso far. Copy the diagram (Figure 5.12), or take a photocopy and draw structures foreach of the functional groups in the appropriate boxes. Then answer the questionswhich follow.

Figure 5.12

Look carefully at the structure of vanillin, whichprovides the flavour in vanilla.


102 TOPIC 5. AMINES

Q19: Which functional group is labelled A?

Q20: Which functional group is labelled B?

Q21: Which functional group is labelled C?

Look carefully at the structure of aspirin,a painkiller.

Q22: Which functional group is labelled A?

Q23: Which functional group is labelled B?

Now, look carefully at the structure oftestosterone, a male sex hormone.

Q24: Which of the following statements is true for testosterone?

a) It is a saturated alcoholb) It is an unsaturated aldehydec) It is a saturated ketoned) It is an unsaturated ketone

Q25: Which of these functional groups does not contain a carbonyl group?

a) esterb) etherc) ketoned) aldehyde


5.7. RESOURCES 103

5.7 Resources• Chemical Storylines: Salters Advanced Chemistry, Heinemann

ISBN 0-435-63106-3

• Chemical Ideas: Salters Advanced Chemistry, Heinemann, ISBN 0-435-63105-5

• Chemistry in Context: Hill and Holman , Nelson ISBN 0-17-438401-7


• A-level chemistry: E.N.Ramsden, Stanley Thornes ISBN 0-85950-154-X

• General Chemistry: Ebbing, Houghton Mifflin, ISBN 0-395-74415-6

• Higher Still Support: Advanced Higher Chemistry- Unit 3:Organic Chemistry. Learning and Teaching Scotland, ISBN 1-85955-873-9

• Nomenclature in the IUPAC system can be investigated at:http://www.acdlabs.com/iupac/nomenclature/




http://www.acdlabs.com/iupac/nomenclature/

104 TOPIC 5. AMINES


105

Topic 6

Aromatics

Contents

6.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 107

6.2 Benzene structure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 107

6.3 Benzene reactions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 110

6.3.1 Halogenation (attack by a halogen) . . . . . . . . . . . . . . . . . . . . . 111

6.3.2 Nitration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 112

6.3.3 Sulphonation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 114

6.3.4 Alkylation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 114

6.4 Acidity of phenol and basicity of phenylamine (aniline) . . . . . . . . . . . . . . 117

6.5 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121

6.6 Resources . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121

6.7 End of Topic test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 122



• describe the bonding in ethene in terms of sigma and pi bonds (Topic 3.2);

• describe benzene as the simplest aromatic compound with distinctive structureand properties (Higher Unit 2);

• explain the stability of benzene and its resistance to addition reactions on the basisof delocalised electrons (Higher Unit 2);

• explain that aromatic compounds can be made by substituting the hydrogen atomsin benzene and that a benzene ring which has one hydrogen substituted byanother group is known as the phenyl group ( -C6H5 ) (Higher Unit 2);

• Explain that using benzene as a feedstock yields a range of consumer products(Higher Unit 2).

Learning Objectives


• describe the stability of aromatic hydrocarbons like benzene in terms of sp2

hydridisation, sigma and pi bonding and electron delocalisation;

106 TOPIC 6. AROMATICS

• explain how benzene resists addition reactions but undergoes electrophilicsubstitutions, giving examples of reactions and describing the mechanisms;

• describe how the delocalised electrons in the phenyl group explain the acidityof phenols and the basicity of aromatic amines compared with their aliphaticequivalents.



6.1 Introduction�

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Learning Objective

To revise some of the main points previously learned in Higher grade.

The word ’aromatic’ was originally applied to hydrocarbons which had a sweet odour,but is now used to classify compounds with benzene-like structures (C6H6) (Figure 6.1).Benzene has a distinctive structural formula and stable nature due to the delocalisedelectrons in the ring. Compounds that have benzene-like structures are quite differentfrom aliphatic structures (Figure 6.1). The systematic name for the family of alkylsubstituted aromatic hydrocarbons is the arenes.

Removing one hydrogen leads to the formation of the phenyl group, C 6H5-, present inmany consumer products such as aspirin, polystyrene, dyestuffs, perfumes and fibres.

Figure 6.1

Despite the fact that benzene is unsaturated, its reactivity is unlike that of the alkenes oralkynes. These unsaturated families undergo addition reactions (see Topic 3.2 section1.5) whereas benzene resists addition reactions but undergoes electrophilic substitutionreactions.

6.2 Benzene structure�

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Learning Objective

To describe the bonding in benzene in terms of sp2 hybridisation, sigma and pi bondsand electron delocalisation and to explain how this can account for its unexpectedstability.

Although known to have a molecular formula C6H6, the structure of benzene remaineda source of controversy for many years until Friedrich Kekulé proposed a hexagonal ringwith alternating double and single bonds. The two possible arrangements are identicaland he proposed that the benzene ring oscillated between the two structures(Figure 6.2(b) and (c)).



Figure 6.2: The structure of benzene

The delocalisation occurs because benzene is a flat symmetrical hexagon with the sixcarbons sp2 hybridised (see Topic 3.2 section 1.3). The carbons are bonded to eachother and to their hydrogen atoms by a sigma (�) bond and each carbon has a p orbitalperpendicular to the plane of the six membered ring. Each p orbital contains a single (�)electron and overlaps sideways with both neighbouring orbitals to form a pi-bond. Thisresults in a doughnut shaped region of negative charge above and below the plane ofthe ring. This is described as a ’delocalised’ �- electron system because the electronsare evenly distributed. Each carbon to carbon bond is neither single nor double, andbenzene is often shown as in Figure 6.2 (d).

Structures Figure 6.2 (b) and (c) show ’extreme’ or ’localised’ arrangements with doubleand single bonds in definite regions. These are called resonance forms of benzeneand Figure 6.2 (d) is a hybrid of these. Kekulé’s ideas of the two structures thereforeanticipated the idea of resonance that was introduced sixty years later. Reactionpathways in benzene chemistry are often shown using the localised arrangementsbecause bond making and breaking are shown more clearly.

A model of benzene gives some idea of the arrangement. (Table 6.1).

Table 6.1

This model of benzene shows the’delocalised’ �- electron system aboveand below the hexagon of carbons.


6.2. BENZENE STRUCTURE 109

Aromatic bonding

10 min

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Learning Objective

To describe the bonding in benzene in terms of sp2 hybridisation, sigma andpi bonding and electron delocalisation and explain how this can account for itsunexpected stability

View the animation of the benzene structure on the website and then answer thequestions that follow.

These pictures (Figure 6.3) show the p orbitals (left) and the ’delocalised’ �- electronsystem (right) in benzene. Use these pictures and the information given previously toanswer the questions.

Figure 6.3: Aromatic bonding

Q1: What word describes the type of bond forming the hexagonal framework?

Q2: What word describes the type of bond formed above and below the framework?

Q3: What word describes the type of bond holding the hydrogens in place?

Q4: How many electrons does each p-orbital contain?

a) 1b) 2c) 3d) 6

Q5: How many electrons are contained in total within the doughnut shaped regions?

a) 1b) 2c) 3d) 6



Q6: What type of hybridisation occurs in benzene?

a) spb) sp2

c) sp3

d) sp4

The cyclic delocalisation of the electrons in benzene makes the molecule more stablethan would be expected with a Kekulé type of structure. The term aromatic has cometo mean any system that contains a ring of atoms stabilised by delocalised � electrons.

Bonding in benzene can be described in terms of sp2 hybridisation, sigma and pi bondsand electron delocalisation. This can explain its unexpected stability.

6.3 Benzene reactions�

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Learning Objective

To explain how benzene resists addition reactions but undergoes electrophilicsubstitutions, giving examples of reactions and describing the mechanisms.

Benzene tends to undergo electrophilic substitution reactions. The stable nature of thebenzene ring system gives benzene the tendency to undergo reactions in which thearomatic character is preserved. The regions of high electron density above and belowthe hexagon attract reagents with a positive charge (electrophiles).

The reaction which takes place is, in the first stage, similar to the electrophilic attack onethene (see Topic 2 section 2.1.5), but differs in the second stage.

The animation on the website shows the general mechanism for electrophilicsubstitutions on benzene with the reagent E+ representing an electrophile with eithera whole or a partial positive charge. This is shown in Figure 6.4.

Figure 6.4: Electrophilic substitution

The electrophile attacks one of the carbons atoms in the ring, forming an intermediateion with a positive charge that can be stabilised by delocalisation around the ring. Theintermediate ion then loses H+ to regain aromatic character and restore the system.

Four specific examples of electrophilic attack on benzene will now be considered.

When writing reaction mechanisms for aromatic systems, it is often useful to representthem as Kekulé structures (the localised arrangements Figure 6.2 (b) and (c)); this will


6.3. BENZENE REACTIONS 111

be done in the four examples shown. It should always be remembered however that thisis not a true picture of the benzene molecule.

6.3.1 Halogenation (attack by a halogen)�

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Learning Objective

To describe the mechanism involved in the electrophilic substitution of benzene by ahalogen in the presence of a suitable catalyst.

Benzene undergoes electrophilic substitution with bromine or chlorine if a catalyst likeiron(III) bromide, iron(III) chloride or aluminium(III) chloride is present.

The catalyst polarises the halogen molecule (here it is bromine) by accepting a pair ofelectrons from one atom and creating an electrophilic centre on the other atom(Figure 6.5).

Figure 6.5: Bromine electrophile production

This partially positive bromine atom attacks one of the carbons on the benzene andcreates a carbocation which is stabilised by delocalisation on the ring. The intermediatethen loses a hydrogen ion by heterolytic fission from the benzene, regaining thearomatic character and forming bromobenzene. The iron(III) bromide is regenerated(Figure 6.6).The same mechanism applies if chlorine is used with a suitable catalyst.

Figure 6.6: Electrophilic substitution of bromine on benzene

Answer these questions concerning the similar reaction of chlorine with benzene in thepresence of aluminium(III) chloride.



Q7: What kind of reaction would take place?

a) Electrophilic additionb) Nucleophilic substitutionc) Electrophilic substitutiond) Nucleophilic addition

Q8: What is generally the best name for the product?

a) Chlorophenylb) Chlorobenzenec) Benzene chlorided) Phenylchloride

Q9: What type of fission does the chlorine molecule undergo in this reaction?

a) Nuclearb) Electrolyticc) Homolyticd) Heterolytic

Q10: What type of organic ion is created as an intermediate in this reaction?

6.3.2 Nitration�

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Learning Objective

To describe the mechanism involved in the electrophilic substitution of benzene by anitronium ion leading to the formation of nitrobenzene.

Benzene will react with nitric acid when a mixture of concentrated nitric acid andconcentrated sulphuric acid (known as a nitrating mixture) is used and the temperatureis kept below 55ÆC (Figure 6.7).

The nitrating mixture generates the nitronium ion, NO2+.

Figure 6.7: Nitrating mixture

The nitronium ion attacks the benzene ring and creates an intermediate ion that isstabilised by the same mechanism as in the halogen reactions.

The ring then regains its aromatic stability by the removal of a hydrogen ion and theproduct molecule nitrobenzene is formed (Figure 6.8).



Figure 6.8: Electrophilic substitution of nitronium on benzene

This is an important industrial reaction since nitrobenzene is used to produce aniline(phenylamine) (Figure 6.14), which is important in the dyestuff industries.

If the temperature is allowed to rise above 55ÆC, further substitution of the ring can takeplace and small amounts of the di- and tri- substituted compounds can result(Figure 6.9).

Figure 6.9: Di- and trinitrobenzene structures

Answer these questions.

Q11: What type of reactant is the nitronium ion?

a) carbanionb) nucleophilec) carbocationd) electrophile

Q12: What type of intermediate is formed?

a) free radicalb) carbocationc) carbaniond) nitronium anion

Q13: Describe how the intermediate is stabilised. Try writing a sentence beforechecking your answer.



Q14: What name would you suggest for structure (x)? (Figure 6.9).

a) 1,1-dinitrobenzeneb) 1,2-dinitrobenzenec) 1,3-dinitrobenzened) 1,4-dinitrobenzene

6.3.3 Sulphonation�

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Learning Objective

To descibe the mechanism involved in the electrophilic substitution of benzene bysulphuric acid leading to the formation of benzenesulphonic acid.

Benzene will react with concentrated sulphuric acid if the reactants are heated togetherunder reflux for several hours. If fuming sulphuric acid is used (sulphuric acid enrichedwith sulphur trioxide) under cold conditions -SO3H substitutes onto the ring. Thissuggests that the active species is sulphur trioxide. The sulphur atom in the sulphurtrioxide molecule carries a partial positive charge and can attack the benzene ring. Themechanism is the same as that shown for nitration (Figure 6.10) .

Figure 6.10: Sulphonation of benzene

Many detergents contain salts of alkylbenzenesulphonic acid that are produced fromarene sulphonic acids. They are an important industrial chemicals.

Q15: In this reaction is the sulphur trioxide acting as a nucleophile or a electrophile?

Q16: Does it react with the benzene in an addition or a substitution reaction?

6.3.4 Alkylation�

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Learning Objective

To describe the mechanism involved in the electrophilic substitution of benzene by analkyl group in the presence of a suitable catalyst.

The aluminium chloride catalyst mentioned in the halogenation reactions can be used toincrease the polarisation of halogen containing organic molecules like halogenoalkanes.This allows electrophilic carbon atom to attack the benzene ring and builds up side-chains. The reaction is called a Friedel-Crafts reaction, after the scientists whodiscovered it.



Figure 6.11: Halogenoalkane and aluminium chloride

The catalyst increases the polarity of the halogenoalkane producing the electrophiliccentre that can attack the benzene ring (Figure 6.11) . The carbocation formed isstabilised by delocalisation and the intermediate so formed regains its stability by lossof a hydrogen ion forming an alkylbenzene (Figure 6.12).

Figure 6.12: Alkylation of benzene

Answer these questions.

Q17: What word describes the role of the aluminium(III) chloride?

Q18: What type of fission does the halogenoalkane undergo in this reaction?

a) nuclearb) homolyticc) heterolyticd) electrolytic

Q19: Name the chloroalkane necessary to produce ethylbenzene in a Friedel-Craftsreaction.

Benzene reactions summary

10 min

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Learning Objective

To give examples of electrophilic substitution reactions on a benzene system, namethe reactants, products and reagents used and describe the mechanism.

Visit the online version of this Topic to try the drag and drop exercise summarising themain electrophilic substitution reactions on the benzene ring system

Answer these questions about Figure 6.13 showing a summary of the main electrophilicsubstitution reactions on the benzene ring system.



Figure 6.13: Benzene electrophilic substitution

Q20: What reagent would be used in reaction 1?

a) Cl2/AlCl3b) RCl/AlCl3c) Br2/FeBr3

d) RCl/H2SO4

Q21: What reagent would be used in reaction 2?

a) Cl2/AlCl3b) RCl/AlCl3c) Br2/FeBr3

d) RCl/H2SO4

Q22: What name would you give to the product in reaction 3? (carried out at roomtemperature).


6.4. ACIDITY OF PHENOL AND BASICITY OF PHENYLAMINE (ANILINE) 117

Q23: What name would you give to the product in reaction 4?

Q24: What name would you give to the product in reaction 6 if the reagent used in 5was iron(III) bromide and bromine?

Benzene undergoes electrophilic substitution reactions. These include chlorination andbromination to produce chlorobenzene and bromobenzene respectively, nitration toproduce nitrobenzene, sulphonation to produce benzenesulphonic acid and alkylationto produce alkylbenzenes.

6.4 Acidity of phenol and basicity of phenylamine (aniline)�

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Learning Objective

To describe the effect that delocalisation of charge over the benzene ring has on theacidity of phenols and the basicity of aromatic amines compared with their aliphaticequivalents.

Phenol

The delocalised electrons present in a phenyl ring have a strong influence onthe properties of functional groups attached to the aromatic system. Phenol andphenylamine (also called aniline) (Figure 6.14) are two compounds which show modifiedpKa values when compared to aliphatic molecules with the same functional groups.

Figure 6.14

Table 6.2 shows a comparison of phenol and ethanol in terms of the pKa value and thereaction with sodium. In each case the sodium is reacting with hydrogen ions to producehydrogen gas. The lower the pKa of an acid, the stronger the acid.(See Unit 2 Topic 4 section 4.3 if revision is needed).

Table 6.2

Alcohol pKaReaction with sodium

ethanol 18.0slow release of

hydrogen

phenol 9.9fast release of

hydrogen



(Note: the phenol reacting with sodium is dissolved in a little solvent, as phenol isnormally solid.)

Q25: Which reacts most rapidly with sodium?

a) ethanolb) phenol

Q26: Which contains the highest concentration of hydrogen ions?

a) ethanolb) phenol

Q27: Does the pKa value given back up your answer?

a) yesb) no

Q28: The pKa value for water is 15.7. Which of these is the strongest acid?

a) ethanolb) phenolc) water

Phenol is acting as a stronger acid than ethanol because ionisation (Figure 6.15)produces the phenoxide ion. The negative charge on the oxygen can be delocalisedover the aromatic � system.

Figure 6.15: Phenol dissociation

The resonance structures that are possible for the phenoxide ion stabilise it and make itless likely to accept a hydrogen ion (Figure 6.16). The equilibrium position for the phenoldissociation therefore lies more to the right.


6.4. ACIDITY OF PHENOL AND BASICITY OF PHENYLAMINE (ANILINE) 119

Figure 6.16

Compared with ethanol, which would have to carry the negative charge withoutdelocalisation when it ionised, phenol is therefore a stronger acid than the aliphaticalcohols. Although weak, phenol can undergo neutralisation reactions and reacts withactive metals in the same way as other acids to give a salt and hydrogen gas.

Phenylamine (aniline)

Phenylamine (also called aniline) is an aromatic amine and, like other amines, is basic.

It is able to accept a hydrogen ion and form the phenylammonium ion (Figure 6.17).

Figure 6.17: Phenylamine acting as a base

The ability of phenylamine to act as a base, and a comparison with other amines andammonia, is illustrated in Table 6.3 which shows values of pKa for the conjugate acids.The lower the value of pKa of the conjugate acid, the weaker the base.(See Unit 2 Topic 4 section 4.4 if revision is needed).

Table 6.3

Base Conjugate acid (ion) pKa

ammonia ammonium 9.3ethylamine ethylammonium 10.7

diethylamine diethylammonium 11.1phenylamine phenylammonium 2.7



Q29: Which conjugate acid ion has the highest value of pKa?

a) ammoniumb) ethylammoniumc) diethylammoniumd) phenylammonium

Q30: Which conjugate acid ion has the lowest value of pKa?

a) ammoniumb) ethylammoniumc) diethylammoniumd) phenylammonium

Q31: Which base is the strongest?

a) ammoniab) ethylaminec) diethylamined) phenylamine

Q32: Which base is the weakest?

a) ammoniab) ethylaminec) diethylamined) phenylamine

Phenylamine(aniline) is considerably less basic than any of the other substances in thetable. This is because the lone pair of electrons on the nitrogen atom is delocalisedaround the aromatic ring. This lone pair is therefore less available to act as a baseand combine with a hydrogen ion. The delocalisation can be represented by theseresonance structures (Figure 6.18).

Figure 6.18


6.5. SUMMARY 121

The equilibrium shown in (Figure 6.17) therefore lies to the left.

6.5 Summary• The unexpectedly stable structure of benzene can be described in a variety of

ways:

1. The Kekulé description involving two different hexagonal forms with doublebonds.

2. A description involving sp2 hybridisation resulting in sigma and pi bondingand delocalisation of the p orbital electrons into a pi bond system above andbelow the hexagon of carbon atoms.

• All of the descriptions of the bonding in benzene can be useful in understandingthe stability of benzene and the reaction mechanisms involved in electrophilicsubstitution reactions.

• Electrophilic substitution in benzene involves the attack of a positive or partlypositive species on the delocalised electron cloud and, in the first instance, theformation of monosubstituted benzene compounds.

• The presence of delocalised electrons creates stabilised carbocationintermediates in the phenyl group during electrophilic substitution.

• The presence of delocalised electrons in the phenyl group can also be used toexplain:

1. The stronger acidic nature of phenol compared with aliphatic alcohols.

2. The weaker basic nature of phenylamine(aniline) compared with otheraliphatic amines.




• Organic Chemistry: J. McMurray, Brooks/Cole Publishing, ISBN 0-534-16218-5


• Chemical Ideas: Salters advanced chemistry, Heinemann, ISBN 0 435 630148

Website for representing molecules in different ways, for hybridisation and reactionmechanisms:





Website for IUPAC naming of organic compounds:

• http://www.acdlabs.com/iupac/nomenclature/





123

Topic 7

Stereoisomers

Contents

7.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 124

7.2 Stereoisomerism . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 125

7.2.1 Geometric isomerism . . . . . . . . . . . . . . . . . . . . . . . . . . . . 126

7.2.2 Optical isomerism. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 135

7.2.3 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 144

7.3 Resources . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 144

7.4 End of Topic test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 145



• define isomers as organic compounds which have the same molecular formula butdifferent structural formulae (Standard Grade).

Learning Objectives


• state that in stereoisomers, the same atoms are bonded together in the same orderbut have a different arrangement in space making them non-superimposable;

• describe geometric isomerism (cis-trans isomerism) in terms of lack of free rotationaround a bond;

• describe some differences in physical and chemical properties between geometricisomers;

• explain how optical isomerism occurs in compounds with four different groupsarranged around a carbon atom and that this usually leads to chiral moleculeswhich are non-superimposable mirror images of each other;

• state that, apart from optical activity, optical isomers have identical physical andchemical properties except when in a chiral environment;

• state that in biological systems only one of a pair of optical isomer is usuallypresent.

124 TOPIC 7. STEREOISOMERS

7.1 Introduction�

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Learning Objective

To revise knowledge of isomers from S.G. and Higher and from earlier Topics.

Isomerism arises whenever there is more than one way to organise a given number ofatoms. It is very common in organic chemistry. Note that when talking about isomerswe are always referring to two or more molecules. One molecule is not an isomer byitself but can be an isomer of one or more other molecules.

Look at the molecules in Figure 7.1. The use of molecular models, in addition to theStudy Guide and the Scholar website, is strongly recommended at all stages of thisTopic.

A B

Figure 7.1: Isomeric molecules

Q1: To what class of compounds does A belong?

Q2: To what class of compounds does B belong?

These two substances are isomers. They both have the same molecular formula,C2H6O. However, the atoms in each are connected in a different order, C-O-C in oneand C-C-O in the other. Isomers like this are known as structural isomers

In this case, the two isomers belong to different homologous series but often thestructural difference is more subtle (Figure 7.2).

Figure 7.2: Structural isomers

Q3: What is the structural difference between these two isomers?


7.2. STEREOISOMERISM 125

Learning Point

Structural isomers have different physical and chemical properties and can evenbelong to different homologous series.

7.2 Stereoisomerism�

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Learning Objective

To state that in stereoisomers, the same atoms are bonded together in the same orderbut have a different arrangement in space making them non-superimposable.

In stereoisomerism, the molecules have the same molecular formula and the samestructural formula (the atoms are connected in the same order in each molecule).However, in each molecule, the atoms have a different three dimensional arrangementin space which makes them non-superimposable. This means that no matter how youtwist and turn the molecules, one isomer cannot fit exactly on top of the other.

There are two main types of stereoisomerism as shown in Figure 7.3.

Figure 7.3: Isomerism

In this Topic, we will concentrate on stereoisomerism in organic compounds, althoughthis phenomenon also occurs in other compounds such as transition metal complexes.



7.2.1 Geometric isomerism�

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Learning Objective

To describe geometric isomerism (cis-trans isomerism) in terms of lack of free rotationaround a bond.

To describe some differences in physical and chemical properties between geometricisomers.

Geometric isomerism can arise when there is lack of free rotation around a bond, oftena C=C bond.

Consider 1,2-dichloroethene. There are two geometric isomers (Figure 7.4).

Figure 7.4: Dichloroethene isomers

They exist because rotation about the C=C bond is very difficult.

In Topic 2, the C=C bond was described as consisting of a � bond and a � bond(Figure 7.5).



Figure 7.5: Carbon to carbon double bond

Rotation is restricted because this would involve breaking the � bond. The two geometricisomers are not superimposable.

Geometric isomers caused by restricted rotation around a bond are distinguished by

using the terms - ’cis’ and ’trans’:

• cis - both groups are on the same side of the double bond,

• trans - the groups are on opposite sides of the double bond (’trans’ means across).

Molecule (a) in Figure 7.4 is trans-1,2-dichloroethene, while molecule (b) in Figure 7.4is cis-1,2-dichloroethene. A similar situation occurs with alkenes containing 4 or morecarbon atoms, e.g.

Figure 7.6 Figure 7.7

3D molecules

2 minExperiment with the ’live’ 3D molecules available on the website.

Q4: Name the molecule in Figure 7.6.

Q5: Name the molecule in Figure 7.7.



Geometric isomerism can also arise in disubstituted cycloalkanes, e.g.

Figure 7.8

Q6: Name the molecule in Figure 7.8(a).

Q7: Name the molecule in Figure 7.8(b).

Physical properties

15 min

Look at the information in the table and answer the questions that follow.

Isomer mp ( ÆC) bp ( ÆC)

cis-but-2-ene -139 +4trans-but-2-ene -106 +1

cis-1,2-dichloroethene -80 +60trans-1,2-dichloroethene -50 +48

cis-pent-2-ene -151 +38trans-pent-2-ene -140 +37

Q8: Look at the melting points. In general, which isomer has the higher melting point?

a) cisb) trans

Q9: Which isomer has the stronger intermolecular forces?

a) cisb) trans

The differences in melting point for each pair of isomers can be explained in terms of theshapes of the molecules. This determines how closely the molecules can pack together.



Q10: What effect will the packing have on the strength of the intermolecular forces?

a) The closer the molecules pack, the weaker the van der Waals forcesb) The closer the molecules pack, the stronger the van der Waals forcesc) The closer the molecules pack, the weaker the repulsive forcesd) The further apart the molecules, the stronger the intermolecular forces

Q11: In general, molecules of which isomer can be more closely packed?

a) cisb) trans

The boiling points of the isomers of 1,2-dichloroethene are significantly different.Consider the types of bond within these molecules.

Q12: Which of these correctly describes the types of bond present?

a) pure covalent onlyb) polar covalent onlyc) pure covalent and polar covalentd) pure covalent and ionic

Q13: By looking at the shape, and considering the bonding within the molecules,suggest an explanation why the cis-isomer has a higher boiling point than thetrans-isomer.

Geometric isomers often display differences in physical properties such as melting pointand boiling point.

Sometimes geometric isomers can show differences in chemical properties althoughthis is much less common than for structural isomers. A simple example involves thegeometric isomers of butenedioic acid (Figure 7.9).

Figure 7.9: Isomers of butenedioic acid



Cis-butenedioic acid readily eliminates water on heating. The product is known as acyclic anhydride (Figure 7.10).

Figure 7.10: Dehydration of cis-butenedioic acid

On the other hand, trans-butenedioic acid cannot eliminate water under the sameconditions since the two carboxyl groups are on opposite sides of the double bond(Figure 7.9).

Case Study - Fats and edible oils

60 min

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Learning Objective

To study examples of relevance of geometric isomerism in consumer products suchas the health issues associated with cis and trans fatty acids

This case study illustrates the differences between fats and edible oils, including thedifferences in physical and chemical properties between cis and trans isomers ofunsaturated fatty acids. The biological importance of cis-fatty acids is also discussed.PLEASE NOTE: Detailed knowledge of the names, their structures and how they workdoes not have to be memorised.

Read through the case study. If you are given it as a homework exercise, follow theinstructions about preparation. If you are only interested in gathering information inorder to complete the ’Fats and oils’ exercise, you may skim through looking for thenecessary information.

Preparation

If you are making a presentation to a group then you will need to prepare:

• a hand-out information sheet.

• audio visual illustrations (blackboard or OHP or data projection).

• a short script for yourself.

Include within your presentation:

• the similarities in structure between fats and edible oils

• the differences between fats and edible oils



• differences in behaviour between saturated and unsaturated fatty acids

• the biological importance of unsaturated fats

Fats and oils

Fats, also called lipids, are the third main class of food type needed in the human diet,the others being proteins and carbohydrates. Oils are simply fats that are liquid at roomtemperature. In foods derived from animals, the main sources of fat are dairy produceand meat although most foods contain some fat. Some of the richer vegetable sourcesof dietary fat are nuts and seeds, soyabeans, olives and peanuts. Fats are an importantcomponent of our diet and at least a minimum intake is required. However, many healthproblems, particularly in the western world, are associated with an excessive intake offat.

The main functions of fat in the body are as an energy reserve and for insulation. Fatscan be burned to release energy when we need it and are not getting enough from thecarbohydrates in our diet. Fatty tissue around internal organs help to protect them fromtrauma and temperature change by providing padding and insulation. They also haveother uses. They are important in transporting other nutrients such as the vitamins A,D, E and K which are not water-soluble. Fats also form an essential part of the cellmembrane. Finally, they are also a source of essential fatty acids.

Structure of Fats

Triglycerides are by far the most common type of fat and make up about 95% of thelipids in food and in our bodies. All triglycerides have a similar structure (Figure 7.11).They are esters formed from the condensation of fatty acids with glycerol(propane-1,2,3-triol). Fatty acids are long, straight chain carboxylic acids which containan even number of carbon atoms, typically between 12 and 20 atoms.

Figure 7.11: Structure of a saturated fat

The three fatty acids present in a triglyceride are generally different, giving rise to a largenumber of possible combinations.

Fats can be classified as saturated or unsaturated according to the nature of the fattyacids present. The fatty acids in turn can be classified as saturated, monounsaturatedor polyunsaturated depending on the presence and number of C=C bonds (Table 7.1).



Table 7.1: Examples of fatty acids

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Physical properties

Saturated fats have higher melting points than unsaturated fats. This can be explainedin terms of differences in the packing of the molecules in the solid state. The longhydrocarbon chains in saturated fat molecules can be packed quite closely togetherwhich maximises the van der Waals forces (Figure 7.11).

Figure 7.12: Structure of a cis-unsaturated fat

Unsaturated fats (oils) contain one or more C=C bond. Almost invariably the C=C bondshave a cis configuration. This introduces a ’kink’ into the hydrocarbon chain since thereis restricted rotation around this bond. This makes close packing much more difficult(Figure 7.12). The intermolecular forces are weaker and so unsaturated fats have lowermelting points than saturated fats.

Figure 7.13: Structure of a trans-unsaturated fat



It is worth pointing out that triglycerides made from trans unsaturated fatty acids canpack together much more closely (Figure 7.13) and so have higher melting points thantheir cis equivalents.

Essential fatty acids

About 40 different fatty acids occur naturally. Table 7.1 shows some of the mostcommon. The most important essential fatty acids are linoleic, linolenic and arachidonicacids, collectively known as vitamin F (Figure 7.14). These are all polyunsaturated fattyacids that are needed by the body. However, they cannot be synthesised from othercompounds in the human body, although if there is sufficient linoleic acid it can beconverted to arachidonic acid. They must be obtained from vegetable oils in the diet.(The term ’essential’ is also used for those amino acids which must be obtained in thediet). Linoleic acid is found in most foods while linolenic acid is common in oily fish.

Figure 7.14: Essential fatty acids

These fatty acids are important for normal growth, especially of the blood vesselsand nerves, and to keep the skin and other tissues youthful and supple through theirlubricating properties.

They are also essential precursors in the biosynthesis of a group of compounds knownas prostaglandins. These are hormone-like substances present in small amounts in allbody tissues and fluids. They are all C20 carboxylic acids that contain a five-memberedring with two long side chains.

Figure 7.15: Prostaglandin

The basic structural similarity between arachidonic acid and the prostaglandin shownin Figure 7.15 is obvious. One step in the conversion involves the formation of a five



membered ring between carbons 8 and 12 in the fatty acid chain. If the configuration ofeither of the two central C=C bonds in arachidonic acid was trans, such a transformationwould be impossible. Prostaglandins have an extraordinarily wide range of biologicaleffects, including controlling blood pressure and controlling inflammation. They arealso involved in blood clotting, kidney function and the reproductive system and arethe subject of a great deal of research to produce new drugs.

Diet

Many health problems are associated with an excessive intake of fat in the diet. Levelsof fat intake are strongly linked to body weight and therefore obesity. Fats provide about42% of the calories in the average American diet. A diet in which about 25% or less oftotal calories is derived from fat would be healthier and help to reduce blood cholesterollevels and risk of blood and heart disease. The type of fat in the diet is also important.Increasing the proportion of unsaturates and polyunsaturates in the diet has been shownto produce a significant reduction in blood cholesterol levels.

Current suggestions for a healthy diet include:

• reduce the total amount of fat in the diet

• specifically reduce the saturated fat intake by reducing consumption of red meatsand dairy produce

• reduce intake of hydrogenated fat found in cooking oils and margarines

• raise the levels of polyunsaturated fats found in vegetable oils such as corn oil andsunflower oil as well as in most seeds and nuts

• eat more fish, particularly oily fish like herring and mackerel rather than red meat

• and of course, keep blood pressure under control, don’t smoke and exerciseregularly.

Summary exercise

Either copy the table below or photocopy it and then complete it. The words and phrasesin the wordbank will help. Some may be used more than once or not at all. Alternativelyvisit the website for a drag and drop version of this exercise.



7.2.2 Optical isomerism.�

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Learning Objective

To explain how optical isomerism can occur in compounds with four different groupsarranged around a carbon atom.

To explain that this leads to chiral molecules that are non-superimposable mirrorimages of each other.

In the same way as your left hand and your right-hand are mirror images of each other,many chemical compounds can exist in two mirror image forms. Right and left-handscannot be superimposed on top of each other so that all the fingers coincide and aretherefore not identical. A right hand glove does not fit a left hand and is said to be chiral.

The hands and the compounds have no centre of symmetry, plane of symmetry or axisof symmetry. Chirality arises from a lack of symmetry. Lack of symmetry is calledasymmetry.

Look at this picture (Figure 7.16) and answer the questions.



Figure 7.16

The lady is holding one hand in front of the mirror and one hand behind.

Q14: Which hand is the lady holding in front of the mirror?

a) leftb) right

Q15: Which hand appears to be in the mirror?

Q16: Which word describes the property exhibited by these two hands?

a) identicalb) symmetricalc) asymmetrical

Q17: Which of these objects could be described as chiral?

a) golf ballb) pencilc) shoed) methane molecule

Carbon atoms and symmetry.

Optical isomerism occurs in other compounds such as transition metal complexes. Inthis Topic however, we are concentrating on carbon based compounds.

The four single bonds around a carbon atom are arranged tetrahedrally. If there arefour different atoms or groups attached to the carbon, it is asymmetric. It can exist intwo isomeric forms. If you have access to molecular models or a kit from which to buildmodels, it is recommended that you build models of the two isomeric forms shown inthe first picture (Picture 1). Use a different coloured sphere for each of the four attachedgroups. This will help in understanding why they are not identical.



Picture 1 shows the two isomers side by side and an attempt to line them up identically isshown in Picture 4. The arrow shows one of the points of mismatch. It is not possible tosuperimpose one on the other. This exercise is developed in the activity ’Optical isomersof alanine’ in the next section of this Topic.

The tetrahedral carbon atom with four different groups is asymmetric. Moleculescontaining one or more asymmetric carbon atoms are usually (but not always) alsoasymmetric. Asymmetric atoms or molecules are described as chiral with the carbonatom being called the chiral centre. They differ from each other only in that theyare mirror images of each other and exhibit optical activity. Such isomers are calledenantiomers or optical isomers.The crystals of enantiomers are mirror images ofeach other. A mixture containing equal amounts of each enantiomer is known as aracemic mixture.

Learning Points

Optical isomerism can occur in compounds with four different groups arranged arounda carbon atom. This leads to chiral molecules that are non-superimposable mirrorimages of each other.

Optical isomers are stereoisomers since the atoms are connected in the same order butthe arrangement of the atoms in space is different.

7.2.2.1 Biological systems.

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Learning Objective

To explain how optical isomers are non-superimposable mirror images of each otherand that in biological systems only one optical isomer of each organic compound isusually present.

Optical isomerism is immensely important in biological systems. In most biologicalsystems only one optical isomer of each organic compound is usually present. Forexample, when the amino acid alanine is synthesised in the laboratory, a mixture of thetwo possible isomers (a racemic mixture) is produced (Figure 7.17). When alanine isisolated from living cells, only one of the two forms is seen.



Figure 7.17

Optical isomers of alanine

10 min

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Learning Objective

To be able to explain how optical isomers are non-superimposable mirror images ofeach other.

This activity involves 3-dimensional representations of amino acids with text andquestions. It will help you understand why only one optical isomer of each organiccompound is usually present in biological systems.

Visit the Scholar website to find an interactive activity in which you can manipulate thestructure of alanine.

This illustration (Figure 7.18) shows one of the enantiomers of alanine, drawn in adifferent way from Figure 7.17.

Figure 7.18

The carbon atom surrounded byfour different groups is the chiralcentre. As well as the hydrogenatom it has an amine group, acarboxyl group and a methyl group.

Q18: Compare Figure 7.18 with the structures shown in Figure 7.17. Which structureis it?

a) Structure 1b) Structure 2

If you have access to molecular models, you should build the molecule on the left ofPicture 1 by starting with a carbon and using different coloured spheres for the fourgroups. (The suggested colours would be red for carboxyl, blue for amine, yellow



for methyl and white for hydrogen.) Try to build the other enantiomer as the exerciseprogresses.

Picture 1 shows two isomers side by side and when one of them is taken away andreplaced with a mirror, the two images in Picture 2 look the same as in Picture 1.

In Picture 3, the isomer that was removed is placed behind the mirror, showing that it isidentical to the image. Picture 4 points out how the two isomers arenon-superimposable because even with the two substituent groups on the left of eachisomer lined up, the two arrowed groups do not match.


Q19: The molecules shown in Picture 4 are

a) superimposableb) non-superimposable

Q20: The two molecules are said to be

a) symmetricb) asymmetric



Q21: The two molecules can be said to be

a) identicalb) chiral

Optical isomers can occur in compounds with four different groups arranged around acarbon atom and this leads to chiral molecules which are non-superimposable mirrorimages of each other. In biological systems only one of a pair of optical isomer of eachorganic compound is usually present.

The proteins in our bodies are built up using only one of the enantiomeric forms ofamino acids. Only the isomer of alanine with structure 2 (Figure 7.19) occurs naturallyin organisms such as humans. Since amino acids are the monomers in proteins suchas enzymes, enzymes themselves will be chiral. The active site of an enzyme will onlybe able to operate on one type of optical isomer. An interesting example of this is theaction of penicillin, which functions by preventing the formation of peptide links involvingthe enantiomer of alanine present in the cell walls of bacteria. This enantiomer is notfound in humans. Penicillin can therefore attack and kill bacteria but not harm the humanhost, as this enantiomer is not present in human cells.

Figure 7.19: Alanine reflections

7.2.2.2 Polarised light

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Learning Objective

To explain that enantiomers have identical physical and chemical properties, exceptwhen in a chiral environment.

To describe the effect enantiomers have on plane polarised light.

Enantiomers behave in the same way in ordinary test tube reactions. Physical propertieslike melting, density and solubility are also identical. But enantiomers behave differentlywhen in the presence of other chiral molecules. For example, enantiomers can tastedifferently due to the chiral nature of the taste bud receptors in the tongue. Oneenantiomeric form of amino acids all taste sweet, the other isomer is sometimes bitter.Chemicals with different enantiomeric forms can smell different. One isomer of limonene(Figure 7.20) smells of oranges, the other of lemons.



Figure 7.20: Limonene enantiomers

Enantiomers have an opposite effect on plane polarised light. This is why they are saidto be optically active. An instrument called a polarimeter can detect this effect.

The polarimeter

10 min

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Learning Objective

To be able to explain how a polarimeter can be used to determine the effect anoptically active isomer has on plane polarised light.

Go to the website to see the animations that illustrate how a polarimeter works and theeffect optical isomers have on plane polarised light

Look at the diagrams and answer the questions which follow.

Figure 7.21: Polarised light

The waves of normal light vibrate in all directions at right angles to the direction of travel.Some substances will allow light vibrating in only a single plane to pass through them,producing ’plane polarised light’. See Figure 7.21. Polaroid sunglasses use this propertyto reduce glare.

Figure 7.22: Polarimeter

A compound which, when inserted into a beam of plane polarised light can rotate it sothat the light vibrates in a different plane, is said to be optically active. Enantiomers havethis property, since solutions of one kind rotate the plane polarised light in a clockwise



direction (measured as +xÆ), and the other enantiomer rotates plane polarised light byexactly the same amount in the opposite direction (Figure 7.22) (measured as -xÆ). Amixture containing equal amounts of each enantiomer (a racemic mixture) is opticallyinactive as the rotations cancel each other.

Q22: The function of the polariser in a polarimeter is to:

a) rotate normal lightb) produce normal lightc) rotate plane polarised lightd) produce plane polarised light

Q23: One enantiomer of a compound is found to give a reading of +13.5 Æ when insertedinto a polarimeter.

What reading would you expect the same amount of the other enantiomer to produce inthe polarimeter?

a) 0Æ

b) +13.5Æ

c) -13.5Æ

d) +27Æ

Q24: What reading would you expect a mixture of equal amounts of the twoenantiomers to produce?

a) 0Æ

b) +27Æ

c) -27Æ

d) 360Æ

Optical isomers (enantiomers) have identical physical and chemical properties, exceptwhen in a chiral environment. However, they have an opposite effect on plane polarisedlight. Mixtures containing equal amounts of both optical isomers are optically inactive.

Many medicines are produced as a mixture of enantiomers, only one of which ispharmacologically active, as it can prove costly to separate the isomers. One of theenantiomers of salbutamol used in the treatment of asthma, is 68 times more effectivethan the other (see Unit 3, Topic10, section 4). Great care has to be taken when usingdrugs with enantiomeric forms as this has led to tragedy in the past, specifically in thecase of thalidomide (Figure 7.23). A mixture of the isomers was used to treat nauseaduring pregnancy, and one enantiomer, which was thought to be inactive, turned outto cause damage to the unborn child. Many handicapped babies were born before thedrug was recognised as being responsible. Screening of pharmaceuticals has to bevery thorough. Regulations were tightened significantly after the thalidomide tragedy toensure that both enantiomeric forms of chiral drugs are tested.



Figure 7.23: Thalidomide

Mechanisms for substitution reactions

In Unit 3, Topic 2, the SN2 mechanism for nucleophilic substitution reactions ofhalogenoalkanes was described (Figure 7.24). The strongest evidence for thismechanism comes from the use of chiral compounds.

Figure 7.24: SN2 mechanism

If this mechanism occurs with a chiral compound, then the product will also be chiral butwith the configuration inverted (a bit like an umbrella being blown inside out by a strongwind).

Figure 7.25: SN1 mechanism, step1

Figure 7.25 shows the mechanism for the SN1 reaction which occurs in two steps.Figure 7.25 shows the first step in the SN1 mechanism. In the second step, a nucleophilewill attack the planar carbocation to form the product.



Q25: If the original halogenoalkane in this SN1 reaction is chiral, what will the productbe?

a) chiral with the same configuration as the originalb) chiral with inverted configurationc) a racemic mixtured) a non-chiral molecule

Q26: Explain your answer to the previous question.

7.2.3 Summary

• Structural isomers are molecules with the same molecular formula but differentstructural formula (in each molecule, the atoms are connected in a different order).

• In stereoisomers, the same atoms are bonded together in the same order but havea different arrangement in space making them non-superimposable.

• Geometric isomers are a type of stereoisomer that arise due to lack of free rotationaround a bond, often a C=C bond.

• Geometric isomers are labelled cis and trans, according to whether thesubstituents are on the same side or opposite sides of the bond. Theynormally display differences in some physical properties and sometimes chemicalproperties.

• Optical isomers are a type of stereoisomer with four different groups or atomsaround a chiral centre. Such a centre is asymmetric (has no symmetry) and givesrise to a pair of isomers which are non superimposable mirror images of eachother. These optical isomers are also called enantiomers.

• Optical isomers have identical physical and chemical properties except when in thepresence of other chiral molecules. They also have an opposite effect on planepolarised light, rotating it by exactly the same amount but in opposite directions.Mixtures containing equal amounts of each enantiomer (a racemic mixture) areoptically inactive as the rotations cancel each other.

• In biological systems only one optical isomer of each organic compound is usuallypresent.




• Organic Chemistry: J. McMurray, Brooks/Cole Publishing, ISBN 0-534-16218-5


• Chemical Ideas: Salters advanced chemistry, Heinemann, ISBN 0 435 630148


7.4. END OF TOPIC TEST 145

For the case study on ’Fats and edible oils’:

• Fats and Oils: Unilever Educational booklet: Advanced Series, Unilever

• http://www.healthy.net/asp/templates/article.asp?id=2099

• http://www.lipid.co.uk/infores/index.htm

Website for animations of stereochemistry::


Website for IUPAC naming of organic compounds:

• http://www.acdlabs.com/iupac/nomenclature/




http://www.healthy.net/asp/templates/article.asp?id=2099

http://www.lipid.co.uk/infores/index.htm



147

Topic 8

Elemental analysis and Massspectrometry

Contents

8.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 148

8.2 Elemental analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 148

8.3 Mass spectrometry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 150

8.3.1 Interpreting mass spectra . . . . . . . . . . . . . . . . . . . . . . . . . . 151

8.3.2 Fragmentation patterns . . . . . . . . . . . . . . . . . . . . . . . . . . . 153

8.4 High resolution mass spectrometry . . . . . . . . . . . . . . . . . . . . . . . . . 160

8.5 Example analysis of an unknown compound . . . . . . . . . . . . . . . . . . . . 161

8.6 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 162

8.7 Resources . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 163

8.8 End of Topic test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 163



• manipulate empirical and molecular formulae (Higher);

• describe stereoisomerism in organic compounds (Topic 7).

Learning Objectives


• calculate empirical formulae from elemental analysis data;

• explain the operation of a mass spectrometer, and be able to interpret simple massspectra.

148 TOPIC 8. ELEMENTAL ANALYSIS AND MASS SPECTROMETRY

8.1 Introduction

Organic compounds can show a very large degree of structural complexity. Remember,it took nearly a century to work out the basic chemical structure of DNA, from its isolationin 1869 by Friedreich Miescher to the double-helical structure of Watson and Crick in1953.

The methods available to these early chemists have now been refined, and thedevelopment of new instrumental techniques and powerful computers means that theworking out of complex organic structures is almost routine today.

Each of the methods described here, and in Topic 3.9, provides specific informationabout an organic compound, with these individual features combining ultimately toprovide a full picture of the structure.

8.2 Elemental analysis�

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Learning Objective

To determine % composition from combustion analysis data, and to work out empiricalformulae.

From a chemical point of view, the most fundamental data about a compound are: whichelements are present and how much of each are in the compound? This informationleads to the simplest empirical formula for the material.

The apparatus

The percentage of carbon, hydrogen, nitrogen and sulphur in a compound is found byburning a known weight of the material in an oxygen atmosphere and measuring theweights of carbon dioxide, water, nitrogen and sulphur dioxide that are produced.

Other elements known to be present (e.g. halogens) have to be measured by othermethods.

The equipment used is shown in a video ’Modern Chemical Techniques’ produced bythe Royal Society of Chemistry.

The output from the analysis gives the % composition for C, H, N and S.

Q1: Combustion analysis cannot give the % oxygen in a compound. Can you suggesta reason why?

The calculation

Knowing the percentage composition of a compound can lead us to what is reallyrequired - the chemical formula.

To convert from a percentage composition by mass to the ratio of the atoms in asubstance, the percentages are divided by the appropriate atomic mass to give a ratioof moles, which are then converted to whole number ratios of atoms.


8.2. ELEMENTAL ANALYSIS 149

Example : Elemental analysis

Analysis of an alcohol shows it to contain 37.5% carbon and 12.5% hydrogen. What isthe formula?

The % carbon + % hydrogen add up to only 50%. The question says that we are dealingwith an alcohol (i.e. contains C,H and O), so we can assume that the remaining 50% ofthe compound is oxygen.

Element Carbon Hydrogen Oxygen

% by weight 37.5 12.5 50.0

Divide byatomic mass

Ratio of atoms37.5/12

(=3.125)

12.5/1

(=12.50)

50.0/16

(=3.125)

Divide to givewhole numbers

Whole numberratio

3.125/3.125=

1

12.50/3.125=

4

3.125/3.125=

1

The formula for this compound is CH4O.

Try the following problem.

Q2: A liquid is found to contain 52.17% carbon and 13.04% hydrogen, with theremainder being oxygen. What is its formula?

Q3: The answer to Q2 gives no information about the structure of this compound. Fromyour knowledge of isomers, suggest two possible structures for this substance.

Not only does the formula above give no indication of structure, but it only indicates theratio of atoms in the compound. It is an empirical formula.

Q4: A compound of C,H and O contains 40.0% carbon and 6.667% hydrogen. Whatis the empirical formula?

The empirical formula CH2O would be obtained for methanal (molecular formulaHCHO) , but would also be the result from the analysis of ethanoic acid, CH3COOH(C2H4O2), glyceraldehyde (C3H9O3, HOCH2CHOHCHO ), glucose (C6H12O6) or manyother compounds.

Obviously, some further information is required to enable a molecular formula to befound.

Q5: A white solid is found to contain 66.667 % carbon, 7.407 % hydrogen and25.936 % nitrogen. What is its empirical formula?

Q6: The compound in the previous question was found to have a molecular mass of108. What is the molecular formula?

Q7: An antibiotic contains C,H,N,S and O. Combustion of 0.3442 g of the compoundin excess oxygen yielded 0.528 g CO2, 0.144 g H2O, 0.128 g SO2 and 0.056 g N2. Whatis its percentage composition?



Q8: What is the antibiotic’s empirical formula?


8.3 Mass spectrometry�

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Learning Objective

To describe the various parts of a simple mass spectrometer, and understand theirfunctions.

Mass spectrometry is a very powerful technique for obtaining information aboutmolecular masses and the structure of molecules. It requires only a very small amountof sample.

The instrument (Figure 8.1) is divided into four main sections:

1. The inlet system. This is maintained at a high temperature, so that any sampleintroduced will be vaporised rapidly. The interior of a mass spectrometer must bemaintained at high vacuum to minimise collisions, so the injector and the rest ofthe instrument are connected to vacuum pumps.

2. The ion source. Some of the vaporised sample enters the ionisation chamberwhere it is bombarded with electrons. Provided the energy of the collision isgreater than the molecular ionisation energy, some positive ions are producedfrom molecules in the sample material.

Some of these molecular ions (sometimes called parent ions) will contain sufficientenergy to break bonds, producing a range of fragments, some of which may alsobe positive ions (daughter ions).

The parent ions and ion fragments from the source are accelerated and focussedinto a thin, fast moving ion beam by passing through a series of focusing slits.

3. The analyser. This stream of ions is then passed through a powerful magneticfield, where the ions experience forces that cause them to adopt curvedtrajectories.

Ions with the same charge (say, 1+) will experience the same force perpendicularto their motion, but the actual path of the ion will depend on its mass. Heavier ionswill be deflected less than lighter ones.

4. The detector. Ions with the same mass/charge (m/z) ratio will have the sametrajectory and can be counted by a detector. As the magnetic field strength isaltered ions with different m/z ratios will enter the detector so that a graph ofabundance (the ion count) against m/z values can be constructed. This is themass spectrum of the sample compound(s).


8.3. MASS SPECTROMETRY 151

Figure 8.1: Diagram of mass spectrometer

You can try a simple experiment at home. Get someone to roll golf balls and table tennisballs (heavy and light ions!) across a table. As the balls pass you, blow gently acrosstheir path. The two types of ball are about the same size, will experience about the samedeflecting force from your breath, but will have different amounts of deflection owing totheir different masses.

8.3.1 Interpreting mass spectra�

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Learning Objective

To understand how molecular ions and fragments are produced in a massspectrometer, and to observe the general features of mass spectra.

When organic molecules are bombarded with energetic electrons, two basic processescan occur:

An electron can be captured to form a negatively charged radical ion,

M + e-� M.-

Or an electron can be removed to form a positively charged radical ion,

M + e-� M.+ + 2e-



(It is conventional in mass spectrometry to call a molecule M)

This latter process is most common (80% of cases), so that positive ion massspectrometry is mostly employed. The positive radical ion has one electron removedfrom a pair in a filled orbital (giving it an odd number of electrons). For simplicity at thelevel of Advanced Higher this radical electron (.) is omitted from the representation ofthese ions. Molecular ions are, therefore, called M+.

When a molecular ion fragments, it can form other ions (with either an odd or evennumber of electrons) and neutral molecules. The molecular ion and associated fragmentions will be measured in the mass spectrometer and the resulting trace is called themass spectrum. The neutral molecules, and any negatively charged ions will not beregistered.

A typical mass spectrum (Figure 8.2) is shown for benzoic acid. It consists of a series ofvertical lines, whose height represents the relative abundance of ions of a particularmass/charge (m/z) ratio. The most abundant peak is called the base peak and isassigned an abundance of 100%. Other abundances are given smaller percentages,relative to the base peak.

The peak with the highest m/z value is often the molecular ion, but in cases where thisis particularly unstable, its abundance might be very small, or undetectable.

Figure 8.2: Mass spectrum of benzoic acid

Q9: In the spectrum of benzoic acid (Figure 8.2), what is the m/z value of the basepeak?

Q10: What is the m/z value of the molecular ion?



8.3.2 Fragmentation patterns�

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Learning Objective

To work out some structural features of compounds from analysis of their massspectra.

Molecular ions, formed by the collision of a molecule with energetic electrons, canbe either stable enough to remain intact during their progress through the massspectrometer, or will have sufficient energy imparted by the collision to break specificbonds and produce a pattern of fragment ions at the detector.

The mass spectrum (Figure 8.3) of an aromatic hydrocarbon naphthalene, shown inFigure 8.4, has a stable molecular ion, C10H8

+, giving a very large response at m/z 128,with only minor fragmentation peaks elsewhere. In this case, it is very clear that thebase peak is the molecular ion and the compound has a molecular mass of 128. Themolecular ion of naphthalene is so stable that a peak of m/z 64 is seen. This is a doublycharged ion, M2+, with m = 128 and z = 2, so that m/z = 64. This type of fragmentationpattern is fairly typical of aromatic compounds, and is further evidence of the stability ofthis group of compounds (see Topic 6).

Figure 8.3: Mass spectrum of naphthalene

Figure 8.4

In contrast, the mass spectrum of 2-methylpentane (C6H14), Figure 8.6, has only a smallabundance of the molecular ion at m/z 86, but many fragment ions. This pattern offragment ions reflects the fact that the lifetime of the molecular ion is so short



(� 10-6 s) that fragmentation occurs in the source. 2-methylpentane has several bondswhich have broken to give fragments with lifetimes � 10-5 s, long enough to havereached the detector. The bonds which break in the molecular ion will always be theweakest, and the fragment abundances reflect their stability (see Topic 3.1) so thatobservation of the fragmentation pattern can give information about the structure ofthe molecule.

Figure 8.5: 2-methylpentane fragments

In Figure 8.5 above, the wavy lines represent broken bonds. The numbers on each sideof the line show the masses of the fragments produced when that bond breaks.

Figure 8.6: Mass spectrum of 2-methylpentane

The fragments of 2-methylpentane

15 min

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Learning Objective

To show that breaking specific bonds in a molecule leads to fragments detected in amass spectrum.

Visit the online version of this Topic, where you are able to click on the fragment andsee the corresponding mass spectrum position. Use Figure 8.6 to answer the followingquestions.

Q11: From the fragmentation pattern in Figure 8.6 , do you think that C - C or C - Hbonds are weaker?

Q12: Does this comparison of bond strengths agree with the bond enthalpy data onpage 9 of the data booklet?



Q13: The m/z 43 ions are most abundant in Figure 8.6. Which bond in2-methylpentane is the weakest?

The breaking of specific bonds in a molecule leads to the various ion abundance peaksin a mass spectrum.

Fragmentation patterns are useful in two ways:

• If the energy of the electron beam is set to a standard value, the fragmentationpattern produced by a particular compound is very reproducible. The mass spectraof many thousands of compounds are now available in digitised format, so that acomputer can search the data base of known mass spectra to identify an unknownsample whose mass spectrum has been obtained using electron-impact ionisation.

• A study of a mass spectrum can give information about the structure of thecompound.

The fragments of 2-ethoxybutane

10 min

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Learning Objective

To show that the mass spectrum is dependent on the molecular structure.

Figure 8.7 shows the six largest peaks in the mass spectrum of 2-ethoxybutane.

Figure 8.7: Simplified mass spectrum of 2-ethoxybutane

The molecular ion of 2-ethoxybutane can fragment as shown in Figure 8.8 to produce



the ions in the mass spectrum. Can you identify which fragments of the structure areresponsible for each of the lines on the mass spectrum?

Figure 8.8: Fragments of 2-ethoxybutane

The fragment m/z values are produced from the various bonds in the molecule breaking.The relative abundances of the fragment ions reflects the stability of the bonds breakingto produce them and the stability of the ions themselves.

The molecular ion (M+) of 2-ethoxybutane has m/z 102. The ion of m/z 87 is 15 m/zunits less than M+, written (M - 15)+. This is a common difference, often due to the lossof a CH3- group. Differences of 15 between m/z values infer the presence of a methylgroup in the compound. Table 8.1 below shows some common mass differences andthe groups they suggest.

Table 8.1: Fragment masses

Mass difference Suggested group

15 CH3

17 OH28 C=O or C2H4

29 C2H5

31 CH3O

45 COOH77 C6H5

Example : Structure from fragmentation pattern

Compounds A and B both have the same molecular formula, C3H6O, but their molecularstructures are different. The mass spectra of A and B are shown below.



A1. For compound A, which group of atoms could be lost when the ion of m/z 43 formsfrom the ion of m/z 58?

A2. Suggest a formula for the ion of m/z 43 in the spectrum of compound A.

A3. Identify A.



B1. For compound B, what atom or groups of atoms could be lost when

i) the molecular ion changes into the ion of m/z 57?

ii) the ion of m/z 57 changes into the ion of m/z 29?

B2. Suggest formulae for the ions of m/z 28,29, and 57 in the spectrum of compound B.

B3. Identify B.

A1. The mass difference from 58 to 43 is 15. Reference to Table 8.1 indicates that amethyl group is likely to be responsible for this.

A2. The ion of m/z 43 is probably due to loss of a methyl group from the molecule, sothat a fragment of formula C2H3O would form the m/z 43 ion. A likely structure for thisis CH3CO.

A3. A is propanone, CH3COCH3, which has major fragments of CH3CO and CH3.

B1. i) A loss of 1 mass unit from 58 to 57 can only be due to loss of a hydrogen atom.The molecule must have a single, ’loose’ H.

ii) The mass difference from 57 to 29 is 28, probably due to loss of C=O.(See Table 8.1)

B2. The ion of m/z 28 is the C=O+ fragment; addition of 1 mass unit ( an H) to this wouldform aCHO+ (aldehyde) as the m/z 29 peak.

The m/z 57 is loss of hydrogen from the molecule, leaving a fragment of formula C3H5O+.We know this contains a C=O group, so is likely to be C2H5CO.

B3. B is propanal, CH3CH2CHO, which loses the aldehyde H, then the CO.

Here is a series of questions for you to try for yourself.

The questions concern three of the four isomers of butanol.

Q14:

Draw the structures for the four isomers of butanol.

The mass spectra X, Y and Z below are for three isomers of butanol, excluding2-methylpropan-1-ol. There is one example of a primary alcohol, a secondary alcoholand a tertiary one, but not necessarily in that order.



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Given that alcohols tend to fragment at the C-C bond adjacent to the C-OH group(i.e. R-CH2OH would produce R+ and CH2OH+ fragments), can you explain, withreasons, which spectrum is for which isomer?

Q15: What is the m/z value for the molecular (M+) ion of all three butanols? Answerwith one integral value.



If you would like to try to link the mass spectra with the appropriate molecular structures,have a go now, by working out the main fragments lost from the M+ ion. Otherwise,the remaining questions will guide you through to an answer, with the conclusionssummarised at the end.

Q16: Spectrum X has a base peak at m/z 59. Using Table 8.1, what fragment has beenlost from the molecular ion to produce this m/z 59 ion?

Q17: Using the fragmentation information, which of the three butanols would be likelyto produce a CH3 fragment most readily?

Q18: Closer examination of spectrum Y shows a base peak at m/z 45. What could bethe structure of this ion?

Q19: When the C-C bond adjacent to the the C-OH bond breaks in butan-1-ol what arethe structures and m/z values of the fragments?

Q20: In which spectrum are these most prominent?

Q21: Summarise, with reasons, which spectrum applies to which structure.

8.4 High resolution mass spectrometry�

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To use very accurate molecular and fragment masses to distinguish between differentformulae.

Some high resolution mass spectrometers provide an accuracy of 1 in 106 atomic massunits, so that very accurate masses can be obtained for the peaks in the mass spectrum.

If you were asked for the molecular masses of, say, ethane (C2H6) and methanal(HCHO), the data booklet information would give 30.0. This is a 1 decimal place figureusing an average of the masses of the isotopes normally present in these compoundson the Earth.

However, a mass spectrometer would separate the individual molecules containingdifferent isotopes; 12C1H2

16O would appear as a different peak from 13C1H216O, one

mass unit higher. Indeed, the first use of mass spectrometry was to measure the massesof isotopes of elements.

In addition, a high resolution mass spectrometer would be able to distinguish betweenthe two compounds above because their accurate masses are not the same.

Q22: Given that the atomic mass for

• 12C is 12.0000

• 1H is 1.0078

• 16O is 15.9949

What is the accurate molecular mass of ethane (12C21H6)?


8.5. EXAMPLE ANALYSIS OF AN UNKNOWN COMPOUND 161

Q23: What is the accurate molecular mass of methanal (12C1H216O)?

So, without further analysis, an accurate mass for these compounds would identify oneagainst the other.

This ability can be extended to give molecular formulae for unknown samples, theaccurate mass being unique for a particular combination of atoms. This, of course,also applies to the fragments in a full mass spectrum, so that mass spectrometry withan accurate mass capability is a powerful technique for determining molecular structure.

8.5 Example analysis of an unknown compound

A summary series of questions

Here is a series of questions to enable you to see how the techniques described in thisTopic can lead to a compound’s structure.

The questions concern analysis of a colourless, pleasant-smelling liquid, containing onlycarbon, hydrogen and oxygen. (Compound L)

Q24: When 0.4080 g of the liquid is completely burned in an excess of oxygen,1.0560 g of CO2 and 0.2160 g of H2O are produced. What is the empirical formula?

Q25: The mass spectrum of the compound is shown below (Figure 8.9).

Figure 8.9: Mass spectrum of L

What is the m/z value of the highest m/z ion?

Q26: Does this ion’s mass agree with the empirical formula; in other words, is this massa simple multiple of the empirical formula?



Q27: Therefore, what is the molecular formula for the compound?

Q28: What is the m/z value of the base peak?

Table 8.1

might help answer the next few questions

Q29: From the mass difference from the molecular ion to base peak, what fragmentmight have been lost?

Q30: What is a likely structure for the ion of m/z 77?

Q31: What is a possible fragment composition lost when the m/z 105 ion forms the m/z77 ion?

Q32: These three fragments (m/z 77 + 28 + 31) fortunately add to the molecular mass(136). So what is a likely structure for the compound?

Q33: Name compound L.

In this example, the evidence from mass spectrometry is good enough to establish thestructure of the compound. However, this is not always the case and Topic 9 will describeother analytical techniques that are employed to determine a compound’s structure.

8.6 Summary• Elemental microanalysis, by combustion in oxygen, can be used to determine the

masses of C, H, S and N in a known weight of an organic compound. Otherelements are determined by other methods. The % composition information canthen be used to find the empirical formula.

• Mass spectrometry can be used to determine accurate molecular masses and toindicate structural features of an organic compound.

• A mass spectrometer consists of:

– an injector, often at a high temperature to vaporise the sample;

– an ion source, where the molecules of sample are bombarded with electronscausing the formation of positive molecular ions. Depending on the stability ofthese ions, and the energy imparted by the collision, bonds in these molecularions may break in a characteristic fashion to produce fragment ions Theparent molecular ion and ion fragments are then accelerated into a fine, fastmoving beam of ions;

– an analyser, consisting of a magnetic field which causes ions to be deflectedin a curved trajectory dependent on their mass/charge ratio;

– a detector, which counts the ions of particular m/z value, to enable a massspectrum of abundance against m/z to be constructed.

• Investigation of the mass spectrum can yield information about the molecular mass(if the parent, M+ ion is present), and about the structure of the sample, from thedistribution of the fragment ions.


8.7. RESOURCES 163

• Where very accurate masses are available from the mass spectrometer, thesealone can often provide chemical formulae for the parent and fragment ions.

8.7 Resources• Chemical Ideas: Salters Advanced Chemistry, Heinemann, ISBN 0-435-630148


• Higher Still Support: Advanced Higher Chemistry - Unit 3:Organic Chemistry, Learning and Teaching Scotland, ISBN 1-85955-873-9

• Organic Spectroscopy: Kemp, Macmillan, ISBN 0-333-18153-0

• Mass Spectrometry Barker Analytical Chemistry by Open Learning series,John Wiley, ISBN 0-471-96764-5

• Interpreting Organic Spectra Whittaker The Royal Society of Chemistry,ISBN 0-85404-601-1

Video: Modern Chemical Techniques, Royal Society of Chemistry, and associated book,ISBN 1-870343-19-0.




165

Topic 9

Infrared and Nuclear magneticresonance spectroscopy and X-raycrystallography

Contents

9.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 167

9.2 Infrared spectroscopy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 167

9.2.1 The infrared spectrometer . . . . . . . . . . . . . . . . . . . . . . . . . . 171

9.2.2 Interpreting IR spectra . . . . . . . . . . . . . . . . . . . . . . . . . . . . 172

9.2.3 Functional group identification . . . . . . . . . . . . . . . . . . . . . . . 174

9.3 Nuclear magnetic resonance spectroscopy . . . . . . . . . . . . . . . . . . . . 178

9.3.1 The NMR spectrometer . . . . . . . . . . . . . . . . . . . . . . . . . . . 179

9.3.2 Interpreting NMR spectra . . . . . . . . . . . . . . . . . . . . . . . . . . 182

9.3.3 Magnetic resonance image (MRI) scanning . . . . . . . . . . . . . . . . 184

9.4 X-ray crystallography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 185

9.4.1 The diffraction of X-rays by crystals . . . . . . . . . . . . . . . . . . . . 185

9.4.2 Electron density maps . . . . . . . . . . . . . . . . . . . . . . . . . . . . 186

9.5 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 188

9.6 Resources . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 189

9.7 End of Topic test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 189


Before starting this Topic, you should be able to:

• manipulate empirical and molecular formulae (Higher);

• describe stereoisomerism in organic compounds (Topic 7).

Learning Objectives

By the end of this Topic, you should be able to:

• explain how infrared spectra are obtained by the interaction of radiation withmolecules, and be able to interpret spectra in terms of functional groups;

166TOPIC 9. INFRARED AND NUCLEAR MAGNETIC RESONANCE SPECTROSCOPYAND X-RAY CRYSTALLOGRAPHY

• state how nuclear magnetic resonance spectra are obtained and be able to usesimple spectra to define the numbers and environments of hydrogen atoms inorganic compounds;

• explain the use of X-ray crystallography to produce electron-density maps oforganic molecules.



9.1 Introduction

The methods described in the previous Topic to determine molecular formulaeand provide structural information about organic compounds are extended in thisTopic to include infrared and nuclear magnetic resonance spectroscopy and X-raycrystallography, which are used to identify functional groups in organic compounds, toinvestigate the different types of hydrogen atom environments in organic compounds,and to reveal the 3D structure, respectively.

9.2 Infrared spectroscopy

When you sit in front of a fire, which gives out infrared (IR) radiation, you feel warm. Thisis because the IR radiation absorbed by molecules in your skin is of a suitable frequencyto increase their vibrational energy. The greater a molecule’s vibrational energy, thehotter it is.

This property is used in "night-vision" equipment, which visualises changes intemperature as shown in Figure 9.1- a man holding a match.

Figure 9.1: An infrared image of a man holding a match

Infrared photography is used to show cloud cover in meteorology. The meteorologicaloffice web site, has a satellite picture of Europe updated every six hours, see(http://www.meto.govt.uk/weather/satellite/smlinfrared.html).

Vibrational energy levels are quantised (see Unit 1, Topic 2.2), and measuring theparticular IR frequencies that are absorbed by different molecules forms the basis ofinfrared spectroscopy.

Units of electromagnetic radiation.

In Unit 1 Topic 1.2, you saw a diagram giving the wavelength and frequency of the entireelectromagnetic spectrum. The range of the infrared frequencies used for spectroscopylies between 1014 to 1013 Hz. This corresponds to wavelengths bewteen 2.5 and 15�m wavelength. In fact many chemists use wavenumber to describe the energies of IRradiation.


http://www.meto.govt.uk/weather/satellite/smlinfrared.html


The units of IR spectroscopy

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To revise the interrelations between frequency, wavelength and wavenumber.

Put the correct numbers from the bank below into the table to show the comparison offrequency, wavelength and wavenumber over the range used in IR spectroscopy.

Frequency (Hz) 1.2 x 1014 6 x 1013

Wavelength(�m) 2.5 10

Wavenumber(cm-1) 4000 200

2000 1000 50

6 x 1012 5.0 3 x 1013

In infrared spectra, the wavelength of the infrared radiation is expressed inwavenumbers, the reciprocal of length in centimetres (cm-1).

How radiation and matter (molecules) interact.

Simple diatomic molecules, for example halogens, have only one bond, which canvibrate only by stretching and compressing. The two halogen atoms can pull apartand then push together.

The frequency of this vibration is determined by the mass of the two atoms and thestrength of the bond. If you think of this system as two spheres attached by a spring, thenthe frequency of the natural, resonant vibration will decrease as the spheres increase inmass, and increase as the spring gets stronger.

Intermediate Slower (heavier spheres) Faster (stronger spring)

Similar principles apply to molecular vibrations. Table 9.1 shows the bond enthalpies forhydrogen halides.

Table 9.1: Bond enthalpies for hydrogen halides

Compound Bond Enthalpy/kJmol-1

HCl 432HBr 366HI 298


9.2. INFRARED SPECTROSCOPY 169

Q1: Which molecule has the strongest bond?

a) HClb) HBrc) HI

Q2: Which molecule has the largest molecular mass?

a) HClb) HBrc) HI

Q3: Which molecule will have the largest vibration frequency?

a) HClb) HBrc) HI

These diatomic molecules have a single, specific frequency at which they vibrate. Whenthe whole range of IR frequencies is passed through a cell containing, say, HCl, only thatfrequency at which this molecule vibrates will be absorbed, leaving most of the radiationto pass through intact. Some of the HCl molecules will then vibrate at a higher quantumstate. These molecules return to the lower state by giving up the energy in collisionswhich alters their velocity.

Table 9.2 below gives the frequency of vibration for hydrogen halides and the IRabsorption in wavenumbers.

Table 9.2: Frequency and wavenumber for hydrogen halide molecules

CompoundVibration

frequency/Hz IR absorption/cm-1

HCl 8.7 x 1013 2886

HBr 7.7 x 1013 2559

HI 6.7 x 1013 2230

(When you see animations of molecular vibrations, remember that they are veryconsiderably slowed down. The actual bond stretching frequency is around 1013 times

per second!)

Most molecules contain more than two atoms and this gives rise to modes of vibrationother than simple stretching of bonds. Consider a water molecule.

Each of the O-H bonds will have a characteristic stretching frequency, but these caninteract and stretch either symmetrically or asymmetrically, giving rise to two slightlydifferent IR absorption frequencies for the O - H bonds in a water molecule.

See Figure 9.2



Symmetric stretch Asymmetric stretch

Figure 9.2: Stretching modes in water molecules

In addition to the stretching mode, molecules can also have bending modes. Thehydrogen atoms in a methylene - (CH2 group) can move in the same plane (either ina scissor motion, or a rocking motion), or can waggle in and out of the plane (again,either twisting or wagging). These are shown in Figure 9.3. All these vibration modeswill have different frequencies.

Scissor bend Twisting deformation

Rocking bend Waggingdeformation

Figure 9.3: Bending modes

Animations of these stretching and bending modes can be seen on the website.

In general, it takes more energy to stretch the molecule than to bend it.

Q4: Two wavenumbers applicable to the C-H bond are given below, one for stretchingand one for bending. Think about which requires most energy, and therefore which willhave the highest frequency. Which wavenumber applies to the C-H stretch?

a) 1460b) 2930

Molecules can vibrate in all of these fashions, and most of these will have characteristicinfrared absorbance wavenumbers.



LEARNING POINTS

Molecules vibrate with a set of specific frequencies dependent on the types of thebonds and the groups or atoms at the ends of these bonds.

The energies associated with these vibrations are quantised.

Infrared radiation causes parts or all of a molecule to vibrate in a higher vibrationalstate.

9.2.1 The infrared spectrometer

Figure 9.4 shows a diagram of an infrared spectrometer.

The source of infrared radiation has its beam split into two.

One part of the beam passes through the sample; the other through a reference cell,which might contain the solvent, for example.

The monochromator grating scans the wavelengths prior to a detector which comparesthe intensity of the two beams.

The amplified signal is plotted as % transmission or absorbance.

Figure 9.4: Infrared spectrometer

There is a description of an IR spectrometer on the video "Modern ChemicalTechniques" from the Royal Society of Chemistry, which describes the preparation ofsamples and the running of the spectrometer.

Q5: In IR spectroscopy, discs of sodium chloride or potassium bromide ("salt flats")are often used to hold a thin film of liquid samples in the IR beam. What property mustNaCl or KCl have that allows them to be used in this way?

Figure 9.5 shows an IR spectrum for liquid paraffin (a mixture of alkanes).



Figure 9.5: IR spectrum of alkane mixture

This spectrum shows the simplest IR spectrum, for an organic compound consisting ofonly two types of bonds ( C - C and C - H). Many other organic compounds will have thiscarbon backbone with the addition of other functional groups.

In an earlier question you were given 2930 cm-1 as the C-H stretch wavenumber, and1460 cm-1 as C-H bend. The complete IR spectrum for liquid paraffin (Figure 9.5) alsohas absorbance at 2880 cm-1, the asymmetric stretch, and a further bending mode at1380 cm-1.

9.2.2 Interpreting IR spectra

The IR spectrum for propanone is shown (Figure 9.6), together with its structure(Figure 9.7).



Figure 9.6: IR spectrum of propanone

Figure 9.7: Structure of propanone

The spectrum contains some wavenumbers from vibrations of the C - H bonds, but thereis another strong absorbance due to the carbonyl group.

Q6: What is its wavenumber? Give your answer to the nearest 100 cm-1.

The C=O stretch absorbance is characteristic of ketones and ocurrs always around thiswavenumber.

The carbonyl group also occurs in other organic compounds, such as aldehydes,carboxylic acids and esters.

The C=O stretch frequency changes slightly when the group is not directly connectedto alkyl or aryl groups and will be found with other wavenumbers characteristic of thatgroup.

Another common group in organic compounds is hydroxyl, found in alcohols and



phenols. Figure 9.8 shows the IR trace of ethanol.

Figure 9.8: IR spectrum of ethanol

Q7: What is the wavenumber of the O -H stretch? Give your answer to the nearest100 cm-1, for the maximum absorption wavenumber.

The infrared correlation table

Because these infrared absorbance wavenumbers are common for a particular group inany compound, tables of values have been built up by observation of a large number ofcompounds. Page 13 of the Data Booklet gives these data. You will notice that thereis a range of values for a particular group, because the value is slightly altered by thesurrounding groups.

Look at the data booklet, particularly the different C=O stretch wavenumbers in thefunctional groups containing the carbonyl group.

You will be using this table extensively in the next section.

9.2.3 Functional group identification

The main use of IR spectra, which can be obtained quickly and cheaply, is to identifythe presence of functional groups and the carbon backbone type in unknown organiccompounds.

Use the correlation table on page 13 of the data booklet to answer the following



questions regarding the IR spectra labelled X, Y and Z.



Q8: Which compound contains an alcohol group?

a) Xb) Yc) Z

Q9: Which compound is not aromatic?

a) Xb) Yc) Z

Q10: Which class of compound is present in X?

Q11: Which of the following could be Z?

a) Benzyl alcohol (C6H5CH2OH)b) Butan-1-ol (C4H9OH)c) Benzonitrile (C6H5CN)d) Acetonitrile (CH3CN)

9.2.3.1 The "fingerprint" region of an IR spectrum.

The region of an IR spectrum from 4000 to 1400 cm-1 contains many absorbancewavenumbers for specific bond types. Below 1400 cm-1 IR spectra typically have anumber of absorbances not assigned to a particular bond. In fact, these relatively lowenergy vibrations are due to complex vibrations which are unique to that molecule.



Comparison of this region for an unknown material with a set of standards run underidentical conditions will allow identification.

Q12: The IR spectrum for an analgesic (pain killer) is shown, together with standardspectra for aspirin, paracetamol, and phenacetin (Taken from the PharmaceuticalCodex). By comparing the spectra, can you tell which analgesic it is?

Analgesic



a) Aspirinb) Paracetamolc) Phenacetin

9.3 Nuclear magnetic resonance spectroscopy

Some atomic nuclei, such as those for 1H and 13C have a nuclear spin, rather likeelectrons (Unit 1, Topic 2.5). When an electrically charged object moves (for example anucleus spins), it will create a magnetic field around it. When these nuclei are in normalconditions the spins and associated magnetic fields will be randomly distributed.

However, if these nuclei are put into a strong magnetic field the nuclei (behaving likea compass needle in the Earth’s magnetic field) will line up along the field. This canhappen in only two ways because of quantum restrictions: some will align with their tinymagnetic fields in the same direction as the large field; others will align in an opposeddirection. This change in condition is illustrated in Table 9.3 .

Table 9.3: Nuclei alignment without and with a magnetic field.

These two states have slightly different energies, with the opposed state (dark spheres)being higher.

If electromagnetic radiation of a suitable frequency is applied to nuclei in a strongmagnetic field some will absorb energy and flip to the higher energy quantum state.This radiation must have the same frequency as the rotating nuclei, so the two are inresonance; hence the term nuclear magnetic resonance. (Much of the NMR data isobtained for 1H, protons, so that the term proton magnetic resonance is sometimes


9.3. NUCLEAR MAGNETIC RESONANCE SPECTROSCOPY 179

used.) The energy required for this transition is in the radio frequency region of thespectrum, typically from 60 MHz to 1000 MHz.

Very strong magnetic fields are required for NMR, so that superconductingelectromagnets cooled in liquid nitrogen are used.

As the nuclei return to the lower state (relax) they emit radiation which can be detected.

9.3.1 The NMR spectrometer

The equipment used to obtain low resolution NMR spectra is shown in Figure 9.9.

Figure 9.9: Diagram of an NMR spectrometer

The sample is dissolved in CDCl3 or CD3COCD3, "deuterated" solvents with hydrogensreplaced by deuterium (D or 2H).

Q13: Why do you think deuterated solvents are used?

a) They are denser than normal solvents so will not fly out of the sample tube so easily.b) They contain no 1H which would swamp the NMR spectrum of the sample with its

signal.c) They can be obtained in a purer state than normal solvents.



Absolute values are difficult to obtain, so the ppm values are obtained by reference to astandard arbitrarily assigned the Æ value 0. The standard is tetramethylsilane (TMS), seeFigure 9.10, which has an NMR signal well separated from those found in most organicmolecules.

Figure 9.10: Structure of tetramethylsilane (TMS)

Q14: Look at the structure of TMS (Figure 9.10). How many different chemicalenvironments are there for the 12 H atoms?

a) 1b) 3c) 4d) 12

The Royal Society of Chemistry video "Modern Chemical Techniques" shows theoperation of an NMR spectrometer.

If the external magnetic field was the only influence on the nuclei, you would be ableto detect the different response frequencies from, say, 1H and 13C, but since all organiccompounds contain carbon and hydrogen, this would not be much use. However, thereis another factor. All nuclei in atoms are surrounded by electrons, which, because theyhave their own spin, will shield the nuclei from the total effect of the magnetic field. Theeffective field experienced by any given nucleus is, therefore, modified by the extent towhich it is surrounded by electrons.

Look at the structure of a methanol molecule (Figure 9.11).

Figure 9.11: Methanol

Q15: Around which hydrogens (methyl or hydroxyl) are the electron clouds denser?(Hint: think about electronegativity)

Q16: So which hydrogens (methyl or hydroxyl) will feel the effects of the external fieldmore?

The proton NMR spectrum of methanol (Figure 9.12) shows two different signals for thetwo different hydrogens in the different magnetic and chemical environments.



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Figure 9.12: NMR spectrum of methanol

The two signals are said to have different chemical shifts. In this case the hydrogenatom in the -OH group is shifted more than those in -CH3.

Observe the relative size of the two signals from CH3 and OH

Q17: How many hydrogen atoms are in the "methyl" environment?

Q18: How many hydrogen atoms are in the "hydroxyl" environment?

Q19: What would you estimate is the ratio of peak areas for methyl H and hydroxyl Hsignals?

a) 1 to 3b) 1 to 1c) 3 to 1

Since it is difficult to measure areas, NMR spectra often have an integration signal,generated electronically, overlaid on the NMR signal. The height of this line isproportional to the area under the NMR response. (See Figure 9.13)

LEARNING POINT

The areas under the signals in an NMR spectrum are in the ratio of hydrogen atoms inthat part of the molecule.

The chemical shift (Æ) values of protons (1H) in different chemical groups in a moleculehave approximately constant values, with respect to tetramethylsilane Æ = 0.

Because the different signals from protons (1H) in a compound reflect their chemicalenvironment, proton nuclear magnetic resonance is one of the most useful instrumentaltechniques for organic chemists.



9.3.2 Interpreting NMR spectra

Determining structure using NMR

The NMR spectrum of a hydrocarbon is shown in Figure 9.13.

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Figure 9.13: NMR spectrum of hydrocarbon W

Q20: From the spectrum, how many different environments are there for hydrogenatoms?

Q21: What is the ratio of areas for these peaks?

Q22: From the correlation table (page 15 of the data booklet), can you identify the typeof hydrogen in the larger peak?

a) RCH3

b) RCH2Rc) ArCH3

d) ArH

Q23: What type of hydrogens are in the smaller peak?

a) RCH3

b) RCH2Rc) ArCH3

d) ArH

You might find it useful to summarise the information in a table.

Table 9.4: NMR table

Peak 1 Peak 2Æ (ppm) 2.3 7.4

Type of H ArCH3 ArHNumber of H atoms 3 5

Group CH3 C6H5



Q24: What is the hydrocarbon?

Q25: The proton NMR spectrum for a substance U, of formula C2H5OBr is shown inFigure 9.14.

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Figure 9.14: NMR spectrum for U.On paper, summarise the information in a table like the one above (Table 9.4), thenname the compound.

An explanation for the answer will be given in the display answer.

The next three questions refer to N,N - diethylphenylamine (C10H15N), shown inFigure 9.15.

Figure 9.15: N,N - diethylphenylamine

Q26: How many different types of H environment are there in the molecule?

a) 1b) 2c) 3d) 5



Q27: What is the ratio of H atoms in these groups?

a) 2:1b) 4:5:6c) 1:2:3d) 3:2:5

Q28: Can you predict the chemical shift (Æ) ranges for this compound?

Q29: Sketch the NMR spectrum for N,N-diethylphenylamine.

9.3.3 Magnetic resonance image (MRI) scanning

Our bodies consist largely of water, which exists in a large number of differentenvironments in each tissue. Just as hydrogen atoms in molecules are shielded todifferent extents depending on the surrounding atoms, the protons in water in tissueswill experience slight differences in a strong external magnetic field, and so will absorbslightly different radiofrequency radiation.

The part of the body being investigated is moved into a strong magnetic field. By usingcomputers to process the absorption data, a series of images of the different watermolecule environments, is built up into a picture of the body’s tissues. It is assumedthat MRI scanning is harmless to health, unlike other imaging processes (such as CATscans) which involve low doses of ionising X-rays.

MRI scanning is particularly good for brain tissue where there is a large amount of fattylipids which provides a different environment for water compared with the other tissues.

The MRI image opposite shows fluidcollection in the region that separates thebrain from the skull. This is a blood clotwhich applies pressure to the brain and isvery dangerous, perhaps even fatal.


9.4. X-RAY CRYSTALLOGRAPHY 185

9.4 X-ray crystallography

Introduction

The previous sections have described methods for investigating the ways in which theatoms in a molecule are connected chemically to form the molecule. (Which functionalgroups are present, and what different types of hydrogens are there?) This sectiondescribes how X-ray crystallography can be used to determine the actual positions ofatoms in molecules.

9.4.1 The diffraction of X-rays by crystals

When monochromatic light passes through a series of evenly-spaced slits on a grating,the transmitted light produces a series of interference lines on a screen.

Crystals of pure chemical compounds consist of very regularly repeating patterns ofatoms, spaced about 100 pm apart. (You can check this approximation by looking atcovalent and ionic radii in the data booklet.) If you then look at Topic 1.2 in Unit 1 youwill see that X-rays have wavelengths about 10-9 - 10-10 m. similar to the interatomicdistances. (100 x 10-12 m = 10-10 m)

By using X-rays, which are diffracted by atoms in much the same way as light is bya diffraction grating, a pattern of diffracted spots is obtained when radiation passesthrough a crystal of material. (See Figure 9.16).

Figure 9.16: X-ray diffraction pattern

Because X-rays are electromagnetic, it is the electrons in the atoms or ions in the crystalwhich are responsible for the scattering. The differing electron densities of atoms areuseful in assigning the diffraction spots to different scattering species.

The next questions are about sodium and potassium chlorides, which both crystallise ina face-centred cubic lattice.

Q30: A sodium chloride crystal consists of an array of sodium and chloride ions. Howmany electrons are there in each sodium ion?

Q31: How many electrons are there in each chloride ion?



Q32: Which ion would you expect to scatter X-rays to the greater extent?

a) Sodiumb) Chloride

Q33: Work out how many electrons each potassium ion has. What do you notice aboutthe numbers of electrons in potassium and chloride ions?

a) More in potassiumb) More in chloridec) Both the same

You would expect the intensities of spots from planes of potassium and chloride ionsto be similar, unlike sodium chloride with different intensities. These differences canbe exploited to help identify the positions giving rise to the reflections, by deliberatelyinserting a heavy atom into a crystal and noting the more intense reflections producedfrom the modified crystal.

Q34: In organic compounds which commonly-found atom would you expect to scatterX-rays least?

9.4.2 Electron density maps

The photograph of the diffraction pattern above (Figure 9.16) would require a veryconsiderable input of time to measure the angles of the different spots and calculatethe position of atoms in the crystal . Nowadays, the process is automated. The crystalis mounted on a frame, which can be rotated in all directions under the control of acomputer. At any particular position, an array of detectors sends information aboutthe scattered X-rays to the computer. The calculation capability of modern computersallows these data to be manipulated to provide an electron density map of the atoms inthe molecules forming the crystal.

The main points of the process are now illustrated.

The X-ray diffraction pattern from a crystal of urea:


9.4. X-RAY CRYSTALLOGRAPHY 187

The electron density map derived from the previous pattern:

Q35: In the compound urea, CO(NH2)2, which atom has the most electrons?

a) carbonb) hydrogenc) nitrogend) oxygen

Q36: Which atom will show the highest electron density pattern?


Look at the electron density map and identify this atom.

Q37: One of the types of atoms below typically shows a very weak response on accountof its small number of electrons. Which element is this?


Identification of carbon and nitrogen by similar means allows the complete 3D structureof urea to be determined.



Arrangement of atoms in urea

For urea this might not seem a great task, but the technique finds invaluable applicationfor larger molecules, particularly, biomolecules.

Learning point

Electron density maps are produced from the positions and intensities of the "spots" ina diffraction pattern, produced when a crystal of an organic compound is exposed toX-rays of a single wavelength.

From the electron density map the precise location of each atom in the molecule canbe determined and since heavier atoms have more electrons than lighter ones eachatom in a molecule can be identified.

Since a hydrogen atom has a low electron density, it is not easily detected by X-rays.

9.5 Summary• All chemical bonds vibrate with quantum energy states that depend on the mass

of the atoms and the strength of the bonds.

• These vibrations, for molecules with more than two atoms, fall into groups ofstretching and bending modes, whose energies are in the infrared (IR) range.These are usually expressed in wavenumbers. The IR wavenumbers involved arefrom 4000 to 400 cm-1.

• Infrared radiation is passed through a sample of the organic compound and thento a detector which measures the intensity of the transmitted radiation at differentwavelengths.

• From correlation tables the different wavenumber absorptions can be assignedto characteristic groups. For example the carbonyl (C=O) group absorbs near1700 cm-1. Additionally the "fingerprint" region below 1400 cm-1, which containsvibrations from the whole molecule, can be used for identification purposes.

• The nuclei of some atoms, in particular 1H, have a spin which means that theybehave as tiny magnets and will align with or against an externally appliedmagnetic field. Those aligned with the field have a lower energy than those


9.6. RESOURCES 189

opposed to it. Absorption of energy in the radiofrequency range "flips" the nucleifrom low to high state.

• 1H atoms in organic molecules have different energies for the low and high statesof nuclear spin which are dependent on their different electron densities.

• A proton NMR spectrum is generated by spinning a sample in the field of a strongmagnet and measuring the emission of radiofrequency radiation.

• The sample to be analysed is normally dissolved in a solvent containingno 1H atoms. e.g. CDCl3 . Deuterium ( 2H) has an NMR frequency a long wayfrom 1H.

• The low resolution NMR spectrum consists of peaks whose chemical shift(Æ, in ppm) from the signal of tetramethylsilane (set at zero) reflects the chemicalgroup containing the protons and whose areas are proportional to the number ofprotons in that environment.

• Magnetic resonance imaging in medicine uses the same technique to visualise thedifferent environments of protons in body tissues, in order to diagnose disease orinjury.

• X-rays are of wavelength comparable to interatomic distances in molecules(around 10-10 m). When X-rays pass through the regular 3D array of atoms ina crystal, a symmetrical arrangement of diffraction spots is produced, which canbe used to determine the precise 3D structure.

9.6 Resources• Chemical Ideas: Salters Advanced Chemistry, Heinemann, ISBN 0-435-630148


• Higher Still Support: Advanced Higher Chemistry - Unit 3:Organic Chemistry, Learning and Teaching Scotland, ISBN 1-85955-873-9

• Organic Spectroscopy: Kemp, Macmillan, ISBN 0-333-18153-0

• Interpreting Organic Spectra: Whittaker The Royal Society of Chemistry,ISBN 0-85404-601-1

Video: Modern Chemical Techniques, Royal Society of Chemistry, and associated book,ISBN 1-870343-19-0.




191

Topic 10

Medicines

Contents

10.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19210.2 Aspirin development . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19210.3 How a medicine functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 194

10.3.1 Enzyme function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19610.3.2 The pharmacophore . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 197

10.4 Case studies . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19910.5 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20710.6 Resources . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20810.7 End of Topic test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 208



• describe enzyme function in relation to the molecular shapes of proteins (HigherUnit 2);

• use systematic names, full and shortened structural formulae, describe functionalgroups and reaction types and mechanisms for all the families of organiccompounds preceeding this Topic in Advanced Higher.

Learning Objectives


• state that medicines are beneficial drugs containing pharmacologically activecompounds;

• describe the historic development of aspirin as an example of a medicinedeveloped from synthesised derivatives of natural compounds found in plantextracts.;

• describe the interaction of biologically active molecules with receptor sites on cellsand enzymes;

• explain the terms pharmacophore, agonist and antagonist and describe thefunctions they perform.

192 TOPIC 10. MEDICINES

10.1 Introduction

Almost without realising it, many of us probably owe the quality of our lives to theready availability of a wide range of medicines. The pharmaceutical industries areamongst Britain’s largest companies and invest huge sums of money in researchand development of medicinal substances. Any substance which alters the normalbiochemical processes in the body is known as a drug. This includes a wide rangeof chemicals from alcohol and caffeine to more medicinal substances like aspirin andantibiotics.

Those drugs which have a beneficial effect on health are called medicines.

Many common pharmacy productscontain aspirin.

Some products have active drugsremoved. Some have them added!

10.2 Aspirin development�

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Learning Objective

To use aspirin as an example of a medicine from a plant source from whichcompounds and derivatives have been identified and synthesised.

Traditional medicines from plant and animal sources can provide strong clues to ’leadin’ chemicals for the pharmaceutical industry. The pharmacologically active compoundsin such plant and animal extracts can be identified and used as medicines. Sometimesthese compounds and their derivatives can be synthesised and modifications made toimprove efficiency and reduce side effects.

Morphine, for example, was isolated as the active principle from the extract of the juiceof the opium poppy. It is a powerful euphoric and analgesic (pain killer). Many syntheticmedicines related to morphine have been made, including codeine and methadone.

Aspirin is a common medicine that has been developed from natural sources.

The Greek physician Hippocrates recommended the use of willow bark for pain reliefduring childbirth and a boiled vinegar extract of willow leaves was recommended forpain relief in Roman times.


10.2. ASPIRIN DEVELOPMENT 193

In 1829, the compound salicin(Figure 10.1) was isolated as theactive principle in willow showing

the analgesic properties.

Hydrolysis of the salicin yielded salicyl alcohol (Figure 10.1), which is metabolised inthe body to give salicylic acid (Figure 10.1). This proved to be useful in treating painand reducing the fever in rheumatic fever (antipyretic) but unfortunately was also foundto cause gastric bleeding. The most effective derivative found was acetylsalicylic acid(Figure 10.1), better known as aspirin.

The sodium and calcium salts of acetylsalicylic acid are soluble.

It has become clear that aspirin is of benefit, not only as an analgesic and antipyreticbut also in the reduction of blood clotting and as such it is involved in treatmentand prevention of heart disease and stroke. (Remember, however, that as with mostmedicines, there are side effects and it should be used with care).

Figure 10.1: Aspirin related compounds

Q1: Which of the chemicals shown is an alcohol?

a) Ab) Bc) Cd) D



Q2: What other product is formed when A is hydrolysed to B?

a) waterb) alcoholc) aspirind) glucose

Q3: What type of reaction takes place when B forms C?

a) esterificationb) reductionc) hydrolysisd) oxidation

Q4: What type of reaction takes place when C forms D?

a) esterificationb) reductionc) hydrolysisd) oxidation

PPA - Preparation of aspirin (UNIT 3 PPA 4)

120 min

Consult with your tutor to find out whether the PPA on ’Preparation of aspirin’ is to becompleted at this stage.

PPA - Aspirin determination (UNIT 3 PPA 5)

120 min

Consult with your tutor to find out whether the PPA on ’Aspirin determination’ is to becompleted at this stage.

10.3 How a medicine functions�

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Learning Objective

To describe how medicines work by binding to receptor sites and functioning asagonist or antagonist molecules.

A living organism is a complicated system with a multitude of chemical reactions withinit. Why, then, do medicines have such a specific effect on some parts of the body andwhere do they interact with the chemical systems?

Many of today’s medicines interact with receptors, which are usually protein moleculesthat are either on the surface of cells or are enzymes that catalyse chemical reactions(catalytic receptors). Receptors normally interact with a specific small biologically activemolecule which can bind reversibly with the site. The forces involved are electrostaticattractions, varying in strength from van der Waals bonding to hydrogen bonding or evenionic bonding. Generally, the functional groups on both have to be in the correct positionto interact and bind the medicine to the receptor.


10.3. HOW A MEDICINE FUNCTIONS 195

Receptor function

5 min

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Learning Objective

To describe how medicines work by binding to receptor sites.

Visit the web version of this Topic to access an animation and a drag and drop exercisethat illustrate the way receptors function.

Play the animation a few times. It illustrates how a receptor initiates activity in a cell. Inthis example, the cell being activated is a muscle cell.

If the correct ’active molecule’ binds, it activates the cell to trigger a biological responseand then reversibly leaves the site unchanged (Figure 10.2).

Figure 10.2: Receptor function

The biological response here is contraction of a muscle cell. A useful analogy mightbe a car engine and its ignition mechanism. The ignition key represents the biologicallyactive molecule, the ignition lock is the receptor and the engine is the cell. The processof starting the engine is the same as initiating the biological response.

Many of today’s medicines interact with receptors, which are usually protein moleculesthat are either on the surface of cells or are enzymes that catalyse chemical reactions(catalytic receptors). Receptors normally interact with a specific small biologically activemolecule which can bind reversibly with the site.

Effective medicines can work by binding to the receptor site and either mimicking theresponse of the active molecule or blocking the effect of the active molecule.

If the medicine mimics the natural active molecule, it will stimulate the same responseand activate the biological response. In this case, the medicine is classified as anagonist. An agonist therefore is a drug that binds to a receptor, triggering or increasing aparticular activity in that cell (Figure 10.3) . This class of medicine is useful in situationswhere there is a shortage of the natural active molecule. For example salbutamol, isan agonist which mimics adrenaline and switches on receptors which lead to dilation(widening) of the airways. It can be used to treat asthma attacks.

The agonist molecule is like a good copy of the original ignition key in the analogy usedpreviously. It fits the lock, switches the ignition and starts the car (it activates the cell)(Figure 10.3).



Figure 10.3: Agonist function

If the medicine binds onto the receptor site and does not switch it on, it prevents theaction of the body’s natural active molecule and is classed as an antagonist. Anantagonist therefore is a drug which binds to a receptor without stimulating cell activityand prevents any other substances from occupying that receptor ( Figure 10.4 ). Thisclass of medicine is useful if there is a surplus of natural messengers or where onewants to block a particular message. For example, propranolol is an antagonist whichblocks the receptors in the heart that are stimulated by adrenaline. It is called a �-blockerand is used to relieve high blood pressure.

Figure 10.4: Antagonist function

The antagonist molecule is like a badly copied key. It fits into the lock but will not turnand is therefore unable to start the engine (unable to activate the cell). Antagonists alsosometimes stick in place, usually due to stronger bonding, preventing any other activemolecule from binding.

10.3.1 Enzyme function�

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Learning Objective

To describe how the catalytic receptor site in enzymes catalyses chemical reactions.

Enzymes are large protein molecules which act as catalysts in biological processes. Theprotein in enzymes is normally folded into a three dimensional structure, part of which isa pocket known as the active site or the catalytic receptor. Since the catalytic receptoris a specific shape, it can only accommodate certain substrate molecules. Once inposition, the substrate is chemically changed and the products formed. The productscould be smaller (a cleavage reaction) (see Table 10.1) or larger (a synthesis reaction).


10.3. HOW A MEDICINE FUNCTIONS 197

Table 10.1: Enzyme action

Medicines can act on enzyme active sites in similar agonistic or antagonistic ways, forexample acting as a plug or inhibitor to switch off specific enzymes. For example,some ’smart drugs’ are designed to improve memory in Alzheimer’s disease by blockingenzyme breakdown of a chemical called acetylcholine, necessary for neurotransmission.

10.3.2 The pharmacophore�

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Learning Objective

To identify the pharmacophore in a group of medicines and explain how it functions inconferring pharmacological activity.

The development of new medicines very often starts with a ’lead in’ clue from a plant oranimal source. Research then seeks to find related compounds with fewer side effectsor greater activity. The first stage in this is isolation of the minimum structural featurethat gives pharmacological activity; this is called the pharmacophore. The shape ofthe pharmacophore complements that of the receptor site, with the functional groups onboth structures correctly positioned to interact and bind them together. Once identified,chemists can design and synthesise related compounds, looking towards improving theperformance of the medicine.

The pharmacophore can often be identified by comparing the structures of medicineswith similar pharmacological activity.

The highly addictive analgesic morphine and some of its close modifications can providean example. (Figure 10.5)

Figure 10.5: Opiate structures

In each structure, the lines are carbon to carbon bonds with a carbon at each cornerand most hydrogens omitted.



The pharmacophore part of these structures is identified in the associated websiteactivity.

Morphine pharmacophore

10 min

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Learning Objective

To identify the pharmacophore in a group of medicines by comparison of theirstructures.

Visit the online activity which shows the slight structural changes made by chemists tomodify the properties of morphine.The pharmacophore can then be identified.

The pharmacophore can be identified by comparing the structures of medicines withsimilar pharmacological activity.

The analgesic medicines shown here ( Figure 10.6) have a common pharmacophore.They are highly addictive and are legally controlled.

Figure 10.6: Morphine derivatives and the pharmacophore.

When the shape, dimensions and functional groups present have been identified,the pharmacophore can be cropped, added to and manipulated by chemists toproduce compounds which are still analgesic but less addictive (Figure 10.7). Thepharmacophore’s shape and dimensions complement the receptor, usually throughfunctional groups that are positioned by the shape and size of the structures to allowinteraction and binding of the medicine to the receptor.


10.4. CASE STUDIES 199

Figure 10.7: Modified opiates

For example, etorphine (a synthetic opiate), (Figure 10.7) is almost 100 times as potentas morphine and is used in veterinary medicine to immobilize large animals.

10.4 Case studies�

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Learning Objective

To describe a common medicine which performs as either an agonist or antagonist inits treatment of a medical condition.

The following four case studies show how medicines function in different commonapplications. Detailed knowledge does not have to be memorised. Your tutor may wishyou to focus on one of the four and give a short presentation to the rest of your group.If you are simply required to be aware of the case studies, read them through, payingparticular attention to the names of drugs, mode of action and medical conditionsthey are effective against. You will then be able to tackle the final activity.

If you are making a presentation to a group, then you will need to prepare:

• a hand-out information sheet;

• audio visual illustrations (blackboard or OHP or data projection);

• a short script for yourself.

Include within your presentation:

• the medical conditions involved;



• formulae, equations, diagrams;

• mode of action;

• side effects.

Read each of the case studies. You may skim them looking for the necessary informationto carry out the ’Medicines in use’ activity at the end of the Topic, where you will be askedto match up the medical problems with the correct medicine and its mode of action.

Case Study 1 Salbutamol:

30 min

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Learning Objective

To be aware of the structure and function of salbutamol as a treatment for asthma.

A case study illustrating the development of salbutamol from pharmacologically activecompounds from natural sources. PLEASE NOTE: Detailed knowledge of themedicines, their structure and how they work does not have to be memorised.

Read each of the case studies. If you are presenting one as a homework exercise, choseone particular case study to focus on and follow the instructions about preparation. Ifyou are only interested in gathering information in order to complete the ’Medicines inuse’ activity, you may skim them looking for the necessary information which is italicised, and match up the medical problems with the correct medicine and its mode of action.

For over 5000 years, the Chinese plant Ma Huang was used to treatasthma andbronchitis. It was boiled in water and the solutions drunk. The chemical responsiblefor the biological activity is ephedrine (Figure 10.8). Pure ephedrine has cardiovascularside effects.

Figure 10.8: Natural bronchodilators

Bronchodilators like ephedrine widen the airways in the lungs and relieve thebreathlessness and wheezing of an asthma attack. One of the natural substancesthat perform this function is adrenaline (Figure 10.8), which is a hormone circulating inthe blood supply. Using adrenaline to treat asthma also increases heart rate and raisesblood pressure.

Research has shown that there is a variety of different receptors on which adrenalinecan act, each producing a different effect. These have been classed as alpha (�) andbeta (�) type receptors, The main effects of adrenaline are shown in Table 10.2.



Table 10.2: Effects of adrenaline

Receptor EffectOutcome forasthmatics

�-activatedincreased blood

pressureunhelpful

�1-activated increased heart rate unhelpful

�2-activated dilation of bronchi helpful

Instead of adrenaline, which activated all these receptors and was quickly broken downin the body, what was required was a �2 specific agonist which lasted longer and wouldonly activate the dilation effect. One of the first was isoprenaline (Figure 10.9) whichhas a bulky group attached to the nitrogen. This seemed to favour � 2 activation.

Figure 10.9

Adding an even bulkier group to the nitrogen, and replacing the 3-hydroxyl group on thering with a hydroxymethyl group, yielded an even more effective medicine, salbutamol(Figure 10.9). The hydroxymethyl group slowed down the breakdown of the drug in thebody so that it lasted about 4 hours. Interestingly, salbutamol has two enantiomericforms (see Topic 3.7), one of which is 68 times more effective than the other (can youthink why?). Salbutamol is a market leader that is delivered by inhalation from an aerosolspray (Figure 10.10).

Figure 10.10

Recent work to extend the time between doses has yielded a drug which is more fatsoluble, and therefore takes longer to be metabolised. It has a duration of 12 hours.



Figure 10.11

Salbutamol is an effective medicine for quick relief in an asthma attack. Salmeterol(Figure 10.11) provides longer spells of relief.

Case Study 2 Propranolol:

30 min

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Learning Objective

To be aware of the structure and function of propranolol as a treatment for asthma.

A case study illustrating the development of propranolol from pharmacologicallyactive compounds from natural sources. PLEASE NOTE: Detailed knowledge of themedicines, their structure and how they work does not have to be memorised.

As before, read each of the case studies. If you are presenting one as a homeworkexercise, chose one particular case study to focus on and follow the instructions aboutpreparation. If you are only interested in gathering information in order to completethe ’Medicines in use’ activity you may skim them looking for the necessary informationwhich is italicised , and match up the medical problems with the correct medicine andits mode of action.

The heart pumps blood through the arteries and veins of our cardiovascular system.Permanently high blood pressure (hypertension) is caused by increased output fromthe heart or from increased resistance to flow in the arteries or both. Hypertension canlead to heart attack and strokes. Another heart condition called angina results fromrestrictions in blood flow through the arteries that supply the heart muscles. This canlead to heart attack if untreated.

Table 10.2 shows that adrenaline activates the �1 type receptors and also increasesblood pressure. If an antagonist for the �1 receptors could be found then it could beused to block the receptor and allows the heart rate and blood pressure to fall.

The search started with isoprenaline (Figure 10.12) which although an agonist, hadshown specific activity to � receptors rather than � receptors. A large number ofmodifications were made to try to find an antagonist from isoprenaline, one of the firstbeing pronethalol.



Figure 10.12

Further research led to the discovery of propranolol (Figure 10.13) which is a pureantagonist and has become the benchmark against which all � blockers are rated.Interestingly, only one of the enantiomeric forms (the S form) is the active ingredient.(Can you suggest why?)

Figure 10.13: Typical beta blockers

A wide variety of these � blockers are now used by people with heart disorders to keepthe heart calm.

Case Study 3 Sulphonamides and penicillins:

30 min

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Learning Objective

To be aware of the structure and function of sulphonamides and penicillins astreatments for bacterial diseases.

A case study illustrating the development of sulphonamides and penicillins frompharmacologically active compounds from natural sources. PLEASE NOTE: Detailedknowledge of the medicines, their structure and how they work does not have to bememorised.


In 1928, Alexander Fleming observed that a plate on which staphylococci bacteria werebeing cultured at St Mary’s Hospital, London had been contaminated with a mould which



seemed to be killing the bacteria close to it. Although he recognised this, he failed toisolate the substance responsible for the effect. The active compound ’penicillin’ waseventually isolated by Chain and Florey in 1940.

Figure 10.14: Penicillin and penicillium mould

A wide variety of penicillins was soon produced. The basic structure of penicillin isshown in Figure 10.15, the structures differing only in the nature of the group R.

Figure 10.15

All penicillins are bactericidal (kill bacteria). During replication bacteria, need to makenew cell wall material by crosslinking strands of polymer chains to give a strong net-like structure. The penicillins block the active site of the enzyme responsible for thiscrosslinking and thus act as antagonists, preventing further cell wall construction. Sincemammal cells do not have cell walls, this enzyme is absent and the penicillin doesn’taffect them.

Antibiotics can therefore be used against bacterial infections like pneumonia,meningitis and septicaemia. It should be noted however that the increased use ofantibiotics over the last 75 years has led to the development of resistance in some typesof bacteria. Use them sparingly and always complete the course!

Before the penicillin antibiotics were widely available, a dyestuff called prontosil(Figure 10.16) had been found to be an effective bacteriostatic agent, stopping thefurther growth of bacteria.



Figure 10.16

Prontosil showed activity when used on bacteria in living tissue but strangely enoughwould not affect bacteria grown in test tubes. The mystery was solved when itwas shown that prontosil was metabolised by the small intestine into sulphanilamide(Figure 10.17) which was the true bacteriostatic agent.

Figure 10.17

Replacing one of the hydrogens on the -SO2NH2 by various groups gives a wide varietyof sulphonamides that interfere with the growth process in bacteria. Cell growth anddivision in bacteria requires the synthesis of a B vitamin called folic acid. Bacteriause 4-aminobenzenecarboxylic acid (Figure 10.17) to synthesise the folic acid and acomparison of their structure shows that sulphanilamide is sufficiently similar to foolthe enzyme involved and inhibit the formation of folic acid. The bacteria stop dividing.Folic acid is obtained by animals in their diet and does not have to be synthesised,so sulphonamides do not affect the host animal. Sulphonamides act as antagonists(inhibitors) by blocking the enzyme site.

Case Study 4 Antihistamines:

30 min

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Learning Objective

To be aware of the structure and function of antihistamines as treatments for allergicconditions and peptic ulcers.

A case study illustrating the development of antihistamines from pharmacologicallyactive compounds from natural sources. PLEASE NOTE: Detailed knowledge of the



medicines, their structure and how they work does not have to be memorised.


Important advances in the treatment of ulcers were made following the developmentof antihistamine drugs in the 30s and 40s. Histamine (Figure 10.18) is found in mosttissues of the body, particularly lungs, skin and gastrointestinal tract. It is released inresponse to many allergies such as hay fever, insect bites, asthma and is a stimulantof gastric acid secretion. Overproduction of gastric acids can lead to the formation ofpeptic ulcers.

Figure 10.18

When drugs such as mepyramine (Figure 10.18) which relieved some of the symptomsof histamine release by blocking the receptor site (acting as antagonists), werediscovered, it was noticed that the mepyramine had no effect on gastric acid secretion.This suggested that there might be more than one receptor site for histamine.

As a result of work by Sir James Black and colleagues in the 70s, two different shapedreceptor sites for histamine, one for allergic type responses (H1-receptor) and one forgastric acid secretions (H2-receptors) were distinguished. Histamine can adopt two ormore different shapes, one to satisfy the H1-receptor and the other for the H2-receptor(Figure 10.19).

Figure 10.19


10.5. SUMMARY 207

Research since the 1970s has led to the development of effective H1-antagonists likechloropheniramine (Figure 10.20) which alleviate hay fever and relieve skin itching,swelling and sneezing. Antagonists for the H2 sites include ranitidine (Figure 10.20)marketed as Zantac, which has few side effects and cures about 75% of gastric ulcers.

Figure 10.20

Medicines in use summary:

10 min

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Learning Objective

To be able to classify medicines as agonists or as antagonists according to whetherthey enhance or block the body’s natural responces.

Read each of the case studies. You may skim them looking for the necessary informationif your tutor does not ask you to look at one in depth. Using the drag and drop activityavailable on the website, match up the medical problems with the correct medicine andits mode of action.

Medicines can be classified as agonists or as antagonists according to whether theyenhance or block the body’s natural responses.

10.5 Summary• All drugs have biochemical effects on the body; some are beneficial, some are

detrimental. Those which have a beneficial effect are called medicines.

• Historically, the first medicines were plant brews and animal extracts and theseprovide modern medicine with ’lead in’ compounds even today.

• Aspirin is an example of a traditional medicine which has been developed byresearching derivatives synthesised from willow bark.

• Most medicines work by binding to receptors. The receptors are usually proteinmolecules either on the surface of cells where they interact with small biologicallyactive molecules or are enzymes that catalyse chemical reactions (catalyticreceptors).



• Many medicines can be classified as agonists, which produce a response froma cell in the same way as the natural active compound, or as antagonists,which produce no response but prevent the actions of the body’s naturally activecompound. Some medicines can act on enzyme receptor sites in similar agonisticor antagonistic ways.

• Structural comparisons of medicines with similar pharmocological activity can leadto the identification of the active fragment called the pharmacophore which bindsand interacts reversibly with the receptor, sometimes through functional groups.

10.6 Resources• Chemical Storylines: Salters Advanced Chemistry, Heinemann ISBN 0-435-

63106-3

• Chemical Ideas, Salters Advanced Chemistry, Heinemann, ISBN 0-435-63105-5


• Pharmacology (3rd edition), Rang, Dale and Ritter, Churchill Livingstone ISBN0-443-050473

• British Medical Association Guide to Medicines and Drugs, Editor Dr JohnHenry CLB Dorling Kindersley ISBN 0-862-839270

• Chemistry - Advanced Higher : Unit 3: Organic Chemistry. Learning andTeaching Scotland, ISBN 1-85955-873-9

• The Organic Chemistry of Drug Design and Drug Action, R.Silverman,Academic Press ISBN 0-126-43730-0




209

Topic 11

End of Unit 3 Test (NAB)

Contents

210 TOPIC 11. END OF UNIT 3 TEST (NAB)

To gain access to the test, click on the words ’Assessment Exercise’ highlighted below.



GLOSSARY 211

Glossary

addition reaction

a reaction in which a small molecule is added across a double bond to give a singleproduct.

agonist

is a drug that binds to a receptor, triggering or increasing a particular activity inthat cell.

alcohols

organic compounds which contain the -OH (hydroxyl) group (except when the OHgroup is attached to a benzene ring when the term phenol is used)

aliphatic

hydrocarbons are nonaromatic compounds such as simple alkanes, alkenes oralkynes.

alkanes

a family of saturated hydrocarbons that fits the general formula, CnH2n+2.

alkanols

a homologous series all members of which fit the general formula, CnH2n+1OH

alkenes

a family of unsaturated hydrocarbons whose molecules contain a carbon to carbondouble bond.

antagonist

is a drug which binds to a receptor without stimulating cell activity and preventsany other substances from occupying that receptor

arenes

are hydrocarbons which contain both aromatic and aliphatic units.

aromatic

compounds are those which contain benzene or benzene type rings.

base peak

The base peak in a mass spectrum is the most abundant peak, assigned anarbitrary abundance of 100%.

carbanions

ions in which a carbon atom carries a full negative charge.

carbocation

an ion in which a carbon atom carries a full positive charge.

chemical properties

those properties of a substance that describe how it behaves chemically.


212 GLOSSARY

chiral

objects are said to be chiral when they are not superimposable on their mirrorimage.

condensation reaction

a reaction in which two molecules react to form a larger molecule by eliminating asmall molecule like water.

dipoles

a dipole arises when there is an uneven distribution of charge in a molecule, suchthat one part has a partial positive charge and another part has an equal butopposite negative charge

electronegative

electronegativity is a measure of the ability of an atom to attract a bonded pair ofelectrons - the more electronegative, the stronger the attraction.

electrophile

a species that is able to accept a pair of electrons from an atom carrying a full orpartial negative charge.

elimination reaction

a reaction in which a small molecule is lost (eliminated) from a larger molecule,resulting in the formation of a multiple bond.

empirical formula

The empirical formula of a compound is the simplest formula defining the ratio ofatoms in that compound. It does not give the total number of atoms in a molecule.For example, the actual formula for glucose is C6H12O6, but the empirical formulais CH2O.

enantiomers

are molecules that are non-superimposable mirror images of each other.

ethers

organic compounds containing molecules which contain an oxygen atom which isbonded to two carbon groups

free radical chain reaction

a reaction that, once initiated, sustains itself in an endlessly repeating cycle ofpropagation steps.

free radicals

atoms or molecule containing unpaired electrons.

functional group

an atom or group of atoms that is part of a larger molecule and has a characteristicchemical reactivity.


GLOSSARY 213

geometric isomers

are stereoisomers that differ in the spatial arrangement of groups around a doublebond or a ring (also called cis-trans isomers).

heterolytic fission

both of the shared electrons go to only one of the two atoms producing ions.

homologous series

a family of compounds that fits a general formula and has similar chemicalproperties.

homolytic fission

the two shared electrons separate equally, one going to each atom.

hybridisation

the mathematical combination of atomic orbitals to generate new orbitals that willmore effectively form bonds.

hybrid orbitals

orbitals that are derived mathematically from a combination of s, p or d orbitals.

hydrolysis

a reaction in which a larger molecule is broken down by reaction with water.

initiation

the first step in a free radical chain reaction in which a weak bond is broken toproduce free radicals.

intermediate

a species that is formed during a multi-step reaction but is not the final product.

isomers

compounds with the same molecular formula but different structural formulae.

markovnikov’s Rule

In the addition of HX to an alkene, the H atom attaches to the carbon atom withfewer alkyl groups attached to it and the X group attaches to the carbon atom withmore alkyl groups attached to it.

molecular formula

The molecular formula of a compound describes the number of atoms of eachelement that are present in one molecule of the compound.

molecular orbital

the region in space in which there is a high probability of finding an electron that ismoving under the influence of two or more nuclei.

monounsaturated

containing only one carbon to carbon double bond in each molecule.


214 GLOSSARY

nucleophile

a species that is able to donate a pair of electrons to an atom carrying a positiveor partial positive charge.

optical isomers

are non-superimposable mirror images of each other and are said to be chiral.

pharmacophore

is the structural fragment of a molecule which confers pharmacological activity onit.

physical properties

those properties which describe what a substance is like but do not involvechemical changes, e.g. boiling point, solubility.

pi bond

a covalent bond formed by sideways overlap of atomic orbitals.

polyunsaturated

containing more than one carbon to carbon double bond in each molecule.

propagation

the steps in a chain reaction that continue the chain. In each step, a productmolecule and a free radical are produced.

racemic mixture

is an equimolar mixture of the optically active enantiomers of a compound, and isoptically inactive.

receptors

are usually protein molecules that are either on the surface of cells or are enzymesthat catalyse chemical reactions (catalytic receptors) Receptors normally interactwith a specific small biologically active molecule which can bind reversibly with thesite.

saturated

contains only single bonds and cannot undergo addition reactions.

sigma bond

a covalent bond formed by end-on overlap of atomic orbitals lying along the axisof the bond.

stereoisomerism

In stereoisomerism, the isomers have their atoms connected in the same order butarranged differently in space.

structural isomers

have the same molecular formula but the atoms in each are connected in adifferent order.


GLOSSARY 215

substitution

a reaction in which an atom of one element is replaced by an atom of anotherelement.

termination

the steps in a free radical chain reaction in which free radicals combine and sostop the chain.

transition state

a hypothetical activated complex formed between reactants that represents thehighest energy point in a reaction curve. The transition state is unstable and canbreak down to give either reactants or products.

unsaturated

contains carbon to carbon double bonds.

wavenumber

Wavenumber is the reciprocal of wavelength and has the units of cm-1

(number of cycles per cm.)


216 FURTHER QUESTIONS

Further questions

Topic 1: Introduction to Organic Chemistry

Nomenclature of alkanes (page 2)

Q5:

a) 3-ethyl-4,6-dimethyloctaneb) 2,5-diethyl-4-methylheptanec) 3,6-diethyl-4-methylheptaned) 2,4,5-trimethyloctane

Topic 2: Hydrocarbons and Halogenoalkanes

Synthesis of alkenes (page 24)

Q40:

Study the following grid for the next two questions.

Which of the above compounds could be starting materials for the synthesis ofpent-1-ene ?

a) B and Db) A and Cc) B onlyd) B,D and Ee) All of them


FURTHER QUESTIONS 217

Products of elimination reactions (page 34)

Q41:


Products of substitution reactions (page 40)

Q42: Compound 2 can be hydrolysed. Draw the structure of the product of hydrolysis.

Q43: Draw the structure of compound 3 and name it.


Q45: What type of reaction is occurring in each case?

Q46: 2-chloropropane can also be converted into propene. What type of reaction wouldthis be and what reagent would be used?

Topic 3: Alcohols and Ethers

Questions continued from page 46.

Q53: Draw the structural formula of a secondary alcohol of formula, C3H6O, and nameit.

Boiling points of alcohols (page 49)

Q54: Using the graph (Figure 3.3), what can you deduce about the relative strengths ofthe intermolecular forces?

Boiling points of ethers (page 51)

Q55: Explain why the boiling points of ethers are less than the corresponding alkanols.


Q56: Is hydrogen bonding possible between a methoxymethane molecule and a watermolecule?

a) yes


218 FURTHER QUESTIONS

b) no

Q57: Is hydrogen bonding possible between an ethanol molecule and a watermolecule?

a) yesb) no

Q58: Predict the solubility of the ether and ethanol in water.

a) both insolubleb) ether soluble, ethanol insolublec) ether insoluble, ethanol solubled) both soluble

Hydration of alkenes (page 54)

Q59: Why is this less likely?

Q60: What role does the oxygen atom of the water molecule play in the second stageof the reaction (Table 3.3)?

Q61: What role does the X- ion play in the final step?

Q62: Overall, what type of reaction has taken place?

Q63: If all the R- groups were CH3- groups, draw structures for the two possibleproducts, name them and state which would be the major product

Preparation from halogenoalkanes (page 55)

Q64:

The molecule opposite (Figure 11.1) is atertiary halogenoalkane and will undergosubstitution by the mechanism shown inFigure 3.8.

Figure 11.1

Suggest two reasons why this mechanism is more likely for a tertiary compound.

Topic 4: Aldehydes, Ketones and Carboxylic Acids

Boiling points (page 69)

Q52: Which family of compounds has the highest boiling points?

a) alkanesb) aldehydesc) ketones


FURTHER QUESTIONS 219

d) alcoholse) carboxylic acids

Q53: Which family of compounds has the strongest intermolecular forces?

a) alkanesb) aldehydesc) ketonesd) alcoholse) carboxylic acids


Q54:

2-methylpropanoic acid can be prepared from 2-bromopropane in two stages. On paper,outline the two reactions involved. State the reagents used in each step and name eachtype of reaction. Then reveal the answer.

Hint: How many carbon atoms are there in the starting material and the final product?

Topic 5: Amines

Topic 6: Aromatics

Topic 7: Stereoisomers

Topic 8: Elemental analysis and Mass spectrometry


Q34: The molecular mass is 172.1, what is the molecular formula?

Topic 9: Infrared and Nuclear magnetic resonance spectroscopy andX-ray crystallography

Topic 10: Medicines

Topic 11: End of Unit 3 Test (NAB)


220 HINTS

Hints for activities

Topic 1: Introduction to Organic Chemistry

Bond fission and carbocations

Hint 1: Use electronegativity values from the data booklet to help decide the polarity ofthe bond.

Topic 4: Aldehydes, Ketones and Carboxylic Acids

Reactions of aldehydes, ketones and carboxylic acids

Hint 1: The compound C4H8O is an aldehyde. Use this information to identifycompounds P and S.


ANSWERS: TOPIC 1 221

Answers to questions and activities

1 Introduction to Organic Chemistry


Q1: c) 2,4-dimethylpentane

Q2: d) 2,4-dimethylhexane

Q3: b) 3-ethyl-2-methylpentane

Q4: d) 2,3,3-trimethylpentane

Further answers


Q5: a) 3-ethyl-4,6-dimethyloctane


222 ANSWERS: TOPIC 2

2 Hydrocarbons and Halogenoalkanes

Answers from page 17.

Q1: 414

Q2: 243

Q3: a) Cl-Cl

Q4: homolytic

Free radical chain reaction (page 19)

Q5: to supply energy to break bonds to start the reaction

Q6: initiation

Q7:

Q8: 3

Q9: Absorption of one photon produces two chlorine atoms each of which then reactsto produce a hydrogen chloride molecule and a free radical. This free radical reactsfurther, propagating the chain and producing more and more hydrogen molecules untilthe chain is terminated. If a chain reaction was not involved each photon absorbedwould produce one product molecule.

Q10:

Q11: These substances would be able to intercept the free radicals and so terminatethe chain.

Bonding in hydrocarbons (page 22)

Q12: c) sp

Q13: c) There are two � bonds and three � bonds.

Q14: linear




Q15: hex-1-ene

Q16: If the temperature is kept above the boiling point of the alkene but below the boilingpoint of the alcohol, the alkene vapour can be removed and condensed elsewhere toproduce a pure product.


Q17: 1

Q18: 2

Q19: 3

Q20: 2


Q21: 1,2-dichloropropane

Q22:


Q23: a) Primary

Q24: b) Secondary

Q25: c) Tertiary

Q26: b) Secondary


Q27:



Q28:

Q29:

Q30:


Q31: b) polar covalent with the carbon partially positive

Q32: a) as an electrophile

Q33: b) a nucleophile

Q34: b) heterolytically


Q35: A dative covalent bond because both electrons of the shared pair originally camefrom the same atom.


Q36: amine

Q37: ammonia



Q38:

Q39: nitrile

Further answers


Q40: d) B,D and E


Q41:


Q42:



Q43:

Q44: ether

Q45: nucleophilic substitution

Q46: This would be an elimination reaction. The reagent would be potassiumhydroxide in solution in ethanol, not potassium hydroxide in aqueous solution.



3 Alcohols and Ethers


Q1:

Isomers of C2H6O:

A B

Isomers of C3H8O:

C D E

Q2:

Group 1

B E

Group 2

AC D

Q3: They all contain a hydroxyl group, -OH.


Q4: Both contain an oxygen atom between two carbon atoms, i.e.




Q5: a) primary

Q6: c) tertiary

Q7: b) secondary

Q8:


Q9: ethoxyethane

Q10: 2-ethoxypropane

Q11: methoxyethane

Q12: 2-methoxy-3-methylbutane


Q13: a) hydrogen bonding � polar-polar attraction � van der Waals forces


Q14: a) van der Waals forces

Q15: b) it increases

Q16:

van der Waals forces arise because a momentary imbalance in the distribution of theelectrons in the molecule produces a temporary dipole.

This induces a dipole in a neighbouring molecule provided that the molecules are closeenough together. This will only happen when the molecules are very slow moving, i.e.at low temperatures. The bigger the molecule, the slower moving it will be at a giventemperature.

So van der Waals forces increase with molecular mass.

Q17: The relative formula mass of methane is 16, that of propane is 44 and that ofchloromethane is 50.5. Since the strength of van der Waals forces depends on therelative formula mass, a fair comparison can only be made between molecules of similarrelative formula mass.



Q18: Any difference cannot be due to van der Waals forces since both compounds havesimilar relative formula mass. However, the chlorine to carbon bond is permanentlypolar (a permanent dipole). There will be a greater attraction between chloromethanemolecules due to the attraction between these permanent dipoles. More energy willbe needed to move the chloromethane molecules apart and so the boiling point will behigher.

Q19: propane

Q20: chloropropane

Q21: Alkanols have higher boiling points than both alkanes and chloroalkanes of similarrelative formula mass.

Q22: c) hydrogen bonding


Q23: b) alcohols

Q24: b) alcohols

Q25: c) Alkanes will have higher boiling points

Q26: c) Alcohols will tend to have higher boiling points


Q27: a) yes

Q28: a) yes

Q29: b) no

Q30: Since hydrogen bonding is not possible between methoxymethane molecules, theintermolecular forces are weak (methoxymethane is only slightly polar). Less energy isneeded to move the ether molecules apart and so the boiling point is low.


Q31: c) The solubilities are roughly the same.

Q32: The first members of the series are soluble but solubility decreases as thenumber of carbon atoms increases. As the non-polar hydrocarbon part of the moleculeincreases, it begins to dominate and the substances become insoluble just like thecorresponding hydrocarbons.


Q33: c) electrophile



Q34: nucleophile

Q35: tertiary

Q36: secondary


Q37: SN1 reaction because it is a substitution, nucleophilic and only one molecule isinvolved in the rate-determining step.

Q38: SN2 reaction because it is a substitution, nucleophilic and two species are involvedin the rate-determining step.

Q39: The C-X (carbon to halogen) bond is polar with a partial positive charge on thecarbon atom which makes it susceptible to attack by any species carrying a negativecharge or a partial negative charge, i.e. a nucleophile.

Q40:


Q41: Alkenes must contain a C=C bond, i.e. at least two carbon atoms. Methanolmolecules contain only one carbon atom. The simplest alcohol that can be made by thismethod is ethanol.


Q42:

Q43: sodium hydroxide



Q44:

The other product is sodium ethoxide.


Q45: b) elimination

Q46:


Q47: d) condensation

Q48:

1. This is done to speed up the reaction. The reaction is catalysed by acid (H+ ions).

2. The concentrated sulphuric acid is a dehydrating agent. Removal of water as it isformed shifts the equilibrium position towards the right and produces more ester.

Q49: methyl ethanoate


Q50: c) substitution


Q51: d) condensation




Q52: The pair of electrons will be almost equally shared and so the bond is likely tobreak homolytically producing highly reactive free radicals.

Further answers


Q53:


Q54: Hydrogen bonds are stronger than polar-polar attractions which in turn arestronger than van der Waals forces.


Q55:

Ethers contain the functional groupAlcohols contain the functional

group

The presence of a hydrogen atom bonded to an oxygen atom in the alcohol means thathydrogen bonds can be formed between alcohol molecules.

Hydrogen bonds cannot be formed between ether molecules because all the hydrogenatoms are bonded to carbon and the C-H bond is almost non-polar.

The ether molecule will be very slightlypolar since oxygen is more electronegativethan carbon but this only makes a slightdifference to the boiling point.




Q56: a) yes

Q57: a) yes

Q58: d) both soluble


Q59: According to Markovnikov’s Rule (see Topic 3.2), the hydrogen atom becomesattached to the carbon atom of the double bond which already has more hydrogenatoms, in this case forming a tertiary carbocation.

Tertiary carbon atoms are more stable than secondary carbocations and so are morelikely to be formed.

Q60: nucleophile

Q61: base

Q62: electrophilic addition

Q63:

2-methylbutan-2-ol 3-methylbutan-2-ol

(major product) (minor product)


Q64:

1. With a tertiary compound, the intermediate will be a tertiary carbocation which ismore stable than a primary or secondary one.

2. From the space-filling model it should be clear that the methyl groups prevent thenucleophile from attacking the carbon from the side opposite to the chlorine atom.So the second mechanism is much less likely.



4 Aldehydes, Ketones and Carboxylic Acids


Q1: c) 2s and two 2p orbitals

Q2: a) electrophilic

Q3: b) nucleophilic

Q4: nucleophile

Q5: electrophile

Q6: base


Q7: c) polar-polar attractions

Q8: When alkanes boil, van der Waals forces are being broken. Polar-polar attractions(also called dipole-dipole attractions) are stronger than van der Waals forces and somore energy is needed to overcome them. Consequently, aldehydes and ketones havehigher boiling points.

Q9: b) hydrogen bonds

Q10: When alcohols boil, hydrogen bonds are being broken. Hydrogen bonds arestronger than polar-polar attractions (also called dipole-dipole attractions) and so moreenergy is needed to overcome them. Consequently, aldehydes and ketones have lowerboiling points than the corresponding alcohols.


Q11: d) all of them

Q12: d) all of them

Q13: Aldehydes and ketones of low molecular mass will be soluble in water (misciblewith water) but solubility will decrease as the chain length increases.

Q14: c) Hexanoic acid is more soluble than ethanoic acid.


Q15: a) ethanoic acid

Q16: b) ethanol




Q17: b) no

Q18: b) carboxyl

Q19: b) alkanoic acids


Q20: c) nucleophile


Q21: d) from trigonal planar to tetrahedral

Q22: primary

Q23: secondary

Q24: propan-1-ol

Q25: 2-methylpentan-3-ol


Q26: condensation


Q27: primary


Q28: b) ketone

Q29: a) aldehyde


Q30: aldehyde

Q31: oxidation

Q32: hydrolysis

Q33: b) B




Q34:

Q35:

Q36:

Q37:


Q38: c) proteins

Q39: peptide link

Q40: Nylon and kevlar are examples of polyamides.


Q41: condensation

Q42:

The correct name is ethyl propanoate.

The first part of the name is derived from the alcohol by removing the ’-anol’ ending andreplacing with ’-yl’.

The second part of the name is derived from the acid by removing the ’-ic’ ending andreplacing with ’-ate’.



Q43: methyl benzoate

Q44: propyl octanoate

Q45: C

Reactions of aldehydes, ketones and carboxylic acids (page 86)

Q46: b) but-1-ene

Q47: a) primary

Q48: R

Q49: butanoic acid

Q50: amide

Q51: LiAlH4

Q52: LiAlH4

Further answers


Q53: e) carboxylic acids

Q54: e) carboxylic acids


Q55:

Stage 1 is the nucleophilic substitution of bromide by cyanide to increase the chainlength by one carbon atom.

Stage 2 is the acid catalysed hydrolysis of the nitrile to form the product carboxylic acid.



5 Amines


Q1: c) diethylamine

Q2: d) 2-aminopropane

Classification and naming of amines (page 93)

Q3: b

Q4: b) diethylmethylamine

Q5: b) secondary

Q6: a) ethyldimethylamine, tertiary


Q7: b) (d)

Q8: b) (d)

Q9: a) (c)

Q10: tertiary

Q11: a) (a)

Q12: a) (a)

Q13: a) (a)


Q14: stronger

Q15: weaker

Q16: c) 8


Q17: ionic

Q18: ethylammonium chloride

Recognising functional groups (page 101)

Q19: aldehyde



Q20: ether

Q21: hydroxyl

Q22: carboxyl

Q23: ester

Q24: d) It is an unsaturated ketone

Q25: b) ether



6 Aromatics

Aromatic bonding (page 109)

Q1: sigma

Q2: pi

Q3: sigma

Q4: a) 1

Q5: d) 6

Q6: b) sp2


Q7: c) Electrophilic substitution

Q8: b) Chlorobenzene

Q9: d) Heterolytic

Q10: carbocation


Q11: d) electrophile

Q12: b) carbocation

Q13: The intermediate is stabilised by delocalisation of the positive charge around thebenzene ring.


Q14: c) 1,3-dinitrobenzene


Q15: electrophile

Q16: substitution


Q17: catalyst

Q18: c) heterolytic

Q19: chloroethane



Benzene reactions summary (page 115)

Q20: b) RCl/AlCl3

Q21: a) Cl2/AlCl3

Q22: nitrobenzene

Q23: benzenesulphonic acid

Q24: bromobenzene


Q25: b) phenol

Q26: b) phenol

Q27: a) yes

Q28: b) phenol


Q29: c) diethylammonium

Q30: d) phenylammonium

Q31: c) diethylamine

Q32: d) phenylamine



7 Stereoisomers


Q1: ether

Q2: alcohol


Q3: In A, the hydroxyl group is attached to the end carbon atom, whereas in B it isattached to the middle carbon atom.


Q4: cis-but-2-ene

Q5: trans-but-2-ene


Q6: cis-1,2-dichlorocyclobutane

Q7: trans-1,2-dichlorocyclobutane

Physical properties (page 128)

Q8: b) trans

Q9: b) trans

Q10: b) The closer the molecules pack, the stronger the van der Waals forces

Q11: b) trans

Q12: c) pure covalent and polar covalent

Q13:



The trans-isomer will be much less polar than the cis-isomer since the polarities of theC-Cl bond oppose each other and will cancel out. In the cis-isomer, both polar bonds areon the same side of the double bond making the molecule polar. The extra polar-polarattractions in the cis-isomer give it a higher boiling point.


Q14: b) right

Q15: left

Q16: c) asymmetrical

Q17: c) shoe

Optical isomers of alanine (page 138)

Q18: a) Structure 1

Q19: b) non-superimposable

Q20: b) asymmetric

Q21: b) chiral

The polarimeter (page 141)

Q22: d) produce plane polarised light

Q23: c) -13.5Æ

Q24: a) 0Æ


Q25: c) a racemic mixture

Q26: The carbocation intermediate is planar. When the nucleophile attacks thecarbocation, in the second step of the mechanism, it can attack from either side. Attackfrom one side will give one enantiomer. Attack from the other side will give its mirrorimage. So a racemic mixture is produced.



8 Elemental analysis and Mass spectrometry


Q1: The organic compound is burned in an excess of oxygen, so that any oxygenpresent in the sample will only add to that used for combustion.


Q2: C2H6O

There must be 100 - 52.17 - 13.04 = 34.79 % oxygen.

Carbon Hydrogen Oxygen

Mass % 52.17 13.04 34.79Atom ratio 4.3475 13.04 2.1744Formula ratio 2 6 1

Q3: C2H5OH (ethanol) and CH3OCH3 (dimethyl ether).


Q4: CH2O


Q5: C3H4N

Q6: C6H8N2

Q7: 41.836 % C; 4.648 % H; 18.594 % S; 16.270 % N; and 18.652 % oxygen.

Q8: C6H8SN2O2


Q9: 105

Q10: 122

The fragments of 2-methylpentane (page 154)

Q11: The carbon backbone breaks because carbon - carbon bonds are weaker thancarbon - hydrogen. If C - H were weaker, the six carbons would remain bonded andhydrogens would be lost to produce ions like C6H13

+, (M - H)+.

Q12: Yes. C-C 346 kJ mol-1, C-H 414 kJ mol-1.

Q13: The bond labelled B is weakest. Breaking this bond gives fragments of m/z 43.If, for example, the terminal C-C bonds were weakest, then the CH3

+ ion abundance atm/z 15 would be high.



This abundance is also a consequence of the (CH3)2-CH+ ion’s relative stabilitycompared with, for example, CH3

+. (See Topic 2 on the stability of carbocations.)


Q14:


Q15: 74


Q16: CH3

Q17: 2-methylpropan-2-ol has three methyl groups attatched to the carbon with the OHgroup attatched, and is likely most easily to lose one of them. Butan-2-ol also has asingle CH3 attatched to the carbon with the OH on, so would be expected to have areduced m/z 59 ion. (Possibly spectrum Y)

Q18: Loss of the CH3CH2 group from the carbon with the OH, in a similar manner to lossof the CH3 above, will leave a fragment of this m/z value, with structure HC(CH3)OH+.This suggests further that spectrum Y is butan-2-ol.

Q19: CH3CH2CH2+ with m/z 43 and CH2OH+ with m/z 31.

Q20: m/z 31 is the base peak in spectrum Z, and m/z 43 is also a major peak. In otherspectra they are less prominent. Since these two fragments are easily produced fromthe structure of butan-1-ol, spectrum Z applies to this.

Q21: Spectrum X is for 2-methylpropan-2-ol, which loses a CH3 adjacent to C-OH toleave the base peak m/z 59.

Spectrum Y is for butan-2-ol, which loses CH3CH2 adjacent to C-OH to leave the basepeak m/z 45.



Spectrum Z is for butan-1-ol, which loses CH3CH2CH2 leaving CH2OH as the m/z 31base peak.


Q22: 30.0468

Q23: 30.0105


Q24: C4H4O

Q25: 136

Q26: Yes, it is twice C4H4O

Q27: C8H8O2

Q28: 105


Q29: CH3O

Q30: C6H5

Q31: CO

Q32: C6H5COOCH3

Q33: methyl benzoate

Further answers


Q34: C6H8SN2O2



9 Infrared and Nuclear magnetic resonance spectroscopy and X-raycrystallography


Q1: a) HCl

Q2: c) HI

Q3: a) HCl


Q4: b) 2930


Q5: NaCl and KBr consist of a lattice of ions held together by ionic attraction. Theyhave no covalent bonds, like those which are present in organic molecules and whichare responsible for IR absorption. At the wavelengths studied, therefore, they aretransparent to the IR.

Glass, for example, absorbs IR light at wavenumbers less than about 2500 cm-1. This isgood for greenhouses since the more energetic (higher wavenumber) infrared from theSun’s radiation can enter, but the lower energy infrared from warm plants etc. cannotescape.


Q6: 1700


Q7: 3400


Q8: b) Y

Q9: b) Y

Q10: aldehyde

Q11: c) Benzonitrile (C6H5CN)


Q12: a) Aspirin




Q13: b) They contain no 1H which would swamp the NMR spectrum of the sample withits signal.


Q14: a) 1


Q15: methyl

Q16: hydroxyl


Q17: 3

Q18: 1

Q19: c) 3 to 1


Q20: 2

Q21: 5:3

Q22: d) ArH

Q23: c) ArCH3


Q24: Toluene, C6H5CH3, with 5 aromatic Hs and 3 methyl Hs.

Q25: It is BrCH2CH2OH, 2-bromoethanol.

Your table should look like:

Peak 1 Peak 2 Peak 3Æ (ppm) 4.1 3.9 3.5

Type of H -OH RCH2O CH2Hal

Number of Hs 1 2 2Group -OH -CH2- -CH2Br

Reasoning - There are 3 types of Hs, at Æ values 4.1, 3.9 & 3.5 in ratio 1:2:2 respectively(since there are 5 Hs in all, these are the actual H atoms in each group).



From the correlation chart, and noting that the formula contains a bromine and anoxygen atom:

the single H atom (Æ 4.1) could be -OH, (range Æ 3.0 - 6.0 for R.O-H);

the two H atoms (Æ 3.9) could be CH2 next to the OH, (range Æ 3.5 - 4.0 for RCH2O-);and

the two H atoms (Æ 3.5) could be CH2Br, (range Æ 2.0 - 4.0 for RCH2Hal.

These suggest BrCH2CH2OH, which agrees with the absence of any methyl Hs(Æ 0.8 - 1.8) unconnected to Br or O atoms.


Q26: c) 3

Q27: b) 4:5:6

Q28: For the 6 CH3, Æ range 0.8 - 1.3 (actually found at 1.2);

for the 4 CH2 - N, Æ range 2.5 - 3.0 (actually found at 3.5);

for the 5 aromatic H, Æ range 6.5 - 8.3 (actually found at 7.2).

Q29:

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Q30: 10

Q31: 18

Q32: b) Chloride

Q33: c) Both the same




Q34: hydrogen


Q35: d) oxygen

Q36: d) oxygen


Q37: b) hydrogen



10 Medicines


Q1: b) B


Q2: d) glucose

Q3: d) oxidation

Q4: a) esterification


SCHOLAR Study Guide SQA Advanced Higher Chemistry Unit 3 ... · SCHOLAR Study Guide Unit 3:...

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