Organic Chm

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ORGANIC CHEMISTRY – SOME BASIC PRINCIPLES AND TECHNIQUES After studying this unit, you will be able to understand reasons for tetravalence of carbon and shapes of organic molecules; write structures of organic molecules in various ways; classify the organic compounds; name the compounds according to IUPAC system of nomenclature and also derive their structures from the given names; understand the concept of organic reaction mechanism; explain the influence of electronic displacements on structure and reactivity of organic compounds; recognise the types of organic reactions; learn the techniques of purification of organic compounds; write the chemical reactions involved in the qualitative analysis of organic compounds; understand the principles involved in quantitative analysis of organic compounds. In the previous unit you have learnt that the element carbon has the unique property called catenation due to which it forms covalent bonds with other carbon atoms. It also forms covalent bonds with atoms of other elements like hydrogen, oxygen, nitrogen, sulphur, phosphorus and halogens. The resulting compounds are studied under a separate branch of chemistry called organic chemistry. This unit incorporates some basic principles and techniques of analysis required for understanding the formation and properties of organic compounds. 12.1 GENERAL INTRODUCTION Organic compounds are vital for sustaining life on earth and include complex molecules like genetic information bearing deoxyribonucleic acid (DNA) and proteins that constitute essential compounds of our blood, muscles and skin. Organic chemicals appear in materials like clothing, fuels, polymers, dyes and medicines. These are some of the important areas of application of these compounds. Science of organic chemistry is about two hundred years old. Around the year 1780, chemists began to distinguish between organic compounds obtained from plants and animals and inorganic compounds prepared from mineral sources. Berzilius, a Swedish chemist proposed that a ‘vital force’ was responsible for the formation of organic compounds. However, this notion was rejected in 1828 when F. Wohler synthesised an organic compound, urea from an inorganic compound, ammonium cyanate. 4 2 2 Heat NH CNO NH CONH ⎯⎯⎯⎯→ Ammonium cyanate Urea The pioneering synthesis of acetic acid by Kolbe (1845) and that of methane by Berthelot (1856) showed conclusively that organic compounds could be synthesised from inorganic sources in a laboratory. UNIT 12

Transcript of Organic Chm

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326 CHEMISTRY

ORGANIC CHEMISTRY – SOME BASIC PRINCIPLESAND TECHNIQUES

After studying this unit, you will beable to

••••• understand reasons fortetravalence of carbon andshapes of organic molecules;

••••• write structures of organicmolecules in various ways;

••••• classify the organic compounds;••••• name the compounds according

to IUPAC system ofnomenclature and also derivetheir structures from the givennames;

••••• understand the concept oforganic reaction mechanism;

••••• explain the influence ofelectronic displacements onstructure and reactivity oforganic compounds;

••••• recognise the types of organicreactions;

••••• learn the techniques ofpurification of organiccompounds;

••••• write the chemical reactionsinvolved in the qualitativeanalysis of organic compounds;

••••• understand the principlesinvolved in quantitative analysisof organic compounds.

In the previous unit you have learnt that the elementcarbon has the unique property called catenation due towhich it forms covalent bonds with other carbon atoms.It also forms covalent bonds with atoms of other elementslike hydrogen, oxygen, nitrogen, sulphur, phosphorus andhalogens. The resulting compounds are studied under aseparate branch of chemistry called organic chemistry.This unit incorporates some basic principles andtechniques of analysis required for understanding theformation and properties of organic compounds.

12.1 GENERAL INTRODUCTION

Organic compounds are vital for sustaining life on earthand include complex molecules like genetic informationbearing deoxyribonucleic acid (DNA) and proteins thatconstitute essential compounds of our blood, muscles andskin. Organic chemicals appear in materials like clothing,fuels, polymers, dyes and medicines. These are some ofthe important areas of application of these compounds.

Science of organic chemistry is about two hundredyears old. Around the year 1780, chemists began todistinguish between organic compounds obtained fromplants and animals and inorganic compounds preparedfrom mineral sources. Berzilius, a Swedish chemistproposed that a ‘vital force’ was responsible for theformation of organic compounds. However, this notionwas rejected in 1828 when F. Wohler synthesised anorganic compound, urea from an inorganic compound,ammonium cyanate.

4 2 2HeatNH CNO NH CONH⎯⎯⎯⎯→

Ammonium cyanate Urea

The pioneering synthesis of acetic acid by Kolbe (1845)and that of methane by Berthelot (1856) showedconclusively that organic compounds could be synthesisedfrom inorganic sources in a laboratory.

UNIT 12

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The development of electronic theory ofcovalent bonding ushered organic chemistryinto its modern shape.

12.2 TETRAVALENCE OF CARBON:SHAPES OF ORGANIC COMPOUNDS

12.2.1 The Shapes of Carbon Compounds

The knowledge of fundamental concepts ofmolecular structure helps in understandingand predicting the properties of organiccompounds. You have already learnt theoriesof valency and molecular structure in Unit 4.Also, you already know that tetravalence ofcarbon and the formation of covalent bondsby it are explained in terms of its electronicconfiguration and the hybridisation of s andp orbitals. It may be recalled that formationand the shapes of molecules like methane(CH4), ethene (C2H4), ethyne (C2H2) areexplained in terms of the use of sp3, sp2 andsp hybrid orbitals by carbon atoms in therespective molecules.

Hybridisation influences the bond lengthand bond enthalpy (strength) in organiccompounds. The sp hybrid orbital containsmore s character and hence it is closer to itsnucleus and forms shorter and strongerbonds than the sp3 hybrid orbital. The sp2

hybrid orbital is intermediate in s characterbetween sp and sp3 and, hence, the lengthand enthalpy of the bonds it forms, are alsointermediate between them. The change inhybridisation affects the electronegativity ofcarbon. The greater the s character of thehybrid orbitals, the greater is theelectronegativity. Thus, a carbon atom havingan sp hybrid orbital with 50% s character ismore electronegative than that possessing sp2

or sp3 hybridised orbitals. This relativeelectronegativity is reflected in severalphysical and chemical properties of themolecules concerned, about which you willlearn in later units.

12.2.2 Some Characteristic Features of πππππBonds

In a π (pi) bond formation, parallel orientationof the two p orbitals on adjacent atoms is

necessary for a proper sideways overlap.Thus, in H

2C=CH

2 molecule all the atoms

must be in the same plane. The p orbitalsare mutually parallel and both the p orbitalsare perpendicular to the plane of themolecule. Rotation of one CH

2 fragment with

respect to other interferes with maximumoverlap of p orbitals and, therefore, suchrotation about carbon-carbon double bond(C=C) is restricted. The electron charge cloudof the π bond is located above and below theplane of bonding atoms. This results in theelectrons being easily available to theattacking reagents. In general, π bonds providethe most reactive centres in the moleculescontaining multiple bonds.

Problem 12.1

How many σ and π bonds are present ineach of the following molecules?

(a) HC≡CCH=CHCH3 (b) CH2=C=CHCH3

Solution

(a) σC – C: 4; σC–H : 6; πC=C :1; π C≡C:2

(b) σC – C: 3; σC–H: 6; πC=C: 2.

Problem 12.2

What is the type of hybridisation of eachcarbon in the following compounds?

(a) CH3Cl, (b) (CH3)2CO, (c) CH3CN,

(d) HCONH2, (e) CH3CH=CHCN

Solution

(a) sp3, (b) sp3, sp2, (c) sp3, sp, (d) sp2, (e)sp3, sp2, sp2, sp

Problem 12.3

Write the state of hybridisation of carbonin the following compounds and shapesof each of the molecules.

(a) H2C=O, (b) CH3F, (c) HC≡N.

Solution

(a) sp2 hybridised carbon, trigonal planar;(b) sp3 hybridised carbon, tetrahedral; (c)sp hybridised carbon, linear.

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12.3 STRUCTURAL REPRESENTATIONSOF ORGANIC COMPOUNDS

12.3.1Complete, Condensed and Bond-lineStructural Formulas

Structures of organic compounds arerepresented in several ways. The Lewisstructure or dot structure, dash structure,condensed structure and bond line structuralformulas are some of the specific types. TheLewis structures, however, can be simplifiedby representing the two-electron covalentbond by a dash (–). Such a structural formulafocuses on the electrons involved in bondformation. A single dash represents a singlebond, double dash is used for double bondand a triple dash represents triple bond. Lone-pairs of electrons on heteroatoms (e.g.,oxygen, nitrogen, sulphur, halogens etc.) mayor may not be shown. Thus, ethane (C2H6),ethene (C2H4), ethyne (C2H2) and methanol(CH3OH) can be represented by the followingstructural formulas. Such structuralrepresentations are called complete structuralformulas.

Similarly, CH3CH2CH2CH2CH2CH2CH2CH3

can be further condensed to CH3(CH2)6CH3.For further simplification, organic chemistsuse another way of representing thestructures, in which only lines are used. Inthis bond-line structural representation oforganic compounds, carbon and hydrogenatoms are not shown and the linesrepresenting carbon-carbon bonds are drawnin a zig-zag fashion. The only atomsspecifically written are oxygen, chlorine,nitrogen etc. The terminals denote methyl(–CH3) groups (unless indicated otherwise bya functional group), while the line junctionsdenote carbon atoms bonded to appropriatenumber of hydrogens required to satisfy thevalency of the carbon atoms. Some of theexamples are represented as follows:

(i) 3-Methyloctane can be represented invarious forms as:

(a) CH3CH2CHCH2CH2CH2CH2CH3

| CH3

These structural formulas can be furtherabbreviated by omitting some or all of thedashes representing covalent bonds and byindicating the number of identical groupsattached to an atom by a subscript. Theresulting expression of the compound is calleda condensed structural formula. Thus, ethane,ethene, ethyne and methanol can be writtenas:

CH3CH3 H2C=CH2 HC≡≡≡≡≡CH CH3OH

Ethane Ethene Ethyne Methanol

Ethane Ethene

Ethyne Methanol

(ii) Various ways of representing 2-bromobutane are:

(a) CH3CHBrCH2CH3 (b)

(c)

(b)

(c)

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In cyclic compounds, the bond-line formulasmay be given as follows:

Cyclopropane

Cyclopentane

chlorocyclohexane

Problem 12.4

Expand each of the following condensedformulas into their complete structuralformulas.

(a) CH3CH2COCH2CH3

(b) CH3CH=CH(CH2)3CH3

Solution

(b)

Solution

Condensed formula:

(a) HO(CH2)3CH(CH3)CH(CH3)2(b) HOCH(CN)2Bond-line formula:

(a)

(b)

Problem 12.5

For each of the following compounds,write a condensed formula and also theirbond-line formula.

(a) HOCH2CH2CH2CH(CH3)CH(CH3)CH3

(b)

(a)

Problem 12.6

Expand each of the following bond-lineformulas to show all the atoms includingcarbon and hydrogen

(a)

(b)

(c)

(d)

Solution

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Framework model Ball and stick model

Space filling model

Fig. 12.2

12.3.2 Three-DimensionalRepresentation of OrganicMolecules

The three-dimensional (3-D) structure oforganic molecules can be represented onpaper by using certain conventions. Forexample, by using solid ( ) and dashed( ) wedge formula, the 3-D image of amolecule from a two-dimensional picturecan be perceived. In these formulas thesolid-wedge is used to indicate a bondprojecting out of the plane of paper, towardsthe observer. The dashed-wedge is used todepict the bond projecting out of the plane ofthe paper and away from the observer. Wedgesare shown in such a way that the broad endof the wedge is towards the observer. Thebonds lying in plane of the paper are depictedby using a normal line (—). 3-D representationof methane molecule on paper has beenshown in Fig. 12.1.

Fig. 12.1 Wedge-and-dash representation of CH4

Molecular Models

Molecular models are physical devices thatare used for a better visualisation andperception of three-dimensional shapes oforganic molecules. These are made of wood,plastic or metal and are commerciallyavailable. Commonly three types of molecularmodels are used: (1) Framework model, (2)Ball-and-stick model, and (3) Space fillingmodel. In the framework model only thebonds connecting the atoms of a moleculeand not the atoms themselves are shown.This model emphasizes the pattern of bondsof a molecule while ignoring the size of atoms.In the ball-and-stick model, both the atomsand the bonds are shown. Balls representatoms and the stick denotes a bond.Compounds containing C=C (e.g., ethene) canbest be represented by using springs in placeof sticks. These models are referred to as ball-and-spring model. The space-filling modelemphasises the relative size of each atombased on its van der Waals radius. Bondsare not shown in this model. It conveys thevolume occupied by each atom in themolecule. In addition to these models,computer graphics can also be used formolecular modelling.

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12.4 CLASSIFICATION OF ORGANICCOMPOUNDS

The existing large number of organiccompounds and their ever -increasingnumbers has made it necessary to classifythem on the basis of their structures. Organiccompounds are broadly classified as follows:

I. Acyclic or open chain compounds

These compounds are also called as aliphaticcompounds and consist of straight orbranched chain compounds, for example:

(homocyclic). Sometimes atoms other thancarbon are also present in the ring(heterocylic). Some examples of this type ofcompounds are:

Cyclopropane Cyclohexane

Cyclohexene Tetrahydrofuran

These exhibit some of the properties similarto those of aliphatic compounds.

Aromatic compounds

Aromatic compounds are special types ofcompounds. You will learn about thesecompounds in detail in Unit 13. These includebenzene and other related ring compounds(benzenoid). Like alicyclic compounds,aromatic comounds may also have heteroatom in the ring. Such compounds are calledhetrocyclic aromatic compounds. Some of theexamples of various types of aromaticcompounds are:

Benzenoid aromatic compounds

Benzene Aniline Naphthalene

Non-benzenoid compound

Tropolone

Isobutane

Acetaldehyde Acetic acid

CH3CH3

Ethane

II Alicyclic or closed chain or ringcompounds

Alicyclic (aliphatic cyclic) compounds containcarbon atoms joined in the form of a ring

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Heterocyclic aromatic compounds

Furan Thiophene Pyridine

Organic compounds can also be classifiedon the basis of functional groups, into familiesor homologous series.

Functional GroupThe functional group may be defined as anatom or group of atoms joined in a specificmanner which is responsible for thecharacteristic chemical properties of theorganic compounds. The examples arehydroxyl group (–OH), aldehyde group (–CHO)and carboxylic acid group (–COOH) etc.

Homologous SeriesA group or a series of organic compounds eachcontaining a characteristic functional groupforms a homologous series and the membersof the series are called homologues. Themembers of a homologous series can berepresented by general molecular formula andthe successive members differ from each otherin molecular formula by a –CH

2 unit. There

are a number of homologous series oforganic compounds. Some of these arealkanes, alkenes, alkynes, haloalkanes,alkanols, alkanals, alkanones, alkanoic acids,amines etc.

12.5 NOMENCLATURE OF ORGANICCOMPOUNDS

Organic chemistry deals with millions ofcompounds. In order to clearly identify them, asystematic method of naming has beendeveloped and is known as the IUPAC(International Union of Pure and AppliedChemistry) system of nomenclature. In thissystematic nomenclature, the names arecorrelated with the structure such that thereader or listener can deduce the structure fromthe name.

Before the IUPAC system of nomenclature,however, organic compounds were assignednames based on their origin or certainproperties. For instance, citric acid is namedso because it is found in citrus fruits and the

acid found in red ant is named formic acidsince the Latin word for ant is formica. Thesenames are traditional and are considered astrivial or common names. Some commonnames are followed even today. For example,Buckminsterfullerene is a common namegiven to the newly discovered C60 cluster(a form of carbon) noting its structuralsimilarity to the geodesic domes popularisedby the famous architect R. BuckminsterFuller. Common names are useful and inmany cases indispensable, particularly whenthe alternative systematic names are lengthyand complicated. Common names of someorganic compounds are given in Table 12.1.Table 12.1 Common or Trivial Names of Some

Organic Compounds

12.5.1 The IUPAC System of Nomenclature

A systematic name of an organic compoundis generally derived by identifying the parenthydrocarbon and the functional group(s)attached to it. See the example given below.

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By further using prefixes and suffixes, theparent name can be modified to obtain theactual name. Compounds containing carbonand hydrogen only are called hydrocarbons. Ahydrocarbon is termed saturated if it containsonly carbon-carbon single bonds. The IUPACname for a homologous series of suchcompounds is alkane. Paraffin (Latin: littleaffinity) was the earlier name given to thesecompounds. Unsaturated hydrocarbons arethose, which contain at least one carbon-carbon double or triple bond.

12.5.2 IUPAC Nomenclature of Alkanes

Straight chain hydrocarbons: The namesof such compounds are based on their chainstructure, and end with suffix ‘-ane’ and carrya prefix indicating the number of carbonatoms present in the chain (except from CH4

to C4H10, where the prefixes are derived fromtrivial names). The IUPAC names of somestraight chain saturated hydrocarbons aregiven in Table 12.2. The alkanes in Table 12.2differ from each other by merely the numberof -CH

2 groups in the chain. They are

homologues of alkane series.

In order to name such compounds, the namesof alkyl groups are prefixed to the name ofparent alkane. An alkyl group is derived froma saturated hydrocarbon by removing ahydrogen atom from carbon. Thus, CH4

becomes -CH3 and is called methyl group. Analkyl group is named by substituting ‘yl’ for‘ane’ in the corresponding alkane. Some alkylgroups are listed in Table 12.3.

Table 12.3 Some Alkyl Groups

Table 12.2 IUPAC Names of Some UnbranchedSaturated Hydrocarbons

Branched chain hydrocarbons: In abranched chain compound small chains ofcarbon atoms are attached at one or morecarbon atoms of the parent chain. The smallcarbon chains (branches) are called alkylgroups. For example:

CH3–CH–CH2–CH3 CH3–CH–CH2–CH–CH3 ⏐ ⏐ ⏐ CH3 CH2CH3 CH3

(a) (b)

Abbreviations are used for some alkylgroups. For example, methyl is abbreviatedas Me, ethyl as Et, propyl as Pr and butyl asBu. The alkyl groups can be branched also.Thus, propyl and butyl groups can havebranched structures as shown below.

CH3-CH- CH3-CH2-CH- CH3-CH-CH2- ⏐ ⏐ ⏐

CH3 CH3 CH3

Isopropyl- sec-Butyl- Isobutyl- CH3 CH3

⏐ ⏐ CH3-C- CH3-C-CH2- ⏐ ⏐

CH3 CH3

tert-Butyl- Neopentyl-

Common branched groups have specifictrivial names. For example, the propyl groupscan either be n-propyl group or isopropylgroup. The branched butyl groups are calledsec-butyl, isobutyl and tert-butyl group. Wealso encounter the structural unit,–CH2C(CH3)3, which is called neopentyl group.

Nomenclature of branched chain alkanes:We encounter a number of branched chainalkanes. The rules for naming them are givenbelow.

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separated from the groups by hyphens andthere is no break between methyl andnonane.]

4. If two or more identical substituent groupsare present then the numbers areseparated by commas. The names ofidentical substituents are not repeated,instead prefixes such as di (for 2), tri(for 3), tetra (for 4), penta (for 5), hexa (for6) etc. are used. While writing the name ofthe substituents in alphabetical order,these prefixes, however, are not considered.Thus, the following compounds arenamed as:

CH3 CH3 CH3 CH3

⏐ ⏐ ⏐ ⏐CH3-CH-CH2-CH-CH3 CH3⎯C⎯CH2⎯CH⎯CH3

⏐ CH3

2,4-Dimethylpentane 2,2,4-Trimethylpentane

H3C H2C CH3

⏐ ⏐ CH3⎯CH2⎯CH⎯C⎯CH2⎯CH2⎯CH3

⏐ CH3

3-Ethyl-4,4-dimethylheptane

5. If the two substituents are found inequivalent positions, the lower number isgiven to the one coming first in thealphabetical listing. Thus, the followingcompound is 3-ethyl-6-methyloctane andnot 6-ethyl-3-methyloctane.

1 2 3 4 5 6 7 8CH3 — CH2—CH—CH2—CH2—CH—CH2 —CH3

⏐ ⏐ CH2CH3 CH3

6. The branched alkyl groups can be namedby following the above mentionedprocedures. However, the carbon atom ofthe branch that attaches to the rootalkane is numbered 1 as exemplifiedbelow.

4 3 2 1CH3–CH–CH2–CH–

⏐ ⏐ CH3 CH3

1,3-Dimethylbutyl-

1. First of all, the longest carbon chain inthe molecule is identified. In the example(I) given below, the longest chain has ninecarbons and it is considered as the parentor root chain. Selection of parent chain asshown in (II) is not correct because it hasonly eight carbons.

1 2 3 4 5 1 2 3 4 5

1 2 3 4 5 6 7

2. The carbon atoms of the parent chain arenumbered to identify the parent alkane andto locate the positions of the carbon atomsat which branching takes place due to thesubstitution of alkyl group in place ofhydrogen atoms. The numbering is donein such a way that the branched carbonatoms get the lowest possible numbers.Thus, the numbering in the above exampleshould be from left to right (branching atcarbon atoms 2 and 6) and not from rightto left (giving numbers 4 and 8 to thecarbon atoms at which branches areattached).

1 2 3 4 5 6 7 8 9C ⎯ C ⎯ C ⎯ C ⎯ C ⎯ C ⎯C ⎯ C ⎯ C ⏐ ⏐ C C⎯C

9 8 7 6 5 4 3 2 1C ⎯ C ⎯ C ⎯ C ⎯ C ⎯ C ⎯ C ⎯ C ⎯ C ⏐ ⏐ C C ⎯ C

3. The names of alkyl groups attached as abranch are then prefixed to the name ofthe parent alkane and position of thesubstituents is indicated by theappropriate numbers. If different alkylgroups are present, they are listed inalphabetical order. Thus, name for thecompound shown above is: 6-ethyl-2-methylnonane. [Note: the numbers are

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The name of such branched chain alkyl groupis placed in parenthesis while naming thecompound. While writing the trivial names ofsubstituents’ in alphabetical order, theprefixes iso- and neo- are considered to bethe part of the fundamental name of alkylgroup. The prefixes sec- and tert- are notconsidered to be the part of the fundamentalname. The use of iso and related commonprefixes for naming alkyl groups is alsoallowed by the IUPAC nomenclature as longas these are not further substituted. In multi-substituted compounds, the following rulesmay aso be remembered:

• If there happens to be two chains of equalsize, then that chain is to be selectedwhich contains more number of sidechains.

• After selection of the chain, numbering isto be done from the end closer to thesubstituent.

5-(2-Ethylbutyl)-3,3-dimethyldecane[and not 5-(2,2-Dimethylbutyl)-3-ethyldecane]

5-sec-Butyl-4-isopropyldecane

5-(2,2-Dimethylpropyl)nonane

Problem 12.7Structures and IUPAC names of somehydrocarbons are given below. Explainwhy the names given in the parenthesesare incorrect.

3-Ethyl-1,1-dimethylcyclohexane(not 1-ethyl-3,3-dimethylcyclohexane)

2,5,6- Trimethyloctane[and not 3,4,7-Trimethyloctane]

3-Ethyl-5-methylheptane[and not 5-Ethyl-3-methylheptane]

Cyclic Compounds: A saturated monocycliccompound is named by prefixing ‘cyclo’ to thecorresponding straight chain alkane. If sidechains are present, then the rules given aboveare applied. Names of some cyclic compoundsare given below.

Solution

(a) Lowest locant number, 2,5,6 is lowerthan 3,5,7, (b) substituents are inequivalent position; lower number isgiven to the one that comes first in thename according to alphabetical order.

12.5.3 Nomenclature of OrganicCompounds having FunctionalGroup(s)

A functional group, as defined earlier, is anatom or a group of atoms bonded together in aunique manner which is usually the site of

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chemical reactivity in an organic molecule.Compounds having the same functional groupundergo similar reactions. For example,CH

3OH, CH

3CH

2OH, and (CH

3)2CHOH — all

having -OH functional group liberate hydrogenon reaction with sodium metal. The presenceof functional groups enables systematisationof organic compounds into different classes.Examples of some functional groups with theirprefixes and suffixes along with someexamples of organic compounds possessingthese are given in Table 12.4.

First of all, the functional group presentin the molecule is identified which determinesthe choice of appropriate suffix. The longestchain of carbon atoms containing thefunctional group is numbered in such a waythat the functional group is attached at thecarbon atom possessing lowest possiblenumber in the chain. By using the suffix asgiven in Table 12.4, the name of the compoundis arrived at.

In the case of polyfunctional compounds,one of the functional groups is chosen as theprincipal functional group and the compound isthen named on that basis. The remainingfunctional groups, which are subordinatefunctional groups, are named as substituentsusing the appropriate prefixes. The choice ofprincipal functional group is made on the basisof order of preference. The order of decreasingpriority for some functional groups is:-COOH, –SO3H, -COOR (R=alkyl group), COCl,-CONH2, -CN,-HC=O, >C=O, -OH, -NH2, >C=C<,-C≡≡≡≡≡C- .The –R, C6H5-, halogens (F, Cl, Br, I), –NO2,alkoxy (–OR) etc. are always prefixsubstituents. Thus, a compound containingboth an alcohol and a keto group is namedas hydroxyalkanone since the keto group ispreferred to the hydroxyl group.

For example, HOCH2(CH2)3CH2COCH3 will benamed as 7-hydroxyheptan-2-one and not as2-oxoheptan -7-ol. Similarly, BrCH2CH=CH2

is named as 3-bromoprop-1-ene and not 1-bromoprop-2-ene.

If more than one functional group of thesame type are present, their number isindicated by adding di, tri, etc. before the class

suffix. In such cases the full name of the parentalkane is written before the class suffix. Forexample CH2(OH)CH2(OH) is named asethane–1,2–diol. However, the ending – ne ofthe parent alkane is dropped in the case ofcompounds having more than one double ortriple bond; for example, CH2=CH-CH=CH2 isnamed as buta–1,3–diene.

Problem 12.8

Write the IUPAC names of the compoundsi-iv from their given structures.

Solution

• The functional group present is analcohol (OH). Hence the suffix is ‘-ol’.

• The longest chain containing -OH haseight carbon atoms. Hence thecorresponding saturated hydrocarbonis octane.

• The -OH is on carbon atom 3. Inaddition, a methyl group is attachedat 6th carbon.Hence, the systematic name of thiscompound is 6-Methyloctan-3-ol.

Solution

The functional group present is ketone(>C=O), hence suffix ‘-one’. Presence oftwo keto groups is indicated by ‘di’,hence suffix becomes ‘dione’. The twoketo groups are at carbons 2 and 4. Thelongest chain contains 6 carbon atoms,hence, parent hydrocarbon is hexane.Thus, the systematic name is Hexane-2,4-dione.

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Table 12.4 Some Functional Groups and Classes of Organic Compounds

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Solution

Here, two functional groups namelyketone and carboxylic acid are present.The principal functional group is thecarboxylic acid group; hence the parentchain will be suffixed with ‘oic’ acid.Numbering of the chain starts fromcarbon of – COOH functional group. Theketo group in the chain at carbon 5 isindicated by ‘oxo’. The longest chainincluding the principal functionalgroup has 6 carbon atoms; hence theparent hydrocarbon is hexane. Thecompound is, therefore, named as5-Oxohexanoic acid.

Solution

The two C=C functional groups arepresent at carbon atoms 1 and 3, whilethe C≡C functional group is present atcarbon 5. These groups are indicated bysuffixes ‘diene’ and ‘yne’ respectively. Thelongest chain containing the functionalgroups has 6 carbon atoms; hence theparent hydrocarbon is hexane. The nameof compound, therefore, is Hexa-1,3-dien-5-yne.

Problem 12.9

Derive the structure of (i) 2-Chlorohexane,(ii) Pent-4-en-2-ol, (iii) 3- Nitrocyclohexene,(iv) Cyclohex-2-en-1-ol, (v) 6-Hydroxy-heptanal.

Solution

(i) ‘hexane’ indicates the presence of6 carbon atoms in the chain. Thefunctional group chloro is present atcarbon 2. Hence, the structure of thecompound is CH3CH2CH2CH2CH(Cl)CH3.

(ii) ‘pent’ indicates that parenthydrocarbon contains 5 carbon atoms inthe chain. ‘en’ and ‘ol’ correspond to thefunctional groups C=C and -OH atcarbon atoms 4 and 2 respectively. Thus,the structure is

CH2=CHCH2CH (OH)CH3.

(iii) Six membered ring containing acarbon-carbon double bond is implied bycyclohexene, which is numbered asshown in (I). The prefix 3-nitro means thata nitro group is present on C-3. Thus,complete structural formula of thecompound is (II). Double bond is suffixedfunctional group whereas NO2 is prefixedfunctional group therefore double bondgets preference over –NO2 group:

(iv) ‘1-ol’ means that a -OH group ispresent at C-1. OH is suffixed functionalgroup and gets preference over C=Cbond. Thus the structure is as shownin (II):

(v) ‘heptanal’ indicates the compound tobe an aldehyde containing 7 carbonatoms in the parent chain. The‘6-hydroxy’ indicates that -OH group ispresent at carbon 6. Thus, the structuralformula of the compound is:CH3CH(OH)CH2CH2CH2CH2CHO. Carbonatom of –CHO group is included whilenumbering the carbon chain.

12.5.4 Nomenclature of SubstitutedBenzene Compounds

For IUPAC nomenclature of substitutedbenzene compounds, the substituent isplaced as prefix to the word benzene asshown in the following examples. However,common names (written in bracket below)of many substituted benzene compoundsare also universally used.

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339ORGANIC CHEMISTRY – SOME BASIC PRINCIPLES AND TECHNIQUES

2-Chloro-4-methylanisole 4-Ethyl-2-methylaniline

1-Chloro-2,4-dinitrobenzene(not 4-chloro,1,3-dinitrobenzene)

If benzene ring is disubstituted, theposit ion of subst i tuents is def inedby numbering the carbon atoms ofthe ring such that the substituentsare located at the lowest numberspossible.For example, the compound(b) isnamed as 1,3-dibromobenzene and not as1,5-dibromobenzene.

Substituent of the base compound isassigned number1 and then the direction ofnumbering is chosen such that the nextsubstituent gets the lowest number. Thesubstituents appear in the name inalphabetical order. Some examples are givenbelow.

2-Chloro-1-methyl-4-nitrobenzene(not 4-methyl-5-chloro-nitrobenzene)

3,4-Dimethylphenol

Methylbenzene Methoxybenzene Aminobenzene (Toluene) (Anisole) (Aniline)

Nitrobenzene Bromobenzene

1,2-Dibromo-benzene

1,3-Dibromo-benzene

1,4-Dibromo-benzene

(a) (b) (c)

In the trivial system of nomenclature theterms ortho (o), meta (m) and para (p) are usedas prefixes to indicate the relative positions1,2- ;1,3- and 1,4- respectively. Thus,1,3-dibromobenzene (b) is named asm-dibromobenzene (meta is abbreviated asm-) and the other isomers of dibromobenzene1,2-(a) and 1,4-(c), are named as ortho (or justo-) and para (or just p-)-dibromobenzene,respectively.

For tri - or higher substituted benzenederivatives, these prefixes cannot be used andthe compounds are named by identifyingsubstituent positions on the ring by followingthe lowest locant rule. In some cases, commonname of benzene derivatives is taken as thebase compound.

When a benzene ring is attached to analkane with a functional group, it isconsidered as substituent, instead of aparent. The name for benzene as substituentis phenyl (C6H5-, also abbreviated as Ph).

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340 CHEMISTRY

different carbon skeletons, these are referredto as chain isomers and the phenomenon istermed as chain isomerism. For example, C5H12

represents three compounds:

Isomerism

Structural isomerism Stereoisomerism

Chainisomerism

Positionisomerism

Functionalgroup

isomerism

Metamerism Geometricalisomerism

Opticalisomerism

12.6 ISOMERISM

The phenomenon of existence of two or morecompounds possessing the same molecularformula but different properties is known asisomerism. Such compounds are called asisomers. The following flow chart showsdifferent types of isomerism.

12.6.1 Structural Isomerism

Compounds having the same molecularformula but different structures (manners inwhich atoms are linked) are classified asstructural isomers. Some typical examples ofdifferent types of structural isomerism are givenbelow:

(i) Chain isomerism: When two or morecompounds have similar molecular formula but

(a) (b)

(c) (d)

Problem 12.10

Write the structural formula of:(a) o-Ethylanisole, (b) p-Nitroaniline,(c) 2,3 - Dibromo -1 - phenylpentane,(d) 4-Ethyl-1-fluoro-2-nitrobenzene.

Solution

CH3

⏐CH3CH2CH2CH2CH3 CH3−CHCH2CH3

Pentane Isopentane(2-Methylbutane)

CH3

⏐CH3⎯ C⎯ CH3

⏐ CH3

Neopentane(2,2-Dimethylpropane)

(ii) Position isomerism: When two or morecompounds dif fer in the position ofsubstituent atom or functional group on thecarbon skeleton, they are called positionisomers and this phenomenon is termed asposition isomerism. For example, themolecular formula C3H8O represents twoalcohols:

OH ⏐

CH3CH2CH2OH CH3−CH-CH3

Propan-1-ol Propan-2-ol

(iii) Functional group isomerism: Two ormore compounds having the same molecularformula but different functional groups arecalled functional isomers and thisphenomenon is termed as functional groupisomerism. For example, the molecularformula C

3H

6O represents an aldehyde and a

ketone:

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341ORGANIC CHEMISTRY – SOME BASIC PRINCIPLES AND TECHNIQUES

O H y x

CH3−C-CH3 CH3−CH2—C= O

Propanone Propanal

(iv) Metamerism: It arises due to different alkylchains on either side of the functional groupin the molecule. For example, C4H10Orepresents methoxypropane (CH3OC3H7) andethoxyethane (C2H5OC2H5).

12.6.2 Stereoisomerism

The compounds that have the sameconstitution and sequence of covalent bondsbut differ in relative positions of their atomsor groups in space are called stereoisomers.This special type of isomerism is called asstereoisomerism and can be classified asgeometrical and optical isomerism.

12.7 FUNDAMENTAL CONCEPTS INORGANIC REACTION MECHANISM

In an organic reaction, the organic molecule(also referred as a substrate) reacts with anappropriate attacking reagent and leads to theformation of one or more intermediate(s) andfinally product(s)

The general reaction is depicted as follows :

understanding the reactivity of organiccompounds and in planning strategy for theirsynthesis.

In the following sections, we shall learnsome of the principles that explain how thesereactions take place.

12.7.1 Fission of a Covalent Bond

A covalent bond can get cleaved either by : (i)heterolytic cleavage, or by (ii) homolyticcleavage.

In heterolytic cleavage, the bond breaksin such a fashion that the shared pair ofelectrons remains with one of the fragments.

After heterolysis, one atom has a sextetelectronic structure and a positive charge andthe other, a valence octet with at least onelone pair and a negative charge. Thus,heterolytic cleavage of bromomethane will give

3CH+

and Br– as shown below.

A species having a carbon atom possessingsextext of electrons and a positive charge iscalled a carbocation (earlier called carboniumion). The C

+H

3 ion is known as a methyl cation

or methyl carbonium ion. Carbocations areclassified as primary, secondary or tertiarydepending on whether one, two or threecarbons are directly attached to the positivelycharged carbon. Some other examples ofcarbocations are: CH3C

+H2

(ethyl cation, aprimary carbocation), (CH3)2C

+H (isopropyl

cation, a secondary carbocation), and (CH3)3C+

(tert-butyl cation, a tertiary carbocation).Carbocations are highly unstable and reactivespecies. Alkyl groups directly attached to thepositively charged carbon stabilise thecarbocations due to inductive andhyperconjugation effects, which you will bestudying in the sections 12.7.5 and 12.7.9.The observed order of carbocation stability is:C+H

3 < CH

3C+H

2 < (CH

3)2C+H < (CH

3)3C+. These

carbocations have trigonal planar shape withpositively charged carbon being sp2

hybridised. Thus, the shape of C+H3 may be

considered as being derived from the overlapof three equivalent C(sp2) hybridised orbitals

Organicmolecule(Substrate)

[Intermediate] Product(s)

AttackingReagent

Byproducts

Substrate is that reactant which suppliescarbon to the new bond and the other reactantis called reagent. If both the reactants supplycarbon to the new bond then choice isarbitrary and in that case the molecule onwhich attention is focused is called substrate.

In such a reaction a covalent bondbetween two carbon atoms or a carbon andsome other atom is broken and a new bond isformed. A sequential account of each step,describing details of electron movement,energetics during bond cleavage and bondformation, and the rates of transformationof reactants into products (kinetics) isreferred to as reaction mechanism. Theknowledge of reaction mechanism helps in

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342 CHEMISTRY

with 1s orbital of each of the three hydrogenatoms. Each bond may be represented asC(sp2)–H(1s) sigma bond. The remainingcarbon orbital is perpendicular to themolecular plane and contains no electrons.(Fig. 12.3).

Fig. 12.3 Shape of methyl cation

The heterolytic cleavage can also give aspecies in which carbon gets the shared pairof electrons. For example, when group Zattached to the carbon leaves without

electron pair, the methyl anion is

formed. Such a carbon species carrying anegative charge on carbon atom is calledcarbanion. Carbanions are also unstable andreactive species. The organic reactions whichproceed through heterolytic bond cleavage arecalled ionic or heteropolar or just polarreactions.

In homolytic cleavage, one of theelectrons of the shared pair in a covalent bondgoes with each of the bonded atoms. Thus, inhomolytic cleavage, the movement of a singleelectron takes place instead of an electronpair. The single electron movement is shownby ‘half-headed’ (fish hook: ) curved arrow.Such cleavage results in the formation ofneutral species (atom or group) whichcontains an unpaired electron. These speciesare called free radicals. Like carbocationsand carbanions, free radicals are alsovery reactive. A homolytic cleavage can beshown as:

Alkyl free radical

Alkyl radicals are classified as primary,secondary, or tertiary. Alkyl radical stabilityincreases as we proceed from primary totertiary:

,Methyl Ethyl Isopropyl Tert-butyl

free free free freeradical radical radical radical

Organic reactions, which proceed byhomolytic fission are called free radical orhomopolar or nonpolar reactions.

12.7.2 Nucleophiles and Electrophiles

A reagent that brings an electron pair is calleda nucleophile (Nu:) i.e., nucleus seeking andthe reaction is then called nucleophilic. Areagent that takes away an electron pair iscalled electrophile (E+) i.e., electron seekingand the reaction is called electrophilic.

During a polar organic reaction, anucleophile attacks an electrophilic centre ofthe substrate which is that specific atom orpart of the electrophile that is electrondeficient. Similarly, the electrophiles attack atnucleophilic centre, which is the electronrich centre of the substrate. Thus, theelectrophiles receive electron pair fromnucleophile when the two undergo bondinginteraction. A curved-arrow notation is usedto show the movement of an electron pair fromthe nucleophile to the electrophile. Someexamples of nucleophiles are the negativelycharged ions with lone pair of electrons suchas hydroxide (HO– ), cyanide (NC–) ions andcarbanions (R3C:–). Neutral molecules such

as etc., can also act asnucleophiles due to the presence of lone pairof electrons. Examples of electrophiles

include carbocations (3CH

+) and neutral

molecules having functional groups likecarbonyl group (>C=O) or alkyl halides(R3C-X, where X is a halogen atom). Thecarbon atom in carbocations has sextetconfiguration; hence, it is electron deficientand can receive a pair of electrons from thenucleophiles. In neutral molecules such asalkyl halides, due to the polarity of the C-Xbond a partial positive charge is generated

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343ORGANIC CHEMISTRY – SOME BASIC PRINCIPLES AND TECHNIQUES

on the carbon atom and hence the carbonatom becomes an electrophilic centre atwhich a nucleophile can attack.

Problem 12.11

Using curved-arrow notation, show theformation of reactive intermediates whenthe following covalent bonds undergoheterolytic cleavage.

(a) CH3–SCH3, (b) CH3–CN, (c) CH3–Cu

Solution

Problem 12.12

Giving justification, categorise thefollowing molecules/ions as nucleophileor electrophile:

Solution

Nucleophiles: ( )2 5 3 23: :HS ,C H O , CH N ,H N− −−

These species have unshared pair ofelectrons, which can be donated andshared with an electrophile.

Electrophi les : 3 3 2BF ,Cl,CH C O,NO+ + +

− = .Reactive sites have only six valenceelectrons; can accept electron pair froma nucleophile.

Problem 12.13Identify electrophilic centre in thefollowing: CH3CH=O, CH3CN, CH3I.

Solution

Among CH3HC*=O, H

3CC*≡N, and

H3C*–I, the starred carbon atoms are

electrophilic centers as they will havepartial positive charge due to polarity ofthe bond.

12.7.3 Electron Movement in OrganicReactions

The movement of electrons in organicreactions can be shown by curved-arrownotation. It shows how changes in bondingoccur due to electronic redistribution duringthe reaction. To show the change in positionof a pair of electrons, curved arrow starts fromthe point from where an electron pair is shiftedand it ends at a location to which the pair ofelectron may move.

Presentation of shifting of electron pair isgiven below :

(i) from π bond to adjacent bond position

(ii) from π bond toadjacent atom

(iii) from atom to adjacentbond position

Movement of single electron is indicatedby a single barbed ‘fish hooks’ (i.e. half headedcurved arrow). For example, in transfer ofhydroxide ion giving ethanol and in thedissociation of chloromethane, the movementof electron using curved arrows can bedepicted as follows:

12.7.4 Electron Displacement Effects inCovalent Bonds

The electron displacement in an organicmolecule may take place either in the groundstate under the influence of an atom or asubstituent group or in the presence of anappropriate attacking reagent. The electrondisplacements due to the influence ofan atom or a substituent group present inthe molecule cause permanent polarlisationof the bond. Inductive ef fect andresonance effects are examples of this type ofelectron displacements. Temporary electrondisplacement effects are seen in a molecule

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344 CHEMISTRY

when a reagent approaches to attack it. Thistype of electron displacement is calledelectromeric effect or polarisability effect. Inthe following sections we will learn about thesetypes of electronic displacements.

12.7.5 Inductive Effect

When a covalent bond is formed betweenatoms of different electronegativity, theelectron density is more towards the moreelectronegative atom of the bond. Such a shiftof electron density results in a polar covalentbond. Bond polarity leads to various electroniceffects in organic compounds.

Let us consider cholorethane (CH3CH2Cl)in which the C–Cl bond is a polar covalentbond. It is polarised in such a way that thecarbon-1 gains some positive charge (δ+

) andthe chlorine some negative charge (δ–

). Thefractional electronic charges on the two atomsin a polar covalent bond are denoted bysymbol δ (delta) and the shift of electrondensity is shown by an arrow that points fromδ+

to δ– end of the polar bond.

δδ+ δ+ δ−

CH3 ⎯→⎯CH

2⎯→⎯⎯→⎯⎯→⎯⎯→⎯⎯→⎯Cl 2 1

In turn carbon-1, which has developedpartial positive charge (δ+

) draws someelectron density towards it from the adjacentC-C bond. Consequently, some positive charge(δδ+

) develops on carbon-2 also, where δδ+

symbolises relatively smaller positive chargeas compared to that on carbon – 1. In otherwords, the polar C – Cl bond induces polarityin the adjacent bonds. Such polarisation ofσ-bond caused by the polarisation of adjacentσ-bond is referred to as the inductive effect.This effect is passed on to the subsequentbonds also but the effect decreases rapidlyas the number of intervening bonds increasesand becomes vanishingly small after threebonds. The inductive effect is related to theability of substituent(s) to either withdraw ordonate electron density to the attached carbonatom. Based on this ability, the substitutentscan be classified as electron-withdrawing orelectron donating groups relative to hydrogen.Halogens and many other groups such as

nitro (- NO2), cyano (- CN), carboxy (- COOH),ester (-COOR), aryloxy (-OAr, e.g. – OC6H5),etc. are electron-withdrawing groups. On theother hand, the alkyl groups like methyl(–CH3) and ethyl (–CH2–CH3) are usuallyconsidered as electron donating groups.

Problem 12.14

Which bond is more polar in the followingpairs of molecules: (a) H3C-H, H3C-Br(b) H3C-NH2, H3C-OH (c) H3C-OH,H3C-SH

Solution

(a) C–Br, since Br is more electronegativethan H, (b) C–O, (c) C–O

Problem 12.15

In which C–C bond of CH3CH2CH2Br, theinductive effect is expected to be theleast?

Solution

Magnitude of inductive effect diminishesas the number of intervening bondsincreases. Hence, the effect is least in thebond between carbon-3 and hydrogen.

12.7.6 Resonance Structure

There are many organic molecules whosebehaviour cannot be explained by a singleLewis structure. An example is that ofbenzene. Its cyclic structurecontaining alternating C–C singleand C=C double bonds shown isinadequate for explaining itscharacteristic properties.

As per the above representation, benzeneshould exhibit two different bond lengths, dueto C–C single and C=C double bonds. However,as determined experimentally benzene has auniform C–C bond distances of 139 pm, avalue intermediate between the C–Csingle(154 pm) and C=C double (134 pm)bonds. Thus, the structure of benzene cannotbe represented adequately by the abovestructure. Further, benzene can berepresented equally well by the energeticallyidentical structures I and II.

Benzene

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345ORGANIC CHEMISTRY – SOME BASIC PRINCIPLES AND TECHNIQUES

Therefore, according to the resonance theory(Unit 4) the actual structure of benzenecannot be adequately represented by any ofthese structures, rather it is a hybrid of thetwo structures (I and II) called resonancestructures. The resonance structures(canonical structures or contributingstructures) are hypothetical andindividually do not represent any realmolecule. They contribute to the actualstructure in proportion to their stability.

Another example of resonance is providedby nitromethane (CH3NO2) which can berepresented by two Lewis structures, (I andII). There are two types of N-O bonds in thesestructures.

However, it is known that the two N–Obonds of nitromethane are of the samelength (intermediate between a N–O singlebond and a N=O double bond). The actualstructure of nitromethane is therefore aresonance hybrid of the two canonicalforms I and II.

The energy of actual structure of themolecule (the resonance hybrid) is lower thanthat of any of the canonical structures. Thedifference in energy between the actualstructure and the lowest energy resonancestructure is called the resonancestabilisation energy or simply theresonance energy. The more the number ofimportant contributing structures, the moreis the resonance energy. Resonance isparticularly important when the contributingstructures are equivalent in energy.

The following rules are applied while writingresonance structures:

The resonance structures have (i) the samepositions of nuclei and (ii) the same number of

unpaired electrons. Among the resonancestructures, the one which has more numberof covalent bonds, all the atoms with octet ofelectrons (except hydrogen which has aduplet), less separation of opposite charges,(a negative charge if any on moreelectronegative atom, a positive charge if anyon more electropositive atom) and moredispersal of charge, is more stable than others.

Problem 12.16

Write resonance structures of CH3COO

and show the movement of electrons bycurved arrows.

Solution

First, write the structure and putunshared pairs of valence electrons onappropriate atoms. Then draw the arrowsone at a time moving the electrons to getthe other structures.

Problem 12.17

Write resonance structures ofCH2=CH–CHO. Indicate relative stability ofthe contributing structures.

Solution

Stability: I > II > III

[I: Most stable, more number of covalentbonds, each carbon and oxygen atom hasan octet and no separation of oppositecharge II: negative charge on moreelectronegative atom and positive chargeon more electropositive atom; III: doesnot contribute as oxygen has positivecharge and carbon has negative charge,hence least stable].

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346 CHEMISTRY

Problem 12.18Explain why the following two structures,I and II cannot be the major contributorsto the real structure of CH3COOCH3.

Solution

The two structures are less importantcontributors as they involve chargeseparation. Additionally, structure Icontains a carbon atom with anincomplete octet.

12.7.7 Resonance Effect

The resonance effect is defined as ‘the polarityproduced in the molecule by the interactionof two π-bonds or between a π-bond and lonepair of electrons present on an adjacent atom’.The effect is transmitted through the chain.There are two types of resonance ormesomeric effect designated as R or M effect.

(i) Positive Resonance Effect (+R effect)In this effect, the transfer of electrons is awayfrom an atom or substituent group attachedto the conjugated system. This electrondisplacement makes certain positions in themolecule of high electron densities. This effectin aniline is shown as :

The atoms or substituent groups, whichrepresent +R or –R electron displacementeffects are as follows :

+R effect: – halogen, –OH, –OR, –OCOR, –NH2,–NHR, –NR2, –NHCOR,

– R effect: – COOH, –CHO, >C=O, – CN, –NO2

The presence of alternate single anddouble bonds in an open chain or cyclicsystem is termed as a conjugated system.These systems often show abnormalbehaviour. The examples are 1,3- butadiene,aniline and nitrobenzene etc. In such systems,the π-electrons are delocalised and the systemdevelops polarity.

12.7.8 Electromeric Effect (E effect)

It is a temporary effect. The organiccompounds having a multiple bond (a doubleor triple bond) show this effect in the presenceof an attacking reagent only. It is defined asthe complete transfer of a shared pair ofπ-electrons to one of the atoms joined by amultiple bond on the demand of an attackingreagent. The effect is annulled as soon as theattacking reagent is removed from the domainof the reaction. It is represented by E and theshifting of the electrons is shown by a curvedarrow ( ). There are two distinct types ofelectromeric effect.(i) Positive Eelctromeric Effect (+E effect) Inthis effect the π−electrons of the multiple bondare transferred to that atom to which thereagent gets attached. For example :

(ii) Negative Resonance Effect (- R effect)This effect is observed when the transfer ofelectrons is towards the atom or substituentgroup attached to the conjugated system. Forexample in nitrobenzene this electrondisplacement can be depicted as :

(ii) Negative Electromeric Effect (–E effect) Inthis effect the π - electrons of the multiplebond are transferred to that atom to whichthe attacking reagent does not get attached.For example:

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When inductive and electromeric effectsoperate in opposite directions, the electomericeffect predominates.

12.7.9 Hyperconjugation

Hyperconjugation is a general stabilisinginteraction. It involves delocalisation ofσ electrons of C—H bond of an alkyl groupdirectly attached to an atom of unsaturatedsystem or to an atom with an unsharedp orbital. The σ electrons of C—H bond of thealkyl group enter into partial conjugation withthe attached unsaturated system or with theunshared p orbital. Hyperconjugation is apermanent effect.

To understand hyperconjugation effect, let

us take an example of 3 2CH CH+

(ethyl cation)in which the positively charged carbon atomhas an empty p orbital. One of the C-H bondsof the methyl group can align in the plane ofthis empty p orbital and the electronsconstituting the C-H bond in plane with thisp orbital can then be delocalised into theempty p orbital as depicted in Fig. 12.4 (a).

In general, greater the number of alkylgroups attached to a positively charged carbonatom, the greater is the hyperconjugationinteraction and stabilisation of the cation.Thus, we have the following relative stabilityof carbocations :

Fig. 12.4(b) Orbital diagram showinghyperconjugation in propene

Fig. 12.4(a) Orbital diagram showinghyperconjugation in ethyl cation

This type of overlap stabilises thecarbocation because electron density from theadjacent σ bond helps in dispersing thepositive charge.

Hyperconjugation is also possible inalkenes and alkylarenes.

Delocalisation of electrons byhyperconjugation in the case of alkene canbe depicted as in Fig. 12.4(b).

There are various ways of looking at thehyperconjugative effect. One of the way is toregard C—H bond as possessing partial ioniccharacter due to resonance.

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The hyperconjugation may also beregarded as no bond resonance.

Problem 12.19Explain why (CH3)3C

+ is more stable than++

3 2 3CH CH andCH is the least stablecation.

SolutionHyperconjugation interaction in (CH3)3C

+

is greater than in +

3 2CH CH as the

(CH3)3C+ has nine C-H bonds. In

+

3CH ,vacant p orbital is perpendicular to theplane in which C-H bonds lie; hence

cannot overlap with it. Thus, +

3CH lackshyperconjugative stability.

12.7.10 Types of Organic Reactions andMechanisms

Organic reactions can be classified into thefollowing categories:(i) Substitution reactions(ii) Addition reactions(iii) Elimination reactions(iv) Rearrangement reactions

You will be studying these reactions inUnit 13 and later in class XII.

12.8 METHODS OF PURIFICATION OFORGANIC COMPOUNDS

Once an organic compound is extracted froma natural source or synthesised in thelaboratory, it is essential to purify it. Variousmethods used for the purification of organiccompounds are based on the nature of thecompound and the impurity present in it.

The common techniques used forpurification are as follows :(i) Sublimation(ii) Crystallisation(iii) Distillation(iv) Differential extraction and(v) Chromatography

Finally, the purity of a compound isascertained by determining its melting orboiling point. Most of the pure compoundshave sharp melting points and boiling points.

New methods of checking the purity of anorganic compound are based on differenttypes of chromatographic and spectroscopictechniques.

12.8.1 Sublimation

You have learnt earlier that on heating, somesolid substances change from solid to vapourstate without passing through liquid state.The purification technique based on the aboveprinciple is known as sublimation and is usedto separate sublimable compounds from non-sublimable impurities.

12.8.2 Crystallisation

This is one of the most commonly usedtechniques for the purification of solid organiccompounds. It is based on the difference inthe solubilities of the compound and theimpurities in a suitable solvent. The impurecompound is dissolved in a solvent in whichit is sparingly soluble at room temperaturebut appreciably soluble at highertemperature. The solution is concentrated toget a nearly saturated solution. On coolingthe solution, pure compound crystallises outand is removed by filtration. The filtrate(mother liquor) contains impurities and smallquantity of the compound. If the compoundis highly soluble in one solvent and very littlesoluble in another solvent, crystallisation canbe satisfactorily carried out in a mixture ofthese solvents. Impurities, which impartcolour to the solution are removed byadsorbing over activated charcoal. Repeatedcrystallisation becomes necessary for thepurification of compounds containingimpurities of comparable solubilities.

12.8.3 Distillation

This important method is used to separate (i)volatile liquids from nonvolatile impurities and(ii) the liquids having sufficient difference intheir boiling points. Liquids having differentboiling points vaporise at dif ferenttemperatures. The vapours are cooled and theliquids so formed are collected separately.Chloroform (b.p 334 K) and aniline (b.p. 457K) are easily separated by the technique ofdistillation (Fig 12.5). The liquid mixture is

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taken in a round bottom flask andheated carefully. On boiling, thevapours of lower boiling componentare formed first. The vapours arecondensed by using a condenser andthe liquid is collected in a receiver. Thevapours of higher boiling componentform later and the liquid can becollected separately.

Fractional Distillation: If thedifference in boiling points of twoliquids is not much, simple distillationcannot be used to separate them. Thevapours of such liquids are formedwithin the same temperature range andare condensed simultaneously. Thetechnique of fractional distillation isused in such cases. In this technique,vapours of a liquid mixture are passedthrough a fractionating column beforecondensation. The fractionatingcolumn is fitted over the mouth of theround bottom flask (Fig.12.6).

Fig.12.5 Simple distillation. The vapours of a substanceformed are condensed and the liquid is collectedin conical flask.

Fig.12.6 Fractional distillation. The vapours of lower boiling fraction reach thetop of the column first followed by vapours of higher boiling fractions.

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Vapours of the liquid with higher boilingpoint condense before the vapours of theliquid with lower boiling point. The vapoursrising up in the fractionating column becomericher in more volatile component. By the timethe vapours reach to the top of thefractionating column, these are rich in themore volatile component. Fractionatingcolumns are available in various sizes anddesigns as shown in Fig.12.7. A fractionatingcolumn provides many surfaces for heatexchange between the ascending vapoursand the descending condensed liquid. Someof the condensing liquid in the fractionatingcolumn obtains heat from the ascendingvapours and revaporises. The vapours thusbecome richer in low boiling component. Thevapours of low boiling component ascend tothe top of the column. On reaching the top,the vapours become pure in low boilingcomponent and pass through the condenserand the pure liquid is collected in a receiver.After a series of successive distillations, theremaining liquid in the distillation flask getsenriched in high boiling component. Eachsuccessive condensation and vaporisationunit in the fractionating column is called a

Fig.12.7 Different types of fractionating columns.

theoretical plate. Commercially, columnswith hundreds of plates are available.

One of the technological applications offractional distillation is to separate differentfractions of crude oil in petroleum industry.

Distillation under reduced pressure: Thismethod is used to purify liquids having veryhigh boiling points and those, whichdecompose at or below their boiling points.Such liquids are made to boil at a temperaturelower than their normal boiling points byreducing the pressure on their surface. Aliquid boils at a temperature at which itsvapour pressure is equal to the externalpressure. The pressure is reduced with thehelp of a water pump or vacuum pump(Fig.12.8). Glycerol can be separated fromspent-lye in soap industry by using thistechnique.

Steam Distillation: This technique isapplied to separate substances which aresteam volatile and are immiscible withwater. In steam distillation, steam from asteam generator is passed through a heatedflask containing the liquid to be distilled.The mixture of steam and the volatileorganic compound is condensed andcollected. The compound is later separatedfrom water using a separating funnel. Insteam distillation, the liquid boils whenthe sum of vapour pressures due to theorganic liquid (p1) and that due to water(p2) becomes equal to the atmosphericpressure (p), i.e. p =p1+ p2. Since p1 islower than p, the organic liquid vaporisesat lower temperature than its boilingpoint.

Thus, if one of the substances in themixture is water and the other, a waterinsoluble substance, then the mixture will boilclose to but below, 373K. A mixture of waterand the substance is obtained which can beseparated by using a separating funnel.Aniline is separated by this technique fromaniline – water mixture (Fig.12.9).

12.8.4 Differential Extraction

When an organic compound is present in anaqueous medium, it is separated by shaking

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Fig.12.8 Distillation under reduced pressure. A liquid boils at a temperature below itsvapour pressure by reducing the pressure.

Fig.12.9 Steam distillation. Steam volatile component volatilizes, the vapours condense inthe condenser and the liquid collects in conical flask.

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352 CHEMISTRY

Fig.12.11 Column chromatography. Differentstages of separation of componentsof a mixture.

Fig.12.10 Differential extraction. Extraction of com-pound takes place based on differencein solubility

it with an organic solvent in which it is moresoluble than in water. The organic solvent andthe aqueous solution should be immisciblewith each other so that they form two distinctlayers which can be separated by separatoryfunnel. The organic solvent is later removedby distillation or by evaporation to get backthe compound. Differential extraction iscarried out in a separatory funnel as shownin Fig. 12.10. If the organic compound is less

mixture get gradually separated from oneanother. The moving phase is called the mobilephase.

Based on the principle involved,chromatography is classified into differentcategories. Two of these are:

(a) Adsorption chromatography, and(b) Partition chromatography.

a) Adsorption Chromatography: Adsor-ption chromatography is based on the factthat different compounds are adsorbed on anadsorbent to different degrees. Commonlyused adsorbents are silica gel and alumina.When a mobile phase is allowed to moveover a stationary phase (adsorbent),the components of the mixture move byvarying distances over the stationaryphase. Following are two main types ofchromatographic techniques based on theprinciple of differential adsorption.

(a) Column chromatography, and(b) Thin layer chromatography.

Column Chromatography: Columnchromatography involves separation of amixture over a column of adsorbent(stationary phase) packed in a glass tube. Thecolumn is fitted with a stopcock at its lowerend (Fig. 12.11). The mixture adsorbed onsoluble in the organic solvent, a very large

quantity of solvent would be required toextract even a very small quantity of thecompound. The technique of continuousextraction is employed in such cases. In thistechnique same solvent is repeatedly used forextraction of the compound.

12.8.5 ChromatographyChromatography is an important techniqueextensively used to separate mixtures intotheir components, purify compounds and alsoto test the purity of compounds. The namechromatography is based on the Greek wordchroma, for colour since the method was firstused for the separation of colouredsubstances found in plants. In this technique,the mixture of substances is applied onto astationary phase, which may be a solid or aliquid. A pure solvent, a mixture of solvents,or a gas is allowed to move slowly over thestationary phase. The components of the

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adsorbent is placed on the top of theadsorbent column packed in a glass tube. Anappropriate eluant which is a liquid or amixture of liquids is allowed to flow down thecolumn slowly. Depending upon the degreeto which the compounds are adsorbed,complete separation takes place. The mostreadily adsorbed substances are retained nearthe top and others come down to variousdistances in the column (Fig.12.11).

Thin Layer Chromatography: Thin layerchromatography (TLC) is another type ofadsorption chromatography, which involvesseparation of substances of a mixture over athin layer of an adsorbent coated on glassplate. A thin layer (about 0.2mm thick) of anadsorbent (silica gel or alumina) is spread overa glass plate of suitable size. The plate isknown as thin layer chromatography plate orchromaplate. The solution of the mixture tobe separated is applied as a small spot about2 cm above one end of the TLC plate. Theglass plate is then placed in a closed jarcontaining the eluant (Fig. 12.12a). As the

eluant rises up the plate, the components ofthe mixture move up along with the eluant todifferent distances depending on their degreeof adsorption and separation takes place. Therelative adsorption of each component of themixture is expressed in terms of itsretardation factor i.e. Rf value (Fig.12.12 b).

Rf = Distance moved by the substance from base line (x)

Distance moved by the solvent from base line (y)

The spots of coloured compounds arevisible on TLC plate due to their originalcolour. The spots of colourless compounds,which are invisible to the eye but fluoresce,can be detected by putting the plate underultraviolet light. Another detection techniqueis to place the plate in a covered jar containinga few crystals of iodine. Spots of compounds,which adsorb iodine, will show up as brownspots. Sometimes an appropriate reagent mayalso be sprayed on the plate. For example,amino acids may be detected by spraying theplate with ninhydrin solution (Fig.12.12b).

Partition Chromatography: Partitionchromatography is based on continuousdifferential partitioning of components of amixture between stationary and mobilephases. Paper chromatography is a type ofpartition chromatography. In paperchromatography, a special quality paperknown as chromatography paper is used.Chromatography paper contains watertrapped in it, which acts as the stationaryphase.

A strip of chromatography paper spottedat the base with the solution of the mixture issuspended in a suitable solvent or a mixtureof solvents (Fig. 12.13). This solvent acts asthe mobile phase. The solvent rises up thepaper by capillary action and flows over thespot. The paper selectively retains differentcomponents according to their differingpartition in the two phases. The paper stripso developed is known as a chromatogram.The spots of the separated colouredcompounds are visible at different heightsfrom the position of initial spot on thechromatogram. The spots of the separatedcolourless compounds may be observed eitherFig.12.12 (b) Developed chromatogram.

Fig.12.12 (a) Thin layer chromatography.Chromatogram being developed.

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354 CHEMISTRY

5H2O + CuSO4 ⎯→ CuSO4.5H2O White Blue

12.9.2 Detection of Other Elements

Nitrogen, sulphur, halogens and phosphoruspresent in an organic compound are detectedby “Lassaigne’s test”. The elements presentin the compound are converted from covalentform into the ionic form by fusing thecompound with sodium metal. Followingreactions take place:

Na + C + N Δ⎯⎯→ NaCN

2Na + S Δ⎯⎯→ Na2S

Na + X Δ⎯⎯→ Na X(X = Cl, Br or I)

C, N, S and X come from organiccompound.

Cyanide, sulphide and halide of sodiumso formed on sodium fusion are extracted fromthe fused mass by boiling it with distilledwater. This extract is known as sodium fusionextract.

(A) Test for NitrogenThe sodium fusion extract is boiled withiron(II) sulphate and then acidified withconcentrated sulphuric acid. The formationof Prussian blue colour confirms the presenceof nitrogen. Sodium cyanide first reactswith iron(II) sulphate and forms sodiumhexacyanoferrate(II). On heating withconcentrated sulphuric acid some iron(II) ionsare oxidised to iron(III) ions which react withsodium hexacyanoferrate(II) to produceiron(III) hexacyanoferrate(II) (ferriferrocyanide)which is Prussian blue in colour.

6CN– + Fe2+ → [Fe(CN)6]

4–

3[Fe(CN)6]4– + 4Fe3+ Fe4[Fe(CN)6]3.xH2O

Prussian blue(B) Test for Sulphur(a) The sodium fusion extract is acidified

with acetic acid and lead acetate is addedto it. A black precipitate of lead sulphideindicates the presence of sulphur.

S2– + Pb2+ ⎯→ PbS Black

(b) On treating sodium fusion extract withsodium nitroprusside, appearance of a

Fig.12.13 Paper chromatography.Chromatography paper in two differentshapes.

under ultraviolet light or by the use of anappropriate spray reagent as discussed underthin layer chromatography.

12.9 QUALITATIVE ANALYSIS OFORGANIC COMPOUNDS

The elements present in organic compoundsare carbon and hydrogen. In addition to these,they may also contain oxygen, nitrogen,sulphur, halogens and phosphorus.

12.9.1 Detection of Carbon and Hydrogen

Carbon and hydrogen are detected by heatingthe compound with copper(II) oxide. Carbonpresent in the compound is oxidised to carbondioxide (tested with lime-water, whichdevelops turbidity) and hydrogen to water(tested with anhydrous copper sulphate,which turns blue).

C + 2CuO Δ⎯⎯→ 2Cu + CO2

2H + CuO Δ⎯⎯→ Cu + H2O

CO2 + Ca(OH)2 ⎯→ CaCO3↓ + H2O

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355ORGANIC CHEMISTRY – SOME BASIC PRINCIPLES AND TECHNIQUES

violet colour further indicates thepresence of sulphur.

S2– + [Fe(CN)5NO]2– ⎯→ [Fe(CN)5NOS]4–

VioletIn case, nitrogen and sulphur both are

present in an organic compound, sodiumthiocyanate is formed. It gives blood redcolour and no Prussian blue since there areno free cyanide ions.

Na + C + N + S ⎯→ NaSCNFe3+ +SCN– ⎯→ [Fe(SCN)]2+

Blood redIf sodium fusion is carried out with excess

of sodium, the thiocyanate decomposes toyield cyanide and sulphide. These ions givetheir usual tests.

NaSCN + 2Na ⎯→ NaCN+Na2S

(C) Test for HalogensThe sodium fusion extract is acidified withnitric acid and then treated with silver nitrate.A white precipitate, soluble in ammoniumhydroxide shows the presence of chlorine, ayellowish precipitate, sparingly soluble inammonium hydroxide shows the presence ofbromine and a yellow precipitate, insolublein ammonium hydroxide shows the presenceof iodine.

X– + Ag+ ⎯→ AgX X represents a halogen – Cl, Br or I.

If nitrogen or sulphur is also present in thecompound, the sodium fusion extract is firstboiled with concentrated nitric acid todecompose cyanide or sulphide of sodiumformed during Lassaigne’s test. These ionswould otherwise interfere with silver nitratetest for halogens.

(D) Test for PhosphorusThe compound is heated with an oxidisingagent (sodium peroxide). The phosphoruspresent in the compound is oxidised tophosphate. The solution is boiled with nitricacid and then treated with ammoniummolybdate. A yellow colouration or precipitateindicates the presence of phosphorus.

Na3PO4 + 3HNO3 ⎯→ H3PO4+3NaNO3

H3PO4 + 12(NH4)2MoO4 + 21HNO3 ⎯→ Ammonium

molybdate

(NH4)3PO4.12MoO3 + 21NH4NO3 + 12H2O Ammoniumphosphomolybdate

12.10 QUANTITATIVE ANALYSISThe percentage composition of elementspresent in an organic compound isdetermined by the methods based on thefollowing principles:

12.10.1 Carbon and Hydrogen

Both carbon and hydrogen are estimated inone experiment. A known mass of an organiccompound is burnt in the presence of excessof oxygen and copper(II) oxide. Carbon andhydrogen in the compound are oxidised tocarbon dioxide and water respectively.CxHy + (x + y/4) O2 ⎯→ x CO2 + (y/2) H2O

The mass of water produced is determinedby passing the mixture through a weighedU-tube containing anhydrous calciumchloride. Carbon dioxide is absorbed inanother U-tube containing concentratedsolution of potassium hydroxide. These tubesare connected in series (Fig.12.14). The

Fig.12.14 Estimation of carbon and hydrogen. Water and carbon dioxide formed on oxidation of substanceare absorbed in anhydrous calcium chloride and potassium hydroxide solutions respectivelycontained in U tubes.

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356 CHEMISTRY

increase in masses of calcium chloride andpotassium hydroxide gives the amounts ofwater and carbon dioxide from which thepercentages of carbon and hydrogen arecalculated.

Let the mass of organic compound bem g, mass of water and carbon dioxideproduced be m1 and m2 g respectively;

Percentage of carbon= 212 10044× ×

×m

m

Percentage of hydrogen = 12 10018× ×

×m

m

Problem 12.20On complete combustion, 0.246 g of anorganic compound gave 0.198g of carbondioxide and 0.1014g of water. Determinethe percentage composition of carbonand hydrogen in the compound.

Solution

12 0.198 100Percentage of carbon =

44 0.246= 21.95%

× ××

Fig.12.15 Dumas method. The organic compound yields nitrogen gas on heating it withcopper(II) oxide in the presence of CO

2 gas. The mixture of gases is collected

over potassium hydroxide solution in which CO2 is absorbed and volume of

nitrogen gas is determined.

Percentage of hydrogen = 2 0.1014 100

18 0.246× ×

×

= 4.58%

12.10.2 Nitrogen

There are two methods for estimation ofnitrogen: (i) Dumas method and (ii) Kjeldahl’smethod.(i) Dumas method: The nitrogen containingorganic compound, when heated with copperoxide in an atmosphere of carbon dioxide,yields free nitrogen in addition to carbondioxide and water.CxHyNz + (2x + y/2) CuO ⎯→

x CO2 + y/2 H2O + z/2 N2 + (2x + y/2) Cu

Traces of nitrogen oxides formed, if any,are reduced to nitrogen by passing thegaseous mixture over a heated copper gauze.The mixture of gases so produced is collectedover an aqueous solution of potassiumhydroxide which absorbs carbon dioxide.Nitrogen is collected in the upper part of thegraduated tube (Fig.12.15).

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357ORGANIC CHEMISTRY – SOME BASIC PRINCIPLES AND TECHNIQUES

Let the mass of organic compound = m gVolume of nitrogen collected = V1 mLRoom temperature = T1K

11

1

273Volume of nitrogen at STP=

760××

p V

T

(Let it be V mL)Where p

1 and V

1 are the pressure and volume

of nitrogen, p1 is dif ferent from the

atmospheric pressure at which nitrogen gasis collected. The value of p

1 is obtained by

the relation;

p1= Atmospheric pressure – Aqueous tension

22400 mL N2 at STP weighs 28 g.

2

28 mL N at STP weighs = g

22400×V

V

Percentage of nitrogen = 28 10022400× ×

×V

m

Problem 12.21In Dumas’ method for estimation ofnitrogen, 0.3g of an organic compoundgave 50mL of nitrogen collected at 300Ktemperature and 715mm pressure.

Calculate the percentage composition ofnitrogen in the compound. (Aqueoustension at 300K=15 mm)

Solution

Volume of nitrogen collected at 300K and715mm pressure is 50 mLActual pressure = 715-15 =700 mm

273 700 50Volume of nitrogen at STP =

300 760= 41.9 mL

× ××

22,400 mL of N2 at STP weighs = 28 g

28 41.941.9 mL of nitrogen weighs= g

22400×

28 41.9 100Percentage of nitrogen =

22400 0.3=17.46%

× ××

(ii) Kjeldahl’s method: The compoundcontaining nitrogen is heated withconcentrated sulphuric acid. Nitrogen in thecompound gets converted to ammoniumsulphate (Fig. 12.16). The resulting acidmixture is then heated with excess of sodium

Fig.12.16 Kjeldahl method. Nitrogen-containing compound is treated with concentrated H2SO

4 to get

ammonium sulphate which liberates ammonia on treating with NaOH; ammonia is absorbedin known volume of standard acid.

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358 CHEMISTRY

Fig. 12.17 Carius method. Halogen containingorganic compound is heated with fumingnitric acid in the presence of silvernitrate.

hydroxide. The liberated ammonia gas isabsorbed in an excess of standard solution ofsulphuric acid. The amount of ammoniaproduced is determined by estimating theamount of sulphuric acid consumed in thereaction. It is done by estimating unreactedsulphuric acid left after the absorption ofammonia by titrating it with standard alkalisolution. The difference between the initialamount of acid taken and that left after thereaction gives the amount of acid reacted withammonia.

Organic compound + H2SO4 ⎯→ (NH4)2SO42

2 4 3 22 2NaOH

Na SO NH H O⎯ →⎯⎯⎯⎯ + +

2NH3 + H2SO4 ⎯→ (NH4)2SO4

Let the mass of organic compound taken = m gVolume of H2SO4 of molarity, M,taken = V mLVolume of NaOH of molarity, M, used fortitration of excess of H2SO4 = V1 mLV1mL of NaOH of molarity M

= V1 /2 mL of H2SO4 of molarity MVolume of H2SO4 of molarity M unused= (V - V1/2) mL

(V- V1/2) mL of H2SO4 of molarity M

= 2(V-V1/2) mL of NH3 solution ofmolarity M.1000 mL of 1 M NH3 solution contains 17gNH3 or 14 g of N2(V-V1/2) mL of NH3 solution of molarity Mcontains:

( )114 M 2 V V /2

g N1000

× × −

( )114 M 2 V-V /2 100Percentage of N=

1000

× ××

m

( )11.4 M 2 - /2=

× × V V

m

Kjeldahl method is not applicable tocompounds containing nitrogen in nitro andazo groups and nitrogen present in the ring(e.g. pyridine) as nitrogen of these compoundsdoes not change to ammonium sulphateunder these conditions.

Problem 12.22

During estimation of nitrogen present inan organic compound by Kjeldahl’smethod, the ammonia evolved from 0.5g of the compound in Kjeldahl’sestimation of nitrogen, neutralized 10 mLof 1 M H2SO4. Find out the percentageof nitrogen in the compound.

Solution

1 M of 10 mL H2SO4=1M of 20 mL NH31000 mL of 1M ammonia contains 14 gnitrogen20 mL of 1M ammonia contains

14 20

1000

× g nitrogen

Percentage of nitrogen =14 20 100× ×

×=

1000 05560%

..

12.10.3 Halogens

Carius method: A known mass of an organiccompound is heated with fuming nitric acidin the presence of silver nitrate contained ina hard glass tube known as Carius tube,(Fig.12.17) in a furnace. Carbon and hydrogen

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359ORGANIC CHEMISTRY – SOME BASIC PRINCIPLES AND TECHNIQUES

present in the compound are oxidised tocarbon dioxide and water. The halogenpresent forms the corresponding silver halide(AgX). It is filtered, washed, dried and weighed.Let the mass of organic compound taken = m gMass of AgX formed = m1 g1 mol of AgX contains 1 mol of X

Mass of halogen in m1g of AgX

1atomic mass of X gmolecular mass of AgX

×=

m

Percentage of halogen

=× ×

×

atomic mas

molecular

s of X

mass of AgX

1m

m

100

Problem 12.23

In Carius method of estimation ofhalogen, 0.15 g of an organic compoundgave 0.12 g of AgBr. Find out thepercentage of bromine in the compound.

Solution

Molar mass of AgBr = 108 + 80 = 188 g mol-1

188 g AgBr contains 80 g bromine

0.12 g AgBr contains 80 012

188

× . g bromine

80 0.12 100Percentage of bromine=

188 0.15= 34.04%

× ××

12.10.4 Sulphur

A known mass of an organic compound isheated in a Carius tube with sodium peroxideor fuming nitric acid. Sulphur present in thecompound is oxidised to sulphuric acid. It isprecipitated as barium sulphate by addingexcess of barium chloride solution in water.The precipitate is filtered, washed, dried andweighed. The percentage of sulphur can becalculated from the mass of barium sulphate.

Let the mass of organic

compound taken = m gand the mass of barium sulphate formed = m1g

1 mol of BaSO4 = 233 g BaSO4 = 32 g sulphur

m1 g BaSO4 contains 132233×m

g sulphur

132 100Percentage of sulphur=

233× ×

×m

m

Problem 12.24

In sulphur estimation, 0.157 g of anorganic compound gave 0.4813 g ofbarium sulphate. What is thepercentage of sulphur in the compound?

Solution

Molecular mass of BaSO4 = 137+32+64 = 233 g

233 g BaSO4 contains 32 g sulphur

0.4813 g BaSO4 contains 32 0.4813233× g

sulphur

32 0.4813 100Percentage of sulphur=

233 0.157 = 42.10%

× ××

12.10.5 Phosphorus

A known mass of an organic compound isheated with fuming nitric acid whereuponphosphorus present in the compound isoxidised to phosphoric acid. It is precipitatedas ammonium phosphomolybdate, (NH4)3PO4.12MoO3, by adding ammonia andammonium molybdate. Alternatively,phosphoric acid may be precipitated asMgNH4PO4 by adding magnesia mixturewhich on ignition yields Mg2P2O7.

Let the mass of organic compound taken= m g and mass of ammonium phosphomolydate = m1g

Molar mass of (NH4)3PO4.12MoO3 = 1877 g

Percentage of phosphorus = 131 1001877× ×

×m

m%

If phosphorus is estimated as Mg2P2O7,

Percentage of phosphorus = 162 100%

222× ×

×m

m

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360 CHEMISTRY

SUMMARY

In this unit, we have learnt some basic concepts in structure and reactivity of organiccompounds, which are formed due to covalent bonding. The nature of the covalent bondingin organic compounds can be described in terms of orbitals hybridisation concept, accordingto which carbon can have sp3, sp2 and sp hybridised orbitals. The sp3, sp2 and sp hybridisedcarbons are found in compounds like methane, ethene and ethyne respectively. Thetetrahedral shape of methane, planar shape of ethene and linear shape of ethyne can beunderstood on the basis of this concept. A sp3 hybrid orbital can overlap with 1s orbital ofhydrogen to give a carbon - hydrogen (C–H) single bond (sigma, σ bond). Overlap of a sp2

orbital of one carbon with sp2 orbital of another results in the formation of a carbon–carbonσ bond. The unhybridised p orbitals on two adjacent carbons can undergo lateral (side-by-side) overlap to give a pi (π) bond. Organic compounds can be represented by various structuralformulas. The three dimensional representation of organic compounds on paper can bedrawn by wedge and dash formula.

Organic compounds can be classified on the basis of their structure or the functionalgroups they contain. A functional group is an atom or group of atoms bonded together in aunique fashion and which determines the physical and chemical properties of the compounds.The naming of the organic compounds is carried out by following a set of rules laid down bythe International Union of Pure and Applied Chemistry (IUPAC). In IUPAC nomenclature,the names are correlated with the structure in such a way that the reader can deduce thestructure from the name.

where, 222 u is the molar mass of Mg2P2O7,m, the mass of organic compound taken, m1,the mass of Mg2P2O7 formed and 62, the massof two phosphorus atoms present in thecompound Mg2P2O7.

12.10.6 Oxygen

The percentage of oxygen in an organiccompound is usually found by differencebetween the total percentage composition(100) and the sum of the percentages of allother elements. However, oxygen can also beestimated directly as follows:

A definite mass of an organic compoundis decomposed by heating in a stream ofnitrogen gas. The mixture of gaseous productscontaining oxygen is passed over red-hot cokewhen all the oxygen is converted to carbonmonoxide. This mixture is passed throughwarm iodine pentoxide (I2O5) when carbonmonoxide is oxidised to carbon dioxideproducing iodine.

Compound heat⎯ →⎯⎯ O2 + other gaseous

products

2C + O2 1373K⎯ →⎯⎯⎯ 2CO

I2O5 + 5CO ⎯→ I2 + 5CO2

The percentage of oxygen can be derivedfrom the amount of carbon dioxide or iodineproduced.Let the mass of organic = m gcompound takenMass of carbon dioxide = m1 g44 g carbon dioxide = 32 g oxygen

m1 g carbon dioxide contains 13244×m

g O2

Percentage of oxygen = 132 100

44× ×

×m

m%

Presently, the estimation of elements inan organic compound is carried out by usingmicroquantities of substances and automaticexperimental techniques. The elements,carbon, hydrogen and nitrogen present in acompound are determined by an apparatusknown as CHN elemental analyser. Theanalyser requires only a very small amountof the substance (1-3 mg) and displays thevalues on a screen within a short time. Adetailed discussion of such methods is beyondthe scope of this book.

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Organic reaction mechanism concepts are based on the structure of the substratemolecule, fission of a covalent bond, the attacking reagents, the electron displacement effectsand the conditions of the reaction. These organic reactions involve breaking and making ofcovalent bonds. A covalent bond may be cleaved in heterolytic or homolytic fashion. Aheterolytic cleavage yields carbocations or carbanions, while a homolytic cleavage givesfree radicals as reactive intermediate. Reactions proceeding through heterolytic cleavageinvolve the complimentary pairs of reactive species. These are electron pair donor known asnucleophile and an electron pair acceptor known as electrophile. The inductive, resonance,electromeric and hyperconjugation effects may help in the polarisation of a bond makingcertain carbon atom or other atom positions as places of low or high electron densities.

Organic reactions can be broadly classified into following types; substitution, addition,elimination and rearrangement reactions.

Purification, qualitative and quantitative analysis of organic compounds are carried outfor determining their structures. The methods of purification namely : sublimation, distillationand differential extraction are based on the difference in one or more physical properties.Chromatography is a useful technique of separation, identification and purification ofcompounds. It is classified into two categories : adsorption and partition chromatography.Adsorption chromatography is based on differential adsorption of various components of amixture on an adsorbent. Partition chromatography involves continuous partitioning of thecomponents of a mixture between stationary and mobile phases. After getting the compoundin a pure form, its qualitative analysis is carried out for detection of elements present in it.Nitrogen, sulphur, halogens and phosphorus are detected by Lassaigne’s test. Carbon andhydrogen are estimated by determining the amounts of carbon dioxide and water produced.Nitrogen is estimated by Dumas or Kjeldahl’s method and halogens by Carius method.Sulphur and phosphorus are estimated by oxidising them to sulphuric and phosphoricacids respectively. The percentage of oxygen is usually determined by difference betweenthe total percentage (100) and the sum of percentages of all other elements present.

EXERCISES

12.1 What are hybridisation states of each carbon atom in the following compounds ?

CH2=C=O, CH3CH=CH2, (CH3)2CO, CH2=CHCN, C6H6

12.2 Indicate the σ and π bonds in the following molecules :

C6H6, C6H12, CH2Cl2, CH2=C=CH2, CH3NO2, HCONHCH3

12.3 Write bond line formulas for : Isopropyl alcohol, 2,3-Dimethyl butanal, Heptan-4-one.

12.4 Give the IUPAC names of the following compounds :

(a) (b) (c)

(d) (e) (f) Cl2CHCH

2OH

12.5 Which of the following represents the correct IUPAC name for the compoundsconcerned ? (a) 2,2-Dimethylpentane or 2-Dimethylpentane (b) 2,4,7-Trimethyloctane or 2,5,7-Trimethyloctane (c) 2-Chloro-4-methylpentane or4-Chloro-2-methylpentane (d) But-3-yn-1-ol or But-4-ol-1-yne.

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362 CHEMISTRY

12.6 Draw formulas for the first five members of each homologous series beginning withthe following compounds. (a) H–COOH (b) CH3COCH3 (c) H–CH=CH2

12.7 Give condensed and bond line structural formulas and identify the functionalgroup(s) present, if any, for :

(a) 2,2,4-Trimethylpentane

(b) 2-Hydroxy-1,2,3-propanetricarboxylic acid

(c) Hexanedial

12.8 Identify the functional groups in the following compounds

(a) (b) (c)

12.9 Which of the two: O2NCH2CH2O– or CH3CH2O

– is expected to be more stable andwhy ?

12.10 Explain why alkyl groups act as electron donors when attached to a π system.

12.11 Draw the resonance structures for the following compounds. Show the electronshift using curved-arrow notation.

(a) C6H5OH (b) C6H5NO2 (c) CH3CH=CHCHO (d) C6H5–CHO (e) 6 5 2C H CH

+

(f) 3 2CH CH CHCH

+

=

12.12 What are electrophiles and nucleophiles ? Explain with examples.

12.13 Identify the reagents shown in bold in the following equations as nucleophiles orelectrophiles:

(a) 3 3 2CH COOH CH COO H O−+ → +–HO

(b) ( ) ( ) ( )3 3 3 2CH COCH CH C CN OH+ →

CN

(c) 6 5 6 5 3C H C H COCH+ →+

3CH CO

12.14 Classify the following reactions in one of the reaction type studied in this unit.

(a)3 2 3 2CH CH Br HS CH CH SH−+ →

(b) ( ) ( )3 2 32 2CH C CH HCl CH ClC= + → −

(c) 3 2 2 2 2CH CH Br HO CH CH H O−+ → = +

(d) ( ) ( )3 2 3 23 2CH C CH OH HBr CH CBrCH CH− + →

12.15 What is the relationship between the members of following pairs of structures ?Are they structural or geometrical isomers or resonance contributors ?

(a)

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363ORGANIC CHEMISTRY – SOME BASIC PRINCIPLES AND TECHNIQUES

(b)

(c)

12.16 For the following bond cleavages, use curved-arrows to show the electron flowand classify each as homolysis or heterolysis. Identify reactive intermediateproduced as free radical, carbocation and carbanion.

(a)

(b)

(c)

(d)

12.17 Explain the terms Inductive and Electromeric effects. Which electron displacementeffect explains the following correct orders of acidity of the carboxylic acids?

(a) Cl3CCOOH > Cl2CHCOOH > ClCH2COOH

(b) CH3CH2COOH > (CH3)2CHCOOH > (CH3)3C.COOH

12.18 Give a brief description of the principles of the following techniques taking anexample in each case.

(a) Crystallisation (b) Distillation (c) Chromatography

12.19 Describe the method, which can be used to separate two compounds with differentsolubilities in a solvent S.

12.20 What is the difference between distillation, distillation under reduced pressureand steam distillation ?

12.21 Discuss the chemistry of Lassaigne’s test.

12.22 Differentiate between the principle of estimation of nitrogen in an organic compoundby (i) Dumas method and (ii) Kjeldahl’s method.

12.23 Discuss the principle of estimation of halogens, sulphur and phosphorus presentin an organic compound.

12.24 Explain the principle of paper chromatography.

12.25 Why is nitric acid added to sodium extract before adding silver nitrate for testinghalogens?

12.26 Explain the reason for the fusion of an organic compound with metallic sodiumfor testing nitrogen, sulphur and halogens.

12.27 Name a suitable technique of separation of the components from a mixture ofcalcium sulphate and camphor.

12.28 Explain, why an organic liquid vaporises at a temperature below its boiling pointin its steam distillation ?

12.29 Will CCl4 give white precipitate of AgCl on heating it with silver nitrate? Givereason for your answer.

Page 39: Organic Chm

364 CHEMISTRY

12.30 Why is a solution of potassium hydroxide used to absorb carbon dioxide evolvedduring the estimation of carbon present in an organic compound?

12.31 Why is it necessary to use acetic acid and not sulphuric acid for acidification ofsodium extract for testing sulphur by lead acetate test?

12.32 An organic compound contains 69% carbon and 4.8% hydrogen, the remainderbeing oxygen. Calculate the masses of carbon dioxide and water produced when0.20 g of this substance is subjected to complete combustion.

12.33 A sample of 0.50 g of an organic compound was treated according to Kjeldahl’smethod. The ammonia evolved was absorbed in 50 ml of 0.5 M H2SO4. The residualacid required 60 mL of 0.5 M solution of NaOH for neutralisation. Find thepercentage composition of nitrogen in the compound.

12.34 0.3780 g of an organic chloro compound gave 0.5740 g of silver chloride in Cariusestimation. Calculate the percentage of chlorine present in the compound.

12.35 In the estimation of sulphur by Carius method, 0.468 g of an organic sulphurcompound afforded 0.668 g of barium sulphate. Find out the percentage of sulphurin the given compound.

12.36 In the organic compound CH2 = CH – CH2 – CH2 – C ≡ CH, the pair of hydridisedorbitals involved in the formation of: C2 – C3 bond is:

(a) sp – sp2 (b) sp – sp3 (c) sp2 – sp3 (d) sp3 – sp3

12.37 In the Lassaigne’s test for nitrogen in an organic compound, the Prussian bluecolour is obtained due to the formation of:

(a) Na4[Fe(CN)6] (b) Fe4[Fe(CN)6]3 (c) Fe2[Fe(CN)6] (d) Fe3[Fe(CN)6]412.38 Which of the following carbocation is most stable ?

(a) (CH3)3C.+

C H2 (b) (CH3)3+

C (c) CH3CH2

+

C H2 (d) CH3

+

C H CH2CH3

12.39 The best and latest technique for isolation, purification and separation of organiccompounds is:

(a) Crystallisation (b) Distillation (c) Sublimation (d) Chromatography

12.40 The reaction:

CH3CH2I + KOH(aq) → CH3CH2OH + KI

is classified as :

(a) electrophilic substitution (b) nucleophilic substitution

(c) elimination (d) addition