UNIT 12

ORGANIC CHEMISTRY - SOME BASIC PRINCIPLES AND TECHNIQUES

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 compounds 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.

$$\underset{\text{Ammonium cyanate}}{\mathrm{NH_4CNO}} \xrightarrow{\text{Heat}} \underset{\text{Urea}}{\mathrm{NH_2CONH_2}}$$

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.

The development of electronic theory of covalent bonding ushered organic chemistry into 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 of molecular structure helps in understanding and predicting the properties of organic compounds. You have already learnt theories of valency and molecular structure in Unit 4. Also, you already know that tetravalence of carbon and the formation of covalent bonds by it are explained in terms of its electronic configuration and the hybridisation of $s$ and $p$ orbitals. It may be recalled that formation and the shapes of molecules like methane $\left(\mathrm{CH}_{4}\right)$, ethene $\left(\mathrm{C}_2 \mathrm{H}_4\right)$, ethyne $\left(\mathrm{C}_2 \mathrm{H}_2\right)$ are explained in terms of the use of $s p^{3}, s p^{2}$ and $s p$ hybrid orbitals by carbon atoms in the respective molecules.

Hybridisation influences the bond length and bond enthalpy (strength) in compounds. The $s p$ hybrid orbital contains more $s$ character and hence it is closer to its nucleus and forms shorter and stronger bonds than the $s p^{3}$ hybrid orbital. The $s p^{2}$ hybrid orbital is intermediate in $s$ character between $s p$ and $s p^{3}$ and, hence, the length and enthalpy of the bonds it forms, are also intermediate between them. The change in hybridisation affects the electronegativity of carbon. The greater the $s$ character of the hybrid orbitals, the greater is the electronegativity. Thus, a carbon atom having an $s p$ hybrid orbital with $50 \% s$ character is more electronegative than that possessing $s p^{2}$ or $s p^{3}$ hybridised orbitals. This relative electronegativity is reflected in several physical and chemical properties of the molecules concerned, about which you will learn in later units.

12.2.2 Some Characteristic Features of $\pi$ Bonds

In a $\pi$ (pi) bond formation, parallel orientation of the two $p$ orbitals on adjacent atoms is necessary for a proper sideways overlap. Thus, in $\mathrm{H_2} \mathrm{C}=\mathrm{CH_2}$ molecule all the atoms must be in the same plane. The $p$ orbitals are mutually parallel and both the $p$ orbitals are perpendicular to the plane of the molecule. Rotation of one $\mathrm{CH_2}$ fragment with respect to other interferes with maximum overlap of $p$ orbitals and, therefore, such rotation about carbon-carbon double bond $(\mathrm{C}=\mathrm{C})$ is restricted. The electron charge cloud of the $\pi$ bond is located above and below the plane of bonding atoms. This results in the electrons being easily available to the attacking reagents. In general, $\pi$ bonds provide the most reactive centres in the molecules containing multiple bonds.

Problem 12.1

How many $\sigma$ and $\pi$ bonds are present in each of the following molecules?

(a) $\mathrm{HC} \equiv \mathrm{CCH}=\mathrm{CHCH_3}$ (b) $\mathrm{CH_2}=\mathrm{C}=\mathrm{CHCH_3}$

Solution

(a) $\sigma_{\mathrm{C}-\mathrm{C}}: 4 ; \sigma_{\mathrm{C}-\mathrm{H}}: 6 ; \pi_{\mathrm{C}=\mathrm{C}}: 1 ; \pi \mathrm{C} \equiv \mathrm{C}: 2$

(b) $\sigma_{\mathrm{C}-\mathrm{C}}: 3 ; \sigma_{\mathrm{C}-\mathrm{H}}: 6 ; \pi_{\mathrm{C}=\mathrm{C}}: 2$.

Problem 12.2

What is the type of hybridisation of each carbon in the following compounds?

(a) $\mathrm{CH_3} \mathrm{Cl}$, (b) $\left(\mathrm{CH_3}\right)_{2} \mathrm{CO}$, (c) $\mathrm{CH_3} \mathrm{CN}$,

(d) $\mathrm{HCONH_2}$, (e) $\mathrm{CH_3} \mathrm{CH}=\mathrm{CHCN}$

Solution

(a) $s p^{3}$, (b) $s p^{3}, s p^{2}$, (c) $s p^{3}, s p$, (d) $s p^{2}$, (e) $s p^{3}, s p^{2}, s p^{2}, s p$

Problem 12.3

Write the state of hybridisation of carbon in the following compounds and shapes of each of the molecules. (a) $\mathrm{H_2} \mathrm{C}=\mathrm{O}$, (b) $\mathrm{CH_3} \mathrm{~F}$, (c) $\mathrm{HC} \equiv \mathrm{N}$.

Solution

(a) $s p^{2}$ hybridised carbon, trigonal planar; (b) $s p^{3}$ hybridised carbon, tetrahedral; (c) $s p$ hybridised carbon, linear.

12.3STRUCTURAL REPRESENTATIONS OF ORGANIC COMPOUNDS

12.3.1 Complete, Condensed and Bond-line

Complete, Condensed and Bond-line Structural Formulas

Structures of organic compounds are represented in several ways. The Lewis structure or dot structure, dash structure, condensed structure and bond line structural formulas are some of the specific types. The Lewis structures, however, can be simplified by representing the two-electron covalent bond by a dash (-). Such a structural formula focuses on the electrons involved in bond formation. A single dash represents a single bond, double dash is used for double bond and a triple dash represents triple bond. Lonepairs of electrons on heteroatoms (e.g., oxygen, nitrogen, sulphur, halogens etc.) may or may not be shown. Thus, ethane $\left(\mathrm{C_2} \mathrm{H_6}\right)$, ethene $\left(\mathrm{C_2} \mathrm{H_4}\right)$, ethyne $\left(\mathrm{C_2} \mathrm{H_2}\right)$ and methanol $\left(\mathrm{CH_3} \mathrm{OH}\right)$ can be represented by the following structural formulas. Such structural representations are called complete structural formulas.

These structural formulas can be further abbreviated by omitting some or all of the dashes representing covalent bonds and by indicating the number of identical groups attached to an atom by a subscript. The resulting expression of the compound is called a condensed structural formula. Thus, ethane, ethene, ethyne and methanol can be written as:

$$ \begin{array}{cccc} \underset{\text{Ethane}}{\mathrm{CH_3CH_3}} & \underset{\text{Ethene}}{\mathrm{H_2C}=\mathrm{CH_2}} & \underset{\text{Ethyne}}{\mathrm{HC}=\mathrm{HC}} & \underset{\text{Methanol}}{\mathrm{CH_3} \mathrm{OH}} \end{array} $$

Similarly, $\mathrm{CH_3} \mathrm{CH_2} \mathrm{CH_2} \mathrm{CH_2} \mathrm{CH_2} \mathrm{CH_2} \mathrm{CH_2} \mathrm{CH_3}$ can be further condensed to $\mathrm{CH_3}\left(\mathrm{CH_2}\right)_{6} \mathrm{CH_3}$. For further simplification, organic chemists use another way of representing the structures, in which only lines are used. In this bond-line structural representation of organic compounds, carbon and hydrogen atoms are not shown and the lines representing carbon-carbon bonds are drawn in a zig-zag fashion. The only atoms specifically written are oxygen, chlorine, nitrogen etc. The terminals denote methyl $\left(-\mathrm{CH_3}\right)$ groups (unless indicated otherwise by a functional group), while the line junctions denote carbon atoms bonded to appropriate number of hydrogens required to satisfy the valency of the carbon atoms. Some of the examples are represented as follows:

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

(ii) Various ways of representing 2-bromo butane are:

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

Problem 12.4

Expand each of the following condensed formulas into their complete structural formulas.

(a) $\mathrm{CH_3} \mathrm{CH_2} \mathrm{COCH_2} \mathrm{CH_3}$

(b) $\mathrm{CH_3} \mathrm{CH}=\mathrm{CH}\left(\mathrm{CH_2}\right)_{3} \mathrm{CH_3}$

Solution

Problem 12.5

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

(a) $\mathrm{HOCH_2} \mathrm{CH_2} \mathrm{CH_2} \mathrm{CH} \left(\mathrm{CH_3} \right) \mathrm{CH} \left(\mathrm{CH_3} \right) \mathrm{CH_3}$

(b)

Solution

Condensed formula:

(a) $\mathrm{HO}\left(\mathrm{CH_2}\right)_3 \mathrm{CH}\left(\mathrm{CH_3}\right) \mathrm{CH}\left(\mathrm{CH_3}\right)_2$

(b) $\mathrm{HOCH}(\mathrm{CN})_{2}$

Bond-line formula:

Problem 12.6

Expand each of the following bond-line formulas to show all the atoms including carbon and hydrogen

Solution

12.3.2Three-Dimensional Representation of Organic Molecules

The three-dimensional (3-D) structure of organic molecules can be represented on paper by using certain conventions. For example, by using solid (-) and dashed ( …IIII) wedge formula, the 3 -D image of a molecule from a two-dimensional picture can be perceived. In these formulas the solid-wedge is used to indicate a bond projecting out of the plane of paper, towards the observer. The dashed-wedge is used to depict the bond projecting out of the plane of the paper and away from the observer. Wedges are shown in such a way that the broad end of the wedge is towards the observer. The bonds lying in plane of the paper are depicted by using a normal line ($-$). 3-D representation of methane molecule on paper has been shown in Fig. 12.1.

Molecular Models

Molecular models are physical devices that are used for a better visualisation and perception of three-dimensional shapes of organic molecules. These are made of wood, plastic or metal and are commercially available. Commonly three types of molecular models are used: (1) Framework model, (2) Ball-and-stick model, and (3) Space filling model. In the framework model only the bonds connecting the atoms of a molecule and not the atoms themselves are shown. This model emphasizes the pattern of bonds of a molecule while ignoring the size of atoms. In the ball-and-stick model, both the atoms and the bonds are shown. Balls represent atoms and the stick denotes a bond. Compounds containing $\mathrm{C}=\mathrm{C}$ (e.g., ethene) can best be represented by using springs in place of sticks. These models are referred to as ball-and-spring model. The space-filling model emphasises the relative size of each atom based on its van der Waals radius. Bonds are not shown in this model. It conveys the volume occupied by each atom in the molecule. In addition to these models, computer graphics can also be used for molecular modelling.

12.4 CLASSIFICATION OF ORGANIC COMPOUNDS

The existing large number of organic compounds and their ever-increasing numbers has made it necessary to classify them on the basis of their structures. Organic compounds are broadly classified as follows:

I. Acyclic or open chain compounds

These compounds are also called as aliphatic compounds and consist of straight or branched chain compounds, for example:

II Cyclic or closed chain or ring compounds

(a) Alicyclic compounds

Alicyclic (aliphatic cyclic) compounds contain carbon atoms joined in the form of a ring (homocyclic).

Sometimes atoms other than carbon are also present in the ring (heterocylic). Tetrahydrofuran given below is an example of this type of compound:

These exhibit some of the properties similar to those of aliphatic compounds.

(b) Aromatic compounds

Aromatic compounds are special types of compounds. You will learn about these compounds in detail in Unit 9. These include benzene and other related ring compounds (benzenoid). Like alicyclic compounds, aromatic comounds may also have hetero atom in the ring. Such compounds are called hetrocyclic aromatic compounds. Some of the examples of various types of aromatic compounds are:

Benzenoid aromatic compounds

Non-benzenoid compound

Heterocyclic aromatic compounds

Organic compounds can also be classified on the basis of functional groups, into families or homologous series.

12.4.1 Functional Group

The functional group is an atom or a group of atoms joined to the carbon chain which is responsible for the characteristic chemical properties of the organic compounds. The examples are hydroxyl group (-OH), aldehyde group (-CHO) and carboxylic acid group $(\mathrm{COOH})$ etc.

12.4.2 Homologous Series

A group or a series of organic compounds each containing a characteristic functional group forms a homologous series and the members of the series are called homologues. The members of a homologous series can be represented by general molecular formula and the successive members differ from each other in molecular formula by a $-\mathrm{CH_2}$ unit. There are a number of homologous series of organic compounds. Some of these are alkanes, alkenes, alkynes, haloalkanes, alkanols, alkanals, alkanones, alkanoic acids, amines etc.

It is also possible that a compound contains two or more identical or different functional groups. This gives rise to polyfunctional compounds.

12.5 NOMENCLATURE OF ORGANIC COMPOUNDS

Organic chemistry deals with millions of compounds. In order to clearly identify them, a systematic method of naming has been developed and is known as the IUPAC (International Union of Pure and Applied Chemistry) system of nomenclature. In this systematic nomenclature, the names are correlated with the structure such that the reader or listener can deduce the structure from the name.

Before the IUPAC system of nomenclature, however, organic compounds were assigned names based on their origin or certain properties. For instance, citric acid is named so because it is found in citrus fruits and the acid found in red ant is named formic acid since the Latin word for ant is formica. These names are traditional and are considered as trivial or common names. Some common names are followed even today. For example, Buckminsterfullerene is a common name given to the newly discovered $\mathrm{C_60}$ cluster (a form of carbon) noting its structural similarity to the geodesic domes popularised by the famous architect R. Buckminster Fuller. Common names are useful and in many cases indispensable, particularly when the alternative systematic names are lengthy and complicated. Common names of some organic compounds are given in Table 12.1.

Table 12.1 Common or Trivial Names of Some Organic Compounds

Compound	Common name
$\mathrm{CH_4}$	Methane
$\mathrm{H_3} \mathrm{CCH_2} \mathrm{CH_2} \mathrm{CH_3}$	$n$-Butane
$\left(\mathrm{H_3} \mathrm{C_2} \mathrm{CHCH_3}\right.$	Isobutane
$\left(\mathrm{H_3} \mathrm{C}\right)_{4} \mathrm{C}$	Neopentane
$\mathrm{H_3} \mathrm{CCH_2} \mathrm{CH_2} \mathrm{OH}$	$n$-Propyl alcohol
$\mathrm{HCHO}$	Formaldehyde
$\left(\mathrm{H_3} \mathrm{C}\right)_{2} \mathrm{CO}$	Acetone
$\mathrm{CHCl_3}$	Chloroform
$\mathrm{CH_3} \mathrm{COOH}$	Acetic acid
$\mathrm{C_6} \mathrm{H_6}$	Benzene
$\mathrm{C_6} \mathrm{H_5} \mathrm{OCH_3}$	Anisole
$\mathrm{C_6} \mathrm{H_5} \mathrm{NH_2}$	Aniline
$\mathrm{C_6} \mathrm{H_5} \mathrm{COCH_3}$	Acetophenone
$\mathrm{CH_3} \mathrm{OCH_2} \mathrm{CH_3}$	Ethyl methyl ether

12.5.1 The IUPAC System of Nomenclature

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

By further using prefixes and suffixes, the parent name can be modified to obtain the actual name. Compounds containing carbon and hydrogen only are called hydrocarbons. A hydrocarbon is termed saturated if it contains only carbon-carbon single bonds. The IUPAC name for a homologous series of such compounds is alkane. Paraffin (Latin: little affinity) was the earlier name given to these compounds. Unsaturated hydrocarbons are those, which contain at least one carboncarbon double or triple bond.

12.5.2 IUPAC Nomenclature of Alkanes

Straight chain hydrocarbons: The names of such compounds are based on their chain structure, and end with suffix ‘-ane’ and carry a prefix indicating the number of carbon atoms present in the chain (except from $\mathrm{CH_4}$ to $\mathrm{C_4} \mathrm{H_10}$, where the prefixes are derived from trivial names). The IUPAC names of some straight chain saturated hydrocarbons are given in Table 8.2. The alkanes in Table 12.2 differ from each other by merely the number of $-\mathrm{CH_2}$ groups in the chain. They are homologues of alkane series.

Table 12.2 IUPAC Names of Some Unbranched Saturated Hydrocarbons

Name	Molecular formula	Name	Molecular formula
Methane	$\mathrm{CH_4}$	Heptane	$\mathrm{C_7} \mathrm{H_16}$
Ethane	$\mathrm{C_2} \mathrm{H_6}$	Octane	$\mathrm{C_8} \mathrm{H_18}$
Propane	$\mathrm{C_3} \mathrm{H_8}$	Nonane	$\mathrm{C_9} \mathrm{H_20}$
Butane	$\mathrm{C_4} \mathrm{H_10}$	Decane	$\mathrm{C_10} \mathrm{H_22}$
Pentane	$\mathrm{C_5} \mathrm{H_12}$	Icosane	$\mathrm{C_20} \mathrm{H_42}$
Hexane	$\mathrm{C_6} \mathrm{H_14}$	Triacontane	$\mathrm{C_30} \mathrm{H_62}$

Branched chain hydrocarbons: In a branched chain compound small chains of carbon atoms are attached at one or more carbon atoms of the parent chain. The small carbon chains (branches) are called alkyl groups. For example:

In order to name such compounds, the names of alkyl groups are prefixed to the name of parent alkane. An alkyl group is derived from a saturated hydrocarbon by removing a hydrogen atom from carbon. Thus, $\mathrm{CH_4}$ becomes $-\mathrm{CH_3}$ and is called methyl group. An alkyl group is named by substituting ’ $y$ l’ for ‘ane’ in the corresponding alkane. Some alkyl groups are listed in Table 12.3.

Table 12.3 Some Alkyl Groups

Alkane		Alkyl group
Molecular formula	Name of alkane	Structural formula	Name of alkyl group
$\mathrm{CH_4}$	Methane	$-\mathrm{CH_3}$	Methyl
$\mathrm{C_2} \mathrm{H_6}$	Ethane	$-\mathrm{CH_2} \mathrm{CH_3}$	Ethyl
$\mathrm{C_3} \mathrm{H_8}$	Propane	$-\mathrm{CH_2} \mathrm{CH_2} \mathrm{CH_3}$	Propyl
$\mathrm{C_4} \mathrm{H_10}$	Butane	$-\mathrm{CH_2} \mathrm{CH_2} \mathrm{CH_2} \mathrm{CH_3}$	Butyl
$\mathrm{C_10} \mathrm{H_22}$	Decane	$-\mathrm{CH_2}\left(\mathrm{CH_2}\right)_{8} \mathrm{CH_3}$	Decyl

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

Common branched groups have specific trivial names. For example, the propyl groups can either be n-propyl group or isopropyl group. The branched butyl groups are called sec-butyl, isobutyl and tert-butyl group. We also encounter the structural unit, $-\mathrm{CH_2} \mathrm{C}\left(\mathrm{CH_3}\right)_{3}$, which is called neopentyl group.

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

1. First of all, the longest carbon chain in the molecule is identified. In the example (I) given below, the longest chain has nine carbons and it is considered as the parent or root chain. Selection of parent chain as shown in (II) is not correct because it has only eight carbons.

2. The carbon atoms of the parent chain are numbered to identify the parent alkane and to locate the positions of the carbon atoms at which branching takes place due to the substitution of alkyl group in place of hydrogen atoms. The numbering is done in such a way that the branched carbon atoms get the lowest possible numbers. Thus, the numbering in the above example should be from left to right (branching at carbon atoms 2 and 6) and not from right to left (giving numbers 4 and 8 to the carbon atoms at which branches are attached).

3. The names of alkyl groups attached as a branch are then prefixed to the name of the parent alkane and position of the substituents is indicated by the appropriate numbers. If different alkyl groups are present, they are listed in alphabetical order. Thus, name for the compound shown above is: 6-ethyl-2methylnonane. [Note: the numbers are separated from the groups by hyphens and there is no break between methyl and nonane.]

4. If two or more identical substituent groups are present then the numbers are separated by commas. The names of identical substituents are not repeated, instead prefixes such as di (for 2), tri (for 3), tetra (for 4), penta (for 5), hexa (for 6) etc. are used. While writing the name of the substituents in alphabetical order, these prefixes, however, are not considered.

Thus, the following compounds are named as:

5. If the two substituents are found in equivalent positions, the lower number is given to the one coming first in the alphabetical listing. Thus, the following compound is 3-ethyl-6-methyloctane and not 6-ethyl-3-methyloctane.

6. The branched alkyl groups can be named by following the above mentioned procedures. However, the carbon atom of the branch that attaches to the root alkane is numbered 1 as exemplified below.

The name of such branched chain alkyl group is placed in parenthesis while naming the compound. While writing the trivial names of substituents’ in alphabetical order, the prefixes iso- and neo- are considered to be the part of the fundamental name of alkyl group. The prefixes sec- and tert- are not considered to be the part of the fundamental name. The use of iso and related common prefixes for naming alkyl groups is also allowed by the IUPAC nomenclature as long as these are not further substituted. In multisubstituted compounds, the following rules may aso be remembered:

• If there happens to be two chains of equal size, then that chain is to be selected which contains more number of side chains.
• After selection of the chain, numbering is to be done from the end closer to the substituent.

Cyclic Compounds: A saturated monocyclic compound is named by prefixing ‘cyclo’ to the corresponding straight chain alkane. If side chains are present, then the rules given above are applied. Names of some cyclic compounds are given below.

Problem 12.7

Structures and IUPAC names of some hydrocarbons are given below. Explain why the names given in the parentheses are incorrect.

Solution

(a) Lowest locant number, 2,5,6 is lower than 3,5,7, (b) substituents are in equivalent position; lower number is given to the one that comes first in the name according to alphabetical order.

12.5.3 Nomenclature of Organic Compounds having Functional Group(s)

A functional group, as defined earlier, is an atom or a group of atoms bonded together in a unique manner which is usually the site of chemical reactivity in an organic molecule. Compounds having the same functional group undergo similar reactions. For example, $\mathrm{CH_3} \mathrm{OH}, \mathrm{CH_3} \mathrm{CH_2} \mathrm{OH}$, and $\left(\mathrm{CH_3}\right)_{2} \mathrm{CHOH}-$ all having - $\mathrm{OH}$ functional group liberate hydrogen on reaction with sodium metal. The presence of functional groups enables systematisation of organic compounds into different classes. Examples of some functional groups with their prefixes and suffixes along with some examples of organic compounds possessing these are given in Table 12.4.

First of all, the functional group present in the molecule is identified which determines the choice of appropriate suffix. The longest chain of carbon atoms containing the functional group is numbered in such a way that the functional group is attached at the carbon atom possessing lowest possible number in the chain. By using the suffix as given in Table 12.4, the name of the compound is arrived at.

In the case of polyfunctional compounds, one of the functional groups is chosen as the principal functional group and the compound is then named on that basis. The remaining functional groups, which are subordinate functional groups, are named as substituents using the appropriate prefixes. The choice of principal functional group is made on the basis of order of preference. The order of decreasing priority for some functional groups is:

$-\mathrm{COOH},-\mathrm{SO_3} \mathrm{H},-\mathrm{COOR} \text{(R=alkyl group)}, \mathrm{COCl}, -\mathrm{CONH_2},-\mathrm{CN},-\mathrm{HC}=\mathrm{O},>\mathrm{C}=\mathrm{O},-\mathrm{OH},-\mathrm{NH_2},> \mathbf{C}=\mathbf{C}<, \quad-\mathbf{C} \equiv \mathbf{C}-$.

The $-\mathrm{R}, \mathrm{C_6} \mathrm{H_5}-$, halogens ( $\left.\mathrm{F}, \mathrm{Cl}, \mathrm{Br}, \mathrm{I}\right),-\mathrm{NO_2}$, alkoxy (-OR) etc. are always prefix substituents. Thus, a compound containing both an alcohol and a keto group is named as hydroxyalkanone since the keto group is preferred to the hydroxyl group.

For example, $\mathrm{HOCH_2}\left(\mathrm{CH_2}\right)_{3} \mathrm{CH_2} \mathrm{COCH_3}$ will be named as 7-hydroxyheptan-2-one and not as 2-oxoheptan -7-ol. Similarly, $\mathrm{BrCH_2} \mathrm{CH}=\mathrm{CH_2}$ is named as 3-bromoprop-1-ene and not 1 -bromoprop-2-ene.

If more than one functional group of the same type are present, their number is indicated by adding di, tri, etc. before the class suffix. In such cases the full name of the parent alkane is written before the class suffix. For example $\mathrm{CH_2}(\mathrm{OH}) \mathrm{CH_2}(\mathrm{OH})$ is named as ethane-1,2-diol. However, the ending - ne of the parent alkane is dropped in the case of compounds having more than one double or triple bond; for example, $\mathrm{CH_2}=\mathrm{CH}-\mathrm{CH}=\mathrm{CH_2}$ is named as buta-1,3-diene.

Problem 12.8

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

Solution

• The functional group present is an alcohol (OH). Hence the suffix is ‘-ol’.
• The longest chain containing $-\mathrm{OH}$ has eight carbon atoms. Hence the corresponding saturated hydrocarbon is octane.
• The $-\mathrm{OH}$ is on carbon atom 3. In addition, a methyl group is attached at $6^{\text {th }}$ carbon.

Hence, the systematic name of this compound is 6-Methyloctan-3-ol.

Solution

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

Table 12.4 Some Functional Groups and Classes of Organic Compounds

Here, two functional groups namely ketone and carboxylic acid are present. The principal functional group is the carboxylic acid group; hence the parent chain will be suffixed with ‘oic’ acid. Numbering of the chain starts from carbon of - $\mathrm{COOH}$ functional group. The keto group in the chain at carbon 5 is indicated by ‘oxo’. The longest chain including the principal functional group has 6 carbon atoms; hence the parent hydrocarbon is hexane. The compound is, therefore, named as 5-Oxohexanoic acid.

(iv) $\underset{6}{\mathrm{CH}} \equiv \underset{5}{\mathrm{C}}-\underset{4}{\mathrm{CH}}=\underset{3}{\mathrm{CH}}-\underset{2}{\mathrm{CH}}=\underset{1}{\mathrm{CH_2}} $

Solution

The two $\mathrm{C}=\mathrm{C}$ functional groups are present at carbon atoms 1 and 3, while the $\mathrm{C} \equiv \mathrm{C}$ functional group is present at carbon 5 . These groups are indicated by suffixes ‘diene’ and ‘yne’ respectively. The longest chain containing the functional groups has 6 carbon atoms; hence the parent hydrocarbon is hexane. The name of compound, therefore, is Hexa-1,3dien-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-Hydroxyheptanal.

Solution

(i) ‘hexane’ indicates the presence of 6 carbon atoms in the chain. The functional group chloro is present at carbon 2. Hence, the structure of the compound is $\mathrm{CH_3} \mathrm{CH_2} \mathrm{CH_2} \mathrm{CH_2} \mathrm{CH}(\mathrm{Cl}) \mathrm{CH_3}$.

(ii) ‘pent’ indicates that parent hydrocarbon contains 5 carbon atoms in the chain. ’en’ and ‘ol’ correspond to the functional groups $\mathrm{C}=\mathrm{C}$ and $-\mathrm{OH}$ at carbon atoms 4 and 2 respectively. Thus, the structure is

$$ \mathrm{CH_2}=\mathrm{CHCH_2} \mathrm{CH}(\mathrm{OH}) \mathrm{CH_3} . $$

(iii) Six membered ring containing a carbon-carbon double bond is implied by cyclohexene, which is numbered as shown in (I). The prefix 3-nitro means that a nitro group is present on $\mathrm{C}-3$. Thus, complete structural formula of the compound is (II). Double bond is suffixed functional group whereas $\mathrm{NO_2}$ is prefixed functional group therefore double bond gets preference over $-\mathrm{NO_2}$ group:

(iv) ‘1-ol’ means that a - $\mathrm{OH}$ group is present at $\mathrm{C}-1$. OH is suffixed functional group and gets preference over $\mathrm{C}=\mathrm{C}$ bond. Thus the structure is as shown in (II):

II (v) ‘heptanal’ indicates the compound to be an aldehyde containing 7 carbon atoms in the parent chain. The ‘6-hydroxy’ indicates that - $\mathrm{OH}$ group is present at carbon 6 . Thus, the structural formula of the compound is: $\mathrm{CH_3} \mathrm{CH}(\mathrm{OH})$ $\mathrm{CH_2} \mathrm{CH_2} \mathrm{CH_2} \mathrm{CH_2} \mathrm{CHO}$. Carbon atom of $\mathrm{CHO}$ group is included while numbering the carbon chain.

12.5.4 Nomenclature of Substituted Benzene Compounds

For IUPAC nomenclature of substituted benzene compounds, the substituent is placed as prefix to the word benzene as shown in the following examples. However, common names (written in bracket below) of many substituted benzene compounds are also universally used.

If benzene ring is disubstituted, the position of substituents is defined by numbering the carbon atoms of the ring such that the substituents are located at the lowest numbers possible. For example, the compound(b) is named as 1,3-dibromobenzene and not as 1,5-dibromobenzene.

In the trivial system of nomenclature the terms ortho $(o)$, meta $(\mathrm{m})$ and para $(p)$ are used as prefixes to indicate the relative positions 1,2;1,3 and 1,4 respectively. Thus, 1,3-dibromobenzene (b) is named as $m$-dibromobenzene (meta is abbreviated as $m$-) and the other isomers of dibromobenzene 1,2-(a) and 1,4-(c), are named as ortho (or just $o^{-}$) and para (or just $p$-)-dibromobenzene, respectively.

For tri - or higher substituted benzene derivatives, these prefixes cannot be used and the compounds are named by identifying substituent positions on the ring by following the lowest locant rule. In some cases, common name of benzene derivatives is taken as the base compound.

Substituent of the base compound is assigned number 1 and then the direction of numbering is chosen such that the next substituent gets the lowest number. The substituents appear in the name in alphabetical order. Some examples are given below.

When a benzene ring is attached to an alkane with a functional group, it is considered as substituent, instead of a parent. The name for benzene as substituent is phenyl $\left(\mathrm{C_6} \mathrm{H_5}{ }^{-} \text{ , also abbreviated as } \mathrm{Ph} \right)$.

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.

8.6 ISOMERISM

The phenomenon of existence of two or more compounds possessing the same molecular formula but different properties is known as isomerism. Such compounds are called as isomers. The following flow chart shows different types of isomerism.

8.6.1 Structural Isomerism

Compounds having the same molecular formula but different structures (manners in which atoms are linked) are classified as structural isomers. Some typical examples of different types of structural isomerism are given below:

(i) Chain isomerism: When two or more compounds have similar molecular formula but different carbon skeletons, these are referred to as chain isomers and the phenomenon is termed as chain isomerism. For example, $\mathrm{C_5} \mathrm{H_12}$ represents three compounds:

(ii) Position isomerism: When two or more compounds differ in the position of substituent atom or functional group on the carbon skeleton, they are called position isomers and this phenomenon is termed as position isomerism. For example, the alcohols:

(iii) Functional group isomerism: Two or more compounds having the same molecular formula but different functional groups are called functional isomers and this phenomenon is termed as functional group isomerism. For example, the molecular formula $\mathrm{C_3} \mathrm{H_6} \mathrm{O}$ represents an aldehyde and a ketone:

(iv) Metamerism: It arises due to different alkyl chains on either side of the functional group in the molecule. For example, $\mathrm{C_4} \mathrm{H_10} \mathrm{O}$ represents methoxypropane $\left(\mathrm{CH_3} \mathrm{OC_3} \mathrm{H_7}\right)$ and ethoxyethane $\left(\mathrm{C_2} \mathrm{H_5} \mathrm{OC_2} \mathrm{H_5}\right)$.

8.6.2 Stereoisomerism

The compounds that have the same constitution and sequence of covalent bonds but differ in relative positions of their atoms or groups in space are called stereoisomers. This special type of isomerism is called as stereoisomerism and can be classified as geometrical and optical isomerism.

12.7 FUNDAMENTAL CONCEPTS IN ORGANIC REACTION MECHANISM

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

The general reaction is depicted as follows :

Substrate is that reactant which supplies carbon to the new bond and the other reactant is called reagent. If both the reactants supply carbon to the new bond then choice is arbitrary and in that case the molecule on which attention is focused is called substrate.

In such a reaction a covalent bond between two carbon atoms or a carbon and some other atom is broken and a new bond is formed. A sequential account of each step, describing details of electron movement, energetics during bond cleavage and bond formation, and the rates of transformation of reactants into products (kinetics) is referred to as reaction mechanism. The knowledge of reaction mechanism helps in understanding the reactivity of organic compounds and in planning strategy for their synthesis.

In the following sections, we shall learn some of the principles that explain how these reactions take place.

12.7.1 Fission of a Covalent Bond

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

In heterolytic cleavage, the bond breaks in such a fashion that the shared pair of electrons remains with one of the fragments.

After heterolysis, one atom has a sextet electronic structure and a positive charge and the other, a valence octet with at least one lone pair and a negative charge. Thus, heterolytic cleavage of bromomethane will give $\stackrel{+}{\mathrm{C}} \mathrm{H_3}$ and $\mathrm{Br}^{-}$as shown below.

A species having a carbon atom possessing sextext of electrons and a positive charge is called a carbocation (earlier called carbonium ion). The $\stackrel{+}{\mathrm{C}} \mathrm{H_3}$ ion is known as a methyl cation or methyl carbonium ion. Carbocations are classified as primary, secondary or tertiary depending on whether one, two or three carbons are directly attached to the positively charged carbon. Some other examples of carbocations are: $\mathrm{CH_3} \stackrel{+}{\mathrm{C}} \mathrm{H_2}$ (ethyl cation, a primary carbocation), $\left(\mathrm{CH_3}\right)_2 \stackrel{+}{\mathrm{C}} \mathrm{H}$ (isopropyl cation, a secondary carbocation), and $\left(\mathrm{CH_3}\right)_3 \stackrel{\perp}{\mathrm{C}}$ (tert-butyl cation, a tertiary carbocation). Carbocations are highly unstable and reactive species. Alkyl groups directly attached to the positively charged carbon stabilise the carbocations due to inductive and hyperconjugation effects, which you will be studying in the sections 12.7.5 and 12.7.9. The observed order of carbocation stability is: $\stackrel{+}{\mathrm{C}} \mathrm{H_3}<\mathrm{CH_3} \stackrel{+}{\mathrm{C}} \mathrm{H_2}<\left(\mathrm{CH_3}\right)_2 \stackrel{+}{\mathrm{C}} \mathrm{H}<\left(\mathrm{CH_3}\right)_3 \stackrel{+}{\mathrm{C}}$. These carbocations have trigonal planar shape with positively charged carbon being $s p^{2}$ hybridised. Thus, the shape of $\stackrel{+}{\mathrm{C}} \mathrm{H_3}$ may be considered as being derived from the overlap of three equivalent $\mathrm{C}\left(s p^{2}\right)$ hybridised orbitals with $1 s$ orbital of each of the three hydrogen atoms. Each bond may be represented as $\mathrm{C}\left(s p^{2}\right)-\mathrm{H}(1 s)$ sigma bond. The remaining carbon orbital is perpendicular to the molecular plane and contains no electrons. [Fig. 12.3(a)].

The heterolytic cleavage can also give a species in which carbon gets the shared pair of electrons. For example, when group $\mathrm{Z}$ attached to the carbon leaves without

electron pair, the methyl anion $\left(\mathrm{H_3} \mathrm{C}^{-}\right)$is formed. Such a carbon species carrying a negative charge on carbon atom is called carbanion. Carbon in carbanion is generally $\mathrm{sp}^{3}$ hybridised and its structure is distorted tetrahedron as shown in Fig. 12.3(b).

Carbanions are also unstable and reactive species. The organic reactions which proceed through heterolytic bond cleavage are called ionic or heteropolar or just polar reactions.

In homolytic cleavage, one of the electrons of the shared pair in a covalent bond goes with each of the bonded atoms. Thus, in homolytic cleavage, the movement of a single electron takes place instead of an electron pair. The single electron movement is shown by ‘half-headed’ (fish hook: $\sim$ ) curved arrow. Such cleavage results in the formation of neutral species (atom or group) which contains an unpaired electron. These species are called free radicals. Like carbocations and carbanions, free radicals are also very reactive. A homolytic cleavage can be shown as:

Alkyl radicals are classified as primary, secondary, or tertiary. Alkyl radical stability increases as we proceed from primary to tertiary:

$ \underset{\substack{\text{Methyl} \\ \substack{ \text{free} \\ \text{radical}}}}{\dot{\mathrm{C}} \mathrm{H_3}} < \underset{\substack{\text{Ethyl} \\ \substack{ \text{free} \\ \text{radical}}}}{\dot{\mathrm{C}} \mathrm{H_2} \mathrm{CH_3}} < \underset{\substack{\text{Isopropyl} \\ \substack{ \text{free} \\ \text{radical}}}}{\dot{\mathrm{C}} \mathrm{H} \left(\mathrm{CH_3} \right)_2} < \underset{\substack{\text{Tert-butyl} \\ \substack{ \text{free} \\ \text{radical}}}}{\dot{\mathrm{C}} \left(\mathrm{CH_3}\right)_3}$

Organic reactions, which proceed by homolytic fission are called free radical or homopolar or nonpolar reactions.

12.7.2 Substrate and Reagent

Ions are generally not formed in the reactions of organic compounds. Molecules as such participate in the reaction. It is convenient to name one reagent as substrate and other as reagent. In general, a molecule whose carbon is involved in new bond formation is called substrate and the other one is called reagent. When carbon-carbon bond is formed, the choice of naming the reactants as substrate and reagent is arbitrary and depends on molecule under observation. Example:

Nucleophiles and Electrophiles

Reagents attack the reactive site of the substrate. The reactive site may be electron deficient portion of the molecule (a positive reactive site) e.g., an atom with incomplete electron shell or the positive end of the dipole in the molecule. If the attacking species is electron rich, it attacks these sites. If attacking species is electron deficient, the reactive site for it is that part of the substrate molecule which can supply electrons, e.g., $\pi$ electrons in a double bond.

A reagent that brings an electron pair to the reactive site is called a nucleophile (Nu:) i.e., nucleus seeking and the reaction is then called nucleophilic. A reagent that takes away an electron pair from reactive site is called electrophile $\left(\mathrm{E}^{+}\right)$i.e., electron seeking and the reaction is called electrophilic.

During a polar organic reaction, a nucleophile attacks an electrophilic centre of the substrate which is that specific atom or part of the substrate which is electron deficient. Similarly, the electrophiles attack at nucleophilic centre, which is the electron rich centre of the substrate. Thus, the electrophiles receive electron pair from the substrate when the two undergo bonding interaction. A curved-arrow notation is used to show the movement of an electron pair from the nucleophile to the electrophile. Some examples of nucleophiles are the negatively charged ions with lone pair of electrons such as hydroxide $\left(\mathrm{HO}^{-}\right)$, cyanide $\left(\mathrm{NC}^{-}\right)$ ions and carbanions $\left(\mathrm{R_3} \mathrm{C} \mathrm{C}^{-}\right)$. Neutral molecules such as $\mathrm{H_2} \ddot{\mathrm{O}}:, \mathrm{R_3} \mathrm{~N}$ :, $\mathrm{R_2} \ddot{\mathrm{N}} \mathrm{H}$ etc., can also act as nucleophiles due to the presence of lone pair of electrons. Examples of electrophiles include carbocations $\left(\stackrel{\stackrel{\rightharpoonup}{C}}{ } \mathrm{H_3}\right)$ and neutral molecules having functional groups like carbonyl group ( $>\mathrm{C}=\mathrm{O})$ or alkyl halides $\left(\mathrm{R_3} \mathrm{C}-\mathrm{X}\right.$, where $\mathrm{X}$ is a halogen atom). The carbon atom in carbocations has sextet configuration; hence, it is electron deficient and can receive a pair of electrons from the nucleophiles. In neutral molecules such as alkyl halides, due to the polarity of the $\mathrm{C}-\mathrm{X}$ bond a partial positive charge is generated on the carbon atom and hence the carbon atom becomes an electrophilic centre at which a nucleophile can attack.

Problem 12.11

Using curved-arrow notation, show the formation of reactive intermediates when the following covalent bonds undergo heterolytic cleavage.

(a) $\mathrm{CH_3}-\mathrm{SCH_3}$, (b) $\mathrm{CH_3}-\mathrm{CN}$, (c) $\mathrm{CH_3}-\mathrm{Cu}$

Solution

(a) $\mathrm{CH_3} — \mathrm{SCH_3} \rightarrow \stackrel{+}{\mathrm{C}}\mathrm{H_3} + \stackrel{-}{\mathrm{S}}\mathrm{CH_3}$

(b) $\mathrm{CH_3} — \mathrm{CN} \rightarrow \stackrel{+}{\mathrm{C}}\mathrm{H_3} + \stackrel{-}{\mathrm{C}}\mathrm{N}$

(b) $\mathrm{CH_3} — \mathrm{Cu} \rightarrow \stackrel{-}{\mathrm{C}}\mathrm{H_3} + \stackrel{+}{\mathrm{Cu}}$

Problem 12.12

Giving justification, categorise the following molecules/ions as nucleophile or electrophile:

$\mathrm{HS}^{-}, \mathrm{BF_3}, \mathrm{C_2} \mathrm{H_5} \mathrm{O}^{-},\left(\mathrm{CH_3}\right)_{3} \mathrm{~N:}$,

Ci+ $, \mathrm{CH_3} \stackrel{+}{\mathrm{C}}=\mathrm{O}, \mathrm{H_2} \mathrm{~N:^-} , \stackrel{+}{\mathrm{N}} \mathrm{O_2}$

Solution

Nucleophiles: $\mathrm{HS}^{-}, \mathrm{C_2} \mathrm{H_5} \mathrm{O}^{-},\left(\mathrm{CH_3}\right)_{3} \mathrm{~N}: \mathrm{H_2} \mathrm{~N}^{-}$ These species have unshared pair of electrons, which can be donated and shared with an electrophile.

Electrophiles: $\mathrm{BF_3}, \mathrm{C} \stackrel{+}{1} \mathrm{H_3}-\stackrel{+}{\mathrm{C}}=\mathrm{O}, \stackrel{+}{\mathrm{N}} \mathrm{O_2}$. Reactive sites have only six valence electrons; can accept electron pair from a nucleophile.

Problem 12.13

Identify electrophilic centre in the following: $\mathrm{CH_3} \mathrm{CH}=\mathrm{O}, \mathrm{CH_3} \mathrm{CN}, \mathrm{CH_3} \mathrm{I}$.

Solution

Among $\mathrm{CH_3} \mathrm{H} \stackrel{}{\mathrm{C}}=\mathrm{O}, \mathrm{H_3} \mathrm{C} \stackrel{}{\mathrm{C}} \equiv \mathrm{N}$, and $\mathrm{H_3} \mathrm{C}^*-\mathrm{I}$,, the starred carbon atoms are electrophilic centers as they will have partial positive charge due to polarity of the bond.

12.7.3 Electron Movement in Organic Reactions

The movement of electrons in organic reactions can be shown by curved-arrow notation. It shows how changes in bonding occur due to electronic redistribution during the reaction. To show the change in position of a pair of electrons, curved arrow starts from the point from where an electron pair is shifted and it ends at a location to which the pair of electron may move.

Presentation of shifting of electron pair is given below :

Movement of single electron is indicated by a single barbed ‘fish hooks’ (i.e. half headed curved arrow). For example, in transfer of hydroxide ion giving ethanol and in the dissociation of chloromethane, the movement of electron using curved arrows can be depicted as follows:

12.7.4 Electron Displacement Effects in Covalent Bonds

The electron displacement in an organic molecule may take place either in the ground state under the influence of an atom or a substituent group or in the presence of an appropriate attacking reagent. The electron displacements due to the influence of an atom or a substituent group present in the molecule cause permanent polarlisation of the bond. Inductive effect and resonance effects are examples of this type of electron displacements. Temporary electron displacement effects are seen in a molecule when a reagent approaches to attack it. This type of electron displacement is called electromeric effect or polarisability effect. In the following sections we will learn about these types of electronic displacements.

12.7.5 Inductive Effect

When a covalent bond is formed between atoms of different electronegativity, the electron density is more towards the more electronegative atom of the bond. Such a shift of electron density results in a polar covalent bond. Bond polarity leads to various electronic effects in organic compounds.

Let us consider cholorethane $\left(\mathrm{CH_3} \mathrm{CH_2} \mathrm{Cl}\right)$ in which the $\mathrm{C}-\mathrm{Cl}$ bond is a polar covalent bond. It is polarised in such a way that the carbon-1 gains some positive charge $\left(\delta^{+}\right)$ and the chlorine some negative charge ( $\delta^{-}$). The fractional electronic charges on the two atoms in a polar covalent bond are denoted by symbol $\delta$ (delta) and the shift of electron density is shown by an arrow that points from $\delta^{+}$to $\delta^{-}$end of the polar bond.

$\stackrel{\delta \delta^+}{\underset{2}{\mathrm{CH_3}}} \xrightarrow{}— \stackrel{\delta^+}{\underset{1}{\mathrm{CH_2}}} \xrightarrow{}— \stackrel{\delta^-}{\mathrm{Cl}}$

In turn carbon-1, which has developed partial positive charge $\left(\delta^{+}\right)$draws some electron density towards it from the adjacent $\mathrm{C}-\mathrm{C}$ bond. Consequently, some positive charge $\left(\delta \delta^{+}\right)$develops on carbon- 2 also, where $\delta \delta^{+}$ symbolises relatively smaller positive charge as compared to that on carbon - 1 . In other words, the polar $\mathrm{C}-\mathrm{Cl}$ bond induces polarity in the adjacent bonds. Such polarisation of $\sigma$-bond caused by the polarisation of adjacent $\sigma$-bond is referred to as the inductive effect. This effect is passed on to the subsequent bonds also but the effect decreases rapidly as the number of intervening bonds increases and becomes vanishingly small after three bonds. The inductive effect is related to the ability of substituent(s) to either withdraw or donate electron density to the attached carbon atom. Based on this ability, the substitutents can be classified as electron-withdrawing or electron donating groups relative to hydrogen. Halogens and many other groups such as nitro $\left(-\mathrm{NO_2}\right)$, cyano (- $\mathrm{CN}$ ), carboxy (- $\left.\mathrm{COOH}\right)$, ester (COOR), aryloxy (-OAr, e.g. $-\mathrm{OC_6} \mathrm{H_5}$ ), etc. are electron-withdrawing groups. On the other hand, the alkyl groups like methyl $\left(-\mathrm{CH_3}\right)$ and ethyl $\left(-\mathrm{CH_2}-\mathrm{CH_3}\right)$ are usually considered as electron donating groups.

Problem 12.14

Which bond is more polar in the following pairs of molecules: (a) $\mathrm{H_3} \mathrm{C}-\mathrm{H}, \mathrm{H_3} \mathrm{C}-\mathrm{Br}$ (b) $\mathrm{H_3} \mathrm{C}-\mathrm{NH_2}, \mathrm{H_3} \mathrm{C}-\mathrm{OH}$ (c) $\mathrm{H_3} \mathrm{C}-\mathrm{OH}$, $\mathrm{H_3} \mathrm{C}$-SH

Solution

(a) $\mathrm{C}-\mathrm{Br}$, since $\mathrm{Br}$ is more electronegative than $\mathrm{H}$, (b) $\mathrm{C}-\mathrm{O}$, (c) $\mathrm{C}-\mathrm{O}$

Problem 12.15

In which $\mathrm{C}-\mathrm{C}$ bond of $\mathrm{CH_3} \mathrm{CH_2} \mathrm{CH_2} \mathrm{Br}$, the inductive effect is expected to be the least?

Solution

Magnitude of inductive effect diminishes as the number of intervening bonds increases. Hence, the effect is least in the bond between carbon-3 and hydrogen.

12.7.6 Resonance Structure

There are many organic molecules whose behaviour cannot be explained by a single Lewis structure. An example is that of benzene. Its cyclic structure containing alternating $\mathrm{C}-\mathrm{C}$ single and $\mathrm{C}=\mathrm{C}$ double bonds shown is inadequate for explaining its Benzene characteristic properties.

As per the above representation, benzene should exhibit two different bond lengths, due to $\mathrm{C}-\mathrm{C}$ single and $\mathrm{C}=\mathrm{C}$ double bonds. However, as determined experimentally benzene has a uniform $\mathrm{C}-\mathrm{C}$ bond distances of $139 \mathrm{pm}$, a value intermediate between the $\mathrm{C}-\mathrm{C}$ single $(154 \mathrm{pm}$ ) and $\mathrm{C}=\mathrm{C}$ double (134 $\mathrm{pm}$ ) bonds. Thus, the structure of benzene cannot be represented adequately by the above structure. Further, benzene can be represented equally well by the energetically identical structures I and II.

Therefore, according to the resonance theory (Unit 4) the actual structure of benzene cannot be adequately represented by any of these structures, rather it is a hybrid of the two structures (I and II) called resonance structures. The resonance structures (canonical structures or contributing structures) are hypothetical and individually do not represent any real molecule. They contribute to the actual structure in proportion to their stability.

Another example of resonance is provided by nitromethane $\left(\mathrm{CH_3} \mathrm{NO_2}\right)$ which can be represented by two Lewis structures, (I and II). There are two types of $\mathrm{N}-\mathrm{O}$ bonds in these structures.

However, it is known that the two $\mathrm{N}-\mathrm{O}$ bonds of nitromethane are of the same length (intermediate between a $\mathrm{N}-\mathrm{O}$ single bond and a $\mathrm{N}=\mathbf{O}$ double bond). The actual structure of nitromethane is therefore a resonance hybrid of the two canonical forms I and II.

The energy of actual structure of the molecule (the resonance hybrid) is lower than that of any of the canonical structures. The difference in energy between the actual structure and the lowest energy resonance structure is called the resonance stabilisation energy or simply the resonance energy. The more the number of important contributing structures, the more is the resonance energy. Resonance is particularly important when the contributing structures are equivalent in energy.

The following rules are applied while writing resonance structures:

The resonance structures have (i) the same positions of nuclei and (ii) the same number of unpaired electrons. Among the resonance structures, the one which has more number of covalent bonds, all the atoms with octet of electrons (except hydrogen which has a duplet), less separation of opposite charges, (a negative charge if any on more electronegative atom, a positive charge if any on more electropositive atom) and more dispersal of charge, is more stable than others.

Problem 12.16

Write resonance structures of $\mathrm{CH_3} \mathrm{COO}^{-}$ and show the movement of electrons by curved arrows.

Solution

First, write the structure and put unshared pairs of valence electrons on appropriate atoms. Then draw the arrows one at a time moving the electrons to get the other structures.

Problem 12.17

Write resonance structures of $\mathrm{CH_2}=\mathrm{CH}-\mathrm{CHO}$. Indicate relative stability of the contributing structures.

Solution

[I: Most stable, more number of covalent bonds, each carbon and oxygen atom has an octet and no separation of opposite charge II: negative charge on more electronegative atom and positive charge on more electropositive atom; III: does not contribute as oxygen has positive charge and carbon has negative charge, hence least stable].

Problem 12.18

Explain why the following two structures, I and II cannot be the major contributors to the real structure of $\mathrm{CH_3} \mathrm{COOCH_3}$.

Solution

The two structures are less important contributors as they involve charge separation. Additionally, structure I contains a carbon atom with an incomplete octet.

12.7.7 Resonance Effect

The resonance effect is defined as ’the polarity produced in the molecule by the interaction of two $\pi$-bonds or between a $\pi$-bond and lone pair of electrons present on an adjacent atom’. The effect is transmitted through the chain. There are two types of resonance or mesomeric effect designated as $\mathrm{R}$ or $\mathrm{M}$ effect.

(i) Positive Resonance Effect ( $+R$ effect)

In this effect, the transfer of electrons is away from an atom or substituent group attached to the conjugated system. This electron displacement makes certain positions in the molecule of high electron densities. This effect in aniline is shown as :

(ii) Negative Resonance Effect (- $\boldsymbol{R}$ effect)

This effect is observed when the transfer of electrons is towards the atom or substituent group attached to the conjugated system. For example in nitrobenzene this electron displacement can be depicted as :

The atoms or substituent groups, which represent $+\mathrm{R}$ or $-\mathrm{R}$ electron displacement effects are as follows :

• +R effect: - halogen, - $\mathrm{OH},-\mathrm{OR},-\mathrm{OCOR},-\mathrm{NH_2}$, $-\mathrm{NHR},-\mathrm{NR_2}$, -NHCOR,

• R effect: $-\mathrm{COOH},-\mathrm{CHO},>\mathrm{C}=\mathrm{O},-\mathrm{CN},-\mathrm{NO_2}$

The presence of alternate single and double bonds in an open chain or cyclic system is termed as a conjugated system. These systems often show abnormal behaviour. The examples are 1,3- butadiene, aniline and nitrobenzene etc. In such systems, the $\pi$-electrons are delocalised and the system develops polarity.

12.7.8 Electromeric Effect (E effect)

It is a temporary effect. The organic compounds having a multiple bond (a double or triple bond) show this effect in the presence of an attacking reagent only. It is defined as the complete transfer of a shared pair of $\pi$-electrons to one of the atoms joined by a multiple bond on the demand of an attacking reagent. The effect is annulled as soon as the attacking reagent is removed from the domain of the reaction. It is represented by $\mathrm{E}$ and the shifting of the electrons is shown by a curved arrow $(\frown)$. There are two distinct types of electromeric effect.

(i) Positive Eelctromeric Effect (+E effect) In this effect the $\pi$-electrons of the multiple bond are transferred to that atom to which the reagent gets attached. For example:

(ii) Negative Electromeric Effect (-E effect) In this effect the $\pi$ - electrons of the multiple bond are transferred to that atom to which the attacking reagent does not get attached. For example:

When inductive and electromeric effects operate in opposite directions, the electomeric effect predominates.

8.7.9 Hyperconjugation

Hyperconjugation is a general stabilising interaction. It involves delocalisation of $\sigma$ electrons of $\mathrm{C}-\mathrm{H}$ bond of an alkyl group directly attached to an atom of unsaturated system or to an atom with an unshared $p$ orbital. The $\sigma$ electrons of $\mathrm{C}-\mathrm{H}$ bond of the alkyl group enter into partial conjugation with the attached unsaturated system or with the unshared $p$ orbital. Hyperconjugation is a permanent effect.

To understand hyperconjugation effect, let us take an example of $\mathrm{CH_3} \stackrel{+}{\mathrm{C}} \mathrm{H_2}$ (ethyl cation) in which the positively charged carbon atom has an empty $p$ orbital. One of the $\mathrm{C}-\mathrm{H}$ bonds of the methyl group can align in the plane of this empty $p$ orbital and the electrons constituting the $\mathrm{C}-\mathrm{H}$ bond in plane with this $p$ orbital can then be delocalised into the empty $p$ orbital as depicted in Fig. 12.4 (a).

This type of overlap stabilises the carbocation because electron density from the adjacent $\sigma$ bond helps in dispersing the positive charge.

In general, greater the number of alkyl groups attached to a positively charged carbon atom, the greater is the hyperconjugation interaction and stabilisation of the cation. Thus, we have the following relative stability of carbocations

Hyperconjugation is also possible in alkenes and alkylarenes.

Delocalisation of electrons by hyperconjugation in the case of alkene can be depicted as in Fig. 12.4(b).

There are various ways of looking at the hyperconjugative effect. One of the way is to regard $\mathrm{C}-\mathrm{H}$ bond as possessing partial ionic character due to resonance.

The hyperconjugation may also be regarded as no bond resonance.

Problem 12.19

Explain why $\left(\mathrm{CH_3}\right)_3 \stackrel{+}{\mathrm{C}}$ is more stable than $\mathrm{CH_3} \stackrel{+}{\mathrm{C}} \mathrm{H_2}$ and $\stackrel{+}{\mathrm{C}} \mathrm{H_3}$ is the least stable cation.

Solution

Hyperconjugation interaction in $\left(\mathrm{CH_3}\right)_3 \stackrel{+}{\mathrm{C}}$ is greater than in $\mathrm{CH_3} \stackrel{+}{\mathrm{C}} \mathrm{H_2}$ as the $\left(\mathrm{CH_3}\right)_3 \stackrel{+}{\mathrm{C}}$ has nine $\mathrm{C}-\mathrm{H}$ bonds. In $\stackrel{+}{\mathrm{C}} \mathrm{H_3}$, vacant $p$ orbital is perpendicular to the plane in which $\mathrm{C}-\mathrm{H}$ bonds lie; hence cannot overlap with it. Thus, $\stackrel{+}{\mathrm{C}} \mathrm{H_3}$ lacks hyperconjugative stability.

8.7.10 Types of Organic Reactions and Mechanisms

Organic reactions can be classified into the following categories:

(i) Substitution reactions

(ii) Addition reactions

(iii) Elimination reactions

(iv) Rearrangement reactions

You will be studying these reactions in Unit 9 and later in class XII.

12.8 METHODS OF PURIFICATION OF ORGANIC COMPOUNDS

Once an organic compound is extracted from a natural source or synthesised in the laboratory, it is essential to purify it. Various methods used for the purification of organic compounds are based on the nature of the compound and the impurity present in it.

The common techniques used for purification are as follows :

(i) Sublimation

(ii) Crystallisation

(iii) Distillation

(iv) Differential extraction and

(v) Chromatography

Finally, the purity of a compound is ascertained by determining its melting or boiling point. Most of the pure compounds have sharp melting points and boiling points. New methods of checking the purity of an organic compound are based on different types of chromatographic and spectroscopic techniques.

12.8.1 Sublimation

You have learnt earlier that on heating, some solid substances change from solid to vapour state without passing through liquid state. The purification technique based on the above principle is known as sublimation and is used to separate sublimable compounds from nonsublimable impurities.

8.8.2 Crystallisation

This is one of the most commonly used techniques for the purification of solid organic compounds. It is based on the difference in the solubilities of the compound and the impurities in a suitable solvent. The impure compound is dissolved in a solvent in which it is sparingly soluble at room temperature but appreciably soluble at higher temperature. The solution is concentrated to get a nearly saturated solution. On cooling the solution, pure compound crystallises out and is removed by filtration. The filtrate (mother liquor) contains impurities and small quantity of the compound. If the compound is highly soluble in one solvent and very little soluble in another solvent, crystallisation can be satisfactorily carried out in a mixture of these solvents. Impurities, which impart colour to the solution are removed by adsorbing over activated charcoal. Repeated crystallisation becomes necessary for the purification of compounds containing impurities of comparable solubilities.

8.8.3 Distillation

This important method is used to separate (i) volatile liquids from nonvolatile impurities and (ii) the liquids having sufficient difference in their boiling points. Liquids having different boiling points vaporise at different temperatures. The vapours are cooled and the liquids so formed are collected separately. Chloroform (b.p $334 \mathrm{~K}$ ) and aniline (b.p. 457 K) are easily separated by the technique of distillation (Fig 8.5). The liquid mixture is taken in a round bottom flask and heated carefully. On boiling, the vapours of lower boiling component are formed first. The vapours are condensed by using a condenser and the liquid is collected in a receiver. The vapours of higher boiling component form later and the liquid can be collected separately.

Fractional Distillation: If the difference in boiling points of two liquids is not much, simple distillation cannot be used to separate them. The vapours of such liquids are formed within the same temperature range and are condensed simultaneously. The technique of fractional distillation is used in such cases. In this technique, vapours of a liquid mixture are passed through a fractionating column before condensation. The fractionating column is fitted over the mouth of the round bottom flask (Fig.12.6).

Vapours of the liquid with higher boiling point condense before the vapours of the liquid with lower boiling point. The vapours rising up in the fractionating column become richer in more volatile component. By the

time the vapours reach to the top of the fractionating column, these are rich in the more volatile component. Fractionating columns are available in various sizes and designs as shown in Fig.12.7. A fractionating column provides many surfaces for heat exchange between the ascending vapours and the descending condensed liquid. Some of the condensing liquid in the fractionating column obtains heat from the ascending vapours and revaporises. The vapours thus become richer in low boiling component. The vapours of low boiling component ascend to the top of the column. On reaching the top, the vapours become pure in low boiling component and pass through the condenser and the pure liquid is collected in a receiver. After a series of successive distillations, the remaining liquid in the distillation flask gets enriched in high boiling component. Each successive condensation and vaporisation unit in the fractionating column is called a theoretical plate. Commercially, columns with hundreds of plates are available.

One of the technological applications of fractional distillation is to separate different crude oil in petroleum industry. Distillation under reduced pressure: This method is used to purify liquids having very high boiling points and those, which decompose at or below their boiling points. Such liquids are made to boil at a temperature lower than their normal boiling points by reducing the pressure on their surface. A liquid boils at a temperature at which its vapour pressure is equal to the external pressure. The pressure is reduced with the help of a water pump or vacuum pump (Fig.8.8). Glycerol can be separated from spent-lye in soap industry by using this technique.

Steam Distillation: This technique is applied to separate substances which are steam volatile and are immiscible with water. In steam distillation, steam from a steam generator is passed through a heated flask containing the liquid to be distilled. The mixture of steam and the volatile organic compound is condensed and collected. The compound is later separated from water using a separating funnel. In steam distillation, the liquid boils when the sum of vapour pressures due to the organic liquid $\left(p_{1}\right)$ and that due to water $\left(p_{2}\right)$ becomes equal to the atmospheric pressure (p), i.e. $p=p_{1}+p_{2}$. Since $p_{1}$ is lower than $p$, the organic liquid vaporises at lower temperature than its boiling point.

Thus, if one of the substances in the mixture is water and the other, a water insoluble substance, then the mixture will boil close to but below, 373K. A mixture of water and the substance is obtained which can be separated by using a separating funnel. Aniline is separated by this technique from aniline - water mixture (Fig.12.9,).

12.8.4 Differential Extraction

When an organic compound is present in an aqueous medium, it is separated by shaking it with an organic solvent in which it is more soluble than in water. The organic solvent and the aqueous solution should be immiscible with each other so that they form two distinct layers which can be separated by separatory funnel. The organic solvent is later removed by distillation or by evaporation to get back the compound. Differential extraction is carried out in a separatory funnel as shown in Fig. 8.10 (Page 282). If the organic compound is less soluble in the organic solvent, a very large quantity of solvent would be required to extract even a very small quantity of the compound. The technique of continuous extraction is employed in such cases. In this technique same solvent is repeatedly used for extraction of the compound.

12.8.5 Chromatography

Chromatography is an important technique extensively used to separate mixtures into their components, purify compounds and also to test the purity of compounds. The

name chromatography is based on the Greek word chroma, for colour since the method was first used for the separation of coloured substances found in plants. In this technique, the mixture of substances is applied onto a stationary phase, which may be a solid or a liquid. A pure solvent, a mixture of solvents, or a gas is allowed to move slowly over the stationary phase. The components of the mixture get gradually separated from one another. The moving phase is called the mobile phase.

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

(a) Adsorption chromatography, and

(b) Partition chromatography.

a) Adsorption Chromatography: Adsorption chromatography is based on the fact that different compounds are adsorbed on an adsorbent to different degrees. Commonly used adsorbents are silica gel and alumina. When a mobile phase is allowed to move over a stationary phase (adsorbent), the components of the mixture move by varying

distances over the stationary phase. Following are two main types of chromatographic techniques based on the principle of differential adsorption.

(a) Column chromatography, and

(b) Thin layer chromatography.

Column Chromatography: Column chromatography involves separation of a mixture over a column of adsorbent (stationary phase) packed in a glass tube. The column is fitted with a stopcock at its lower end (Fig. 8.11). The mixture adsorbed on adsorbent is placed on the top of the adsorbent column packed in a glass tube. An appropriate eluant which is a liquid or a mixture of liquids is allowed to flow down the column slowly. Depending upon the degree to which the compounds are adsorbed, complete separation takes place. The most readily adsorbed substances are retained near the top and others come down to various distances in the column (Fig.12.11).

Thin Layer Chromatography: Thin layer chromatography (TLC) is another type of adsorption chromatography, which involves separation of substances of a mixture over a thin layer of an adsorbent coated on glass plate. A thin layer (about $0.2 \mathrm{~mm}$ thick) of an adsorbent (silica gel or alumina) is spread over a glass plate of suitable size. The plate is known as thin layer chromatography plate or chromaplate. The solution of the mixture to be separated is applied as a small spot about $2 \mathrm{~cm}$ above one end of the TLC plate. The glass plate is then placed in a closed jar containing the eluant (Fig. 8.12a). As the eluant rises up the plate, the components of the mixture move up along with the eluant to different distances depending on their degree of adsorption and separation takes place. The relative adsorption of each component of the mixture is expressed in terms of its retardation factor i.e. $\mathbf{R_f}$ value (Fig.8.12 b).

$R_f = \frac{\text{Distance moved by the substance from base line(x)}}{\text{Distance moved by the solvent from base line(y)}}$

The spots of coloured compounds are visible on TLC plate due to their original colour. The spots of colourless compounds, which are invisible to the eye but fluoresce in ultraviolet light, can be detected by putting the plate under ultraviolet light. Another detection technique is to place the plate in a covered jar containing a few crystals of iodine. Spots of compounds, which adsorb iodine, will show up as brown spots. Sometimes an appropriate reagent may also be sprayed on the plate. For example, amino acids may be detected by spraying the plate with ninhydrin solution (Fig.8.12b).

Partition Chromatography: Partition chromatography is based on continuous differential partitioning of components of a mixture between stationary and mobile phases. Paper chromatography is a type of partition chromatography. In paper chromatography, a special quality paper known as chromatography paper is used. Chromatography paper contains water trapped in it, which acts as the stationary phase.

A strip of chromatography paper spotted at the base with the solution of the mixture is suspended in a suitable solvent or a mixture of solvents (Fig. 8.13). This solvent acts as the mobile phase. The solvent rises up the paper by capillary action and flows over the spot. The paper selectively retains different components according to their differing partition in the two phases. The paper strip so developed is known as a chromatogram. The spots of the separated coloured compounds are visible at different heights from the position of initial

spot on the chromatogram. The spots of the separated colourless compounds may be observed either under ultraviolet light or by the use of an appropriate spray reagent as discussed under thin layer chromatography.

12.9 QUALITATIVE ANALYSIS OF ORGANIC COMPOUNDS

The elements present in organic compounds are 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 heating the compound with copper(II) oxide. Carbon present in the compound is oxidised to carbon dioxide (tested with lime-water, which develops turbidity) and hydrogen to water (tested with anhydrous copper sulphate, which turns blue).

$\mathrm{C}+ \mathrm{2CuO} \xrightarrow{\Delta} 2\mathrm{Cu}+ \mathrm{CO_2}$

$2\mathrm{H}+ \mathrm{CuO} \xrightarrow{\Delta} \mathrm{Cu}+ \mathrm{H_2O}$

$\mathrm{CO_2}+ \mathrm{Ca(OH)_2} \rightarrow \mathrm{CaCO_3}\downarrow+ \mathrm{H_2O}$

$5\mathrm{H_2O}+ \underset{\text{White}}{\mathrm{CuSO_4}} \rightarrow \underset{\text{Blue}}{\mathrm{CuSO_4.5H_2O}}$

12.9.2 Detection of Other Elements

Nitrogen, sulphur, halogens and phosphorus present in an organic compound are detected by “Lassaigne’s test”. The elements present in the compound are converted from covalent form into the ionic form by fusing the compound with sodium metal. Following reactions take place:

$$ \begin{array}{lll} \mathrm{Na}+\mathrm{C}+\mathrm{N} & \xrightarrow{\Delta} & \mathrm{NaCN} \\ 2 \mathrm{Na}+\mathrm{S} & \xrightarrow{\Delta} & \mathrm{Na_2} \mathrm{~S} \\ \mathrm{Na}+\mathrm{X} & \xrightarrow{\Delta} & \mathrm{Na} \mathrm{X} \\ & & (\mathrm{X}=\mathrm{Cl}, \text { Br or } \mathrm{I}) \end{array} $$

C, N, S and X come from organic compound.

Cyanide, sulphide and halide of sodium so formed on sodium fusion are extracted from the fused mass by boiling it with distilled water. This extract is known as sodium fusion extract.

(A) Test for Nitrogen

The sodium fusion extract is boiled with iron(II) sulphate and then acidified with concentrated sulphuric acid. The formation of Prussian blue colour confirms the presence of nitrogen. Sodium cyanide first reacts with iron(II) sulphate and forms sodium hexacyanidoferrate(II). On heating with concentrated sulphuric acid some iron(II) ions are oxidised to iron(III) ions which react with sodium hexacyanidoferrate(II) to produce iron(III) hexacyanidoferrate(II) (ferriferrocyanide) which is Prussian blue in colour.

$6 \mathrm{CN}^{-}+\mathrm{Fe}^{2+} \rightarrow\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{4-}$

$3\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{4-}+4 \mathrm{Fe}^{3+} \xrightarrow{\mathrm{xH_2} \mathrm{O}} \underset{\text{Prussian Blue}}{\mathrm{Fe_4}\left[\mathrm{Fe}(\mathrm{CN})_6\right]_3 \cdot \mathrm{xH_2} \mathrm{O}}$

(B) Test for Sulphur

(a) The sodium fusion extract is acidified with acetic acid and lead acetate is added to it. A black precipitate of lead sulphide indicates the presence of sulphur.

$$ \begin{aligned} & \mathrm{S}^{2-}+\mathrm{Pb}^{2+} \longrightarrow \underset{\text{Black}}{\mathrm{PbS}} \end{aligned} $$

(b) On treating sodium fusion extract with sodium nitroprusside, appearance of a violet colour further indicates the presence of sulphur.

$$ \mathrm{S}^{2-}+\left[\mathrm{Fe}(\mathrm{CN})_5 \mathrm{NO}\right]^{2-} \longrightarrow \quad \underset{\text { Violet }}{\left[\mathrm{Fe}(\mathrm{CN})_5 \mathrm{NOS}\right]^{4-}} $$

In case, nitrogen and sulphur both are present in an organic compound, sodium thiocyanate is formed. It gives blood red colour and no Prussian blue since there are no free cyanide ions.

$$ \mathrm{Na}+\mathrm{C}+\mathrm{N}+\mathrm{S} \longrightarrow \quad \mathrm{NaSCN} $$

$\mathrm{Fe}^{3+}+\mathrm{SCN}^{-} \quad \longrightarrow \quad \underset{\text{Blood Red}}{[\mathrm{Fe}(\mathrm{SCN})]^{2+}}$

If sodium fusion is carried out with excess of sodium, the thiocyanate decomposes to yield cyanide and sulphide. These ions give their usual tests.

$$ \mathrm{NaSCN}+2 \mathrm{Na} \longrightarrow \mathrm{NaCN}+\mathrm{Na_2} \mathrm{~S} $$

(C) Test for Halogens

The sodium fusion extract is acidified with nitric acid and then treated with silver nitrate. A white precipitate, soluble in ammonium hydroxide shows the presence of chlorine, a yellowish precipitate, sparingly soluble in ammonium hydroxide shows the presence of bromine and a yellow precipitate, insoluble in ammonium hydroxide shows the presence of iodine.

$$ \mathrm{X}^{-}+\mathrm{Ag}^{+} \longrightarrow \mathrm{AgX} $$

$\mathrm{X}$ represents a halogen $-\mathrm{Cl}, \mathrm{Br}$ or $\mathrm{I}$.

If nitrogen or sulphur is also present in the compound, the sodium fusion extract is first boiled with concentrated nitric acid to decompose cyanide or sulphide of sodium formed during Lassaigne’s test. These ions would otherwise interfere with silver nitrate test for halogens.

(D) Test for Phosphorus

The compound is heated with an oxidising agent (sodium peroxide). The phosphorus present in the compound is oxidised to phosphate. The solution is boiled with nitric acid and then treated with ammonium molybdate. A yellow colouration or precipitate indicates the presence of phosphorus.

$$\mathrm{Na_3} \mathrm{PO_4}+3 \mathrm{HNO_3} \longrightarrow \mathrm{H_3} \mathrm{PO_4}+3 \mathrm{NaNO_3}$$

$$\mathrm{H_3} \mathrm{PO_4}+ \underset{\substack{\text{Ammonium} \\ \text{molybdate}}}{12\left(\mathrm{NH_4}\right)_2 \mathrm{MoO_4}} +21 \mathrm{HNO_3} \longrightarrow $$

$$\underset{\substack{\text{Ammonium} \\ \text{phosphomolybdate}}}{\left(\mathrm{NH_4}\right)_{3} \mathrm{PO_4} \cdot 12 \mathrm{MoO_3}} +21 \mathrm{NH_4} \mathrm{NO_3}+12 \mathrm{H_2} \mathrm{O} $$

12.10 QUANTITATIVE ANALYSIS

Quantitative analysis of compounds is very important in organic chemistry. It helps chemists in the determination of mass per cent of elements present in a compound. You have learnt in Unit-1 that mass per cent of elements is required for the determination of emperical and molecular formula.

The percentage composition of elements present in an organic compound is determined by the following methods:

12.10.1 Carbon and Hydrogen

Both carbon and hydrogen are estimated in one experiment. A known mass of an organic compound is burnt in the presence of excess of oxygen and copper(II) oxide. Carbon and hydrogen in the compound are oxidised to carbon dioxide and water respectively.

$\mathrm{C_\mathrm{x}} \mathrm{H_\mathrm{y}}+(\mathrm{x}+\mathrm{y} / 4)_2 \mathrm{O_2} \longrightarrow \mathrm{xCO_2}+(\mathrm{y} / 2) \mathrm{H_2O}$

The mass of water produced is determined by passing the mixture through a weighed U-tube containing anhydrous calcium chloride. Carbon dioxide is absorbed in another U-tube containing concentrated solution of potassium hydroxide. These tubes are connected in series (Fig. 12.14). The increase in masses of calcium chloride and potassium hydroxide gives the amounts of water and carbon dioxide from which the percentages of carbon and hydrogen are calculated.

Let the mass of organic compound be $\mathrm{m} g$, mass of water and carbon dioxide produced be $m_{1}$ and $m_{2}$ g respectively;

Percentage of carbon $=\frac{12 \times \mathrm{m_2} \times 100}{44 \times \mathrm{m}}$

Percentage of hydrogen $=\frac{2 \times m_{1} \times 100}{18 \times m}$

Problem 12.20

On complete combustion, $0.246 \mathrm{~g}$ of an organic compound gave $0.198 \mathrm{~g}$ of carbon dioxide and $0.1014 \mathrm{~g}$ of water. Determine the percentage composition of carbon and hydrogen in the compound.

Solution

Percentageof carbon $=\frac{12 \times 0.198 \times 100}{44 \times 0.246}$

$$ =21.95 \% $$

Percentage of hydrogen $=\frac{2 \times 0.1014 \times 100}{18 \times 0.246}$

$$ =4.58 \% $$

12.10.2 Nitrogen

There are two methods for estimation of nitrogen: (i) Dumas method and (ii) Kjeldahl’s method.

(i) Dumas method: The nitrogen containing organic compound, when heated with copper oxide in an atmosphere of carbon dioxide, yields free nitrogen in addition to carbon dioxide and water.

$$ \begin{aligned} & \mathrm{C_\mathrm{x}} \mathrm{H_\mathrm{y}} \mathrm{N_\mathrm{z}}+(2 \mathrm{x}+\mathrm{y} / 2) \mathrm{CuO} \longrightarrow \\ & \quad \mathrm{xCO_2}+\mathrm{y} / 2 \mathrm{H_2} \mathrm{O}+\mathrm{z} / 2 \mathrm{~N_2}+(2 \mathrm{x}+\mathrm{y} / 2) \mathrm{Cu} \end{aligned} $$

Traces of nitrogen oxides formed, if any, are reduced to nitrogen by passing the gaseous mixture over a heated copper gauze. The mixture of gases so produced is collected over an aqueous solution of potassium hydroxide which absorbs carbon dioxide. Nitrogen is collected in the upper part of the graduated tube (Fig.8.15).

Let the mass of organic compound $=\mathrm{mg}$

Volume of nitrogen collected $=V_{1} \mathrm{~mL}$

Room temperature $=T_{1} \mathrm{~K}$

V olume of nitrogen atSTP $=\frac{P_{1} V_{1} \times 273}{760 \times T_{1}}$

(Let it be $\mathrm{V} \mathrm{mL}$ )

Where $p_{1}$ and $V_{1}$ are the pressure and volume of nitrogen, $p_{1}$ is different from the atmospheric pressure at which nitrogen gas is collected. The value of $p_{1}$ is obtained by the relation;

$p_{1}=$ Atmospheric pressure - Aqueous tension $22400 \mathrm{~mL} \mathrm{~N_2}$ at STP weighs $28 \mathrm{~g}$.

$\mathrm{V} \mathrm{mL} \mathrm{N}{ _2}$ atSTP weighs $=\frac{28 \times \mathrm{V}}{22400} \mathrm{~g}$

Percentage of nitrogen $=\frac{28 \times \mathrm{V} \times 100}{22400 \times \mathrm{m}}$

Problem 12.21

In Dumas’ method for estimation of nitrogen, $0.3 \mathrm{~g}$ of an organic compound gave $50 \mathrm{~mL}$ of nitrogen collected at $300 \mathrm{~K}$ temperature and $715 \mathrm{~mm}$ pressure. Calculate the percentage composition of nitrogen in the compound. (Aqueous tension at $300 \mathrm{~K}=15 \mathrm{~mm}$ )

Solution

Volume of nitrogen collected at $300 \mathrm{~K}$ and $715 \mathrm{~mm}$ pressure is $50 \mathrm{~mL}$

Actual pressure $=715-15=700 \mathrm{~mm}$

Volume of nitrogen atSTP $\frac{273 \times 700 \times 50}{300 \times 760}$ $41.9 \mathrm{~mL}$

$22,400 \mathrm{~mL}$ of $\mathrm{N_2}$ at STP weighs $=28 \mathrm{~g}$

$$ \begin{aligned} & 41.9 \mathrm{~mL} \text { of nitrogen weighs }=\frac{28 \times 41.9}{22400} \mathrm{~g} \\ & \begin{aligned} \text { Percentage of nitrogen } & =\frac{28 \times 41.9 \times 100}{22400 \times 0.3} \\ & =17.46 \% \end{aligned} \end{aligned} $$

(ii) Kjeldahl’s method: The compound containing nitrogen is heated with concentrated sulphuric acid. Nitrogen in the compound gets converted to ammonium sulphate (Fig. 8.16). The resulting acid mixture is then heated with excess of sodium hydroxide. The liberated ammonia gas is absorbed in an excess of standard solution of sulphuric acid. The amount of ammonia produced is determined by estimating the amount of sulphuric acid consumed in the reaction. It is done by estimating unreacted sulphuric acid left after the absorption of ammonia by titrating it with standard alkali solution. The difference between the initial amount of acid

taken and that left after the reaction gives the amount of acid reacted with ammonia.

Organic compound $+\mathrm{H_2} \mathrm{SO_4} \longrightarrow\left(\mathrm{NH_4}\right)_{2} \mathrm{SO_4}$

$$ \begin{aligned} & \xrightarrow{2 \mathrm{NaOH}} \mathrm{Na_2} \mathrm{SO_4}+2 \mathrm{NH_3}+2 \mathrm{H_2} \mathrm{O} \\ & 2 \mathrm{NH_3}+\mathrm{H_2} \mathrm{SO_4} \longrightarrow\left(\mathrm{NH_4}\right)_{2} \mathrm{SO_4} \end{aligned} $$

Let the mass of organic compound taken $=\mathrm{mg}$

Volume of $\mathrm{H_2} \mathrm{SO_4}$ of molarity, M,

taken $=V \mathrm{~mL}$

Volume of $\mathrm{NaOH}$ of molarity, M, used for titration of excess of $\mathrm{H_2} \mathrm{SO_4}=V_{1} \mathrm{~mL}$

$V_{1} \mathrm{~mL}$ of $\mathrm{NaOH}$ of molarity $\mathrm{M}$

$$ =V_{1} / 2 \mathrm{~mL} \text { of } \mathrm{H_2} \mathrm{SO_4} \text { of molarity } \mathrm{M} $$

Volume of $\mathrm{H_2} \mathrm{SO_4}$ of molarity $\mathrm{M}$ unused $=\left(V-V_{1} / 2\right) \mathrm{mL}$

$\left(V-V_{1} / 2\right) \mathrm{mL}$ of $\mathrm{H_2} \mathrm{SO_4}$ of molarity $\mathrm{M}$

molarity M.

$=2\left(V-V_{1} / 2\right) \mathrm{mL}$ of $\mathrm{NH_3}$ solution of

$1000 \mathrm{~mL}$ of $1 \mathrm{M} \mathrm{NH_3}$ solution contains $17 \mathrm{~g} \mathrm{NH_3}$ or $14 \mathrm{~g}$ of $\mathrm{N}$

$2\left(V-V_{1} / 2\right) \mathrm{mL}$ of $\mathrm{NH_3}$ solution of molarity M contains:

$$ \frac{14 \times \mathrm{M} \times 2\left(\mathrm{~V}-\mathrm{V_1} / 2\right)}{1000} \mathrm{gN} $$

Percentage of $\mathrm{N}=\frac{14 \times \mathrm{M} \times 2\left(\mathrm{~V}-\mathrm{V_1} / 2\right)}{1000} \times \frac{100}{\mathrm{~m}}$

$$ =\frac{1.4 \times \mathrm{M} \times 2(\mathrm{~V}-\mathrm{V} / 2)}{\mathrm{m}} $$

Kjeldahl method is not applicable to compounds containing nitrogen in nitro and azo groups and nitrogen present in the ring (e.g. pyridine) as nitrogen of these compounds does not change to ammonium sulphate under these conditions.

Problem 12.22

During estimation of nitrogen present in an organic compound by Kjeldahl’s method, the ammonia evolved from $0.5 \mathrm{~g}$ of the compound in Kjeldahl’s estimation of nitrogen, neutralized $10 \mathrm{~mL}$ of $1 \mathrm{M} \mathrm{H_2} \mathrm{SO_4}$. Find out the percentage of nitrogen in the compound.

Solution

$1 \mathrm{M}$ of $10 \mathrm{~mL} \mathrm{H_2} \mathrm{SO_4}=1 \mathrm{M}$ of $20 \mathrm{~mL} \mathrm{NH_3}$ $1000 \mathrm{~mL}$ of $1 \mathrm{M}$ ammonia contains $14 \mathrm{~g}$ nitrogen

$20 \mathrm{~mL}$ of $1 \mathrm{M}$ ammonia contains

$$ \frac{14 \times 20}{1000} \mathrm{~g} \text { nitrogen } $$

Percentage of nitrogen $=\frac{14 \times 20 \times 100}{1000 \times 0.5}=56.0 \%$

12.10.3 Halogens

Carius method: A known mass of an organic compound is heated with fuming nitric acid in the presence of silver nitrate contained in a hard glass tube known as Carius tube, (Fig.12.17)

in a furnace. Carbon and hydrogen present in the compound are oxidised to carbon dioxide and water. The halogen present forms the corresponding silver halide $(\operatorname{AgX})$. It is filtered, washed, dried and weighed.

Let the mass of organic compound taken $=\mathrm{m} \mathrm{g}$

Mass of $\mathrm{AgX}$ formed $=\mathrm{m_1} \mathrm{~g}$ $1 \mathrm{~mol}$ of $\mathrm{AgX}$ contains $1 \mathrm{~mol}$ of $\mathrm{X}$

Mass of halogen in $\mathrm{m_1} \mathrm{~g}$ of $\mathrm{AgX}$

$$ =\frac{\text { atomic mass of } \mathrm{X} \times \mathrm{m_1} \mathrm{~g}}{\text { molecular mass of } \mathrm{AgX}} $$

Percentage of halogen

$$ =\frac{\text { atomic mass of } X \times \mathrm{m_1} g}{\text { molecular mass of } \mathrm{AgX}} $$

Problem 12.23

In Carius method of estimation of halogen, $0.15 \mathrm{~g}$ of an organic compound gave $0.12 \mathrm{~g}$ of $\mathrm{AgBr}$. Find out the percentage of bromine in the compound.

Solution

Molar mass of $\mathrm{AgBr}=108+80$

$$ =188 \mathrm{~g} \mathrm{~mol}^{-1} $$

$188 \mathrm{~g}$ AgBr contains $80 \mathrm{~g}$ bromine

$0.12 \mathrm{~g}$ AgBr contains $\frac{80 \times 0.12}{188} \mathrm{~g}$ bromine

$$ \begin{aligned} \text { Percentage of bromine } & =\frac{80 \times 0.12 \times 100}{188 \times 0.15} \\ & =34.04 \% \end{aligned} $$

12.10.4 Sulphur

A known mass of an organic compound is heated in a Carius tube with sodium peroxide or fuming nitric acid. Sulphur present in the compound is oxidised to sulphuric acid. It is precipitated as barium sulphate by adding excess of barium chloride solution in water. The precipitate is filtered, washed, dried and weighed. The percentage of sulphur can be calculated from the mass of barium sulphate.

Let the mass of organic

$$ \text { compound taken }=\mathrm{mg} $$

and the mass of barium

$$ \text { sulphate formed }=\mathrm{m_1} \mathrm{~g} $$

$1 \mathrm{~mol} \text{ of } \mathrm{BaSO_4}=233 \mathrm{~g} \mathrm{BaSO_4}=32 \mathrm{~g}$ sulphur

$\mathrm{m_1} \mathrm{~g} \mathrm{BaSO_4}$ contains $\frac{32 \times m_{1}}{233}$ g sulphur

$$ \text { Percentage of sulphur }=\frac{32 \times m_{1} \times 100}{233 \times m} $$

Problem 12.24

In sulphur estimation, $0.157 \mathrm{~g}$ of an organic compound gave $0.4813 \mathrm{~g}$ of barium sulphate. What is the percentage of sulphur in the compound?

Solution

Molecular mass of $\mathrm{BaSO_4}=137+32+64$

$$ =233 \mathrm{~g} $$

$233 \mathrm{~g} \mathrm{BaSO_4}$ contains $32 \mathrm{~g}$ sulphur

$0.4813 \mathrm{~g} \mathrm{BaSO_4}$ contains $\frac{32 \times 0.4813}{233} \mathrm{~g}$ $\mathrm{~g}$ sulphur

Percentage of sulphur $=\frac{32 \times 0.4813 \times 100}{233 \times 0.157}$

$$ =42.10 \% $$

12.10.5 Phosphorus

A known mass of an organic compound is heated with fuming nitric acid whereupon phosphorus present in the compound is oxidised to phosphoric acid. It is precipitated as ammonium phosphomolybdate, $\left(\mathrm{NH_4}\right)_{3}$ $\mathrm{PO_4} \cdot 12 \mathrm{MoO_3}$, by adding ammonia and ammonium molybdate. Alternatively, phosphoric acid may be precipitated as $\mathrm{MgNH_4} \mathrm{PO_4}$ by adding magnesia mixture which on ignition yields $\mathrm{Mg_2} \mathrm{P_2} \mathrm{O_7}$.

Let the mass of organic compound taken $=\mathrm{mg}$ and mass of ammonium phospho molydate $=\mathrm{m_1} \mathrm{~g}$

Molar mass of $\left(\mathrm{NH_4}\right)_{3} \mathrm{PO_4} \cdot 12 \mathrm{MoO_3}=1877 \mathrm{~g}$

Percentage of phosphorus $=\frac{31 \times m_{1} \times 100}{1877 \times m} \%$

If phosphorus is estimated as $\mathrm{Mg_2} \mathrm{P_2} \mathrm{O_7}$,

Percentage of phosphorus $=\frac{62 \times m_{1} \times 100}{222}$ :

where, $222 \mathrm{u}$ is the molar mass of $\mathrm{Mg_2} \mathrm{P_2} \mathrm{O_7}$, $m$, the mass of organic compound taken, $m_{1}$, the mass of $\mathrm{Mg_2} \mathrm{P_2} \mathrm{O_7}$ formed and 62, the mass of two phosphorus atoms present in the compound $\mathrm{Mg_2} \mathrm{P_2} \mathrm{O_7}$.

12.10.6 Oxygen

The percentage of oxygen in an organic compound is usually found by difference between the total percentage composition (100) and the sum of the percentages of all other elements. However, oxygen can also be estimated directly as follows:

A definite mass of an organic compound is decomposed by heating in a stream of nitrogen gas. The mixture of gaseous products containing oxygen is passed over red-hot coke when all the oxygen is converted to carbon monoxide. This mixture is passed through warm iodine pentoxide $\left(\mathrm{I_2} \mathrm{O_5}\right)$ when carbon monoxide is oxidised to carbon dioxide producing iodine.

Compound $\xrightarrow{\text { heat }} \mathrm{O_2}+ \text{other gaseous product} $

$2 \mathrm{C}+\mathrm{O_2} \xrightarrow{1373 \mathrm{~K}} 2 \mathrm{CO}] 5 \quad \quad \quad \quad [A]$

$\mathrm{I_2} \mathrm{O_5}+5 \mathrm{CO} \longrightarrow \mathrm{I_2}+5 \mathrm{CO_2}] 2 \quad \quad \quad \quad [B]$

On making the amount of $\mathrm{CO}$ produced in equation (A) equal to the amount of $\mathrm{CO}$ used in equation (B) by multiplying the equations (A) and (B) by 5 and 2 respectively; we find that each mole of oxygen liberated from the compound will produce two moles of carbondioxide.

Thus $88 \mathrm{~g}$ carbon dioxide is obtained if $32 \mathrm{~g}$ oxygen is liberated.

Let the mass of organic compound taken be $m g$ Mass of carbon dioxide produced be $m_{1} g$

$\therefore \mathrm{m_1} \mathrm{~g}$ carbon dioxide is obtained from $\frac{32 \times m_{1}}{88} \mathrm{~g} \mathrm{O_2}$

$\therefore$ Percentage of oxygen $=\frac{32 \times m_{1} \times 100}{88 \times m} \%$

The percentage of oxygen can be derived from the amount of iodine produced also.

Presently, the estimation of elements in an organic compound is carried out by using microquantities of substances and automatic experimental techniques. The elements, carbon, hydrogen and nitrogen present in a compound are determined by an apparatus known as CHN elemental analyser. The analyser requires only a very small amount of the substance $(1-3 \mathrm{mg})$ and displays the values on a screen within a short time. A detailed discussion of such methods is beyond the scope of this book.

SUMMARY

In this unit, we have learnt some basic concepts in structure and reactivity of organic compounds, which are formed due to covalent bonding. The nature of the covalent bonding in organic compounds can be described in terms of orbitals hybridisation concept, according to which carbon can have $s p^{3}, s p^{2}$ and $s p$ hybridised orbitals. The $s p^{3}, s p^{2}$ and $s p$ hybridised carbons are found in compounds like methane, ethene and ethyne respectively. The tetrahedral shape of methane, planar shape of ethene and linear shape of ethyne can be understood on the basis of this concept. A $s p^{3}$ hybrid orbital can overlap with $1 s$ orbital of hydrogen to give a carbon - hydrogen (C-H) single bond (sigma, $\sigma$ bond). Overlap of a $s p^{2}$ orbital of one carbon with $s p^{2}$ orbital of another results in the formation of a carbon-carbon $\sigma$ bond. The unhybridised $p$ orbitals on two adjacent carbons can undergo lateral (sideby-side) overlap to give a pi ( $\pi$ ) bond. Organic compounds can be represented by various structural formulas. The three dimensional representation of organic compounds on paper can be drawn by wedge and dash formula.

Organic compounds can be classified on the basis of their structure or the functional groups they contain. A functional group is an atom or group of atoms bonded together in a unique 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 by the 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 the structure from the name.

Organic reaction mechanism concepts are based on the structure of the substrate molecule, fission of a covalent bond, the attacking reagents, the electron displacement effects and the conditions of the reaction. These organic reactions involve breaking and making of covalent bonds. A covalent bond may be cleaved in heterolytic or homolytic fashion. A heterolytic cleavage yields carbocations or carbanions, while a homolytic cleavage gives free radicals as reactive intermediate. Reactions proceeding through heterolytic cleavage involve the complimentary pairs of reactive species. These are electron pair donor known as nucleophile and an electron pair acceptor known as electrophile. The inductive, resonance, electromeric and hyperconjugation effects may help in the polarisation of a bond making certain 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 out for determining their structures. The methods of purification namely : sublimation, distillation and differential extraction are based on the difference in one or more physical properties. Chromatography is a useful technique of separation, identification and purification of compounds. It is classified into two categories : adsorption and partition chromatography. Adsorption chromatography is based on differential adsorption of various components of a mixture on an adsorbent. Partition chromatography involves continuous partitioning of the components of a mixture between stationary and mobile phases. After getting the compound in 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 and hydrogen 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 phosphoric acids respectively. The percentage of oxygen is usually determined by difference between the 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 ? $\mathrm{CH_2}=\mathrm{C}=\mathrm{O}, \mathrm{CH_3} \mathrm{CH}=\mathrm{CH_2}$, $\left(\mathrm{CH_3}\right)_{2} \mathrm{CO}, \mathrm{CH_2}=\mathrm{CHCN}, \mathrm{C_6} \mathrm{H_6}$

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Answer

(i) $\stackrel{1}{C} H_2=\stackrel{2}{C}=0$

Câ€" 1 is $s p^{2}$ hybridised.

Câ€" 2 is sp hybridised.

(ii)

$\stackrel{1}{C} H_3-\stackrel{2}{C} H=\stackrel{3}{C} H_2$

Câ€" 1 is $s p^{3}$ hybridised.

Câ€" 2 is $s p^{2}$ hybridised.

Câ€" 3 is $s p^{2}$ hybridised.

(iii)

C–1 and C–3 are sp³ hybridised.

C— 2 is $s p^{2}$ hybridised.

(iv)

C— 1 is $s p^{2}$ hybridised.

Câ€" 2 is $s p^{2}$ hybridised.

Câ€" 3 is $s p$ hybridised.

(v) $C_6 H_6$

All the 6 carbon atoms in benzene are $s p^{2}$ hybridised.

12.2 Indicate the $\sigma$ and $\pi$ bonds in the following molecules : $\mathrm{C_6} \mathrm{H_6}, \mathrm{C_6} \mathrm{H_12}, \mathrm{CH_2} \mathrm{Cl_2}, \mathrm{CH_2}=\mathrm{C}=\mathrm{CH_2}, \mathrm{CH_3} \mathrm{NO_2}, \mathrm{HCONHCH_3}$

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Answer

(i) $C_6 H_6$

There are six C–C sigma ( ${ }^{\sigma _{C-C}}$ ) bonds, six C–H sigma ( ${ }^{\sigma _{C-H}}$ ) bonds, and three $C=C$ pi $({ }^{C-C})$ resonating bonds in the given compound.

(ii) $C_6 H _{12}$

There are six C–C sigma ( $\sigma _{C-C}$ ) bonds and twelve C–H sigma ( $\sigma _{C-H}$ ) bonds in the given compound.

(iii) $CH_2 Cl_2$

There two C–H sigma ( $\sigma _{C-H}$ ) bonds and two Câe"Cl sigma ( ${ }^{\sigma _{C-Cl}}$ ) bonds in the given compound.

(iv) $CH_2=C=CH_2$

There are two CâE"C sigma ( ${ }^{\sigma _{C-C}}$ ) bonds, four Câe"H sigma ( ${ }^{\sigma _{C-H}}$ ) bonds, and two $C=C$ pi ( ${ }^{\pi _{C-C}}$ ) bonds in the given compound.

(v) $CH_3 NO_2$

There are three C–H sigma ( $.\sigma _{C-H})$ bonds, one C–N sigma ( $.\sigma _{C-N})$ bond, one N–O sigma ( $.\sigma _{N-0})$ bond, and one $N=O$ pi ( $.\pi _{N-O})$ bond in the given compound.

(vi) $HCONHCH_3$

There are two C–N sigma ( $\sigma _{C-N}$ ) bonds, four C–H sigma ( $\sigma _{C-H}$ ) bonds, one N–H sigma bond, and one $C=O$ pi ( $\pi _{C-C}$ ) bond in the given compound.

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

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Answer

The bond line formulae of the given compounds are:

(a) Isopropyl alcohol

$\Rightarrow$

(b) 2, 3-dimethyl butanal

$\Rightarrow$

(c) Heptan-4-one

$ H_3 C-CH_2-CH_2-\overbraceC^{O}-CH_2-CH_2-CH_3 $

$\Rightarrow$

12.4 Give the IUPAC names of the following compounds :

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Answer

(a)

1-phenyl propane

(b)

3-methylpentanenitrile

(c)

2, 5-dimethyl heptane

(d)

3-bromo-3-chloroheptane

(e)

3-chloropropanal

(f) $Cl_2 CHCH_2 OH$

1, 1-dichloro-2-ethanol

12.5 Which of the following represents the correct IUPAC name for the compounds concerned ?

(a) 2,2-Dimethylpentane or 2-Dimethylpentane

(b) 2,4,7-Trimethyloctane or 2,5,7-Trimethyloctane

(c) 2-Chloro-4-methylpentane or 4-Chloro-2-methylpentane

(d) But-3-yn-1-ol or But-4-ol-1-yne.

Show Answer

Answer

(a) The prefix diin the IUPAC name indicates that two identical substituent groups are present in the parent chain. Since two methyl groups are present in the C-2 of the parent chain of the given compound, the correct IPUAC name of the given compound is 2, 2-dimethylpentane.

(b) Locant number 2, 4, 7 is lower than 2, 5, 7. Hence, the IUPAC name of the given compound is 2, 4, 7trimethyloctane.

(c) If the substituents are present in the equivalent position of the parent chain, then the lower number is given to the one that comes first in the name according to the alphabetical order. Hence, the correct IUPAC name of the given compound is 2-chloro-4-methylpentane.

(d) Two functional groups - alcoholic and alkyne - are present in the given compound. The principal functional group is the alcoholic group. Hence, the parent chain will be suffixed with ol. The alkyne group is present in the C-3 of the parent chain. Hence, the correct IUPAC name of the given compound is But-3-yn-1-ol.

12.6 Draw formulas for the first five members of each homologous series beginning with the

following compounds. (a) $\mathrm{H}-\mathrm{COOH}$ (b) $\mathrm{CH_3} \mathrm{COCH_3}$ (c) $\mathrm{H}-\mathrm{CH}=\mathrm{CH_2}$

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Answer

The first five members of each homologous series beginning with the given compounds are shown as follows:

(a)

$H-COOH$ : Methanoic acid

$CH_3-COOH$ : Ethanoic acid

$CH_3-CH_2-COOH$ : Propanoic acid

$CH_3-CH_2-CH_2-COOH$ : Butanoic acid

$CH_3-CH_2-CH_2-CH_2-COOH$ : Pentanoic acid

(b)

$CH_3 COCH_3$ : Propanone

$CH_3 COCH_2 CH_3$ : Butanone

$CH_3 COCH_2 CH_2 CH_3$ : Pentan-2-one

$CH_3 COCH_2 CH_2 CH_2 CH_3$ : Hexan-2-one

$CH_3 COCH_2 CH_2 CH_2 CH_2 CH_3$ : Heptan-2-one

(c)

$H-CH=CH_2$ : Ethene

$CH_3-CH=CH_2$ : Propene

$CH_3-CH_2-CH=CH_2:$ 1-Butene

$CH_3-CH_2-CH_2-CH=CH_2$ : 1-Pentene

$CH_3-CH_2-CH_2-CH_2-CH=CH_2: 1-Hexene$

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

(a) 2,2,4-Trimethylpentane

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

(c) Hexanedial

Show Answer

Answer

(a) 2, 2, 4-trimethylpentane

Condensed formula:

$(CH_3)_2 CHCH_2 C(CH_3)_3$

Bond line formula:

(b) 2-hydroxy-1, 2, 3-propanetricarboxylic acid Condensed Formula:

$(COOH) CH_2 C(OH)(COOH) CH_2(COOH)$

Bond line formula:

The functional groups present in the given compound are carboxylic acid (-COOH) and alcoholic (-OH) groups.

(c) Hexanedial

Condensed Formula:

$(CHO)(CH_2)_4(CHO)$

Bond line Formula:

The functional group present in the given compound is aldehyde $(-CHO)$.

12.8 Identify the functional groups in the following compounds

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Answer

The functional groups present in the given compounds are:

(a) Aldehyde (–CHO),

Hydroxyl(–OH),

Methoxy (–OMe),

$C=C$ double bond $(-\stackrel{1}{C}=\stackrel{1}{C}-)$

(b) Amino (— $NH_2$ ); primary amine,

Ester (-O-CO-),

Triethylamine $(N(C_2 H_5)_2)$; tertiary amine

(c) Nitro (– $NO_2$ ),

$C=C$ double bond $(-C=C-)$

12.9 Which of the two: $\mathrm{O_2} \mathrm{NCH_2} \mathrm{CH_2} \mathrm{O}^{-}$or $\mathrm{CH_3} \mathrm{CH_2} \mathrm{O}^{-}$is expected to be more stable and why?

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Answer

$NO_2$ group is an electron-withdrawing group. Hence, it shows -I effect. By withdrawing the electrons toward it, the $NO_2$ group decreases the negative charge on the compound, thereby stabilising it. On the other hand, ethyl group is an electron-releasing group. Hence, the ethyl group shows +1 effect. This increases the negative charge on the compound, thereby destabilising it. Hence, $O_2 NCH_2 CH_2 O$ is expected to be more stable than $CH_3 CH_2 O$.

12.10 Explain why alkyl groups act as electron donors when attached to a $\pi$ system.

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Answer

When an alkyl group is attached to a Ãâ, $\neg$ system, it acts as an electron-donor group by the process of hyperconjugation. To understand this concept better, let us take the example of propene.

In hyperconjugation, the sigma electrons of the $C-H$ bond of an alkyl group are delocalised. This group is directly attached to an atom of an unsaturated system. The delocalisation occurs because of a partial overlap of a $s p^{3}$-ssigma bond orbital with an empty $p$ orbital of the Ãâ, $\neg$ bond of an adjacent carbon atom.

The process of hyperconjugation in propene is shown as follows:

This type of overlap leads to a delocalisation (also known as no-bond resonance) of the Ãâ, $\neg$ electrons, making the molecule more stable.

IV

III

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

(a) $\mathrm{C_6} \mathrm{H_5} \mathrm{OH}$

(b) $\mathrm{C_6} \mathrm{H_5} \mathrm{NO_2}$

(c) $\mathrm{CH_3} \mathrm{CH}=\mathrm{CHCHO}$

(d) $\mathrm{C_6} \mathrm{H_5}-\mathrm{CHO}$

(e) $\mathrm{C_6} \mathrm{H_5}-\stackrel{+}{\mathrm{C}} \mathrm{H_2}$

(f) $\mathrm{CH_3} \mathrm{CH}=\mathrm{CH} \stackrel{+}{\mathrm{C}} \mathrm{H_2}$

Show Answer

Answer

(a) The structure of $C_6 H_5 OH$ is:

The resonating structures of phenol are represented as:

(b) The structure of $C_6 H_5 NO_2$ is:

The resonating structures of nitro benzene are represented as:

(c) $CH_3 CH=CH-CHO$

The resonating structures of the given compound are represented as:

1

II

(d) The structure of $C_6 H_5 CHO$ is:

The resonating structures of benzaldehyde are represented as:

(e) $C_6 H_5 CH_2^{\text{AंA “c }}$

The resonating structures of the given compound are:

(f) $CH_3 CH=CHCH_2^{\tilde{A A} A^{n} c}$

The resonating structures of the given compound are:

12.12 What are electrophiles and nucleophiles ? Explain with examples.

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Answer

An electrophile is a reagent that takes away an electron pair. In other words, an electron-seeking reagent is called an electrophile $(E^{+})$. Electrophiles are electron-deficient and can receive an electron pair.

Carbocations $(CH_3 CH_2^{+})$and neutral molecules having functional groups such as carbonyl group $(/ C=0$ ) are examples of electrophiles.

A nulceophile is a reagent that brings an electron pair. In other words, a nucleus-seeking reagent is called a nulceophile (Nu:).

For example: $OH^{ate}, NC^{a} e^{e}$, carbanions $(R_3 C^{a \epsilon})$, etc.

Neutral molecules such as $H_2 \tilde{A} f \hat{€} \in$ " and ammonia also act as nulceophiles because of the presence of a lone pair.

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

(a) $\mathrm{CH_3} \mathrm{COOH}+\mathbf{H O}^{-} \rightarrow \mathrm{CH_3} \mathrm{COO}^{-}+\mathrm{H_2} \mathrm{O}$

(b) $\mathrm{CH_3} \mathrm{COCH_3}+\overline{\mathbf{C N}} \rightarrow\left(\mathrm{CH_3}\right)_{2} \mathrm{C}(\mathrm{CN})(\mathrm{OH})$

(c) $\mathrm{C_6} \mathrm{H_6}+\mathbf{C H_3} \stackrel{+}{\mathbf{C}} \mathbf{O} \rightarrow \mathrm{C_6} \mathrm{H_5} \mathrm{COCH_3}$

Show Answer

Answer

Electrophiles are electron-deficient species and can receive an electron pair. On the other hand, nucleophiles are electron-rich species and can donate their electrons.

(a) $CH_3 COOH+HO^{-} \longrightarrow CH_3 COO^{-}+H_2 O$

Here, $HO^{\hat{E}} e$ acts as a nucleophile as it is an electron-rich species, i.e., it is a nucleus-seeking species.

(b) $CH_3 COCH_3+\stackrel{-}{\mathbf{C}} \mathbf{N} \longrightarrow(CH_3)_2 C(CN)+(OH)$

Here, ${ }^{a} e^{\prime} CN$ acts as a nucleophile as it is an electron-rich species, i.e., it is a nucleus-seeking species. (c) $C_6 H_5+\mathbf{C H}_3 \stackrel{+}{\mathbf{C}} O \longrightarrow C_6 H_5 COCH_3$

Here, $CH_3 \stackrel{+}{C} O$ acts as an electrophile as it is an electron-deficient species.

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

(a) $\mathrm{CH_3} \mathrm{CH_2} \mathrm{Br}+\mathrm{HS}^{-} \rightarrow \mathrm{CH_3} \mathrm{CH_2} \mathrm{SH}+\mathrm{Br}^{-}$

(b) $\left(\mathrm{CH_3}\right)_2 \mathrm{C}=\mathrm{CH_2}+\mathrm{HCI} \rightarrow\left(\mathrm{CH_3}\right)_2 \mathrm{CIC}-\mathrm{CH_3}$

(c) $\mathrm{CH_3} \mathrm{CH_2} \mathrm{Br}+\mathrm{HO}^{-} \rightarrow \mathrm{CH_2}=\mathrm{CH_2}+\mathrm{H_2} \mathrm{O}+\mathrm{Br}^{-}$

(d) $\left(\mathrm{CH_3}\right)_3 \mathrm{C}-\mathrm{CH_2} \mathrm{OH}+\mathrm{HBr} \rightarrow\left(\mathrm{CH_3}\right)_2 \mathrm{CBrCH_2} \mathrm{CH_2} \mathrm{CH_3}+\mathrm{H_2} \mathrm{O}$

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Answer

(a) It is an example of substitution reaction as in this reaction the bromine group in bromoethane is substituted by the -SH group.

(b) It is an example of addition reaction as in this reaction two reactant molecules combine to form a single product.

(c) It is an example of elimination reaction as in this reaction hydrogen and bromine are removed from bromoethane to give ethene.

(d) In this reaction, substitution takes place, followed by a rearrangement of atoms and groups of atoms.

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

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Answer

(a) Compounds having the same molecular formula but with different structures are called structural isomers. The given compounds have the same molecular formula but they differ in the position of the functional group (ketone group).

In structure I, ketone group is at the C-3 of the parent chain (hexane chain) and in structure II, ketone group is at the $C-2$ of the parent chain (hexane chain). Hence, the given pair represents structural isomers.

(b) Compounds having the same molecular formula, the same constitution, and the sequence of covalent bonds, but with different relative position of their atoms in space are called geometrical isomers.

In structures I and II, the relative position of Deuterium (D) and hydrogen $(H)$ in space are different. Hence, the given pairs represent geometrical isomers.

(c) The given structures are canonical structures or contributing structures. They are hypothetical and individually do not represent any real molecule. Hence, the given pair represents resonance structures, called resonance isomers.

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

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Answer

(a) The bond cleavage using curved-arrows to show the electron flow of the given reaction can be represented as

It is an example of homolytic cleavage as one of the shared pair in a covalent bond goes with the bonded atom. The reaction intermediate formed is a free radical.

(b) The bond cleavage using curved-arrows to show the electron flow of the given reaction can be represented as

It is an example of heterolytic cleavage as the bond breaks in such a manner that the shared pair of electrons remains with the carbon of propanone. The reaction intermediate formed is carbanion.

(c) The bond cleavage using curved-arrows to show the electron flow of the given reaction can be represented as

It is an example of heterolytic cleavage as the bond breaks in such a manner that the shared pair of electrons remains with the bromine ion. The reaction intermediate formed is a carbocation.

(d) The bond cleavage using curved-arrows to show the electron flow of the given reaction can be represented as

It is a heterolytic cleavage as the bond breaks in such a manner that the shared pair of electrons remains with one of the fragments. The intermediate formed is a carbocation.

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

(a) $\mathrm{Cl_3} \mathrm{CCOOH}>\mathrm{Cl_2} \mathrm{CHCOOH}>\mathrm{ClCH_2} \mathrm{COOH}$

(b) $\mathrm{CH_3} \mathrm{CH_2} \mathrm{COOH}>\left(\mathrm{CH_3}\right)_2 \mathrm{CHCOOH}>\left(\mathrm{CH_3}\right)_3 \mathrm{C} . \mathrm{COOH}$

Show Answer

Answer

Inductive effect

The permanent displacement of sigma ( $\tilde{A} E^{\prime}$ ) electrons along a saturated chain, whenever an electron withdrawing or electron donating group is present, is called inductive effect.

Inductive effect could be $+I$ effect or $-I$ effect. When an atom or group attracts electrons towards itself more strongly than hydrogen, it is said to possess - I effect. For example,

$F \longrightarrow CH_2 \longleftarrow CH_2 \backsim CH_2 \backsim CH_3$

When an atom or group attracts electrons towards itself less strongly than hydrogen, it is said to possess +1 effect. For example,

$CH_3 \longrightarrow CH_2 \to Cl$

Electrometric effect

It involves the complete transfer of the shared pair of Ãâ, $\urcorner$ electrons to either of the two atoms linked by multiple bonds in the presence of an attacking agent. For example,

Electrometric effect could be $+E$ effect or $-E$ effect.

• E effect: When the electrons are transferred towards the attacking reagent

• E effect: When the electrons are transferred away from the attacking reagent

(a) $Cl_3 CCOOH>Cl_2 CHCOOH>ClCH_2 COOH$

The order of acidity can be explained on the basis of Inductive effect (- I effect). As the number of chlorine atoms increases, the - I effect increases. With the increase in - I effect, the acid strength also increases accordingly.

(b) $CH_3 CH_2 COOH>(CH_3)_2 CHCOOH>(CH_3)_3 C . COOH$

The order of acidity can be explained on the basis of inductive effect (+ I effect). As the number of alkyl groups increases, the $+I$ effect also increases. With the increase in $+I$ effect, the acid strength also increases accordingly.

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

(a) Crystallisation

(b) Distillation

(c) Chromatography

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Answer

(a) Crystallisation

Crystallisation is one of the most commonly used techniques for the purification of solid organic compounds.

Principle: It is based on the difference in the solubilites of the compound and the impurities in a given solvent. The impure compound gets dissolved in the solvent in which it is sparingly soluble at room temperature, but appreciably soluble at higher temperature. The solution is concentrated to obtain a nearly saturated solution. On cooling the solution, the pure compound crystallises out and is removed by filtration.

For example, pure aspirin is obtained by recrystallising crude aspirin. Approximately $2-4 g$ of crude aspirin is dissolved in about $20 mL$ of ethyl alcohol. The solution is heated (if necessary) to ensure complete dissolution. The solution is then left undisturbed until some crystals start to separate out. The crystals are then filtered and dried.

(b) Distillation

This method is used to separate volatile liquids from non-volatile impurities or a mixture of those liquids that have a sufficient difference in their boiling points.

Principle: It is based on the fact that liquids having different boiling points vapourise at different temperatures. The vapours are then cooled and the liquids so formed are collected separately.

For example, a mixture of chloroform ( $b . p=334 K$ ) and aniline ( $b . p=457 K$ ) can be separated by the method of distillation. The mixture is taken in a round bottom flask fitted with a condenser. It is then heated. Chloroform, being more volatile, vaporizes first and passes into the condenser. In the condenser, the vapours condense and chloroform trickles down. In the round bottom flask, aniline is left behind.

(c) Chromatography

It is one of the most useful methods for the separation and purification of organic compounds.

Principle: It is based on the difference in movement of individual components of a mixture through the stationary phase under the influence of mobile phase.

For example, a mixture of red and blue ink can be separated by chromatography. A drop of the mixture is placed on the chromatogram. The component of the ink, which is less adsorbed on the chromatogram, moves with the mobile phase while the less adsorbed component remains almost stationary.

12.19 Describe the method, which can be used to separate two compounds with different solubilities in a solvent $\mathrm{S}$.

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Answer

Fractional crystallisation is the method used for separating two compounds with different solubilities in a solvent $S$. The process of fractional crystallisation is carried out in four steps.

(a) Preparation of the solution: The powdered mixture is taken in a flask and the solvent is added to it slowly and stirred simultaneously. The solvent is added till the solute is just dissolved in the solvent. This saturated solution is then heated.

(b) Filtration of the solution: The hot saturated solution is then filtered through a filter paper in a China dish.

(c) Fractional crystallisation: The solution in the China dish is now allowed to cool. The less soluble compound crystallises first, while the more soluble compound remains in the solution. After separating these crystals from the mother liquor, the latter is concentrated once again. The hot solution is allowed to cool and consequently, the crystals of the more soluble compound are obtained.

(d) Isolation and drying: These crystals are separated from the mother liquor by filtration. Finally, the crystals are dried.

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

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Answer

The differences among distillation, distillation under reduced pressure,

and steam distillation are given in the following table.

Distillation	Distillation under reduced pressure	Steam distillation
1.It is used for the purification of compounds that are associated with non-volatile impurities or those liquids, which do not decompose on boiling. In other words, distillation is used to separate volatile liquids from non-volatile impurities or a mixture of those liquids that have sufficient difference in boiling points.	This method is used to purify a liquid that tends to decompose on boiling. Under pressure, the liquid will boil at a low temperature than its boiling point and will, therefore, not decompose.	compound, which is steam volatile and immiscible in water. On passing steam, the compound gets condensed to water. After some time, the mixture of water and liquid starts to boil and passes through the condenser. This condensed mixture of water and liquid is then separated by using a separating funnel.
2.	Mixture of petrol and kerosene is separated by this method.	Glycerol is purified by this method. It boils with decomposition at a temperature of 593 K. At a reduced pressure, it boils at 453 K without decomposition.	A mixture of water and aniline is separated by steam distillation.

12.21 Discuss the chemistry of Lassaigne’s test.

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Answer

Lassaigne’s test

This test is employed to detect the presence of nitrogen, sulphur, halogens, and phosphorous in an organic compound. These elements are present in the covalent form in an organic compound. These are converted into the ionic form by fusing the compound with sodium metal.

$ \begin{aligned} & Na+C+N \xrightarrow{\Delta} NaCN \\ & 2 Na+S \xrightarrow{\Delta} Na_2 S \\ & Na+X \xrightarrow{\Delta} NaX \\ & (X=Cl, Br, I) \end{aligned} $

The cyanide, sulphide, and halide of sodium formed are extracted from the fused mass by boiling it in distilled water. The extract so obtained is called Lassaigne’s extract. This Lassaigne’s extract is then tested for the presence of nitrogen, sulphur, halogens, and phosphorous.

(a) Test for nitrogen

Prussian blue colour

(Ferriferro cyanide)

Chemistry of the test

In the Lassaigne’s test for nitrogen in an organic compound, the sodium fusion extract is boiled with iron (II) sulphate and then acidified with sulphuric acid. In the process, sodium cyanide first reacts with iron (II) sulphate and forms sodium hexacyanoferrate (II). Then, on heating with sulphuric acid, some iron (II) gets oxidised to form iron (III) hexacyanoferrate (II), which is Prussian blue in colour. The chemical equations involved in the reaction can be represented as

$ \begin{aligned} & 6 CN^{-}+Fe^{2+} \longrightarrow[Fe(CN)_6]^{4-} \\ & 3[Fe(CN)_6]^{4-}+4 Fe^{3+} \xrightarrow{x H_2 O} Fe_4[Fe(CN)_6]_3 x H_2 O \end{aligned} $

Prussian blue colour

(b) Test for sulphur

(i) Lassaigne’s extract + Lead acetate $\xrightarrow{\text{ acetic acid }}$ Black precipitate

Chemistry of the test

In the Lassaigne’s test for sulphur in an organic compound, the sodium fusion extract is acidified with acetic acid and then lead acetate is added to it. The precipitation of lead sulphide, which is black in colour, indicates the presence of sulphur in the compound.

$S^{2-}+Pb^{2+} \longrightarrow PbS$

(Black)

(ii) Lassaigne’s extract + Sodium nitroprusside $\longrightarrow$ Violet colour

Chemistry of the test

The sodium fusion extract is treated with sodium nitroprusside. Appearance of violet colour also indicates the presence of sulphur in the compound.

$ \begin{aligned} & S^{2-}+[.Fe(CN)_5 NO]^{2-} \longrightarrow \\ &(\text{ Violet) } \end{aligned} $

If in an organic compound, both nitrogen and sulphur are present, then instead of $NaCN$, formation of NaSCN takes place.

$Na+C+N+S$ Ã

This NaSCN (sodium thiocyanate) gives a blood red colour. Prussian colour is not formed due to the absence of free cyanide ions.

$ Fe^{3+}+SCN^{-} \longrightarrow[Fe(SCN)]^{2+} $

(Blood red)

(c) Test for halogens

Chemistry of the test

In the Lassaigne’s test for halogens in an organic compound, the sodium fusion extract is acidified with nitric acid and then treated with silver nitrate.

$ \begin{aligned} & X^{-}+Ag^{+} \longrightarrow AgX \\ &(X=Cl, Br, I) \end{aligned} $

If nitrogen and sulphur both are present in the organic compound, then the Lassaigne’s extract is boiled to expel nitrogen and sulphur, which would otherwise interfere in the test for halogens.

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

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Answer

In Dumas method, a known quantity of nitrogen containing organic compound is heated strongly with excess of copper oxide in an atmosphere of carbon dioxide to produce free nitrogen in addition to carbon dioxide and water. The chemical equation involved in the process can be represented as

$ C x HyNz+(2 x+y / 2) CuO \longrightarrow x CO_2+y / 2 H_2 O+z / 2 N_2+(2 x+y / 2) Cu $

The traces of nitrogen oxides can also be produced in the reaction, which can be reduced to dinitrogen by passing the gaseous mixture over a heated copper gauge. The dinitrogen produced is collected over an aqueous solution of potassium hydroxide. The volume of nitrogen produced is then measured at room temperature and atmospheric pressure.

On the other hand, in Kjeldahl’s method, a known quantity of nitrogen containing organic compound is heated with concentrated sulphuric acid. The nitrogen present in the compound is quantitatively converted into ammonium sulphate. It is then distilled with excess of sodium hydroxide. The ammonia evolved during this process is passed into a known volume of $H_2 SO_4$. The chemical equations involved in the process are

The acid that is left unused is estimated by volumetric analysis (titrating it against a standard alkali) and the amount of ammonia produced can be determined. Thus, the percentage of nitrogen in the compound can be estimated. This method cannot be applied to the compounds, in which nitrogen is present in a ring structure, and also not applicable to compounds containing nitro and azo groups.

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

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Answer

Estimation of halogens

Halogens are estimated by the Carius method. In this method, a known quantity of organic compound is heated with fuming nitric acid in the presence of silver nitrate, contained in a hard glass tube called the Carius tube, taken in a furnace. Carbon and hydrogen that are present in the compound are oxidized to form $CO_2$ and $H_2 O$ respectively and the halogen present in the compound is converted to the form of $AgX$. This $AgX$ is then filtered, washed, dried, and weighed.

Let the mass of organic compound be $m g$.

Mass of $AgX$ formed $=m_1 g$

1 mol of Agx contains 1 mol of $X$.

Therefore,

Mass of halogen inm $I_1 g$ of $AgX$

$ =\frac{\text{ Atomic mass of } X \times m_1 g}{\text{ Molecular mass of } AgX} $

Thus, $%$ of halogen will be $=\frac{\text{ Atomic mass of } X \times m_1 \times 100}{\text{ Molecular mass of } AgX \times m}$

Estimation of Sulphur

In this method, a known quantity of organic compound is heated with either fuming nitric acid or sodium peroxide in a hard glass tube called the Carius tube. Sulphur, present in the compound, is oxidized to form sulphuric acid. On addition of excess of barium chloride to it, the precipitation of barium sulphate takes place. This precipitate is then filtered, washed, dried, and weighed.

Let the mass of organic compound be $m g$.

Mass of $BaSO_4$ formed $=m_1 g$

$1 mol^{2}$ of $BaSO_4=233 g BaSO_4=32 g$ of Sulphur

Therefore, $m_1 g$ of $BaSO_4$ contains $\frac{32 \times m_1}{233} g$ of sulphur.

Thus, percentage of sulphur $=\frac{32 \times m_1 \times 100}{233 \times m}$

Estimation of phosphorus

In this method, a known quantity of organic compound is heated with fuming nitric acid. Phosphorus, present in the compound, is oxidized to form phosphoric acid. By adding ammonia and ammonium molybdate to the solution, phosphorus can be precipitated as ammonium phosphomolybdate.

Phosphorus can also be estimated by precipitating it as $MgNH_4 PO_4$ by adding magnesia mixture, which on ignition yields $Mg_2 P_2 O_7$.

Let the mass of organic compound be $m g$.

Mass of ammonium phosphomolybdate formed $=m_1 g$

Molar mass of ammonium phosphomolybdate $=1877 g$

Thus, percentage of phosphorus $=\frac{31 \times m_1 \times 100}{1877 \times m} %$

If $P$ is estimated as $Mg_2 P_2 O_7$,

Then, percentage of phosphorus $=\frac{62 \times m_1 \times 100}{222 \times m} %$

12.24 Explain the principle of paper chromatography.

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Answer

In paper chromatography, chromatography paper is used. This paper contains water trapped in it, which acts as the stationary phase. On the base of this chromatography paper, the solution of the mixture is spotted. The paper strip is then suspended in a suitable solvent, which acts as the mobile phase. This solvent rises up the chromatography paper by capillary action and in the procedure, it flows over the spot. The components are selectively retained on the paper (according to their differing partition in these two phases). The spots of different components travel with the mobile phase to different heights. The paper so obtained (shown in the given figure) is known as a chromatogram.

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

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Answer

While testing the Lassaigne’s extract for the presence of halogens, it is first boiled with dilute nitric acid. This is done to decompose $NaCN$ to $HCN$ and $Na_2 S$ to $H_2 S$ and to expel these gases. That is, if any nitrogen and sulphur are present in the form of $NaCN$ and $Na_2 S$, then they are removed. The chemical equations involved in the reaction are represented as

$ \begin{aligned} & NaCN+HNO_3 \longrightarrow NaNO_3+HCN \\ & Na_2 S+2 HNO_3 \longrightarrow 2 NaNO_3+H_2 S \end{aligned} $

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

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Answer

Nitrogen, sulphur, and halogens are covalently bonded in organic compounds. For their detection, they have to be first converted to ionic form. This is done by fusing the organic compound with sodium metal. This is called “Lassaigne’s test”. The chemical equations involved in the test are

$ \begin{aligned} & Na+C+N \longrightarrow NaCN \\ & Na+S+C+N \longrightarrow NaSCN \\ & 2 Na+S \longrightarrow Na_2 S \\ & Na+X \longrightarrow NaX \\ &(X=Cl, Br, I) \end{aligned} $

Carbon, nitrogen, sulphur, and halogen come from organic compounds.

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

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Answer

The process of sublimation is used to separate a mixture of camphor and calcium sulphate. In this process, the sublimable compound changes from solid to vapour state without passing through the liquid state. Camphor is a sublimable compound and calcium sulphate is a non-sublimable solid. Hence, on heating, camphor will sublime while calcium sulphate will be left behind.

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

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Answer

In steam distillation, the organic liquid starts to boil when the sum of vapour pressure due to the organic liquid $(p_1)$ and the vapour pressure due to water $(p_2)$ becomes equal to atmospheric pressure $(p)$, that is, $p=p_1+p_2$

Since $p_1<p_2$, organic liquid will vapourise at a lower temperature than its boiling point.

12.29 Will $\mathrm{CCl_4}$ give white precipitate of $\mathrm{AgCl}$ on heating it with silver nitrate? Give reason for your answer.

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Answer

$CCl_4$ will not give the white precipitate of $AgCl$ on heating it with silver nitrate. This is because the chlorine atoms are covalently bonded to carbon in $CCl_4$. To obtain the precipitate, it should be present in ionic form and for this, it is necessary to prepare the Lassaigne’s extract of $CCl_4$.

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

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Answer

Carbon dioxide is acidic in nature and potassium hydroxide is a strong base. Hence, carbon dioxide reacts with potassium hydroxide to form potassium carbonate and water as $2 KOH+CO_2 \longrightarrow K_2 CO_3+H_2 O$

Thus, the mass of the U-tube containing $KOH$ increases. This increase in the mass of U-tube gives the mass of $CO_2$ produced. From its mass, the percentage of carbon in the organic compound can be estimated.

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

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Answer

Although the addition of sulphuric acid will precipitate lead sulphate, the addition of acetic acid will ensure a complete precipitation of sulphur in the form of lead sulphate due to common ion effect. Hence, it is necessary to use acetic acid for acidification of sodium extract for testing sulphur by lead acetate test.

12.32 An organic compound contains $69 \%$ carbon and $4.8 \%$ hydrogen, the remainder being oxygen. Calculate the masses of carbon dioxide and water produced when $0.20 \mathrm{~g}$ of this substance is subjected to complete combustion.

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Answer

Percentage of carbon in organic compound $=69 %$

That is, $100 g$ of organic compound contains $69 g$ of carbon.

$\therefore 0.2 g$ of organic compound will contain $100=0.138 g$ of $C$

$ =\frac{69 \times 0.2}{100}=0.138 g \text{ of } C $

Molecular mass of carbon dioxide, $CO_2=44 g$

That is, $12 g$ of carbon is contained in $44 g$ of $CO_2$.

$ \text{ Therefore, } 0.138 g \text{ of carbon will be contained in } \frac{44 \times 0.138}{12}=0.506 g \text{ of } CO_2 $

Thus, $0.506 g$ of $CO_2$ will be produced on complete combustion of $0.2 g$ of organic compound.

Percentage of hydrogen in organic compound is 4.8.

i.e., $100 g$ of organic compound contains $4.8 g$ of hydrogen.

Therefore, $0.2 g$ of organic compound will contain

$ \frac{4.8 \times 0.2}{100}=0.0096 g \text{ of } H $

It is known that molecular mass of water $(H_2 O)$ is $18 g$.

Thus, $2 g$ of hydrogen is contained in $18 g$ of water. $\therefore 0.0096 g$ of hydrogen will be contained in $\frac{18 \times 0.0096}{2}=0.0864 g$ of water

Thus, $0.0864 g$ of water will be produced on complete combustion of $0.2 g$ of the organic compound.

12.33 A sample of $0.50 \mathrm{~g}$ of an organic compound was treated according to Kjeldahl’s method. The ammonia evolved was absorbed in $50 \mathrm{ml}$ of $0.5 \mathrm{M} \mathrm{H_2} \mathrm{SO_4}$. The residual acid required $60 \mathrm{~mL}$ of $0.5 \mathrm{M}$ solution of $\mathrm{NaOH}$ for neutralisation. Find the percentage composition of nitrogen in the compound.

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Answer

Given that, total mass of organic compound $=0.50 g$

$60 mL$ of $0.5 M$ solution of $NaOH$ was required by residual acid for neutralisation.

$60 mL$ of $0.5 M NaOH$ solution $=\frac{60}{2} mL$ of $0.5 M H_2 SO_4=30 mL$ of $0.5 M H_2 SO_4$

$ =\frac{60}{2} mL \text{ of } 0.5 M $

$\therefore$ Acid consumed in absorption of evolved ammonia is (50âє"30) mL $=20 mL$

Again, $20 mL$ of $0.5 MH_2 SO_4=40 mL$ of $0.5 MNH_3$

Also, since $1000 mL$ of $1 MNH_3$ contains $14 g$ of nitrogen,

$\therefore 40 mL$ of $0.5 M NH_3$ will contain $\frac{14 \times 40}{1000} \times 0.5=0.28 g$ of N

Therefore, percentage of nitrogen in $0.50 g$ of organic compound $=\frac{0.28}{0.50} \times 100=56 %$

12.34 $ 0.3780 \mathrm{~g}$ of an organic chloro compound gave $0.5740 \mathrm{~g}$ of silver chloride in Carius estimation. Calculate the percentage of chlorine present in the compound.

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Answer

Given that,

Mass of organic compound is $0.3780 g$.

Mass of $AgCl$ formed $=0.5740 g$

$1 mol$ of $AgCl$ contains $1 mol$ of $Cl$.

Thus, mass of chlorine in $0.5740 g$ of $AgCl$ $=\frac{35.5 \times 0.5740}{143.32}$

$=0.1421 g$

$\therefore$ Percentage of chlorine $=\frac{0.1421}{0.3780} \times 100=37.59 %$

Hence, the percentage of chlorine present in the given organic chloro compound is $37.59 %$

12.35 In the estimation of sulphur by Carius method, $0.468 \mathrm{~g}$ of an organic sulphur compound afforded $0.668 \mathrm{~g}$ of barium sulphate. Find out the percentage of sulphur in the given compound.

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Answer

Total mass of organic compound $=0.468 g$ [Given]

Mass of barium sulphate formed $=0.668 g$ [Given]

$1 mol$ of $BaSO_4=233 g$ of $BaSO_4=32 g$ of sulphur

Thus, $0.668 g$ of $BaSO_4$ contains $\frac{32 \times 0.668}{233} g$ of sulphur $=0.0917 g$ of sulphur

Therefore, percentage of sulphur $=\frac{0.0197}{0.468} \times 100=19.59 %$

Hence, the percentage of sulphur in the given compound is $19.59 %$.

12.36 In the organic compound $\mathrm{CH_2}=\mathrm{CH}-\mathrm{CH_2}-\mathrm{CH_2}-\mathrm{C} \equiv \mathrm{CH}$, the pair of hydridised orbitals involved in the formation of: $\mathrm{C_2}-\mathrm{C_3}$ bond is:

(a) $s p-s p^{2}$

(b) $s p-s p^{3}$

(c) $s p^{2}-s p^{3}$

(d) $s p^{3}-s p^{3}$

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Answer

$\stackrel{6}C_2=\stackrel{5}{C} H-\stackrel{4}C_2-\stackrel{3}C_2-\stackrel{2}{C} \equiv \stackrel{1}{C} H$

In the given organic compound, the carbon atoms numbered as 1, 2, 3, 4, 5, and 6 are $s p, s p, s p^{3}, s p^{3}, s p^{2}$, and $s p^{2}$ hybridized respectively. Thus, the pair of hybridized orbitals involved in the formation of $C_2-C_3$ bond is $s p \hat{a} \epsilon^{\prime \prime} s p^{3}$.

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

(a) $\mathrm{Na_4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]$

(b) $\mathrm{Fe_4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]_3$

(c) $\mathrm{Fe_2}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]$

(d) $\mathrm{Fe_3}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]_4$

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Answer

In the Lassaigne’s test for nitrogen in an organic compound, the sodium fusion extract is boiled with iron (II) sulphate and then acidified with sulphuric acid. In the process, sodium cyanide first reacts with iron (II) sulphate and forms sodium hexacyanoferrate (II). Then, on heating with sulphuric acid, some iron (II) gets oxidised to form iron (III) hexacyanoferrate (II), which is Prussian blue in colour. The chemical equations involved in the reaction can be represented as

$ \begin{matrix} 6 CN^{-}+Fe^{2+} \longrightarrow[Fe(CN)_6]^{4-} \\ 3[Fe(CN)_6]^{4-}+4 Fe^{3+} \xrightarrow{x H_2 O} Fe_4[Fe(CN)_6]_3 x H_2 O \\ \text{ Prussian blue } \end{matrix} $

Hence, the Prussian blue colour is due to the formation of $Fe_4[Fe(CN)_6]_3$.

12.38 Which of the following carbocation is most stable?

(a) $\left(\mathrm{CH_3}\right)_3 \mathrm{C} \cdot \stackrel{+}{\mathrm{C}} \mathrm{H_2}$

(b) $\left(\mathrm{CH_3}\right)_3 \stackrel{+}{\mathrm{C}}$

(c) $\mathrm{CH_3} \mathrm{CH_2} \stackrel{+}{\mathrm{C}} \mathrm{H_2}$

(d) $\mathrm{CH_3} \stackrel{+}{\mathrm{C}} \mathrm{H} \mathrm{CH_2} \mathrm{CH_3}$

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Answer

$(CH_3)_3 \stackrel{+}{C}$

is a tertiary carbocation. A tertiary carbocation is the most stable carbocation due to the electron releasing effect of three methyl groups. An increased + I effect by three methyl groups stabilizes the positive charge on the carbocation.

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

(a) Crystallisation

(b) Distillation

(c) Sublimation

(d) Chromatography

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Answer

Chromatography is the most useful and the latest technique of separation and purification of organic compounds. It was first used to separate a mixture of coloured substances.

12.40 The reaction:

$\mathrm{CH_3} \mathrm{CH_2} \mathrm{I}+\mathrm{KOH}(\mathrm{aq}) \rightarrow \mathrm{CH_3} \mathrm{CH_2} \mathrm{OH}+\mathrm{KI}$

is classified as :

(a) electrophilic substitution

(b) nucleophilic substitution

(c) elimination

(d) addition

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Answer

$CH_3 CH_2 I+KOH _{(a q)} \longrightarrow CH_3 CH_2 OH+KI$

It is an example of nucleophilic substitution reaction. The hydroxyl group of $KOH(OH^{\text{af }})$ with a lone pair of itself acts as a nucleophile and substitutes iodide ion in $CH_3 CH_2$ to form ethanol.