Organic Chemistry  Some Basic Principles and Techniques

Objectives

After studying this unit, you will be able to

- understand reasons for tetravalence of carbon and shapes of organic molecules;

• write structures of organic molecules in various ways:

- classify the organic compounds:

- name the compounds according to I.U.P.A.C system of nomenclature and also derive their structures from the given names;

- understand the concept of organic reaction mechanism:

• explain the influence of electronic displacements on structure and reactivity of organic compounds;

- recognise the types of organic reactions:

- learn the techniques of purification of organic compounds;

• write the chemical reactions involved in the qualitative analysis of organic compounds;

- understand the principles involved in quantitative analysis of organic compounds.

In the previous unit you have learnt that the element carbon has the unique property called catenation due to which it forms covalent bonds with other carbon atoms. It also forms covalent bonds with atoms of other elements like hydrogen, oxygen, nitrogen, sulphur, phosphorus and halogens. The resulting compounds are studied under a separate branch of chemistry called organic chemistry . This unit incorporates some basic principles and techniques of analysis required for understanding the formation and properties of organic compounds.

8.1 General Introduction

Organic compounds are vital for sustaining life on earth and include complex molecules like genetic information bearing deoxyribonucleic acid (D.N.A) 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.

Math summary: This expression describes the chemical synthesis of urea from ammonium cyanate. The process is performed by applying heat to the inorganic ammonium cyanate input to produce the organic urea output.

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.

8.2 Tetravalence of Carbon: Shapes of Organic Compounds

8.2.1 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 (C.H 4) , ethene (C_2H 4) , ethyne (C_2H 2) are explained in terms of the use of sp ^3 , sp ^2 and sp hybrid orbitals by carbon atoms in the respective molecules.

Hybridisation influences the bond length and bond enthalpy (strength) in compounds. The sp hybrid orbital contains more s character and hence it is closer to its nucleus and forms shorter and stronger bonds than the sp hybrid orbital. The sp hybrid orbital is intermediate in s character between sp and sp 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 sp hybrid orbital with 50% s character is more electronegativity than that possessing sp or sp 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.

8.2.2 Some Characteristic Features of pi Bonds

In a pi bond formation, parallel orientation of the two p orbitals on adjacent atoms is necessary for a proper sideways overlap.

?: Thus, in H 2 C equals C H 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 C H 2 fragment with respect to other interferes with maximum overlap of p orbitals and, therefore, such rotation about carbon-carbon double bond ( C equals 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 8.1

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

: Math summary: This expression identifies the chemical structures of two different molecules. Part b specifies three carbon to carbon sigma bonds, six carbon to hydrogen sigma bonds, and two carbon to carbon pi bonds.

Solution

Math summary: This expression lists the counts of specific chemical bonds in a molecule. It specifies four carbon to carbon sigma bonds, six carbon to hydrogen sigma bonds, one carbon to carbon pi bond, and two carbon to carbon triple bond pi bonds.

(b) sigma c minus c: 3; sigma c minus h: 6; pi c equals c: 2.

Problem 8.2

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

C H 3 Cl, (b) (C H 3) 2 C O, (c) C H 3 C N,

(d) H C O N H 2, (e) C H 3 C H equals C H C N

Solution

(a) sp cubed, (b) sp cubed, sp squared, (c) sp cubed, sp, (d) sp squared, (e) sp cubed, sp squared, sp squared, sp

Problem 8.3

Write the state of hybridisation of carbon in the following compounds and shapes of each of the molecules.

:Math summary: This expression lists the chemical formulas for formaldehyde, fluoromethane, and hydrogen cyanide. It identifies the carbon hybridization and molecular shapes as trigonal planar for the first, tetrahedral for the second, and linear for the third.

Solution

(a) s p squared hybridised carbon, trigonal planar; (b) s p cubed hybridised carbon, tetrahedral; (c) s p hybridised carbon, linear.

8.3 Structural Representations of Organic Compounds

8.3.1 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. Lone-pairs of electrons on heteroatoms (e.g., oxygen, nitrogen, sulphur, halogens etcetera) may or may not be shown. Thus, ethane ( C_2H 6 ), ethene ( C_2H 4 ), ethylene ( C_2H 2 ) and methanol ( C.H 3O.H ) can be represented by the following structural formulas. Such structural representations are called complete structural formulas.

Image summary: Chemical structures showing the molecular formulas for ethane and ethene. Ethane is depicted with a single bond between two carbon atoms, while ethene is shown with a double bond between two carbon atoms. The figure illustrates the difference between a saturated hydrocarbon and an unsaturated hydrocarbon.

Image summary: Chemical structural formulas for ethyne and methanol. Ethyne is shown as two carbons with a triple bond between them, each bonded to a hydrogen atom. Methanol is shown in two equivalent representations, depicting a carbon atom bonded to three hydrogen atoms and one hydroxyl group. The figure illustrates the basic molecular structures of these two compounds.

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:

Math summary: This expression displays the condensed structural formulas for four different organic chemical compounds. It maps each chemical formula to its corresponding name: ethane, ethene, ethyne, and methanol.

Similarly, C.H_3C.H_2C.H_2C.H_2C.H_2C.H_2C.H 3 can be further condensed to C.H 3(C.H 2)_6C.H 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 etcetera The terminals denote methyl ( -C.H 3 ) 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:

Image summary: Chemical structural formulas for two organic molecules, labeled (a) and (b). Both structures represent isomers of an alkane, showing different arrangements of carbon and hydrogen atoms. The figures illustrate how the same chemical formula can result in different structural configurations.

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

Image summary: A chemical formula for 2-bromobutane, written as CH3CHBrCH2CH3. The image depicts the structural formula of a brominated organic compound.

Image summary: A chemical structure diagram of 2-bromobutane, showing a central carbon atom bonded to a bromine atom, a methyl group, a hydrogen atom, and an ethyl group. This diagram identifies the molecular arrangement of a secondary alkyl bromide.

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

Image summary: A chemical diagram showing a structural formula of cyclopropane, composed of three methylene groups arranged in a triangle, and its equivalence to a simple triangle symbol. The figure demonstrates the shorthand skeletal representation of a three-carbon ring.

।Image summary: A chemical diagram showing two representations of the same molecule: a structural formula explicitly labeling the five carbon and hydrogen groups in a ring, and a skeletal formula represented as a simple pentagon. The equality sign between them indicates that the skeletal structure is a shorthand abbreviation for the full chemical structure of cyclopentane.

Image summary: A chemical structural diagram showing the equivalence between two representations of chlorocyclohexane. On the left is a condensed structural formula explicitly labeling the carbon and hydrogen atoms in the six-membered ring, and on the right is the skeletal structure showing the same molecule as a hexagon with a chlorine atom substituent. The point is to show that these two notations represent the same chemical compound.

Figure c summary: A chemical structure diagram of a zig-zag hydrocarbon chain, with arrows indicating that the terminal ends of the structure represent methyl groups. This represents a simplified way of depicting a saturated hydrocarbon chain.

Problem 8.4

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

(a) C H 3 C H 2 C O C H 2 C H 3

(b) C.H 3 C.H equals C.H C.H 2 3 C.H 3

Solution

Image summary: Two chemical structural formulas showing organic molecules. The top molecule is a ketone with a carbonyl group on the third carbon of a five-carbon chain. The bottom molecule is an alkene with a double bond between the second and third carbons of a seven-carbon chain. The image depicts two different types of functional groups within hydrocarbon chains.

Problem 8.5

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

(a) H O C H 2 C H 2 C H 2 C H ( C H 3 ) C H ( C H 3 ) C H 3

Image summary: A chemical structure diagram depicting a molecule with a central carbon atom bonded to a hydrogen atom, a hydroxyl group (OH), and two cyano groups (CN) on opposite sides. This represents the molecule malononitrile derivatives or specifically 2-hydroxymalononitrile.

(b) nitrogen triple bond carbon minus carbon hydrogen minus carbon triple bond nitrogen

Condensed formula:

(a) H O, open parenthesis, C H 2, close parenthesis 3, C H, open parenthesis, C H 3, close parenthesis, C H, open parenthesis, C H 3, close parenthesis 2

(b) H.O.C.H(C.N) 2

Bond-line formula:

(a)

(b)

Image summary: A chemical structure diagram of a branched organic molecule featuring a hydroxyl group (HO) attached to a carbon chain with two methyl branches. This depicts the molecular structure of a specific alcohol.

Image summary: A chemical structure diagram showing a central carbon atom bonded to one hydroxyl group (OH) and two cyano groups (NC and CN). This depicts the molecule 2-hydroxypropanedinitrile, often referred to as cyanohydrin.

Problem 8.6

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

(b)

Image summary: A skeletal chemical structure showing a six-membered carbon ring attached to an isopropyl group. This depicts the molecule isopropylcyclohexane.

Image summary: A simple line drawing depicting a continuous zigzag or wave pattern consisting of alternating upward and downward diagonal lines. It is a graphic representing a repetitive, oscillating sequence.

(c)

(d)

Image summary: A chemical structure diagram of a molecule featuring a terminal alkyne group, a short carbon chain, and a hydroxyl group attached to a secondary carbon. This depicts a pentyne derivative with an alcohol functional group, specifically 4-pentyn-2-ol.

Image summary: A simple line drawing consisting of a horizontal central segment with two diverging lines extending from each end, forming a symmetrical shape resembling a wide, stylized letter M or a flat-bottomed valley. It is a geometric figure with no data or analytical result to report.

Solution

(a)

Image summary: A chemical structural formula showing a cyclohexane ring where one of the carbon atoms is substituted with an isopropyl group. The takeaway is that this molecule is isopropylcyclohexane.

Image summary: A set of three chemical structure diagrams labeled (b), (c), and (d). Diagram (b) shows a long straight-chain alkane, (c) shows a shorter chain containing a triple bond and a hydroxyl group, and (d) shows a branched molecule with multiple methyl groups. The figure depicts different types of organic molecules, ranging from a simple saturated hydrocarbon to an unsaturated alcohol and a branched hydrocarbon.

8.3.2 Three-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 (…) 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 figure 8.1

Figure 8.1 summary: A diagram of a methane molecule using wedge-and-dash notation to represent its three-dimensional structure. It shows two hydrogen bonds in the plane of the paper, one solid wedge representing a bond pointing toward the observer, and one dashed wedge representing a bond pointing away from the observer. The purpose of the diagram is to illustrate how 2D representations are used to depict the tetrahedral geometry of a molecule.

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 C=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.

Figure 8.2 summary: Three molecular models showing different representations of the same molecule: a framework model showing only the bonds, a ball-and-stick model showing atoms as spheres and bonds as rods, and a space-filling model showing the relative sizes of the atoms. These diverse representations allow for different ways of visualizing molecular geometry, connectivity, and steric bulk.

8.4 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:

Image summary: A classification diagram of organic compounds. Organic compounds are first divided into acyclic or open chain compounds and cyclic or closed chain or ring compounds. Cyclic compounds are further split into alicyclic compounds, which branch into homocyclic/carbocyclic and non-aromatic heterocyclic compounds, and aromatic compounds, which are divided into benzenoid and non-benzenoid compounds, the latter of which includes heterocyclic aromatic compounds. The diagram provides a hierarchical structure for categorizing organic molecules based on their chain and ring properties.

1. Acyclic or open chain compounds

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

Image summary: A collection of four chemical structures with their corresponding names: ethane, isobutane, acetaldehyde, and acetic acid. The image serves as a reference for comparing simple organic molecules, ranging from alkanes to an aldehyde and a carboxylic acid.

I.I 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).

Image summary: Three chemical structures showing cyclopropane as a triangle, cyclohexane as a hexagon, and cyclohexene as a hexagon with one double bond. The figure serves to illustrate the structural differences between a small strained ring, a saturated six-membered ring, and an unsaturated six-membered ring.

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

Tetrahydrofuran

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 compounds may also have heteroatom in the ring. Such compounds are called hetrocyclic aromatic compounds. Some of the examples of various types of aromatic compounds are:

Benzenoid aromatic compounds

Image summary: A diagram showing the chemical structures of four organic compounds. Benzene, Aniline, and Naphthalene are grouped together as benzenoid compounds, while Tropone is labeled as a non-benzenoid compound. The figure serves to illustrate the structural difference between benzenoid and non-benzenoid aromatic systems.

Heterocyclic aromatic compounds

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

Furan

Thiophene

Pyridine

8.4.1 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 (-O.H), aldehyde group (-C.H.O) and carboxylic acid group (-C.O.O.H) etcetera

8.4.2 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 -C.H 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 etcetera

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

8.5 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 I.U.P.A.C (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 I.U.P.A.C 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 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 8.1.

Table 8.1 summary: A list of common or trivial names for various organic compounds and their corresponding chemical formulas. Examples include Methane for CH4, n-Butane for H3CCH2CH2CH3, and Benzene for C6H6. The list also covers diverse functional groups, such as Formaldehyde for HCHO, Acetic acid for CH3COOH, and Aniline for C6H5NH2.

8.5.1 The I.U.P.A.C 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.

Image summary: A structural diagram of an organic molecule showing a horizontal parent chain of six carbon atoms. It identifies a methyl group (H3C) as a branch and two other groups, a bromine atom (Br) and a hydroxyl group (OH), as functional groups. The purpose of the diagram is to illustrate the basic components used in naming and classifying organic molecules.

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 I.U.P.A.C 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 carbon-carbon double or triple bond.

8.5.2 I.U.P.A.C 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 C.H 4 to C 4H 10 , where the prefixes are derived from trivial names). The I.U.P.A.C names of some straight chain saturated hydrocarbons are given in Table 8.2. The alkanes in Table 8.2 differ from each other by merely the number of -C.H 2 groups in the chain. They are homologues of alkane series.

Table 8.2 summary: IUPAC names and molecular formulas for unbranched saturated hydrocarbons, ranging from simple alkanes like Methane with CH4 and Ethane with C2H6 to longer chains such as Decane with C10H22, Icosane with C20H42, and Triacontane with C30H62.

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:

:Math summary: This expression illustrates the chemical structures of two branched chain hydrocarbons. Each structure consists of a main parent chain of carbon atoms with smaller methyl or ethyl alkyl groups attached at specific positions.

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, C.H 4 becomes -C.H 3 and is called methyl group.

An alkyl group is named by substituting 'yl' for 'ane' in the corresponding alkane. Some alkyl groups are listed in Table 8.3.

Table 8.3 summary: A list of common alkanes and their corresponding alkyl groups. Each entry pairs an alkane's molecular formula and name with the structural formula and name of its derived alkyl group. Examples include methane with the methyl group, ethane with the ethyl group, propane with the propyl group, butane with the butyl group, and decane with the decyl group.

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.

Image summary: A diagram showing the chemical structures of five common alkyl groups: isopropyl, sec-butyl, isobutyl, tert-butyl, and neopentyl. Each structure depicts the arrangement of carbon and hydrogen atoms to illustrate the different branching patterns of the alkyl chains.

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, minus C H 2 C parenthesis C H 3 parenthesis cubed, 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 (2) is not correct because it has only eight carbons.

Math summary: This process identifies the longest carbon chain in a molecule to determine the parent alkane. It compares a nine carbon chain numbered one through nine against an incorrect eight carbon chain to ensure the branching positions receive the lowest possible numbers.

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).

Math summary: This expression represents a chemical structure for the compound six ethyl two methylnonane. It shows a primary chain of nine carbon atoms with branching substituents located at the second and sixth carbon positions.

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-2-methylnonane. [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) etcetera 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:

Image summary: A diagram showing the chemical structures and IUPAC names for three branched alkanes: 2,4-dimethylpentane, 2,2,4-trimethylpentane (isooctane), and 3-ethyl-4,4-dimethylheptane. The structures include carbon chain numbering to illustrate how the names correspond to the positions of the methyl and ethyl substituents.

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

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.

Math summary: This expression illustrates the chemical structure for three ethyl six methyloctane. It shows an eight carbon chain with an ethyl group at the third position and a methyl group at the sixth position.

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.

Math summary: This expression illustrates the numbering system for a one three dimethylbutyl group. It identifies the carbon atom attaching to the root alkane as number one and sequences the remaining chain up to number four.

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 I.U.P.A.C nomenclature as long as these are not further substituted. In multi-substituted compounds, the following rules may also 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.

Image summary: A chemical structure diagram of 5-(2-ethylbutyl)-3,3-dimethyldecane, with numbers labeling the carbon atoms of the main ten-carbon chain and the substituents. The diagram identifies the correct numbering for the parent chain and the ethylbutyl branch to distinguish it from an incorrect naming convention. The point is to demonstrate the proper IUPAC nomenclature for this specific branched alkane.

: Image summary: A chemical structure diagram of 5-(2-ethylbutyl)-3,3-dimethyldecane, showing a ten-carbon main chain numbered 1 to 10. The structure features an isopropyl group at carbon 3 and a 2-ethylbutyl group at carbon 5. The purpose of the figure is to illustrate the correct connectivity and naming of this specific branched alkane.

5-sec-Butyl-4-isopropyldecane 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.

Image summary: A chemical structure diagram showing a long carbon chain with numbered positions 1 through 9, featuring a branched side chain attached at position 5. The side chain consists of a CH2 group connected to a quaternary carbon atom bonded to three methyl groups. This diagram illustrates the structural arrangement of 5-(2,2-dimethylpropyl)nonane.

Image summary: Two chemical structures displaying IUPAC nomenclature rules. The left shows cyclopentane, a simple five-membered ring. The right shows 1-methyl-3-propylcyclohexane, a six-membered ring with methyl and propyl substituents, with numbering labels 1, 2, and 3 indicating that the methyl group receives the lower number based on alphabetical priority. This demonstrates the application of alphabetical order when determining the numbering of substituted cycloalkanes.

Cyclopentane 1-Methyl-3-propylcyclohexane

Image summary: A chemical structure of 3-ethyl-1,1-dimethylcyclohexane, with carbon positions 1, 2, and 3 labeled on the ring. The accompanying text explains that more branched carbons are assigned lower numbers. The point is to illustrate the IUPAC numbering rule for substituents on a cyclohexane ring, where the more substituted carbon is designated as position 1.

Problem 8.7

Structures and I.U.P.A.C names of some hydrocarbons are given below. Explain why the names given in the parentheses are incorrect.

Image summary: A chemical structure diagram of an organic molecule featuring an eight-carbon main chain with three methyl group branches. This structural formula represents an isomer of an alkane, specifically 3,5,6-trimethyloctane, showing the connectivity of the carbon atoms and their associated hydrogen atoms.

2,5,6-Trimethyloctane

[and not 3,4,7-Trimethyloctane]

Image summary: A structural chemical formula showing a hydrocarbon chain with substituents, identified as 3-ethyl-1,1-dimethylcyclohexane. This diagram clarifies the correct chemical nomenclature for the depicted molecule, explicitly distinguishing it from 1-ethyl-3,3-dimethylcyclohexane.

3-Ethyl-5-methylheptane

[and not 5-Ethyl-3-methylheptane]

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.

8.5.3 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, C H 3 O H , C H 3 C H 2 O H , and ( C H 3 ) squared C H O H — all having O H 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 8.4.

Table 8.4 summary: A reference guide for organic compound classes and their associated IUPAC naming conventions. It maps functional group structures to their respective prefixes and suffixes. For example, hydrocarbons include Alkanes with the suffix -ane, Alkenes with -ene, and Alkynes with -yne. Oxygen-containing groups include Alcohols, characterized by the hydroxy- prefix and -ol suffix, Aldehydes with the -al suffix, and Ketones with the -one suffix. Carbonyl-based compounds include Carboxylic acids with the -oic acid suffix, Esters with the -oate suffix, and Amides with the -amide suffix. Nitrogenous compounds include Nitriles with the nitrile suffix and Amines with the -amine suffix. Other listed classes include Halides, Arenes, Ethers, Carboxylate ions, Acyl halides, Nitro compounds, and Sulphonic acids, providing a specific example for each, such as Butane for Alkanes and Benzene for Arenes.

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 8.4, 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:

-C.O.O.H, minus S O 3 H, minus C O O R (R equals alkyl group), C.O.C.L, minus C O N H 2, minus C N, minus H C equals O, greater than C equals O, minus O H, minus N H 2, greater than C equals C less than, minus C triple bond C minus.

The -R, C_6H 5^-, halogens (F, Cl, Br, 1), -N.O 2 , alkoxy (-O.R) etcetera 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, H.O.C.H 2(C.H 2) 3C.H 2C.O.C.H 3 will be named as 7-hydroxyheptan-2-one and not as 2-oxoheptan-7-ol. Similarly, BrCH 2C.H equals C.H 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, etcetera before the । In such cases the full name of the parent alkane is written before the class suffix. For example C H 2 open parenthesis O H close parenthesis C H 2 open parenthesis O H close parenthesis 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, C H 2 equals C H dash C H equals C H 2 is named as buta-1,3-diene.

Problem 8.8

Write the I.U.P.A.C names of the compounds i-4 from their given structures.

Math summary: This expression presents a structural formula for an organic compound to be named using IUPAC conventions. It depicts an eight carbon chain with an alcohol group at the third position and a methyl group at the sixth position.

Solution

- The functional group present is an alcohol (O.H). Hence the suffix is '-ol'.

- The longest chain containing -O.H has eight carbon atoms. Hence the corresponding saturated hydrocarbon is octane.

- The -O.H is on carbon atom 3. In addition, a methyl group is attached at 6 superscript th carbon.

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

Image summary: A chemical structure diagram of a diketone molecule with carbon atoms numbered from 1 to 6. Carbonyl groups are located at positions 2 and 4. This diagram identifies the numbering sequence of the carbon chain for naming the compound.

Solution

The functional group present is ketone ( greater than C=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.

(3)

: Math summary: This expression represents the chemical structure of a six carbon fatty acid. It labels the carbon chain from one to six, identifying a carboxyl group at the first position and a ketone group at the fifth position.

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 -C.O.O.H 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.

Equation iv summary: This expression represents the structural formula and carbon numbering for a chemical compound named Hexa-1,3-dien-5-yne. It shows a six carbon chain with double bonds at the first and third positions and a triple bond at the fifth position.

Solution

The two C=C functional groups are present at carbon atoms 1 and 3, while the C≡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-1,3-dien-5-yne.

Problem 8.9

Derive the structure of (1) 2-Chlorohexane, (2) Pent-4-en-2-ol, (3) 3-Nitrocyclohexene, (4) Cyclohex-2-en-1-ol, (5) 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 C H 3 C H 2 C H 2 C H 2 C H parenthesis C l parenthesis C H 3.

(2) 'pent' indicates that parent hydrocarbon contains 5 carbon atoms in the chain. 'en' and 'ol' correspond to the functional groups C=C and -O.H at carbon atoms 4 and 2 respectively. Thus, the structure is

Math summary: This expression represents the structural formula of a chemical compound. It describes a five carbon chain containing a double bond at the fourth carbon and a hydroxyl group at the second carbon.

(3) 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 C-3. Thus, complete structural formula of the compound is (2). Double bond is suffixed functional group whereas N.O 2 is prefixed functional group therefore double bond gets preference over -N.O 2 group:

(4) '1-ol' means that a -O.H group is present at C-1. O.H is suffixed functional group and gets preference over C=C bond. Thus the structure is as shown in (2):

(v) 'heptanal' indicates the compound to be an aldehyde containing 7 carbon atoms in the parent chain. The '6-hydroxy' indicates that -O.H group is present at carbon 6. Thus, the structural formula of the compound is: C H 3 C H O H C H 2 C H 2 C H 2 C H 2 C H O. Carbon atom of -C.H.O group is included while numbering the carbon chain.

Figure I summary: A chemical structure diagram of cyclohexene, showing a six-membered carbon ring with one double bond. The carbon atoms are numbered 1 through 6 around the ring. This diagram depicts the basic molecular structure and numbering convention for a cyclohexene molecule.

Figure II summary: Two chemical structures of substituted cyclohexene rings. The left structure shows a nitro group (NO2) attached to the ring at position 3, while the right structure shows a hydroxyl group (OH) attached at position 1. The purpose of the figure is to illustrate two different functional groups attached to a cyclohexene scaffold.

8.5.4 Nomenclature of Substituted Benzene Compounds

For I.U.P.A.C 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.

Image summary: A chemical structure diagram showing a benzene ring with a nitro group (NO2) attached to the top position. Although the caption identifies it as bromobenzene, the image actually depicts nitrobenzene.

Image summary: A chemical structure diagram of bromobenzene, consisting of a benzene ring with a single bromine atom attached. The image depicts the molecular structure of the compound bromobenzene.

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.

Image summary: A chemical structure diagram of a benzene ring with two bromine atoms attached to adjacent carbon atoms, labeled 1 and 2. This depicts the molecule 1,2-dibromobenzene.

Image summary: A chemical structure diagram of 1,4-dibromobenzene featuring a benzene ring with bromine atoms attached at the 1 and 4 positions. The image depicts the molecular structure of this specific halogenated aromatic compound.

In the trivial system of nomenclature the terms ortho (o), meta (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 number1 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.

Image summary: A chemical structure diagram of a benzene ring with a chlorine atom at position 2 and nitro groups at positions 1 and 4. The figure depicts the molecular structure of 2-chloro-1,4-dinitrobenzene.

Image summary: A chemical structure diagram showing a benzene ring with three substituents: a nitro group (NO2) at the top, a chlorine atom (Cl) at the 3-position, and a methyl group (CH3) at the 4-position. This depicts the molecular structure of 2-chloro-4-methylnitrobenzene.

Image summary: A chemical structure diagram of 2-chloro-4-methylanisole, showing a benzene ring with a methoxy group at position 1, a chlorine atom at position 2, and a methyl group at position 4. This layout depicts the specific arrangement of substituents on the aromatic ring.

: Image summary: A chemical structure diagram of 4-Ethyl-2-methylaniline, showing a benzene ring with an amino group at position 1, a methyl group at position 2, and an ethyl group at position 4. The diagram identifies the numbering of the carbon atoms in the ring to specify the arrangement of these functional groups.

Image summary: A chemical structure diagram of 3,4-Dimethylphenol, consisting of a benzene ring with a hydroxyl group at position 1 and methyl groups at positions 3 and 4. The figure serves to illustrate the molecular structure and numbering of the atoms in this organic compound.

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 (C 6H 5-, also abbreviated as Ph).

Methylbenzene (Toluene)

Methoxybenzene (Anisole)

Nitrobenzene

Aminobenzene (Aniline)

1,2-Dibromo- benzene

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

Figure b summary: A chemical structure diagram of 1,3-dibromobenzene, showing a benzene ring with bromine atoms attached at the 1 and 3 positions. This illustrates the structure of the specific organic molecule 1,3-dibromobenzene.

Problem 8.10

Write the structural formula of:

(a) o-Ethylanisole, (b) p-Nitroaniline,

(c) 2,3-Dibromo-1-phenylpentane,

(d) 4-Ethyl-1-fluoro-2-nitrobenzene.

Solution

Image summary: A grid of four chemical structure diagrams labeled (a) through (d). Structure (a) shows a benzene ring with methoxy and ethyl groups; (b) shows a benzene ring with amino and nitro groups in para positions; (c) shows a phenyl group attached to a brominated alkyl chain; and (d) shows a benzene ring with fluorine, nitro, and ethyl substituents. The image serves to provide the solutions for naming or identifying these specific organic molecules.

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, C 5H 12 represents three compounds:

: Image summary: Chemical structures showing the three isomers of pentane: the straight-chain pentane, the branched isopentane (2-methylbutane), and the highly branched neopentane (2,2-dimethylpropane). The figure illustrates how the same molecular formula can be arranged in different structural configurations.

- (2) 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 molecular formula C_3H 8O represents two alcohols:

Image summary: Two chemical structural formulas for propanol isomers. The left image shows propan-1-ol as a straight chain with the hydroxyl group at the end, and the right image shows propan-2-ol with the hydroxyl group attached to the middle carbon. The figure illustrates the difference between a primary and a secondary alcohol isomer.

- (3) 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 C_3H 6O represents an aldehyde and a ketone:

Image summary: A hierarchical diagram of isomerism, which is divided into two main categories: structural isomerism and stereoisomerism. Structural isomerism is further subdivided into chain, position, functional group, and metamerism, while stereoisomerism is split into geometrical and optical isomerism. The diagram serves to classify the different types of isomers based on their structural and spatial arrangements.

Image summary: Chemical structural formulas for propanone and propanal. Propanone is shown as a ketone with a central carbonyl group bonded to two methyl groups, while propanal is shown as an aldehyde with a carbonyl group at the end of the three-carbon chain bonded to a hydrogen atom. The figure illustrates the difference between a ketone and an aldehyde with the same number of carbon atoms.

(4) Metamerism: It arises due to different alkyl chains on either side of the functional group in the molecule. For example, C 4 H 10 O represents methoxypropane ( C.H 3 O C 3 H 7 ) and ethoxyethane ( C 2 H 5 O C 2 H 5 ).

8.6.2 Stereooisomerism

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.

8.7 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 :

Image summary: A chemical reaction diagram showing an organic molecule substrate being acted upon by an attacking reagent to form an intermediate. This intermediate then branches to produce either final products or byproducts. The diagram illustrates the general pathway of a chemical reaction from substrate to final products and side-products.

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.

8.7.1 Fission of a Covalent Bond

A covalent bond can get cleaved either by : (1) heterolytic cleavage, or by (2) 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 {C.H}_{3} and {Br superscript minus} as shown below.

Math summary: This expression describes the heterolytic cleavage of bromomethane. The process breaks the bond between the methyl group and bromine to produce a positively charged methyl cation and a negatively charged bromide ion.

A species having a carbon atom possessing sextext of electrons and a positive charge is called a carbocation (earlier called carbonium ion). The {C}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: C H 3 C H 2 (ethyl cation, a primary carbocation), (C H 3) 2 C H (isopropyl cation, a secondary carbocation), and (C H 3) 3 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 8.7.5 and 8.7.9. The observed order of carbocation stability is: C H 3 less than C H 3 C H 2 less than (C H 3) 2 C H less than (C H 3) 3 C. These carbocations have trigonal planar shape with positively charged carbon being s p 2 hybridised. Thus, the shape of C H 3 may be considered as being derived from the overlap of three equivalent C (s p 2) hybridised orbitals with 1s orbital of each of the three hydrogen atoms. Each bond may be represented as C sp 2 to H 1 s sigma bond. The remaining carbon orbital is perpendicular to the molecular plane and contains no electrons. [Fig. 8.3(a)].

Figure 8.3 summary: A chemical diagram depicting the shape of a methyl carbocation. It shows a central carbon atom bonded to three hydrogen atoms in a planar arrangement, with a vacant p-orbital represented by two lobes aligned perpendicular to the plane of the H-C-H bonds. This structure illustrates the trigonal planar geometry and the electronic configuration of the carbocation.

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

Math summary: This expression describes a heterolytic cleavage process where a methyl group attached to a group Z is split. The process results in the formation of a negatively charged methyl anion and a positively charged Z species.

electron pair, the methyl anion (H 3 C minus) is formed. Such a carbon species carrying a negative charge on carbon atom is called carbanion. Carbon in carbanion is generally sp 3 hybridised and its structure is distorted tetrahedron as shown in figure 8. 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: 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:

Math summary: This expression illustrates a homolytic cleavage process for an alkyl free radical. It shows an input molecule splitting into two separate radicals when triggered by heat or light.

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

Math summary: This expression establishes a hierarchy of stability for four different alkyl free radicals. Stability increases sequentially from the methyl radical to the ethyl radical, followed by the isopropyl radical, and peaking with the tert-butyl radical.

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

8.7.2 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:

Equation i summary: This expression shows a chemical reaction where a substrate and a reagent combine to form a product. Specifically, ethene reacts with bromine to produce one two bromine ethylene molecule.

।Image summary: A chemical equation showing the reaction between a benzene ring (substrate) and chloromethane (reagent) to produce toluene and hydrogen chloride. The diagram illustrates a Friedel-Crafts alkylation process where a methyl group is added to the benzene ring.

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) for example, 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, for example, pi electrons in a double bond.

A reagent that brings an electron pair to the reactive site is called a nucleophile (Nu colon) that is, nucleus seeking and the reaction is then called nucleophilic. A reagent that takes away an electron pair from reactive site is called electrophile (E plus) that is, 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 (H.O minus), cyanide (NC minus) ions and carbanions (R 3 C minus). Neutral molecules such as H 2 O, R 3 N, R 2 N.H etcetera, can also act as nucleophiles due to the presence of lone pair of electrons. Examples of electrophiles include carbocations (C.H 3 plus) and neutral molecules having functional groups like carbonyl group (C equals O) or alkyl halides (R 3 C minus X, where 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 C-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 8.11

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

(a) C.H 3 minus S C.H 3, (b) C.H 3 minus C.N, (c) C.H 3 minus Cu

Solution

Equations a, b, c summary: This process determines whether specific molecules act as nucleophiles or electrophiles by analyzing their bond cleavage. For the first two cases, the bond breaks to leave a positive charge on the methyl group and a negative charge on the sulfur or nitrogen atoms, while the third case results in a negative charge on the methyl group and the copper atom.

Problem 8.12

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

Math summary: This expression lists a series of chemical species and ions to be categorized as nucleophiles or electrophiles. The group includes species such as hydrosulfide, boron trifluoride, ethoxide, trimethylamine, chlorine, an acyl anion, ammonia, and nitrogen dioxide.

Solution

Nucleophiles: H S minus, C 2 H 5 O minus, (C H 3) 3 N, H 2 N. These species have unshared pair of electrons, which can be donated and shared with an electrophile.

Electrophiles: B F 3, Cl minus H 3 dash C equals O, N O 2. Reactive sites have only six valence electrons; can accept electron pair from a nucleophile.

Problem 8.13

Identify electrophilic centre in the following: C.H 3 C.H equals O, C.H 3 C.N, C.H 3 I.

Solution

Among C.H 3 H.C star equals O, H 3 C.C star triple bond N, and H 3C minus I, the starred carbon atoms are electrophilic centers as they will have partial positive charge due to polarity of the bond.

8.7.3 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 :

Math summary: This expression outlines a system for depicting the shifting of electron pairs using directional arrows. It defines three specific movements: from one bond to an adjacent bond position, from one bond to an adjacent atom, and from an atom to an adjacent entity.

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:

Math summary: This expression depicts a chemical reaction showing the transfer of a hydroxide ion to a methyl bromide molecule. The process results in the formation of ethanol and a bromide ion through the movement of electron pairs indicated by curved arrows.

Image summary: A diagram showing two chemical reaction mechanisms. The top reaction shows a nucleophilic substitution where a hydroxide ion attacks a methyl bromide molecule, resulting in methanol and a bromide ion. The bottom reaction shows homolytic bond cleavage of a methyl chloride molecule, resulting in a methyl radical and a chlorine radical. The image illustrates the difference between heterolytic bond cleavage involving electron pair movement and homolytic cleavage involving single electron movement.

Math summary: This expression describes the dissociation of chloromethane into two separate radicals. The process involves the breaking of a carbon iodine bond to produce methyl and iodine radicals.

8.7.4 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 polarisation 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.

8.7.5 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 (C H 3 C H 2 Cl) in which the C–Cl bond is a polar covalent bond. It is polarised in such a way that the carbon-1 gains some positive charge (delta plus) and the chlorine some negative charge (delta minus). The fractional electronic charges on the two atoms in a polar covalent bond are denoted by symbol delta and the shift of electron density is shown by an arrow that points from delta plus to delta minus end of the polar bond.

Image summary: A chemical diagram of an ethyl chloride molecule showing the flow of electron density. Arrows point from the methyl group (CH3) to the methylene group (CH2) and then to the chlorine atom (Cl), with corresponding partial charge labels indicating a progression from delta positive to delta negative. This depicts the inductive effect, where the electronegative chlorine atom pulls electron density away from the carbon chain.

In turn carbon-1, which has developed partial positive charge ( delta^+ ) draws some electron density towards it from the adjacent C-C bond. Consequently, some positive charge ( deltadelta^+ ) develops on carbon-2 also, where deltadelta^+ symbolises relatively smaller positive charge as compared to that on carbon - 1. In other words, the polar C - 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 substitutes can be classified as electron-withdrawing or electron donating groups relative to hydrogen. Halogens and many other groups such as nitro (minus N O 2), cyano (minus C N), carboxy (minus C O O H), ester (C O O R), aryloxy (-OAr, for example minus O C 6 H 5), etcetera are electron-withdrawing groups. On the other hand, the alkyl groups like methyl (minus C H 3) and ethyl (minus C H 2 dash C H 3) are usually considered as electron donating groups.

Problem 8.14

Which bond is more polar in the following pairs of molecules: (a) H 3 C dash H, H 3 C dash Br (b) H 3 C dash N H 2, H 3 C dash O H (c) H 3 C dash O H, H 3 C dash S H

Solution

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

Problem 8.15

In which C–C bond of CH3CH2CH2Br, 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.

8.7.6 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 C-C single and C=C double bonds shown is inadequate for explaining its characteristic properties.

As per the above representation, benzene should exhibit two different bond lengths, due to C–C single and C=C double bonds. However, as determined experimentally benzene has a uniform C–C bond distances of 139 pm, a value intermediate between the C–C single (154 pm) and C=C double (134 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 2.

Image summary: A chemical diagram showing two resonance structures of benzene, represented as hexagonal rings with alternating single and double bonds. A double-headed arrow between the two structures indicates that the actual molecule is a hybrid of these two forms. The point is to illustrate the resonance stabilization of the benzene ring.

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 2) 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 (C H 3 N O 2) which can be represented by two Lewis structures, (I and 2). There are two types of N-O bonds in these structures.

Image summary: A diagram showing the resonance structures of a chemical species, labeled I and II, connected by a double-headed arrow. Curved arrows indicate the movement of electrons between the nitrogen atom and the two oxygen atoms, shifting the double bond and formal charges. The figure illustrates how electron delocalization creates resonance stability between these two equivalent forms.

However, it is known that the two N-O bonds of nitromethane are of the same length (intermediate between a N-O single bond and a N=O double bond). The actual structure of nitromethane is therefore a resonance hybrid of the two canonical forms I and 2.

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 (1) the same positions of nuclei and (2) 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 8.16

Write resonance structures of C H 3 C O O minus 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.

Image summary: A chemical diagram showing two resonance structures of an acetate ion connected by a double-headed arrow. Curved arrows indicate the movement of electrons from a lone pair on the negatively charged oxygen to form a double bond, while the existing double bond shifts its electrons to the other oxygen. This illustrates that the negative charge is delocalized across both oxygen atoms.

Problem 8.17

Write resonance structures of C H 2 equals C H dash C H O. Indicate relative stability of the contributing structures.

Solution

Image summary: A chemical diagram showing three resonance structures, labeled I, II, and III, for an organic molecule consisting of carbon, hydrogen, and oxygen atoms. Structure I shows a carbonyl group adjacent to a vinyl group, structure II shifts the double bond and places a positive charge on the terminal carbon and a negative charge on oxygen, and structure III places the negative charge on the terminal carbon and a positive charge on oxygen. The accompanying text indicates that the stability of these structures follows the order I > II > III, identifying structure I as the most stable contributor.

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

Problem 8.18

Explain why the following two structures, I and I.I cannot be the major contributors to the real structure of C H 3 C O O C H 3 .

Image summary: A chemical diagram showing two resonance structures, labeled I and II, separated by a double-headed arrow. Structure I shows a positively charged carbon atom bonded to two methyl groups, an oxygen anion, and an ether group, while structure II shows a carbon-oxygen double bond and a positively charged oxygen atom. This depicts the resonance stabilization of a carbocation through electron donation from an adjacent oxygen atom.

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

8.7.7 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 R or 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:

Image summary: A chemical diagram showing the resonance structures of an aniline-like species, using curved arrows to illustrate the movement of electron pairs from the nitrogen atom into the aromatic ring. The sequence shows the lone pair on nitrogen forming a double bond, which pushes the ring's pi electrons to create negative charges at the ortho and para positions. This demonstrates how the nitrogen atom donates electron density into the ring, making it more electron-rich.

(2) Negative Resonance Effect (-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:

Image summary: A series of four chemical resonance structures for a nitrobenzene-like molecule, connected by double-headed arrows. The sequence shows the movement of electrons from the nitrogen-oxygen bond into the aromatic ring, shifting the positive charge and double bonds throughout the ring system. The purpose is to illustrate the delocalization of electrons and the resonance stabilization of the molecule.

The atoms or substituent groups, which represent +R or -R electron displacement effects are as follows:

Math summary: This expression lists substituent groups that create positive and negative Reffect electronic displacement effects. Positive effects are attributed to halogen, hydroxyl, alkoxy, amino, and other specified groups, while negative effects are attributed to carboxyl, formyl, carbonyl, cyano, and nitro groups.

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 etcetera In such systems, the pi -electrons are delocalised and the system develops polarity.

8.7.8 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 E and the shifting of the electrons is shown by a curved arrow ( ). 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:

: Image summary: A chemical equation showing the reaction of an alkene with a hydrogen ion (H+), denoted as the attacking reagent. A curved arrow indicates the double bond attacking the proton, resulting in a carbocation where the positive charge is located on one carbon and the hydrogen is bonded to the other. This depicts the mechanism of electrophilic addition to form a carbocation intermediate.

(2) 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:

Image summary: A chemical reaction diagram showing the addition of a cyanide ion (CN-) to a carbon-carbon double bond. The arrow indicates that the cyanide reagent attacks one of the double-bonded carbons, resulting in a single bond between the two carbons and the addition of the nitrile group to one of them. This depicts the mechanism of a nucleophilic addition reaction.

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 C—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 C—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 CH_3CH 2 (ethyl cation) in which the positively charged carbon atom has an empty p orbital. One of the C-H bonds of the methyl group can align in the plane of this empty p orbital and the electrons constituting the C-H bond in plane with this p orbital can then be delocalised into the empty p orbital as depicted in figure 8.4 (a).

Figure 8.4 summary: An orbital diagram depicting hyperconjugation in an ethyl cation. The figure shows the interaction between a Csp3-H1s sigma bond and an empty 2p orbital of a neighboring positively charged carbon, with dotted lines indicating the orbital overlap. This interaction allows for the delocalization of electrons, which stabilizes the carbocation.

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:

Math summary: This expression represents a series of chemical species showing the relative stability of different carbocations. It ranks the stability from a tertiary carbon center down to a methyl cation, illustrating a decrease in stability across the sequence.

Hyperconjugation is also possible in alkenes and alkylarenes.

Delocalisation of electrons by hyperconjugation in the case of alkene can be depicted as in figure 8.4(b).

There are various ways of looking at the hyperconjugative effect. One of the way is to regard C—H bond as possessing partial ionic character due to resonance.

The hyperconjugation may also be regarded as no bond resonance.

Problem 8.19

Explain why (C H 3) 3 C is more stable than C H 3 C H 2 and C H 3 is the least stable cation.

Solution

Hyperconjugation interaction in C methyl 3 dot C is greater than in C methyl dot C H 2 as the C methyl 3 dot C has nine C-H bonds. In dot C H 3, vacant p orbital is perpendicular to the plane in which C-H bonds lie; hence cannot overlap with it. Thus, dot C 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

(2) Addition reactions

(3) Elimination reactions

(4) Rearrangement reactions

You will be studying these reactions in Unit 9 and later in class 12.

8.8 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

(2) Crystallisation

(3) Distillation

(4) 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.

8.8.1 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 (2) 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 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.

: Figure 8.5 summary: A diagram of a simple distillation setup showing a round-bottom flask heated by a Bunsen burner, a vertical thermometer for temperature monitoring, and a water-cooled condenser. Vapors rise from the flask, travel through the condenser where they are cooled back into a liquid, and are collected in a conical flask. The purpose of this apparatus is to separate components of a mixture by evaporating a substance and condensing the resulting vapors into a pure liquid.

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.8.6, page 280).

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 figure8.7. 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 fractions of 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.

: Figure 8.8 summary: A diagram showing the apparatus for distillation under reduced pressure, featuring a heating source, a boiling flask with a stopcock and thermometer, a condenser with a water inlet, and a collection flask for distilled liquid. The system is connected to a vacuum pump and a manometer to lower the internal pressure. This setup allows a liquid to boil at a temperature below its normal boiling point by reducing the pressure within the system.

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 ( p 1 ) and that due to water ( p 2 ) becomes equal to the atmospheric pressure (p), that is 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.8.9, Page 282).

Figure 8.6 summary: A diagram of a fractional distillation apparatus featuring a heat source and oil bath to warm a flask of liquid, a fractionating column with packing to separate vapours, and a condenser that cools vapours back into a collected liquid in a conical flask. The setup is designed to separate a mixture based on boiling points, as vapours of lower boiling fractions reach the top of the column and the condenser before those of higher boiling fractions.

Figure 8.7 summary: A diagram comparing two types of fractionating columns. The simple packed column on the left fills the interior with glass beads, while the bubble plate column on the right uses a series of horizontal plates with openings. Both designs serve to increase the surface area and contact between rising vapor and falling liquid to improve separation efficiency during distillation.

8.8.4 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 figure 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.

: Figure 8.10 summary: Two diagrams of a separatory funnel illustrating differential extraction. Before extraction, an organic compound is located in the lower aqueous layer; after extraction, the compound has moved into the upper solvent layer. This demonstrates that a compound can be separated from a mixture based on its difference in solubility between two immiscible liquids.

The technique of continuous extraction is employed in such cases. In this technique same solvent is repeatedly used for extraction of the compound.

8.8.5 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.

Figure 8.11 summary: A diagram of column chromatography showing three sequential stages of separation. A mixture of compounds (a+b+c) is placed atop a stationary adsorbent phase within a glass column. As solvent is added, the mixture separates into distinct bands based on differing affinities for the adsorbent, with compound a moving fastest and compound c moving slowest. The process demonstrates how a complex mixture is resolved into its individual components over time.

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. 8.11).

Thin Layer Chromatography: Thin layer chromatography (T.L.C) 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 millimeters 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 chromatplate. The solution of the mixture to be separated is applied as a small spot about 2 centimeters above one end of the T.L.C 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 that is R f value (Fig.8.12 b).

The spots of coloured compounds are visible on T.L.C 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.

Figure 8.13 summary: Two diagrams illustrating paper chromatography setups, one using a cylindrical rolled sheet of chromatography paper and the other using a flat strip. Both setups show a spot of sample placed on a base line, with the paper positioned inside a jar containing a solvent. The diagrams demonstrate how different shapes of chromatography paper can be used to separate components of a mixture through solvent migration.

Figure 8.9 summary: A schematic diagram of a steam distillation setup showing a water flask and a flask containing the compound to be distilled, both heated by Bunsen burners. Water vapor travels from the first flask into the second, carrying the volatile component through a water-cooled condenser and into a collecting conical flask. This process allows for the separation of steam-volatile components by condensing them into a collected liquid mixture.

Figure 8.12 summary: A two-part diagram illustrating thin layer chromatography. Part (a) shows the initial setup with a sample dot on an adsorbent-coated glass plate placed in a jar containing solvent. Part (b) shows the developed chromatogram, where the solvent has moved up the plate to a solvent front and the sample has separated into a spot. The diagram defines the retardation factor, Rf, as the ratio of the distance the substance moved from the base line (x) to the distance the solvent moved from the base line (y).

8.9 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.

8.9.1 Detection of Carbon and Hydrogen

Carbon and hydrogen are detected by heating the compound with copper(I.I) 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).

Math summary: These chemical equations describe the processes for detecting carbon and hydrogen in an organic compound using copper oxide. Carbon is oxidized to carbon dioxide and tested with lime water, while hydrogen is oxidized to water which turns anhydrous copper sulphate blue.

8.9.2 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:

Math summary: These equations describe the chemical transformations used in Lassaigne's test to convert covalent elements into ionic forms. Sodium metal reacts under heat with carbon and nitrogen to form sodium cyanide, with sulfur to form sodium sulfide, and with halogens like chlorine, bromine, or iodine to form sodium halides.

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(I.I) 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(I.I) sulphate and forms sodium hexacyanidoferrate(I.I). On heating with concentrated sulphuric acid some iron(I.I) ions are oxidised to iron(I.I.I) ions which react with sodium hexacyanidoferrate(I.I) to produce iron(I.I.I) hexacyanidoferrate(I.I) (ferriferrocyanide) which is Prussian blue in colour.

Math summary: This expression describes the chemical reaction process used to produce Prussian blue to confirm the presence of nitrogen. Six cyanide ions combine with one iron two plus ion to form hexacyanidoferrate two, which then reacts with four iron three plus ions and water to yield the final Prussian blue complex.

(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.

Math summary: This expression represents a chemical reaction that identifies the presence of sulphur. It shows sulphide ions reacting with lead ions to produce a black precipitate of lead sulphide.

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

Math summary: This expression represents a chemical reaction used to indicate the presence of sulphur when treating a sodium fusion extract with sodium nitroprusside. A sulphur ion reacts with a nitroprusside complex to produce a violet colored output complex.

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.

Math summary: This process describes the formation of sodium thiocyanate from sodium, carbon, nitrogen, and sulfur. The resulting sodium thiocyanate then reacts with iron three plus ions to produce a blood red iron thiocyanate complex.

Blood red

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

Math summary: This expression describes a sodium fusion chemical reaction where sodium thiocyanate reacts with sodium. The process uses two parts sodium to decompose the thiocyanate into sodium cyanide and sodium sulfide.

(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.

Math summary: This expression represents a chemical reaction used to identify a halogen. It shows a halide ion and a silver ion combining to produce a silver halide precipitate.

X represents a halogen - Cl, Br or 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.

Math summary: This chemical sequence describes a two step process to detect phosphorus by first converting sodium phosphate into phosphoric acid using nitric acid. The phosphoric acid then reacts with ammonium molybdate and more nitric acid to produce a yellow precipitate of ammonium phosphomolybdate, ammonium nitrate, and water.

8.10 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 empirical and molecular formula.

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

8.10.1 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(I.I) oxide. Carbon and hydrogen in the compound are oxidised to carbon dioxide and water respectively.

: Math summary: This expression represents a chemical reaction where an organic compound containing carbon and hydrogen reacts with oxygen. The process transforms the organic input into a calculated output of carbon dioxide and water based on the proportions of carbon and hydrogen present.

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. 8.14). 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.

Figure 8.14 summary: A schematic diagram of an apparatus for estimating carbon and hydrogen. Pure dry oxygen passes through a combustion tube containing a sample in a platinum boat and CuO pellets, where oxidation occurs. The resulting gases flow through two U-tubes: the first containing anhydrous calcium chloride to absorb water, and the second containing a KOH solution to absorb carbon dioxide, with excess oxygen exiting the system. The setup is designed to isolate and quantify the water and carbon dioxide produced from the oxidation of a substance.

Let the mass of organic compound be m g, mass of water and carbon dioxide produced be m 1 and m 2 g respectively;

Math summary: These expressions calculate the percentage composition of carbon and hydrogen in an organic compound. The process involves multiplying the mass of carbon dioxide and water by specific atomic constants and one hundred, then dividing by the product of molar masses and the total mass of the compound.

Problem 8.20

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

Solution

:Math summary: This computation determines the percentage of carbon and hydrogen within a sample. The process scales the atomic weights of carbon and hydrogen by their respective masses and divides by a common denominator to yield results of twenty-one point nine five percent for carbon and four point five eight percent for hydrogen.

8.10.2 Nitrogen

There are two methods for estimation of nitrogen: (1) Dumas method and (2) 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.

Math summary: This chemistry expression describes the reaction used in the Dumas method to decompose a nitrogen containing organic compound. It shows the compound reacting with copper oxide to produce carbon dioxide, water, nitrogen, and copper.

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 = m g

Volume of nitrogen collected equals V 1 milliliters

Math summary: This computation determines the volume of nitrogen at standard temperature and pressure. It multiplies the initial pressure, initial volume, and two hundred seventy three, then divides that result by the product of seven hundred sixty and the room temperature in kelvin.

(Let it be V 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 equals Atmospheric pressure minus Aqueous tension

22400 mL N 2 at S.T.P weighs 28 g.

Figure 8.15 summary: A diagram of the Dumas method for determining nitrogen content. An organic compound is heated within a furnace containing copper(II) oxide gauze and coarse CuO under a stream of CO2 gas, releasing nitrogen. The resulting gas mixture is passed through a mercury seal into a nitrometer containing potassium hydroxide solution, which absorbs the CO2 and allows the volume of evolved nitrogen gas to be measured.

Math summary: This computation determines the weight of nitrogen at standard temperature and pressure and its subsequent percentage in a compound. It calculates the weight by multiplying the volume by twenty eight and dividing by twenty two thousand four hundred, then finds the percentage by multiplying that weight by one hundred and dividing by the total mass.

Problem 8.21

In Dumas' method for estimation of nitrogen, 0.3g of an organic compound gave 50 mL of nitrogen collected at 300K temperature and 715 millimeters pressure. Calculate the percentage composition of nitrogen in the compound. (Aqueous tension at 300K=15 millimeters)

Solution

Volume of nitrogen collected at 300K and 715 millimeters pressure is 50 mL

Math summary: This calculation determines the actual pressure of nitrogen. It subtracts fifteen from seven hundred fifteen to reach a final result of seven hundred millimeters.

Volume of nitrogen at S.T.P 273 times 700 times 50 divided by 300 times 760

41.9 mL

22,400 mL of N₂ at S.T.P weighs = 28 g

Math summary: This computation determines the weight and percentage of nitrogen using Kjeldahl's method. It first calculates the nitrogen weight by multiplying twenty eight by forty one point nine and dividing by twenty two thousand four hundred, then divides that result by zero point three and multiplies by one hundred to reach a final value of seventeen point four six percent.

(2) 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.

Figure 8.16 summary: A diagram of the Kjeldahl method showing two stages. In the first stage, an organic compound is heated in Kjeldahl's flask with concentrated sulfuric acid and copper sulfate for digestion. In the second stage, the digest is treated with sodium hydroxide and heated to distill ammonia, which passes through a Kjeldahl's trap and is collected in a flask containing a known volume of standard acid. The process ensures nitrogen-containing compounds are converted to ammonia and then absorbed for quantification.

Math summary: This expression details a chemical sequence to determine the amount of acid reacted with ammonia. It follows a process where an organic compound reacts with sulfuric acid to form ammonium sulfate, which is then processed with two parts sodium hydroxide into sodium sulfate, ammonia, and water, before the resulting ammonia reacts again with sulfuric acid to yield ammonium sulfate.

Let the mass of organic compound taken = m g

Volume of H 2 S O 4 of molarity, M, taken = V mL

Volume of N-A-O-H of molarity, M, used for titration of excess of H 2 S O 4 equals V 1 mL

V 1 milliliters

equals V 1 divided by 2 milliliters of H 2 S.O 4 of molarity M

Volume of H 2 S.O 4 of molarity M unused equals open parenthesis V minus V 1 divided by 2 close parenthesis milliliters

greater than (V minus V 1 divided by 2) milliliters of H 2 S O 4 of molarity M equals 2 times the quantity V minus V 1 divided by 2 milliliters of N.H 3 solution of molarity M.

1000 mL of 1 molar N.H 3 solution contains 17g N.H 3 or 14 g of N

2(V-V1/2) mL of NH3 solution of molarity M contains:

:Math summary: This expression computes the percentage of nitrogen using the Kjeldahl method. It calculates this by multiplying fourteen, the molarity, and twice the difference between the volume and half of the first volume, then dividing that result by the total mass.

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 8.22

During estimation of nitrogen present in an organic compound by Kjeldahl's method, the ammonia evolved from 0.5 g of the compound in Kjeldahl's estimation of nitrogen, neutralized 10 mL of 1 M H 2SO 4 . Find out the percentage of nitrogen in the compound.

Solution

1 M of 10 mL H2SO4=1M of 20 mL NH3

1000 mL of 1M ammonia contains 14 g nitrogen

20 mL of 1M ammonia contains 14 times 20 divided by 1000 grams nitrogen

Math summary: This calculation determines the percentage of nitrogen using the Carius method. It multiplies fourteen by twenty and one hundred, then divides that product by the result of one thousand multiplied by zero point five to arrive at fifty six percent.

8.10.3 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.8.17)

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 (AgX). It is filtered, washed, dried and weighed.

Let the mass of organic

Math summary: This expression calculates the mass of a compound taken and the mass of silver halide formed. It defines these quantities as the product of a mass value and the acceleration due to gravity.

1 mol of AgX contains 1 mol of X

Math summary: This computation determines the mass and percentage of a halogen within a sample of silver halide. It multiplies the atomic mass of the halogen by the mass of silver halide and divides that product by the molecular mass of silver halide.

Figure 8.17 summary: A schematic diagram of the Carius method, depicting a 50 cm sealed capillary tube containing a halogen-containing organic compound, fuming HNO3, and AgNO3. The setup is designed to heat these components together to analyze the halogen content of the organic compound.

Problem 8.23

In Carius method of estimation of halogen, 0.15 g of an organic compound gave 0.12 g of AgBr. Find out the percentage of bromine in the compound.

Solution

Math summary: This expression calculates the molar mass of silver bromide. It sums the values three hundred eight and eight hundred to arrive at a final result of one hundred eighty eight grams per mole.

188 g AgBr contains 80 g bromine

: Math summary: This calculation determines the mass of bromine in a sample and the overall percentage of bromine in a compound. It multiplies the sample mass of zero point one two grams by eighty and divides by one hundred eighty eight to find the bromine mass, then scales this by one hundred and divides by zero point one five to reach a final result of thirty four point zero four percent.

8.10.4 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

Math summary: This expression defines the mass of the compound taken as being measured in milligrams. It specifies that this value represents the mass of barium.

sulphate formed equals m 1 g

1 mol of BaSO 4 equals 233 g BaSO 4 equals 32 g sulphur m 1 g BaSO 4 contains 32 times m 1 divided by 233 g sulphur

Math summary: This expression calculates the percentage of sulphur in an organic compound. The process multiplies the mass of barium sulphate by thirty-two and one hundred, then divides that result by the product of two hundred thirty-three and the total mass of the compound.

Problem 8.24

In sulphur estimation, 0.157 g of an organic compound gave 0.4813 g of barium sulphate. What is the percentage of sulphur in the compound?

Molecular mass of barium sulfate equals 137 plus 32 plus 64 equals 233 grams

233 g Barium Sulfate contains 32 g sulphur

0.4813 g BaSO 4 contains 32 times 0.4813 divided by 233 g

Percentage of sulphur equals 32 times 0.4813 times 100 divided by 233 times 0.157 equals 42.10 percent

8.10.5 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, ammonium 3 phosphate dot 12 molybdenum oxide 3, by adding ammonia and ammonium molybdate. Alternatively, phosphoric acid may be precipitated as magnesium ammonium phosphate by adding magnesia mixture which on ignition yields magnesium 2 phosphorus 2 oxygen 7.

Let the mass of organic compound taken = m g and mass of ammonium phospho molydate = m 1g

Molar mass of ammonium phosphate twelve molybdenum trioxide equals 1877 grams

Percentage of phosphorus equals 31 times m 1 times 100 divided by 1877 times m percent

If phosphorus is estimated as magnesium 2 phosphorus 2 oxygen 7,

Percentage of phosphorus equals 62 times m 1 times 100 divided by 222 times where, 222 u is the molar mass of Mg 2 P 2 O 7 , m, the mass of organic compound taken, m 1, the mass of Mg 2 P 2 O 7 formed and 62, the mass of two phosphorus atoms present in the compound Mg 2 P 2 O 7.

8.10.6 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 (I_2O 5) when carbon monoxide is oxidised to carbon dioxide producing iodine.

Equation A summary: This process describes the decomposition of an organic compound using heat to produce oxygen and other gaseous products. The oxygen is then converted to carbon monoxide using coke at one thousand three hundred seventy three Kelvin, and subsequently oxidized to carbon dioxide using iodine pentoxide to determine the molar relationship between liberated oxygen and produced carbon dioxide.

On making the amount of C-O produced in equation (A) equal to the amount of C-O 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 g carbon dioxide is obtained if 32 g oxygen is liberated.

Let the mass of organic compound taken be m g

Mass of carbon dioxide produced be m 1 g

Therefore, m 1 grams carbon dioxide is obtained from

32 times m 1 divided by 88 grams of O 2

Math summary: This expression calculates the percentage of oxygen. It divides the product of thirty two, the mass of the sample, and one hundred by the product of eighty eight and the total mass.

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 C,H,N elemental analyser. The analyser requires only a very small amount of the substance (1 to 3 milligrams) 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 sp cubed , sp squared and sp hybridised orbitals. The sp cubed , sp squared and sp 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 sp cubed hybrid orbital can overlap with 1s orbital of hydrogen to give a carbon - hydrogen (C-H) single bond (sigma, sigma bond). Overlap of a sp squared orbital of one carbon with sp squared orbital of another results in the formation of a carbon-carbon sigma bond. The unhybridised p orbitals on two adjacent carbons can undergo lateral (side-by-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 (I.U.P.A.C). In I.U.P.A.C 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

8.1 What are hybridisation states of each carbon atom in the following compounds? C H 2 equals C equals O, C H 3 C H equals C H 2, open parenthesis C H 3 close parenthesis squared C O, C H 2 equals C H C N, C 6 H 6

8.2 Indicate the sigma and pi bonds in the following molecules:

C 6 H 6, C 6 H 12, C H 2 Cl 2, C H 2 equals C equals C H 2, C H 3 N O 2, H C O N H C H 3

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

8.4 Give the I.U.P.A.C names of the following compounds :

Image summary: A chemical structure diagram showing an ethyl group attached to a benzene ring, representing the molecule 2-phenylethane (ethylbenzene). It is a structural formula used to depict the molecular arrangement of this organic compound.

(d)

(b)

(e)

Image summary: A chemical skeletal structure of a molecule featuring a carbon chain with a methyl branch and a nitrile group (CN) at the terminus. This depicts the structure of 2-methylbutanenitrile.

Image summary: A chemical structure diagram showing a four-carbon chain with a chlorine atom at one end and an aldehyde group (a carbonyl carbon double-bonded to oxygen and single-bonded to hydrogen) at the other. It depicts the molecule 4-chlorobutanal.

Image summary: A skeletal chemical structure showing a branched hydrocarbon chain with several methyl groups. The image depicts a specific organic molecule, illustrating its connectivity and branched architecture.

(c)

(f) Cl 2 C.H C.H 2 O.H

8.5 Which of the following represents the correct I.U.P.A.C 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.

8.6 Draw formulas for the first five members of each homologous series beginning with the following compounds. (a) H-C.O.O.H (b) C H 3 C O C H 3 (c) H-C.H=C.H 2

8.7 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

8.8 Identify the functional groups in the following compounds

Figure 8.8 summary: Two chemical structure diagrams. Structure (a) shows a benzene ring substituted with an aldehyde group, a hydroxy group, and a methoxy group. Structure (b) shows a benzene ring substituted with an amino group and an ester group containing a tertiary amine. These diagrams provide the molecular structures of two different organic compounds.

(a)

(b)

(c)

Image summary: A chemical structure diagram showing a benzene ring attached to a nitroethenyl group, specifically a CH=CHNO2 chain. This depicts the molecule trans-beta-nitrostyrene.

8.9 Which of the two: O 2 N C H 2 C H 2 O minus or C H 3 C H 2 O minus is expected to be more stable and why?

8.10 Explain why alkyl groups act as electron donors when attached to a pi system.

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

(a) C 6 H 5 O.H (b) C 6 H 5 N.O 2 (c) C.H 3 C.H equals C.H C.H.O (d) C 6 H 5 dash C.H.O (e) C 6 H 5 dash C.H 2 (f) C.H 3 C.H equals C.H

8.12 What are electrophiles and nucleophiles? Explain with examples.

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

Math summary: This chemical equation describes a reaction between acetic acid and a hydroxide ion. The process shows these two reactants combining to produce an acetate ion and water.

(b)

Math summary: This expression describes two separate chemical reactions. In the first reaction, acetone and cyanide react to form a cyanohydrin, while the second reaction involves benzene and acetyl chloride combining to produce acetophenone.

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

(a) C.H 3 C.H 2 Br plus H.S minus arrow C.H 3 C.H 2 S.H plus Br minus

(b) C H 3 in parentheses squared C equals C H 2 plus H C l arrow C H 3 in parentheses squared C I C dash C H 3

(c) C.H 3 C.H 2 Br plus H.O minus arrow C.H 2 equals C.H 2 plus H 2 O plus Br

open parenthesis C H 3 close parenthesis 3 C dash C H 2 O H plus H B r arrow open parenthesis C H 3 close parenthesis 2 C B r C H 2 C H 2 C H 3 plus H 2 O

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

(a)

(b)

(c)

Figure 8.15 summary: Two skeletal chemical structures of ketones. The top image shows 3-pentanone, with a carbonyl group at the third carbon of a five-carbon chain. The bottom image shows 2-pentanone, with the carbonyl group at the second carbon. The figure illustrates two constitutional isomers of pentanone.

Image summary: Two chemical structures showing the addition of a hydroxyl group to different isomers of deuterated ethene. In the first case, the hydroxyl group attaches to a carbon bonded to deuterium and hydrogen, while in the second case, it attaches to a carbon bonded to two hydrogens. This illustrates the regioselectivity of the reaction based on the arrangement of deuterium and hydrogen atoms.

Image summary: Two chemical structures showing the transition from a neutral molecule to a positively charged ion. The structure on the left shows a carbon atom double-bonded to an OH group and single-bonded to a hydrogen and another OH group; the structure on the right shows the same molecule but with the double bond replaced by a single bond, resulting in a positive charge on the carbon. The image depicts the protonation of a carbonyl group to form a carbocation.

8.16 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.

(a)

Math summary: This expression describes a chemical reaction where a dimethyl peroxide molecule decomposes. The process splits the initial molecule into two separate methoxy fragments.

(b)

Math summary: This expression describes a chemical transformation where a starting component is combined with others to form a final product. The process shows a change from a state involving an initial substance and a specific combination of inputs to a result consisting of that initial substance and water.

(c)

(d)

Image summary: A chemical equation showing a tert-butyl bromide molecule dissociating into a tert-butyl cation and a bromide ion. This depicts the heterolytic cleavage of a carbon-bromine bond to form a stable tertiary carbocation.

Figure 8.16 summary: A chemical reaction diagram showing a benzene ring reacting with an electrophile E+ to form a positively charged arenium ion intermediate. This depicts the first step of an electrophilic aromatic substitution reaction, where the aromaticity of the ring is temporarily broken to create a carbocation.

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

(a) Cl 3 C C O O H is greater than Cl 2 C H C O O H is greater than Cl C H 2 C O O H

(b) C.H 3 C.H 2 C.O.O.H is greater than (C.H 3) 2 C.H C.O.O.H is greater than (C.H 3) 3 C C.O.O.H

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

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

8.19 Describe the method, which can be used to separate two compounds with different solubilities in a solvent S.

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

8.21 Discuss the chemistry of Lassaigne's test.

8.22 Differentiate between the principle of estimation of nitrogen in an organic compound by (1) Dumas method and (2) Kjeldahl's method.

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

8.24 Explain the principle of paper chromatography.

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

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

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

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

8.29 Will CCl 4 give white precipitate of A-G C-L on heating it with silver nitrate? Give reason for your answer.

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

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

8.32 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 g of this substance is subjected to complete combustion.

8.33 A sample of 0.50 g of an organic compound was treated according to Kjeldahl's method. The ammonia evolved was absorbed in 50 ml of 0.5 M H2SO4. The residual acid required 60 mL of 0.5 M solution of N-A-O-H for neutralisation. Find the percentage composition of nitrogen in the compound.

8.34 0.3780 g of an organic chloro compound gave 0.5740 g of silver chloride in Carius estimation. Calculate the percentage of chlorine present in the compound.

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

8.36 In the organic compound C H 2 equals C H minus C H 2 minus C H 2 minus C triple bond C H, the pair of hydridised orbitals involved in the formation of: C 2 minus C 3 bond is: (a) s p minus s p squared (b) s p minus s p cubed (c) s p squared minus s p cubed (d) s p cubed minus s p cubed

8.37 In the Lassaigne's test for nitrogen in an organic compound, the Prussian blue colour is obtained due to the formation of: (a) sodium 4 ferrocyanide (b) iron 4 ferricyanide 3 (c) iron 2 ferricyanide (d) iron 3 ferricyanide 4

8.38 Which of the following carbocation is most stable? (a) (C.H 3) 3C dot C.H 2 (b) (C.H 3) 3C (c) C.H 3C.H 2C.H 2 (d) C.H 3C.H dot C.H 2C.H 3

8.39 The best and latest technique for isolation, purification and separation of organic compounds is: (a) Crystallisation (b) Distillation (c) Sublimation (d) Chromatography

(a) Crystallisation (b) Distillation (c) Sublimation (d) Chromatography 8.40 The reaction: C.H 3 C.H 2 I plus K.O.H aqueous equals C.H 3 C.H 2 O.H plus K.I is classified as : (a) electrophilic substitution (b) nucleophilic substitution (c) elimination (d) addition