Organic Chemistry Carbon - Honourscheminnerweb.ukzn.ac.za/Files/Class Slides with Solutions...

10/4/2011

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CHEM 120 CHEMICAL REACTIVITY

ORGANIC CHEMISTRY

19 Lectures5 Tutorials

2 Quiz1 Test

Dr. Vincent O. Nyamori

http://cheminnerweb.ukzn.ac.za/Firstyear/chemonetwenty.aspx

worksheets

Textbooks

Recommended books

H. Hart, L.E. Craine, D.J. Hart and C.M. Hadad, Organic Chemistry - A

Short Course, 12th Edition, Houghton-Mifflin (students not intending to

continue with Chemistry).

P.Y. Bruice, Organic Chemistry, 6th Edition,

Pearson/Prentice Hall (Chemistry major students).

Any other relevant Organic textbook - Library

Prescribed book

T.L. Brown, H.E. LeMay Jr, B.E. Bursten, C.J. Murphy,

S. Langford and D. Sagatys, Chemistry: The Central

Science: A Broad Perspective, 2nd Edition, Pearson,

Australia, 2010.

Organic Chemistry• The study of carbon‐containing

compounds and their properties.

• The vast majority of organic

compounds contain chains or rings of

carbon atoms.

3

SurfactantC17H35COO

‐Ascorbic acidHC6H7O6

GlucoseC6H12O6

Carbon1s22s22p2

Periodic Table4

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2

Structure of Carbon Compounds

C C C C C C

Three hybridization states

sp3 sp2 sp

each satisfies the octet rule for each carbon!

1.54 Å1.20 Å

1.33 Å

“Chemistry: The Central Science: A Broad Perspective”, 2nd

Edition, by LeMay et al. Chapter 21.1 & Chapter 23.1 pg 852 5

“Organic Chemistry: A Short Course” 12th

Edition, by Hart et al. Chapter 1.14 ‐ 1.18

Geometries of carbon compounds

sp3

Tetrahedral

sp2

Trigonal planar

spLinear

C

C

6

Hydrogens are in a tetrahedral

arrangement around the sp3

hybridized carbon atom.

Hydrogens bond to the carbon sp3

orbitals with 1s orbitals.

Methane: CH4

Energy

sp3

2p

2s

1s

Hybridization

sp3 Hybridizationcarbon

7

Energy

sp2

2p

2s

1s

Hybridization

sp

2p

2s

1s

Hybridization

Energy

Ethene: C2H4

Ethyne: C2H2

sp2 Hybridization

sp Hybridizationcarbon

carbon

8

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HYDROCARBONS

• compounds composed of only carbon and hydrogen

• chain of carbon atoms bonded to enough hydrogen atoms

to satisfy the octet rule for each carbon

• chain is bent because of the 109.5° C–C–C tetrahedral angle

C

CC

CC

CC

H

H

H

H

HH

H

H

HH

CH3

CHHC

CH2

CC

H3C

e.g.

sp3

sp

sp2

Line notation

180°

109.5°

120°

9

Π = θ =

3 bonds16 bonds

How many bonds are 10

• They contain the maximum number of hydrogen atoms

• UNSATURATED hydrocarbons

HYDROCARBONS

• Hydrocarbons with all single carbon‐carbon bonds(no double or triple bonds)

• Alkanes are SATURATED

‐ they ARE NOT ALKANES

• contain carbon‐carbon multiple bonds

Alkanes

Alkenes, alkynes and aromatic compounds

“Chemistry: The Central Science: A Broad Perspective” ‐ Chapter 21

“Chemistry: The Central Science: A Broad Perspective”

‐ Chapter 23 & 24

Past exam Question

C‐1: _____ C‐3: _____

C‐2: _____ C‐4: _____

sp2 sp

sp3 sp3

11

Example

CC

C

CC

CC

C

O

CN

F H

O

ClH

HBr

H

H

H

12

3

46

78

910

1. Indicate the hybridization for carbons 1 – 10 and theirrespective geometry. Include bond angles in your answer.

Solution

Carbon 1, 2 and 6:

Carbon 3, 4 and 9: sp2

Carbon 7, 8 and 10:

sp3

sp

Hybridization Geometry Bond angles

Tetrahedral 109.5°

Trigonal planar 120°

Linear 180°12

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Hydrocarbons

• Four basic types:

‐ Alkenes

‐ Alkynes

‐ Aromatic hydrocarbons

‐ AlkanesCnH2n+2

C2H6 Ethane

CnH2n

C2H4 Ethene

C2H2 Ethyne

CnHn

CnH2n‐2

HH

H

H H

H

C6H6 Benzeneor

120°

Organic Nomenclature

• Three parts to a compound name:

1 2 3

1. Prefix

2. Base

3. Suffix

Chapter 2: “Organic Chemistry”, 5th Edition , Bruice P. Y.

Chapter 21: pg. 807‐9: “Chemistry: The Central Science: A

Broad Perspective”, 2nd Edition, LeMay et al.


Suffix:

Tells what type of compound it is.

suffix

What family?

15


Base/parent:

Tells how many carbons are in the longest continuous chain.

base suffix

How many carbons?

What family?

16

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Organic Nomenclature ‐ IUPAC Rules

Tells what substituent(s) are attached, if any.

prefix base suffix

What substituent? How many

carbons?

What family?

Prefix:

17

Alkanes

• Only van der Waals force: London force.

• Boiling point increases with length of chain.

• Combust to give mainly CO2 and H2O

• Nomenclature suffix “‐ane” E1

Properties

To Name a Compound…1. Determine what type of

compound it is. ‐ Suffix

2. Find the longest chain in the

molecule. ‐ Base

3. Number the chain from the

end nearest the first

substituent encountered.

4. List the substituents as a prefix

in alphabetical order along

with the number(s) of the

carbon(s) to which they are

attached.

CH2CH

CH2H2C

CH2

H3C

H3C

CH32‐Ethylhexane3‐Methylheptane Name?? 19

Example

If there is more than one type of

substituent in the molecule, list them

alphabetically i.e. name of substituent,

not prefix for frequency e.g. di, tri,

tetra, etc...are not considered.

CH3

CH2CH

CH CH

CHCH3

CH2 CH3

CH3

CH3

CH3

3‐Ethyl‐2,4,5‐trimethylheptane

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Cycloalkanes

• Carbon can also form ringed structures.

• Five‐ and six‐membered rings are most stable.

– Can take on conformation in which angles are

very close to tetrahedral angle.

– Smaller rings are quite strained.

cyclohexane cyclopentane cyclopropane

General formulaCnH2n

CYCLIC ALKANES

How do we name….CH3

CH2CH3

methyl

ethylEthylmethylcyclohexane

1

23

4

56

1‐Ethyl‐3‐methylcyclohexane

22

Unsaturated Hydrocarbons

Carbon can form multiple bonds with itself or other

atoms, e.g. N or O.

Multiple bonds affect, physical and chemical

properties (reactivity of organic molecules).

Hydrocarbons that contain one or more double or

triple bonds are called unsaturated hydrocarbons.

C

CC

C

CC

C C

C C

H

H

H

H

H H

H

H H

H

H

H

CH3C

O

CH3

CH3C

N

CH3

CH3C

S

CH3

H

Alkenes

• Contain at least one carbon–carbon double bond

• Unsaturated

– Have fewer than maximum number of hydrogens

– The C atoms on double bond are sp2 hybridized

VSEPR Theory

120°

Chapter 23: “Chemistry: The Central Science: A Broad Perspective”

2nd Edition, by LeMay et al.

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Structure of Alkenes

• Unlike alkanes, alkenes cannot rotate

freely about the double bond.

– Side‐to‐side overlap makes this

impossible without breaking ‐bond.

C C

25

Structure of Alkenes

This creates geometric isomers

difference in the spatial arrangement of groups about the double bond

Cis‐ isomer “Z”‐isomer

Trans‐ isomer “E”‐isomer

Z‐2‐Pentene

E‐2‐Pentene

26

27

NAMING ALKENES

1. Find the longest unbranched carbon chain containing the

double bond.

2. Number the carbon atoms in the main chain.

• Name chain according to number of carbon atoms.

add ‐ene as a suffix

• Start from the end of the chain that is closest to the

double bond.

location of the double bond is numbered with the

lowest‐numbered carbon in the double bond.28

1. The longest unbranched chain containing the double

bond is seven carbons long, so this is a heptene.

Carbon #2 is the lowest‐numbered carbon in the

double bond, so this is a 2‐heptene or hept‐2‐ene

1 23 4

5 6 7Name this alkene

Example

C

H

CHCH3 C

H

CH

CH2

CH2

CH

CH3

CH3

H2C

2. The chain numbering starts closest to the double bond.

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3. There are two substituent groups on this alkene:

C

H

CHCH3 C

H

CH

CH2

CH2

CH

CH3

CH3

H2C

1 23 4 5 6 7

5‐methyl

4‐cyclopropyl

hept the alkene is:

Compose the name…..

4‐cyclopropyl‐5‐methyl‐2‐ ene

• Add the substituent groups alphabetically to the alkene name

• Specify the position of each group on the main chain

29

Properties of Alkenes

Structure also affects physical properties of alkenes

2‐Methyl‐1‐propenebp. ‐7 ⁰C

1‐Butenebp ‐6 ⁰C

Cis‐2‐Butenebp +4 ⁰C

Trans‐2‐Butenebp +1 ⁰C

Example: C4H8

Can we have more than one double bond? 1,3‐butadiene

Chapter 22.2: “Chemistry: The Central Science: A Broad Perspective” 2nd Edition

• Identical substituents on

opposite sides of the double

bond

CCH3

H

CH

CH3

C

CH3

HC

H

CH3

Alkenes exhibit cis‐trans isomerism.

GEOMETRICAL ISOMERS

• Identical substituents on

same side of the double

bondtrans‐ cis‐

Stick diagram

Trans‐ isomer “E‐” Cis‐isomer “Z‐”

31

Priorities are assigned by the atomic numbers of the atoms

bonded to the carbon in the double bond.

“Z‐”“E‐”

Geometric isomers of Alkenes

• Cis‐alkenes have similar higher priority elements or group in

the chain on the same side of the molecule (or Z‐isomer i.e.

have higher priority elements but not necessarily the same on

the same side of the molecule)

• Trans‐alkenes have similar higher priority elements or group in

the chain on opposite sides of the molecule (or E‐isomer i.e.

have higher priority elements but not necessarily the same on

opposite sides of the molecule).

2‐Pentene 32

2‐Pentene

1‐Pentene

(Z)‐

(E)‐

e.g.

C5H10

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Examples1. Name the following alkenes and determine whether there

are geometric isomers are either Trans‐ (E‐) or Cis‐ (Z‐)isomers if applicable.

H

a)

3‐Methylhex‐3‐ene

Br

Hb)

3‐Fluoro‐2‐methylpent‐2‐eneF

c)

(E)‐

1 2

1

1

1

2 2

2

Equal priority

No geometric isomers formed.

3‐Bromopent‐2‐ene

33

(Z)‐

34

Each functional group is specified by a suffix or prefix

Certain groups of atoms give a molecule a.... FUNCTION

Acidic, basic, alcohol, etc…

The GROUPS are called functional groups.

depicted on the nomenclature of the organic molecule

FUNCTIONAL GROUPS

34

Functional groups are given an order of priority to decide

on which is the suffix.

HOMEWORK!

• “Chemistry: The Central Science: A Broad Perspective”, 2nd Edition, by LeMay et

al. Chapter 24.1, 25.1, 26.1 & 26.3

• “Organic Chemistry” 5th Edition, by Bruice

• Hart et al. “Organic Chemistry: A Short Course” 12th Edition

Hint: Please refer to your textbooks !!

Group / Family

FormulaStructural Formula

Prefix Suffix Example

Alkane RH alkyl- -ane

Alkene R2C=CR2 alkenyl- -ene

Alkyne RC≡CR' alkynyl- -yne

Benzene derivative

RC6H5

RPhphenyl- -benzene

Ethane

2-phenylpropaneisopropylbenzene

Ethene

Ethyne

Group / Family



Haloalkane RX halo-alkyl

halide

Fluoroalkane RF fluoro-alkyl

fluoride

Chloroalkane RCl chloro-alkyl

chloride

Bromoalkane RBr bromo-alkyl

bromide

Iodoalkane RI iodo-alkyl

iodide

Chloroethane

Ethyl chloride

Fluoromethane

Methyl fluoride

Chloromethane

Methyl chloride

Bromomethane

Methyl bromide

Iodomethane

Methyl iodide

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Group / Family

Group FormulaStructural Formula


Amines

Primary amine

(1º)RNH2 amino- -amine

Secondary amine

(2º)R2NH amino- -amine

Tertiary amine

(3º)R3N amino- -amine

Quaternaryammonium

ionR4N+X-

ammonio--ammonium

H3C

N

H CH3

CH3

Cl

Methylamine

Dimethylamine

Trimethylamine

Trimethyl-ammonium

chloride

Group / Family



Alcohol ROH hydroxy- -ol

Ketone RCOR' keto-, oxo- -one

Aldehyde RCHO aldo- -al

Methanol

2-ButanoneMethyl-

ethyl ketone

EthanalAcetaldehyde

Group / Family



Carboxylic acid

RCOOH carboxy- -oic acid

Acyl halide RCOX haloformyl- -oyl halide

Ether ROR' alkoxy-alkyl alkyl

ether

Ester RCOOR'alkyl

alkanoate

Ethanoic acidAcetic acid

Ethanoyl chlorideAcetyl chloride

Ethoxy ethaneDiethyl ether

Ethyl butanoateEthyl butyrate

Primary (1°) alcohols and amines

General structure

AlcoholR1 C

H

H

OH

Amine

Example

1° alcohols

1° amine

C

H

H

OHCH3CH2

N

H

H

CH3CH2R1 N

H

HEthylamine

Propan‐1‐ol

1‐Propanol

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Secondary (2°) alcohols and amines

General structure

AlcoholR1 C

H

R2

OH

Amine

Example

2° Alcohol

2° Amine

C

H

CH3

OHCH3CH2

N

H

CH3

CH3CH2R1 N

H

R2

Ethylmethylamine

Butan‐2‐ol

2‐Butanol

Tertiary (3°) alcohols and amines

General structure

AlcoholR1 C

R3

R2

OH

Amine

Example

3° alcohols

3° Amine

C

CH3

CH3

OHCH3CH2

N

CH3

CH3

CH3CH2R1 N

R3

R2

Ethyldimethylamine

2‐Methylbutan‐2‐ol

2‐Methyl‐2‐butanol

42

Quaternary amines

General structure

AmineR4 N

R2

R1

R3+ N

CH3

CH3

CH3CH3CH2

+

Example

Ethyltrimethylammonium ion

43

Exercise

1. Draw the structures of the following alcohols and amines

and classify them as either 1°, 2°, 3° or quaternary

a) Pentan‐1‐ol

b) Dimethylamine

c) 3‐Ethylhexan‐3‐ol

d) Diethylmethylamine

e) Butan‐2‐ol

f) Triethylmethyl ammonium ion

44

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CC

CC

CHO

H

HH

H

H

HH

H

H

HH

a) Pentan‐1‐ol

Solutions

N

H

CH3

CH3

b) Dimethylamine

OH

CH3

CH2 C

CH2

CH2

CH3CH2

CH3

c) 3‐Ethylhexan‐3‐ol

N

CH2CH3

CH2CH3

H3C

d) Diethylmethylamine

C

OH

CH3

H CH2CH3

e) Butan‐2‐ol

N

CH2CH3

CH2CH3

H3C

CH3CH2

f) Triethylmethylammonium ion

1° alcohol 2° amine3° alcohol

3° amine 2° alcoholQuaternary amine

A45 46

Naming Hydrocarbons with Functional Groups

Name the other substituent groups, using the

prefixes for alkyl groups and the prefixes for any

other functional groups

Specify the position of each group on the main chain.

Add the substituent groups alphabetically to the name

of the alkane (or alkene or alkyne) along with the

frequency of each group.

Example…..

47

Example: Name this organic molecule:

CH3 CHCH

OH

CH3

NO2

48


1. This molecule contains a hydroxyl and a nitro group

CH3 CHCH

OH

CH3

NO2

Only the hydroxyl group has priority.

So this is an ALCOHOL…..

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49


1. This molecule contains a hydroxyl group and a nitro

group

longest carbon chain containing the hydroxyl

group has four carbons.

Therefore its a “butanol”.

CH3 CHCH

OH

CH3

NO2

this is an ALCOHOL.

An alcohol suffix is ‐OL

50

3. Number the carbons, starting NEAREST the

functional group.

CH3 CHCH

OH

CH3

NO2 Butanol

1 2 3 4

The hydroxyl group is on position 2

…so this is a 2‐butanol or butan‐2‐ol

4. This molecule has one substituent

A nitro group on position 3

3‐nitro‐2‐butanol 3‐nitrobutan‐2‐olor

51

Example: Name this organic molecule

CH3 CH CH CH

CH3

C

CH2

CH2C

CH2

CH3

CH3

CH3

OH

O

H2N

1. Identify functional groups

carboxyl group amino group

52


CH3 CH CH CH

CH3

C

CH2

CH2C

CH2

CH3

CH3

CH3

OH

O

H2N

1. Molecule contains a carboxyl group and an amino group.

a carboxylic acid

Highest priority

Now determine longest carbon chain

The carboxyl group has highest priority, hence

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53


CH3 CH CH CH

CH3

C

CH2

CH2C

CH2

CH3

CH3

CH3

OH

O

H2N

The longest carbon chain has 8 carbons.

BUT………………...

THIS CHAIN DOES NOT CONTAIN THE ‐CO2H group

54

Example: Name this organic molecule

CH3 CH CH CH

CH3

C

CH2

CH2C

CH2

CH3

CH3

CH3

OH

O

H2N

The longest carbon CONTAINING the ‐CO2H

chain has 7 carbons.

so this molecule is based on a heptane.

55

Highest priority functional group is a carboxyl group

CH3 CH CH CH

CH3

C

CH2

CH2C

CH2

CH3

CH3

CH3

OH

O

H2N

Suffix ‐OIC ACIDHence… Heptanoic acid

Now number chain…….

3. Number the carbons starting with the functional group

123

4

5 6 7

CH3 CH CH CH

CH3

C

CH2

CH2C

CH2

CH3

CH3

CH3

OH

O

H2N

Heptanoic acid

The carboxyl group is on position 1,

do not include in the name because

the carboxyl group is always a terminal group.56

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57

Compose the name

2‐ethyl3‐isopropyl

5‐amino

5‐methyl

CH3 CH CH CH

CH3

C

CH2

CH2C

CH2

CH3

CH3

CH3

OH

O123

4

5 6 7H2N

heptanoic acid5‐methyl3‐isopropyl‐2‐ethyl‐5‐amino‐

ALPHABETICAL LIST

INTERPETING AN IUPAC NAME…...

Heptanoic acid

4. This molecule has four substituents

i.e. at carbons 2, 3 and two at 5

58

WHAT IS THE STRUCTURAL FORMULA OF

BUTANONE ?

4 CARBONS

KETONE GROUP

CH3 CH2C

O

CH3

CANNOT BE AT END!!!! WHY?????

C CH2CH2

O

CH3H

BUTANAL

Examples

Give the correct IUPAC name for the following compounds.

(Z)‐3‐methylhex‐2‐ene

2,3,7‐trimethyloctane

Br

H

6‐bromo‐4‐ethyl‐2,3,7‐trimethyloctane

3,3‐dimethylbutan‐2‐ol

OH

Alkane

BromoalkaneAlkylbromide

Alkene

Alcohol(2°)

a)

c)

b)

d)

59

O

H

O

CH3

O

Cl

OH

2‐chloropentanoic acid ethylbutanoate

3‐methylhexan‐2‐one2‐ethylpentanal

Examples

Give the correct IUPAC name for the following compounds.

f) g)

h) i) O

O

Aldehyde Ketone

Carboxylic acid Ester

60

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Examples

Draw structural formulae for the following compounds:

a) 2,3,5‐trimethylhexane

b) (Z)‐3‐chlorohept‐2‐ene

c) 3‐ethylnonanol

d) 2,3‐dimethylpentanoic acid

e) methyl hexanoate

f) 3‐iodohexanal

g) pentan‐2‐one

h) 3‐aminopentane 61

Solutions

2,3,5‐trimethylhexane

Cl

H

(Z)‐3‐chlorohept‐2‐ene

OH

3‐ethylnonanol

O

OH

2,3‐dimethylpentanoic acid

a) b)

c) d)

62

Solutions

methyl hexanoate 3‐iodohexanal

pentan‐2‐one

e) f)

g) h)

O

O

OI

H

OH2N

3‐aminopentane

63

ISOMERS

(a) Structural isomers

CONSTITUTIONAL ISOMERS

Constitutional isomers have different properties:

These are…..?

e.g. butane (C4H10) has 2 structural isomers

Molecules with the same chemical formula but different

bonds between the atoms

Now called...

CH3 CH2 CH2 CH3 CH3 CH CH3

CH3

n‐Butane: C4H10 2‐methylpropane: C4H10

bp = ‐12 °C mp = ‐159 °Cbp = 0 °C mp = ‐138 °C

Two types: (a) Structural isomers (b) Stereoisomers

64

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65

CH2CH3 CH2 CH2 CH3

n‐pentane

Solution:

2‐methylbutane

1) 2) CH3 CH CH2

CH3

CH3

CCH3 CH3

CH3

CH3

2,2‐dimethylpropane

3)

ExampleHow many constitutional isomers are formed from C5H12? Draw their structures.

THREE CONSTITUTIONAL ISOMERS

CONSTITUTIONAL (STRUCTURAL) ISOMERS

The general formula for ALKANES is…... 22 nnHC

The number of ISOMERS increases with n…..

n = 1, 2 and 3 1 ISOMER

n = 4 2 ISOMERS

n = 5 3 ISOMERS

n = 6 5 ISOMERS

n = 7 9 ISOMERS

n = 8 18 ISOMERS

n = 9 35 ISOMERS

n = 10 75 ISOMERS

n = 20 366,319 ISOMERS

n = 40 62,491,178,805,831 ISOMERS

HOMEWORK: DRAW THE ISOMERS OF C40H82 !!! 66

Constitutional isomers for multibonds

(E,E)‐2,4‐hexadiene

(E)‐1,3‐hexadiene

(Z,E)‐2,4‐hexadiene

(Z)‐1,3‐hexadiene

(Z,Z)‐2,4‐hexadiene

C6H10

Example: Alkene

1,5‐hexadiene67 68

Constitutional isomers for multibonds

Example: Alkyne

C6H6

H3C C C C C CH3

C C C C CH2

CH3

H

1,3‐hexadiyne

hexa‐1,3‐diyne

2,4‐hexadiyne

hexa‐2,4‐diyne

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69

OPTICAL ISOMERISM

Optical isomerism arises when molecules have a

structure such that the mirror image is not

superimposable on the original molecule.

occurs whenever there are four different groups

bound to the same tetrahedral carbon atom.

Some terminology…….

70

OPTICAL ACTIVITY IN ORGANIC COMPOUNDS

The carbon involved is called a chiral carbon or

stereogenic carbon and the molecule is known as a chiral

molecule.

Stereogenic centre has four different groups attached

to a tetrahedral carbon atom

Example: 2‐butanol

ZX

W

Y

C

Stereogenic centre

Chiral carbon atom

*

C

CH2CH3

OH

H

H3C

2‐ButanolDash shows bond going

backwards from the viewer

*

Solid wedge represents a bond extending out towards the viewer

Simple line for Bonds aligned to the asymmetric

center in the plane

Chiral centre

Perspective formula

71

Wedge‐dash notation

CH2CH3HO

H

CH3

72

2‐Butanol

*

Fischer projection

Bonds aligned to the asymmetric

center in the plane

Chiral centre

Bond going backwards from

the viewer

Bond extending out towards the

viewer

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(non‐superimposable mirror images of each other)

View in 3‐DView in 3‐D“mirror”

• 2‐butanol has two optical isomers.

• A pair of isomers called enantiomers ‐

C

CH2CH3

OH

H

H3C

C

OH

HCH3CH3CH2

ENANTIOMERISM in ORGANIC CHEMISTRY


E

Identify the chiral carbon (stereogenic centre) and draw the

structural formula for each of the following molecules:

Solution

(a) 1‐chloroethanol (b) 2,3,5‐trimethylhexane

(c) Methylcyclohexane (d) 1,3‐dimethlycyclopentane

C CH3

Cl

HO

H

(a) 1‐chloroethanol

*

*

* *No chiral carbon

Question

(b) 2,3,5‐trimethylhexane

(d) 1,3‐dimethlycyclopentane(c) methylcyclohexane

74

75

NAMING OPTICAL ISOMERS

Solution: Use R‐S nomenclature system for designating

the configuration

Stereogenic centre creates twomolecular optical isomers

We assign priorities as in the E, Z system……...

How do we name these isomers??

CH

Cl

OHH3C

CH

Cl

HO CH3

Two configurations

Enantiomers* *

1. Assign relative priorities to each of the four groups on

the stereogenic carbon to describe the configuration.

The priorities are given by rules:

• Higher atomic numbers are given higher priorities.

• If necessary, the second atom in each substituent is

used to determine the priorities.

CAHN‐PRELOG‐INGOLD R,S‐NOMENCLATURE


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20

2. Draw the molecule with the lowest‐priority group

pointing directly into the page….

and the other three groups pointing out of the page in an

arrangement like a steering wheel orMercedes‐Benz symbol

Example: 2‐Butanol…..

CAHN‐PRELOG‐INGOLD R,S‐NOMENCLATURE

CH2CCH3

OH

CH3

H

Draw molecule as a wedge and dashed line diagram

C

78

H3CCH2CH3

C

OH

HH3CH2C

C

OH

CH3

H

2‐ butanol

DO NOT FORGET THE OTHER ISOMER……….

CH2CCH3

OH

CH3

H

THESE ARE THE TWO…. ENANTIOMERS...or ....OPTICAL ISOMERS

p

Chapter 22; pg. 829‐839. “Chemistry: The Central Science: A Broad Perspective” 2th Edition.

79

1. Assign priorities:

O > CMe = CEt > H

C,H,H > H,H,H

CEt > CMe

2‐ butanol

C

CH3

CH2CH3

OH

H

80

CH3

CH2CH3

C

OH

H

1

2

3

4

H3C CH2CH3

C

OH1

2

3

2. Redraw the molecule with the lowest priority group facing in.

Now what?????

O > CEt > CMe > H

1. Assign priorities to each group:

2‐Butanol

10/4/2011

21

81

3. Look at the direction in which the priorities decrease.

If they decrease in a clockwise direction, the stereogenic centre is called “R” or rectus

which is Latin for “right.”Or….

H3C CH2CH3

OH

1

2

3

C

2‐Butanol

(R)‐2‐Butanol

CH3CH3CH2

OH

1

2 3

If the priorities decrease in a counter‐clockwise

direction the stereogenic centre is called “S” or

sinister, which is Latin for “left.”

C

2‐Butanol

(S)‐2‐Butanol

82

A83

Example:

CH3

CH2CH3

C

NH2

HCH3C

CH2CH3H

NH2

(S)‐2‐butanamine(R)‐2‐butanamine

H3C CH2CH3

C

NH2

3 2

1CH3

CH2CH3

C

H2N

1 2

3

N > CEt > CMe > H

What are the configurations of the following chiral molecules?

84

Example: Give the configuration of the stereogenic

centre in each of the following molecules:

CH

CH2OH

ClC OH

O

CHCH2OH

Cl

HO C

O

THEY LOOK DIFFERENT BUT ARE THEY???

(A) (B)

2‐chloro‐3‐hydroxy‐propanoic acid

10/4/2011

22

85

A)

CH

CH2OH

ClC OH

O


Cl > Calc = Ccarb > H

O,O,O > O,H,H

Go to next atoms on tied carbons

Ccarbox > Calc

Cl > Ccarb > Calc > H

86

A)

CH

CH2OH

ClC OH

O1

3

24


Cl > Ccarb > Calc > H

86

ClC

1

2

3 CH2OH

C OH

O

The priorities decrease anti‐clockwise, so this centre is “S”

(S)‐2‐chloro‐3‐hydroxy‐propanoic acid

2. Redraw the molecule with the

lowest priority group facing in.

3. Look which way the priorities decrease.

1. Assign priorities to each group

Cl > Ccarb > Calc > H CHCH2OH

Cl

HO C

O

B)

2

1

4

3

2. Redraw the molecule with the

lowest priority group facing in.

C

32

Cl

CH2OHHO C

O

1

3. Direction of the priorities

decrease?

The priorities decrease anti‐clockwise, so this centre is “S”

(S)‐2‐chloro‐3‐hydroxy‐propanoic acid 88

Example: Give the configuration of the stereogenic

centre in each of the following molecules:

THEY LOOK DIFFERENT BUT ARE THEY???

2‐chloro‐3‐hydroxy‐propanoic acid

WE JUST FOUND OUT THAT THEY ARE IDENTICAL!!!!!

WHAT IF?????

CHCH2OH

Cl

HO C

O

CH

CH2OH

ClC OH

O

B)A)

10/4/2011

23

If we flip the COOH and H ???????

CH

CH2OH

ClC OH

O

A)

C H

CH2OH

Cl

C

O

HO

A’)

ClC

1

2

3 CH2OH

C OH

O

C

2

1

3 CH2OH

C

O

HO Cl

“S” configuration “R” configuration89 90

What is the structural formula of (R)‐2‐chloro‐2‐butanol?Hint: Make use of perspective formula or dash‐wedge notation.

DRAW MOLECULAR STRUCTURE

CH2CH3C

OH

CH3

Cl

ASSIGN PRIORITIES

1

3

2

4

EXAM QUESTION

Solution:

91

(R)‐2‐chloro‐2‐butanol

CH2CH3C

OH

CH3

Cl

DRAW MOLECULE…..

1

3

2

4

Draw steering wheel with lowest priority group pointing in…….

MAKE SURE PRIORITY GOES

CLOCKWISE FOR “R” configuration

Cl CH2CH3

OH

1

2

3

C

Assign prioritiesDRAW MOLECULE TO SEE

ALL GROUPS…..

Make sure priority goes clockwise for “R”

Cl

OH

H3CCH2CH3

C

Cl CH2CH3

OH

1

2

3

C

(R)‐2‐chloro‐2‐butanol

Remember lowest priority group has a “dashed” bond

92

Perspective formulaor

Dash‐wedge notation

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24

enantiomeric pairs enantiomeric pairs

[2R,3R] [2S,3S] [2R,3S] [2S,3R]

• Enantiomeric pair differ only in optical activity

a b c d

R

R

C

C

CH3

CH3

H

Cl

Br

H

S

S

C

C

CH3

CH3

Br

H

H

Cl

R

S

C

C

CH3

CH3

H

H

Br

ClR

SC

C

CH3

CH3

Br

Cl

H

H

CHCH3 CH

Br Cl

CH3* *

93Chapter 22.2, page 838‐9: “Chemistry: The Central Science: A Broad Perspective”

R

S

C

C

CH3

CH3

H

H

Br

ClR

SC

C

CH3

CH3

Br

Cl

H

H

R

R

C

C

CH3

CH3

H

Cl

Br

H

S

S

C

C

CH3

CH3

Br

H

H

Cl

a b c d

Diastereomers ‐ are distinct chemical compounds, differing not only in optical activity but also in mp, bp., , solubility etc…

Diastereomers b & ca & c

b & da & d

molecule with n stereogenic centres may exist inmaximum of 2n stereisomeric forms, with maximum of2n/2 enantiomeric pairs 94

meso compound ‐ an achiral (optically inactive)

diastereomer of compound with stereogenic centres

arises because 4 different groups making each of C‐2 & C‐3

stereogenic are same 4 different groups…(!)

Meso compoundCHHOC CH COH

OH OH

OO* *

R

R

C

C

CO2H

CO2H

OH

H

H

HO

S

S

C

C

CO2H

CO2H

H

OH

HO

H

R

S

C

C

CO2H

CO2H

OH

OH

H

HR

SC

C

CO2H

CO2H

H

H

HO

HO

170 °C 170 °C 140 °C

+12° ‐12° 0°

95

Meso compounds

Enantiomeric pairs

possess plane of symmetry bisecting central C‐C bond

Meso compound

R

R

C

C

CO2H

CO2H

OH

H

H

HO

S

S

C

C

CO2H

CO2H

H

OH

HO

H

R

S

C

C

CO2H

CO2H

OH

OH

H

HR

SC

C

CO2H

CO2H

H

H

HO

HO-------------------- -------------------- -------------------- --------------------

Enantiomers, Chiral Identical, achiral, Meso form

96

10/4/2011

25

Shows arrangements in space

Newman projections

e.g. Ethane C2H6

H

H

H

HH

H

“dash‐wedge”

H H

HHH H

“dash‐wedge”

HH

HH

HH

H

HHH

H H

“sawhorse”

“sawhorse”

60°

0°

Newman

Newman

Staggeredconformation

Eclipsed conformation

H

H

H

H

H

H

HH

HHH

H

97

a

e

e

ae

a

aa

e

e

a

e

"flip"16

5

43

2

a

e

e

a e

a

aa

e

e

a

e

16

5

43

2

H

H H

HH

H

HH

H

HH

H

Chair conformations

Boat conformation

Cyclohexane conformations

Cyclohexane

98

Examples1,2‐Dimethylcyclopentane

CH3 CH3

H H

H

H

H

H

H H

CH3

CH3H

H H

H

H

H

H H

Cis‐1,2‐Dimethylcyclopentane Trans‐1,2‐Dimethylcyclopentane

99

Different bond pattern

isomers

Structural (constitutional)

isomer

Stereoisomer

Interconvertible by single bond rotation

Not interconvertibleby bond rotation

Same bond pattern

Conformers (rotamers)

Configurational isomers

Summary of isomerism

Hart et al. “Organic Chemistry: A Short Course” 12th Edition, page 52‐54

100

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26

Exercise

1. Draw the structures of

a) (Z)‐3‐methyl‐2‐pentene c) (E,Z)‐2,4‐heptadiene

b) (S)‐2‐bromopropan‐1‐ol d) (2S,3R)‐3‐bromobutan‐2‐ol

2. Using the Newman projection draw the structure of a

staggered conformation of butane.

3. Using the Fischer projection draw a structure of

(S)‐2‐methylpentanoic acid.

4. Draw the structure for the cis‐ and trans‐isomers of

1‐bromo‐2‐chlorocyclopropane 101

Solutions

1. (a) (Z)‐3‐methyl‐2‐pentene

CH3

H2C CH3

H3C

H

(b) (S)‐2‐bromopropanol

CC

CH3

Br

OHH

H

H

C

C

CH3

Br

H3C

H

HHO

(c) (E,Z)‐2,4‐heptadiene (d) (2S,3R)‐3‐bromobutan‐2‐ol

102

CH 2C H2CH 3

COOH

H

H 3C

CH3

Staggered conformation of butane

Solutions

2.

60°Newman projectionH

H

CH3

H

H

Fischer projection

3.

*

(S)‐2‐methylpentanoic acid

103

(4) 1‐bromo‐2‐chlorocyclopropane

H

H

Br

H

Cl

H

H

H

Br

H

H

Cl

Cis‐1‐bromo‐2‐chlorocyclopropane

Trans‐1‐bromo‐2‐chlorocyclopropane

104


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Substitution reaction: Substitution of one atom or group of

atoms by another atom or group of atoms.

Addition reaction: Addition of one molecule or atom to another

to give a new molecule . normally on double or triple bond.

Elimination reaction: Elimination of two atoms or group of atoms

from a molecule (reverse of the addition reaction).

Oxidation and reduction reactions: Involve the loss or gain of

electron density by a carbon respectively.

Rearrangement reaction: Conversion of one structure to an

isomeric structure.

Five important organic reactions

Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia

Uses

shows redistribution of valence electrons a molecule

defining resonance

chemical reaction direction

defining a reaction mechanism.

Two common types of electron redistribution:

From a bond to an adjacent atom, and

From an atom to an adjacent bond.

Arrow Notation

Homolytic bond breaking Heterolytic bond breaking

Bond forming

A B BA + A B B-A+ +

Homolytic bond forming Heterolytic bond forming

BA + A B

Bond breaking

A BB-A+ +


Terminology

Nucleophile (Nu):- electron rich species and/or negatively

charged species

examples

-OH, Br-, -NH2,

Electrophile (E+):- electron poor species and/or positively

charged species

examples

H+, M+, H3O+

O

C

O

O

CC C

10/4/2011

28

PowerPoint to accompany

Chapter 21

Alkanes: Scaffolds for

Organic Chemistry

Reactions of Alkanes

Because of its structural role, alkanes are relatively

unreactive.

They make excellent nonpolar solvents.

Forcing conditions are required for a reaction

Combustion: give mainly H2O and CO2

Free-radical reactions: chain reactions

In order to be able to predict the product of a reaction, it is

important to be able to distinguish between carbon (and

hydrogen) types in an alkane.

Primary, secondary, tertiary and quaternary carbon atoms.

Primary, secondary and tertiary hydrogens.

Figure 21.25

Classification

hydrogen and carbon atoms in alkanes

Combustion

Combustion of alkane gives CO2 and H2O.

Combustion reactions are exothermic.

2C2H6(g) + 7O2(g) 4CO2(g) + 6H2O(l) ∆H = -2855 kJ

e.g. ethane

Write a balance equation for the combustion of butane

Question

Solution

C4H10(g) + O2(g) CO2(g) H2O(l)+6½ 4 5

2C4H10(g) + O2(g) CO2(g) H2O(l)13 8 10+

Homework Try combustion of pentane C5H12 ??

10/4/2011

29


Combustion - Napalm

The commonly quoted composition is 21% benzene,

33% gasoline, and 46% polystyrene. This mixture is

difficult to ignite.

Movie – Vietnam.


Free-Radical Reactions

Consider the following halogenation reactions:

Reaction 1

Reaction 2

CH4 + Cl2 no reaction

CH4 + Cl2 CH3Cl + CH2Cl2 + CHCl3 + CCl4

dark

Light

λ

• This type of reaction is called a substitution reaction.


The free-radical halogenation is also an example

of a radical chain reaction.

Such reactions are characterised by three steps

Initiation

Propagation

Termination

Initiation produces the first free-radicals by a

homolytic bond cleavage.

Cl Cl λ

Cl2


Propagation steps - a series of new reactions

which generate other radicals

e.g.

radical

radical

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30


Termination steps: stop the formation of free-radicals

neutral molecular species formed

e.g.

radical pair

radical pair

Free-Radical Reactions

Given the opportunity, free-radical reactions are regiospecific.

More so for bromination than chlorination.

3°> 2° > 1°most stable least stable

R

C R = akly group - EDG

1°> 2° > 3°most stable least stable

X

C e.g X = halide - EWG


Chapter 23

Alkenes and Alkynes

Electrophilic addition and Substitution reactions

The presence of carbon-carbon double or triple bonds in a

compound markedly increases its chemical reactivity.

Most characteristic reactions of alkene and alkynes are addition

reactions:

Halogenation

Hydrohalogenation

Hydration

The electron-rich nature of a multiple bond leads to enhanced

reactivity with electrophiles (electron-loving species).

10/4/2011

31


Alkenesreactions

Figure 23.20

Mechanism for HX Addition rxn

Two-steps mechanism

1. First step is slow, rate-determining step.

2. Second step is fast.

slow

H Br

Fast

carbocation

carbocation


Addition of HX

In the first step, -bond breaks and new C—H bond

and cation form.

Figure 23.22

In the second step, a new bond forms between negative

bromide ion and positive carbon. Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia

Regioselectivity

A occurs when a non-symmetric alkene is used.

The preference based on intermediate carbocation stability:

Primary < Secondary < Tertiary

(Less Stable) (Most Stable)

The regioselectivity is stated as Markovnikov’s Rule:

“In the addition of HX to an alkene, the hydrogen adds to the

carbon atom of the double bond bearing the greater number of

hydrogen atoms bonded directly to it.”

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32


Addition of HX

H Br

C C

H

H CH3

CH31 2

Major product

Minor product

Exercise

H 3CC

C H 3

C H 3

CH 2

C H 3

C H 2

+

+

+

+

(A ) (B )

(C ) (D )

1. Which of the following is the most stable carbocation and why?

Soln:

• It is a tertiary (3˚)

carbocation.

• Inductive effect from

electron rich groups

2. What will be the major product from the reaction by reflux in dark

of 2-methylbut-2-ene with hydrochloride in a carbon tetrachloride

solution? Name and draw the structure of the product.

(D)

C C

H3C CH3

H3C H

2-methylbut-2-ene

HCl

CCl4

C C

CH3 CH3

CH3 H

ClH (A)

C C

CH3 CH3

CH3 H

HCl (B)

Soln:

2-chloro-3-methylbutane

2-chloro-2-methylbutane

minor product

major product

Addition of H2O

The hydration of alkenes and alkynes requires the addition of a

strong acid catalyst e.g. H2SO4

The first step is rate determining.

Leads to the formation of a carbocation.

hydronium ion(electrophile)

10/4/2011

33

Second step is fast

Carbocation reaction with excess H2O (nucleophile)

leads to the formation of an oxonium ion.

Loss of H+ from the oxonium ion leads to formation of an alcohol

and regenerates the hydronium ion, hence it is catalytic reaction.

butan-2-ol2-butanol

hydronium ionregenerated

oxonium ion

Addition of H2O (Reversible)

The fact that the reaction is always described by equilibria

means that the overall reaction is reversible.

That is, in the presence of H2SO4, alcohols can be dehydrated to

yield alkenes.


Formation of Alkenes

Previously, we have discussed the formation of alcohols from alkenes.

The reverse process is also possible.

This dehydration reaction is also known as an elimination reaction.

Requires the elimination of a good leaving group.

I- > H2O > Br- > Cl- >> F- > CH3COO- > OH-

Formation of Alkenes

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34

Exercise

1. Draw and name the major product in the following reactions

(a)

(b)

H

OH

H+ / H2O

pentan-2-ol2-pentanol

OH H2SO4

Heatcyclohexene


Halogenation

One of the easiest and most

dramatic reactions is the

addition of Br2 to an alkene.

Colour change

Useful test for alkenes

Figure 23.23

Reactions of Alkynes

Many similar reactions as alkenes.

Reactions involve the replacement of -bonds with -bonds.

C CH3C CH3

Example

Br2

CH2Cl2

C C

H3C

CH3Br

Br

Br2

CH2Cl2C C

CH3

CH3

Br Br

Br

Br

(E)-2,3-dibromobut-2-ene

2,2,3,3-tetrabromobutane

Problems1. What will be the product for a reaction between hex-2-yne and one

mole of chlorine in dichloromethane under anhydrous condition? Draw

the structure of the product and provide it IUPAC name.

2. How would you form 1,1,2,2-tetrabromo-1,2-difluoroethane using

bromine as one of your reagent?

Solutions

C C

F

F

Br Br

Br

Br

C C

F

FBr

BrBr2

CH2Cl2 C CF FBr2

CH2Cl2

1,2-difluoroethyne(E)-1,2-dibromo-1,2-difluoroethene

1,1,2,2-tetrabromo-1,2-difluoroethane

H3C C C CH2

H2C CH3(1)

(E)-2,3-dichlorohex-2-ene

Cl2

CH2Cl2

(2)

H3C

C C

H2C CH2

CH3

Cl

Cl

hex-2-yne

10/4/2011

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Catalytic Hydrogenation

Reaction requires molecular hydrogen (H2) and a metal

catalyst

Ni, Pd, Pt, Ru typical metals used

Because a gas, liquid and solid metal are used, this

reaction is known as a heterogeneous catalysis

Problems

H2

Pd/C

3-ethylpent-1-ene

(1)

3-ethylpentane

(2)

(3)

Br2

CH2Cl2

Br2

H2O

Br

Br

1,2-dibromo-3-ethylpentane

HO

Br

1-bromo-3-ethylpentan-2-ol

Provide the major organic product(s) in the reactions below

3-ethylpent-1-ene

3-ethylpent-1-ene


Chapter 24

Alcohols, Ethers and Haloalkanes


Conventions in Organic Chemistry

Figure 24.1

10/4/2011

36

Alcohol Properties

The difference in electronegativity between oxygen and

hydrogen allows alcohols to hydrogen bond. Compared to

alkanes they:

Are more soluble in water

Have higher boiling and melting points

Figure 24.4

Alcohol Solubility

There are differences in solubility:

Like dissolves like Table 24.2


Ether Properties

Ethers are more polar than alkanes but are unable to hydrogen bond.

They are chemically inert.

Both these properties mean that ethers are excellent solvents!

Ether Nomenclature

Figure 24.10

10/4/2011

37

Reactions of Alcohols

Figure 24.12

Fill in the reagents andname type of reaction !!Fill in the reagents andname type of reaction !!

Homework

Reactions of Alcohols

Alcohols react with sodium metal to form

alkoxides and H2(g).

Alkoxides are excellent nucleophiles and

strong bases.

Figure 24.13

2 CH3CH2O H + 2 Na 2 CH3CH2O-Na+ + H2(g)

Sodium ethoxide

2 CH3CH2O H + 2 Na 2 CH3CH2O-Na+ + H2(g)

Sodium ethoxide

Basicity of Alcohols

An alcohol can act as a weak base, generating an oxonium ion

under acidic conditions, which then loses water to form

carbocation intermediate.

The stability of the carbocation is an enormous driving force for

the nature of the reactivity of alcohols.

CH3CH2 H + H+O CH3CH2 O

H

H

+

CH3CH2+ + H2O

carbocation

oxonium ion

10/4/2011

38

Alcohols to Haloalkanes

Haloalkanes can be prepared from alcohols.

Haloalkanes are important for introducing alkyl groups in

chemical reactions.

Two reactions are common

Depend on the type of alcohol

Dehydration of Alcohols Dehydration of alcohols is an acid catalysed process. The rate-

determining step is the elimination of water to form the corresponding

carbocation. Removal of a small group from a larger molecule is called

elimination reaction.

Water is classed as a good leaving group because of its stability and

relatively unreactive nature. Hydroxide ion, on the other hand, does not

eliminate easily and so is classed as a poor leaving group.

Elimination reactionoxonium ion

Order of leaving group:

I- > H2O > Br- > Cl- >> F- >

CH3COO- > OH- > NH2-

Nucleophilic Substitution (Haloalkanes)

Nucleophilic substitution reactions of haloalkanes proceed by attack of the

nucleophile at the carbon bearing the halogen.

Because of the difference in electronegativity between carbon and any of

the halides, this bond is polarised such that the carbon becomes mildly

electrophilic.

This is a SN2 or biomolecular (2), nucleophilic (N), substitution (S)

reaction.

Figure 24.15

Transition state

SN2Mechanism

Example

1. Show the reaction mechanism between sodium hydroxide

and 1-bromoethane.

CCH3

BrH

H

NaOH

Soln

Transition state1-bromoethane. ethanol

mechanism

+

10/4/2011

39

2. Use reaction mechanisms to explain the why the inversion of the

carbon reaction centre is observed when (S)-2-bromobutane is

reacted with NaOH.

C

H3C Br

H

CH2CH3

Soln

HO-

HO C Br

CH2CH3

CH3

_

(S)-2-bromobutan

C

CH3

H

CH2CH3

HO

(R)-butan-2-ol

Inversion:- the (S)-configuration now becomes (R)-configuration.

Transition state


University of KwaZulu-Natal, Westville Campus, Durban

Examinations : November 2010

CHEMICAL REACTIVITY - CHEM120

Duration: 3 hours Total Marks for Examination: 100

Internal Examiners: V.O. Nyamori, H.G. Kruger, M.D. Bala, B.S. Martincigh,

P.G. Ndungu and G.D. Dawson

QUESTION 6Primary organic halides react with nucleophiles through a SN2mechanism. In the mechanism inversion of the carbon reaction centre isobserved. Provide the mechanism of the SN2 mechanism below.

(4)

C

H

H2CH3C

H

BrHO

SN2

(a) (b) (c)

starting material transition state product(s)

Four Factors affecting the SN2 reaction rate

Substrate

The nucleophile attacks from the back of the substrate and hence to

maximise the rate of the SN2 reaction, the back of the substrate must

be as unhindered as possible.

Order of reactivity: Primary > secondary > tertiary substrates

(Most reactive) (not reactive)

Nucleophile

Steric hindrance affects the nucleophile's strength e.g. methoxide

anion is better nucleophile compared to tert-butoxide.

Nucleophile strength is also affected by charge and electronegativity:

nucleophilicity increases with increasing negative charge and

decreasing electronegativity, e.g. nucleophile OH- > H2O and I- > Br-

(in polar protic solvents).

Solvent.

Solvents may or may not surround a nucleophile, thus hindering or

not hindering its approach to the carbon atom.

Polar aprotic solvents because polar protic solvents will be solvated

by the solvent hydrogen bonding to the nucleophile and thus

hindering it from attacking the carbon with the leaving group.

Leaving group

The more stable the leaving group is, the more likely that it will take

the two electrons of its carbon-leaving group bond with it when the

nucleophile attacks the carbon.

Therefore, the weaker the leaving group is as a conjugate base,

and thus the stronger its corresponding acid, the better the leaving

group. Examples of good leaving groups are therefore the halides

(except fluoride) and tosylate, whereas HO- and H2N- are not

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University of KwaZulu-Natal, Westville Campus, DurbanExaminations : November 2010

CHEM120 - CHEMICAL REACTIVITYSection B

QUESTION 6Primary organic halides react with nucleophiles through a SN2 mechanism.In the mechanism inversion of the carbon reaction centre is observed.Provide the mechanism of the SN2 reaction below. (4)

C

H

H2CH3C

H

BrHO

SN2

(a) (b) (c)

starting material transition state product(s)

Soln

Example

Nucleophilic Substitution (Haloalkanes)

Table 24.4

SN1 reaction mechanism

C R2

LG

R1

R3Nu + CR2

R1

R3

Nu

CNu

R2

R1

R3C

Nu

R2

R1

R3

and

Racemic mixture: Both (R) and (S)-configurationenantiomeric mixtures

carbocation

or

This allows two different avenues forthe nucleophilic attack, one on eitherside of the planar molecule.

If the reaction takes place at astereocenter yielding a racemicmix of enantiomers.

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41

Example

Step 1

Step 2

Step 3

Separation of the leaving group

Nucleophilic attack

Deprotonation

Carbocation

oxonium ion

Product : 2-methylpropan-2-ol

The SN1 mechanism tends to dominate when the central carbon

atom is surrounded by bulky groups. SN2 reaction hindered by steric

factor.

Bulky substituents on the central carbon favour carbocation

formation because of the relief of steric strain that occurs.

The resultant carbocation is also stabilized by both inductive

stabilization and hyperconjugation from attached alkyl groups.

The SN1 mechanism dominates in reactions at tertiary alkyl centers

and is further observed at secondary alkyl centers in the presence

of weak nucleophiles.

The carbocation intermediate formed in the reaction's rate limiting

step is an sp2 hybridized carbon with trigonal planar molecular

geometry.

Scope of the reaction

Haloalkanes to Alkenes: -Elimination

The process of elimination of a good leaving group.

Haloalkanes undergo elimination under basic conditions. This

process is called -elimination reaction. It involves the

dehydrohalogenation of the haloalkane.

For the reaction to proceed, a hydrogen atom on the carbon atom

adjacent to the carbon bearing the halide must exist and be

removed.

The -elimination reaction is a E2 or biomolecular (2) elimination

(E) reaction.

When there are two sets of -hydrogens present in the

haloalkane, there is only one product formed and it is determined

by the Zaitsev’s Rule:

The major alkene formed by dehydrohalogenation is the more

substituted one.

majorminor

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Figure 24.16

monosubstituted

disubstituteddisubstituted disubstituted

trisubstituted

tetrasubstituted

ethene

C C

H

H

H

X

HH

X = Cl, Br, I

C2H5O-Na+

C2H5O-Na+

C2H5O-Na+

C2H5O-Na+

C C

H

H

H

X

RR

C C

R

H

R

X

HR

C C

R

H

R

X

RR

C C

R

H

H

X

HR

HX +

HX +

HX +

HX +

C C

H

H

H

X

HR

C2H5O-Na+Increasing degree of

substitution

+HX

Substitution vs Elimination

Both Substitution reactions (SN1 & SN2) and elimination are important,

however, each of these reactions does not work in isolation.

E.g. the addition of strong base to an alkyl halide yields an alkene:

This elimination reaction competes with a substitution reaction which,

in this case, forms ethers:

Substitution vs Elimination

Mechanism shows a trigonal planar intermediate formed by

concerted bond breaking and bond making processes

There are usually stereochemical consequences

Elimination vs Substitution

Thus, there is a competition between elimination and substitution reactions and the result is dependant on:

Size of the alkoxide

The accessibility of the carbon bearing the halide

The accessibility of the -hydrogen

Figure 24.17

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Elimination vs Substitution

Figure 24.19

We looked at the

hydration of alkenes

The reverse reaction

is also possible!

The presence of other

nucleophiles add another

dimension to the reaction.


Chapter 25

Aldehydes and Ketones


Carbonyl Compounds

Contain C=O double bond.

Include many classes of compounds:

aldehydes, ketones, carboxylic acids, esters, amides.

C

O

CR H

O

CR R'

O

CR O

O

HC

R O

O

R'

CR N

O

R'

R''Aldehyde Ketone Carboxylic acid Ester Amide


Aldehydes

Characterised by the functional group CHO.

Contains a carbonyl group Both carbon and oxygen are sp2 hybridised.

Figure 25.1

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Aldehyde Nomenclature Name the longest chain containing the carbonyl group

and replace ane with al.

Must, by definition, occur at one of the ends of a

molecule. Remember that aldehyde group takes priority

over both methyl and hydroxyl groups.


Ketones

Name the longest chain containing the carbonyl

group and replace ane with one.

Carbonyl group is bonded to alkyl groups on either

side.

O

O

H

H

H

progesterone

O

camphor

O

OH

H

H

H

testosterone


Ketone Nomenclature

Longest chain containing the carbonyl group.

Replace ane with one.

Constitutional isomers occur for C5 or greater.

Preparation of Aldehydes and Ketones

Oxidation of alcohols using a range of reagents

• Most are chromium(VI) compounds (CrO3).

Again, different types of alcohols yield different products

• Primary alcohols can lead to aldehydes and carboxylic acids.

• Secondary alcohols lead to ketones.

• Tertiary alcohols do not oxidise.

Figure 25.2

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Preparation of Aldehydes and Ketones Preparation of Aldehydes and Ketones

Using pyridinium chlorochromate (PCC)

Preparation of Aldehydes and Ketones

Secondary alcohols can be oxidised by PCC, Jones

Reagent (CrO3) or K2Cr2O7 to yield the corresponding

ketone.

Aldehydes and ketones can also be prepared in good yields by the

reaction of alkenes with ozone, followed by a mild reduction using

dimethyl sulfide. This reaction is called ozonolysis

ozonolysis

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Reactions Aldehydes and Ketones

In general

Reduction reaction

Reduction Reactions

Aldehydes and ketones can be reduced to give alcohols

Catalytic hydrogenation is one way


Examples

H

O

O

O

O

H

H2

H2

H2

H2

Pd/C or Pt

Pd/C or Pt

Pd/C or Pt

Pd/C or Pt

(A)

(C)

(B)

(D)

H

OH

OH

OH

OH

H

QuestionProvide the structure of the reaction products from the followingequations.

Reduction Reactions

Hydrides:

Figure 25.5

NaBH4 LiAlH4

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ExamplesQuestion

Provide the structure of the reaction products from the followingequations.

H

O

O

O

O

H

LiAlH4(A)

(C)

(B)

(D)

LiAlH4

LiAlH4

LiAlH4

H

OH

OH

OH

OH

H

Tautomerism in Aldehydes and Ketones

Aldehydes and ketones bearing -hydrogen are able to undergo

tautomerism to form enols (i.e. compounds bearing an alkene and an

alcohol). Enols are constitutional isomers of their respective keto form

(i.e. aldehyde or ketone).

Tautomerism is the process by which two isomers are interconverted by

a formal movement of an atom or group bearing in mind valency rules.

Hence, an enol and its keto form are tautomers of the same structure.

Aldehydes and ketones are in equilibrium with their enol forms and this

equilibrium is acid catalysed. The process involves two separate proton

transfers:

The extent of tautomerisation is dependent on the aldehyde or ketone:

Halogenation of Aldehydes and Ketones

Aldehydes and ketones undergo a substitution reaction at the α-carbon in the presence of halogen in good yield.

This reaction is regiospecific in that only the α-hydrogen issubstituted under these conditions. The rate determining step inthis reaction is the conversion of the ketone to its enol isomer,e.g.

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Chapter 26

Carboxylic Acids and their

Derivatives

Carboxylic Acids Have hydroxyl group bonded to carbonyl group.

Carboxylic acids are weak acids.

Figure 26.3

Carboxylic Acids

Figure 26.4

Figure 26.5

Carboxylic Acids - Acidity

Table 26.2

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49


Question

Which of the following organic acids is expected to be the

strongest and why? Provide an explanation

O

CC O H

ethanoic acid

H

H

H

O

CC O H

F

F

F

O

CC O H

CH3

H3C

CH3

2,2,2-trifluoroacetic acid

2,2,2-trifluoroethanoic acid

acetic acid pivalic acid

2,2-dimethylpropanoic acid

Soln

(A) (B) (C)

(B)

Inductive effect from the substituents on the α-carbon.

F atom is the most electronegative, hence, stabilizes thecarboxylate anion generated

O

CC O-

F

F

F

+ H+carboxylate anion

Carboxylic Acids - Acidity

Stronger acid than alcohols because of resonance stabilisation.

pKa = 4.8

Figure 26.7

Carboxylic Acids - Reactions with Bases


ExamplesQuestion

Provide the structure of the reaction products from the followingequations.

HO

O

O

OH

NaOH(A)

(B)

NaOH

+Na-O

O

O

O-Na+

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Esters and Esterification

Products of reaction between carboxylic acids and alcohols.

Found in many fruits, perfumes and many other everyday

products:

Figure 26.9

O

CR O

+ HO R'H

O

CR O R'

+ OHH

Esters

Table 26.3

Examples

A

Ester Hydrolysis

Saponification = hydrolysis of ester using aqueous base.

Figure 26.12


Carboxylic Acid Derivatives

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Acid Chlorides

Formed by treatment of a carboxylic acid with thionyl chloride.


Acid Anhydrides

Formed by the reaction of an acid chloride and carboxylic acid or by dehydration.

Questions

(a) Provide the structures for the reaction products from the followingequations.

(b) Provide appropriate starting compounds or reagents used to form theproducts in the following equations.

CH2

CH3

O

Cl

C CH2

CH3O

H3CC

O

O

SOCl3

? ?

O

OH

+ HO

O

O

HO

C CH2

CH3OH3C

C

O

OH

(i)

(ii)

O

O F Dilute NaOH

H2OHO F

O

OH +


Chapter 28

Nitrogen-Containing

Compounds of Biological

Relevance

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Biological Chemistry

Research at the interface of biology and

chemistry.

The principles of chemistry are applied to

understand the molecular basis of biological

processes.

Figure 28.1


Biological Chemistry

We will investigate: Amines and amides

Amino acids

Peptides, proteins and enzymes

Vitamins

DNA


Amines

General formula:

Can be aliphatic, heterocyclic or aromatic:• Next slide shows a number of simple

amines.

• Classified as 1o, 2o, 3o

• Quaternary (4o) ammonium salts are also likely.

Figure 28.3

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Figure 28.4

Simple amines are named using the same guidelines for alcohols.

In many instances, the amino group is named as a substituent.

Amine Nomenclature

Amine Nomenclature Reactivity of Amines

Amines react with acids to form ammonium salts.

Amines react with alkyl halides to form ammonium salts.

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Amides

Amides are derivatives of carboxylic acids.

They are classified as 1o, 2o and 3o

O

CH3C N

CH3

H

O

CH3C N

H

H

O

CH3C N

CH3

CH3

acetamide N-methylacetamide N,N-dimethylacetamide

2°1° 3°

Formation of Amides

Amides are formed by the reaction of amines with “activated”

carboxylic acids.

C

O

H3C Cl NCH2CH3

H

H

+ C

O

H3C N

H

CH2CH3

+ HCl

CHEM 120 CHEMICAL REACTIVITY

ORGANIC CHEMISTRY

19 Lectures5 Tutorials

2 Quiz1 Test

Dr. Vincent O. Nyamori


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216

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