Organic Chemistry – Some Basic Principles and Techniques: Class 11 Chemistry · Chapter 8

Chemistry Part 2 · Companion notes, simulations and practice · Official NCERT chapter PDF ↗

Published 22 Oct 2026 · Updated 22 Oct 2026

115 min
46 practice questions
41 MCQs

What you will learn in Organic Chemistry – Some Basic Principles and Techniques

Explain how sp³, sp² and sp hybridisation fix the shapes of carbon compounds, and the special features of π bonds
Draw and interconvert complete, condensed, bond-line and wedge–dash representations of organic molecules
Classify organic compounds by skeleton and by functional group, and recognise homologous series
Name compounds by IUPAC rules, including functional groups and substituted benzenes, and draw structures from names
Identify the types of structural isomerism and stereoisomerism
Describe bond fission, reactive intermediates, nucleophiles and electrophiles, and use curved arrows
Use the inductive, resonance, electromeric and hyperconjugation effects to explain stability and reactivity
Choose purification techniques, describe the tests for elements and calculate percentage composition from analytical data

Original companion text aligned to NCERT Chemistry Class 11, Chapter 8, pp. 256–294. Not affiliated with or endorsed by NCERT.

Press a paracetamol tablet out of its strip and you are handling several carbon compounds at once: the drug, the PVC blister and the lacquer on the foil. No other element builds such long, stable chains and rings out of its own atoms (catenation), and carbon dresses these skeletons with hydrogen, oxygen, nitrogen and a handful of other elements. Millions of compounds result, and their study is organic chemistry.

This chapter is the toolkit for the rest of organic chemistry: how bonds set the shape of a molecule, how to draw and name molecules, how electrons move in reactions, and how chemists purify a compound and find which elements it contains, and how much of each.

Look inside a living cell and almost every working part is an organic molecule: enzymes that speed up reactions, fats that form membranes, DNA that carries instructions. Outside the body, organic chemistry gives us petrol and LPG, nylon and polyester, paints, detergents and most medicines.

8.1 General Introduction

The word is older than the science. Chemists once sorted compounds by origin, from living things or from minerals, and Berzelius held that organic ones needed a vital force found only in living cells. In 1828 Wöhler made urea, a compound of urine, simply by heating ammonium cyanate:

Wöhler's synthesis of urea

\begin{aligned} N H_{4} C N O \\ \overset{heat}{\to} N H_{2} C O N H_{2} \end{aligned}

(8.1)

Others soon went further: Hermann Kolbe built acetic acid from carbon disulphide in 1845, and Marcellin Berthelot made methane in the laboratory in 1856. Today 'organic' simply means a compound of carbon, apart from a few simple ones such as $C O_{2}$ and the carbonates. What lets us predict how these compounds behave is the electron-pair bond, the thread running through this whole chapter.

8.2 Tetravalence of Carbon: Shapes of Organic Compounds

8.2.1 The Shapes of Carbon Compounds

In Unit 4 you met carbon's four valence electrons and the hybrid orbitals it forms. These decide its shape. A carbon with four single bonds is sp³ and tetrahedral, as in methane. A carbon with one double bond is sp² and trigonal planar, as in ethene. A carbon with a triple bond (or two double bonds) is sp and linear, as in ethyne. Fig. 8.1 compares the three.

2/3Three σ bonds in a plane at 120°; one p orbital is left for a π bond.

Fig. 8.1More s character pulls bonding electrons closer to carbon.NCERT p. 257

A carbon atom in three states beside a data card. Step 1, sp³: four bonds (plain, wedge, hashed), 109.5°, C–H 109.4 pm, s-character bar at 25 per cent. Step 2, sp²: three bonds in a dashed plane at 120°, one p orbital above and below, C–H 108.7 pm, bar at 33 per cent. Step 3, sp: two bonds in a line at 180°, two p orbitals, C–H 106.3 pm, bar at 50 per cent.

Look down the C–H row of Table 8.1: the bond shrinks from 109.4 pm to 106.3 pm as carbon goes from sp³ to sp. The reason is s character (25 %, 33 %, 50 %). Electrons in an s orbital spend more time near the nucleus than those in a p orbital, so an s-rich hybrid holds its bonding pair tightly: the bond is shorter and stronger, and the carbon behaves as if it were more electronegative. That is why a strong base can pull a proton off ethyne but not off ethane (Unit 9).

Table 8.1How hybridisation changes carbon's bonds

	sp³	sp²	sp
s character	25 %	33 %	50 %
Shape at carbon	tetrahedral	trigonal planar	linear
Bond angle	109.5°	120°	180°
C–H length (pm)	109.4 (ethane)	108.7 (ethene)	106.3 (ethyne)
C–C length (pm)	154 (C–C)	134 (C=C)	120 (C≡C)

8.2.2 Some Characteristic Features of π Bonds

Lay two pencils side by side and they touch along their whole length; cross them and they meet at one point. A π bond is the side-by-side overlap of two p orbitals, so it forms only when the p orbitals on neighbouring carbons are parallel. In ethene each carbon's spare p orbital points straight out of the molecular plane, and the π electrons occupy two lobes, one above and one below that plane (Fig. 8.2).

2/3Side-by-side overlap puts π electron density above and below the plane.

Fig. 8.2Parallel p orbitals overlap; a 90° twist breaks the π bond.NCERT p. 257

Two carbons joined by a σ bond, each with two hydrogens in a dashed plane. Step 1: an upright p orbital on each carbon, parallel to the other. Step 2: the lobes merge into a π cloud above and below the C–C axis. Step 3: the right CH₂ is turned 90 degrees, its p orbital lies flat, and a cross marks that there is no overlap.

Now try to twist one $C H_{2}$ group about the C=C axis. The p orbitals turn away from each other, the overlap falls to zero at 90°, and the π bond is gone. Twisting therefore costs about the energy of a π bond, so the ends of a double bond stay locked (restricted rotation) and all six atoms of ethene lie in one plane; this locking gives the geometrical isomers of Section 8.6.2. And because π electrons sit away from the C–C axis, less tightly held than σ electrons, electron-seeking reagents reach them easily: multiple bonds are usually where reactions happen.

Example 8.1NCERT Problem 8.1 · p. 257

Counting σ and π bonds

How many σ and π bonds are there in each molecule?

$C H_{2} = C H - C \equiv N$ (acrylonitrile, used for acrylic fibres)
$H C \equiv C - C \equiv C H$
$C H_{3} - C H = C H - C H O$

Try it yourself first. Then reveal the solution one step at a time.

Example 8.2NCERT Problems 8.2, 8.3 · p. 257

Hybridisation and shape

Give the hybridisation of each carbon, left to right, in (i) $C H_{2} = C H - C H_{2} O H$ , (ii) $H C \equiv C - C H = C H_{2}$ and (iii) $C H_{3} C O O H$ . Then give the hybridisation and shape at carbon in (iv) $C O C l_{2}$ and (v) $C S_{2}$ .

Try it yourself first. Then reveal the solution one step at a time.

8.3 Structural Representations of Organic Compounds

8.3.1 Complete, Condensed and Bond-line Structural Formulas

A molecular formula such as $C_{4} H_{10} O$ does not say how the atoms are joined, and several compounds share it, so chemists draw structures. Replace each shared pair of a Lewis structure by a dash and you have the complete structural formula; lone pairs on O, N, S or halogens may be shown or left out.

Condensed formula. Write each carbon with its hydrogens as one group and drop most dashes: butan-2-ol becomes $C H_{3} C H (O H) C H_{2} C H_{3}$ . Brackets mark a branch, and a repeated unit takes a subscript, so hexane is $C H_{3} (C H_{2})_{4} C H_{3}$ .

Bond-line formula. Draw only the carbon skeleton as a zig-zag. Every corner and line end is a carbon with enough hydrogens to make four bonds; atoms other than C and H are written, with any H on them. A bare line end is a $C H_{3}$ group, and rings become polygons: cyclohexanol is a hexagon with OH on one corner. Fig. 8.3 shows all three styles for butan-2-ol.

3/3Only the skeleton: corners and ends are carbons; OH is written.

Fig. 8.3One molecule, three levels of detail.NCERT p. 258

Butan-2-ol three ways. Step 1: the complete formula, four carbons in a row with every bond drawn as a dash and OH above the second carbon. Step 2: the condensed formula CH₃CH(OH)CH₂CH₃. Step 3: the bond-line formula, a three-line zig-zag with OH at the second corner and small labels naming the group at each corner and end.

Example 8.4NCERT Problems 8.4, 8.5 · p. 259

Complete, condensed and bond-line formulas

(i) Write the complete structural formula of propanal, $C H_{3} C H_{2} C H O$ . (ii) Write a compact condensed formula for $C H_{3} C H_{2} C H_{2} C H (C H_{3}) C H_{2} O H$ and describe its bond-line formula.

Try it yourself first. Then reveal the solution one step at a time.

Example 8.6NCERT Problem 8.6 · p. 260

Reading bond-line formulas

Write the condensed and molecular formula of each compound in Fig. 8.4.

Fig. 8.4Bond-line formulas for Example 8.6.NCERT p. 260

Four skeletal drawings. (i) A pentagon with one short line from a corner. (ii) A zig-zag of five lines with a double line at the left end. (iii) A zig-zag of three lines with a double bond to O at the second corner. (iv) A zig-zag with CN written at the right end and a short branch line on the second corner from the left.

Try it yourself first. Then reveal the solution one step at a time.

8.3.2 Three-Dimensional Representation of Organic Molecules

Paper is flat; molecules are not. The wedge–dash convention brings back the third dimension. A plain line lies in the plane of the page, a solid wedge comes out of the page towards you (broad end nearest you), and a hashed wedge goes behind the page, away from you. Step 1 of Fig. 8.5 draws bromochlorofluoromethane, $C H B r C l F$ , in this style.

3/4Atoms as balls, bonds as sticks: angles are easy to see.

Fig. 8.5Each model shows a different feature of the same molecule.NCERT Figs. 8.1, 8.2 · p. 260

CHBrClF shown four ways. Step 1: wedge–dash drawing, H and Br in the page, Cl on a solid wedge towards you, F on a hashed wedge away. Step 2: framework model, four rods only. Step 3: ball-and-stick model with labelled balls of different sizes. Step 4: space-filling model of overlapping spheres scaled to van der Waals radii, H smallest and Br largest.

Which model should you pick? To see which atom is joined to which, sticks alone will do: a framework model. To read bond angles, add balls for the atoms: a ball-and-stick model (springs stand in for double bonds). To judge whether two molecules can fit together, as a drug must fit its target, you need atoms at their true relative sizes, overlapping as in the real molecule: a space-filling model built from van der Waals radii. Free molecule-viewing software draws all three and lets you rotate them.

Quick check

1 of 4

In a bond-line formula, a corner where three lines meet stands for

a CH groupa

C H_{2}

groupa carbon with no Ha

C H_{3}

group

Show answer

Answer (A) a CH group

Three bonds are drawn, so one H completes the four. Two lines mean

C H_{2}

; a line end is

C H_{3}

8.4 Classification of Organic Compounds

With millions of known compounds, chemists sort them twice: once by the shape of the carbon skeleton and once by the functional group. Fig. 8.6 builds the skeleton tree.

3/3Aromatic: benzenoid, non-benzenoid, or with a heteroatom in the ring.

Fig. 8.6Sorting by skeleton: chains, ordinary rings and aromatic rings.NCERT p. 261

A classification tree in three steps. Step 1: organic compounds split into acyclic (example propanone) and cyclic (example cyclobutane). Step 2: alicyclic rings split into homocyclic (cyclopentene) and heterocyclic (oxane, one ring oxygen). Step 3: aromatic rings split into benzenoid (toluene), non-benzenoid (azulene, fused five- and seven-membered rings) and heterocyclic (pyrrole).

Acyclic (open-chain) compounds, also called aliphatic compounds, have straight or branched chains, like the butane in a gas lighter, the ethanol in a sanitiser and the citric acid in lemon juice.

Cyclic (closed-chain) compounds contain at least one ring. Alicyclic rings behave much like aliphatic chains. Most are homocyclic (carbocyclic), with only carbon in the ring, such as cyclobutane, methylcyclopentane and cyclooctane. Some are heterocyclic, with O, N or S in the ring: oxane (one O in a six-membered ring) and piperidine (one N) are non-aromatic heterocycles.

Aromatic compounds have a special stability (Unit 9). Benzenoid ones contain benzene rings, like toluene, phenol and anthracene (three fused rings). Azulene, a deep-blue hydrocarbon with fused five- and seven-membered rings, is aromatic without a benzene ring (non-benzenoid). With a heteroatom in the aromatic ring the compound is heterocyclic aromatic: pyrrole has an NH in a five-membered ring, and pyrimidine, parent of the DNA bases cytosine and thymine, has two N in a six-membered ring.

The second sorting uses functional groups, which divides compounds into families called homologous series.

8.4.1 Functional Group

Definition

Functional group

An atom or group of atoms, bonded to the carbon skeleton, that gives a compound its characteristic chemical reactions.

Example–OH (hydroxyl), >C=O (carbonyl), –COOH (carboxyl), $- N H_{2}$ (amino).

Because the functional group, not the rest of the skeleton, does most of the chemistry, compounds that share one react alike. Methanoic, ethanoic and propanoic acids all contain –COOH, and all three fizz with baking soda (sodium hydrogencarbonate), giving off carbon dioxide.

8.4.2 Homologous Series

Definition

Homologous series

A family of compounds with the same functional group and one general formula, in which each member differs from the next by a $C H_{2}$ unit (14 u).

ExampleAlkanols, $C_{n} H_{2 n + 1} O H$ : $C H_{3} O H$ , $C_{2} H_{5} O H$ , $C_{3} H_{7} O H$ , …

The members (homologues) share chemical properties, while physical properties change steadily with size: methanol, ethanol, propan-1-ol and butan-1-ol boil at 65, 78, 97 and 118 °C. Besides the alkanols, you will meet the amines, the alkanoic acids, the carbonyl series (alkanals and alkanones), the haloalkanes, and the hydrocarbon series: alkanes, alkenes and alkynes.

A molecule may carry two or more functional groups, alike or different; it is then polyfunctional. Lactic acid in curd, $C H_{3} C H (O H) C O O H$ , has a hydroxyl and a carboxyl group; citric acid in lemons has one hydroxyl and three carboxyl groups.

8.5 Nomenclature of Organic Compounds

A name is useful only if it points to exactly one structure and the structure can be rebuilt from it. The rules of IUPAC (the International Union of Pure and Applied Chemistry) do exactly that, so a chemist in Chennai and one in Chile draw the same structure for the same name.

Older common (trivial) names usually record where a compound was found: lactic acid in sour milk (Latin lac), malic acid in apples (malum), butyric acid in rancid butter (butyrum). Cubane, $C_{8} H_{8}$ , is named after its cube-shaped skeleton. Common names survive because they are short.

Table 8.2Common and IUPAC names of some compoundsNCERT Table 8.1 · p. 262

Formula	Common name	IUPAC name
$H C \equiv C H$	acetylene	ethyne
$(C H_{3})_{2} C H O H$	isopropyl alcohol	propan-2-ol
$C H_{3} C H O$	acetaldehyde	ethanal
$C H_{3} C O C H_{2} C H_{3}$	methyl ethyl ketone	butan-2-one
$C H_{3} C H_{2} O C H_{2} C H_{3}$	diethyl ether	ethoxyethane
$H O C H_{2} C H (O H) C H_{2} O H$	glycerol	propane-1,2,3-triol
$C H_{3} C O O C H_{2} C H_{3}$	ethyl acetate	ethyl ethanoate
$C_{6} H_{5} C H_{3}$	toluene	methylbenzene

8.5.1 The IUPAC System of Nomenclature

An IUPAC name is built from a word root for the number of carbons in the parent chain, a primary suffix for saturation (-ane, -ene, -yne), a secondary suffix for the main functional group, and prefixes, each with a locant, for everything else. In 5-chloro-4-methylhexan-2-one, 'hex' is the six-carbon chain, '-an-' says every C–C bond is single, '-2-one' puts a ketone C=O on C2, and the prefixes put Cl on C5 and $C H_{3}$ on C4: $C H_{3} C O C H_{2} C H (C H_{3}) C H C l C H_{3}$ .

Compounds of carbon and hydrogen only are hydrocarbons: saturated if every C–C bond is single (the alkanes; their old name, paraffins, comes from the Latin parum affinis, 'lacking affinity', because they react so little) and unsaturated if they contain C=C or C≡C.

8.5.2 IUPAC Nomenclature of Alkanes

Unbranched alkanes are named by a root for the chain length plus -ane. The first four roots are historical; from five carbons on they come from Greek numbers. Removing one hydrogen from an alkane gives an alkyl group, named by changing -ane to -yl and written R in general formulas. Alkanes form a homologous series with the general formula $C_{n} H_{2 n + 2}$ .

Table 8.3Roots, unbranched alkanes and alkyl groupsNCERT Tables 8.2, 8.3 · p. 263

C atoms	Root	Alkane	Formula	Alkyl group
1	meth	methane	$C H_{4}$	methyl (Me)
2	eth	ethane	$C_{2} H_{6}$	ethyl (Et)
3	prop	propane	$C_{3} H_{8}$	propyl (Pr)
4	but	butane	$C_{4} H_{10}$	butyl (Bu)
5	pent	pentane	$C_{5} H_{12}$	pentyl
6	hex	hexane	$C_{6} H_{14}$	hexyl
7	hept	heptane	$C_{7} H_{16}$	heptyl
8	oct	octane	$C_{8} H_{18}$	octyl
9	non	nonane	$C_{9} H_{20}$	nonyl
10	dec	decane	$C_{10} H_{22}$	decyl
20	icos	icosane	$C_{20} H_{42}$	icosyl

Branched alkyl groups. Three- and four-carbon groups can be attached through different carbons. IUPAC allows their common names; the systematic names number the group from the carbon attached to the main chain.

Table 8.4Branched alkyl groupsNCERT p. 263

Group	Structure	Systematic name	Attached C
isopropyl	$(C H_{3})_{2} C H^{-}$	1-methylethyl	2°
sec-butyl	$C H_{3} C H_{2} C H (C H_{3})^{-}$	1-methylpropyl	2°
isobutyl	$(C H_{3})_{2} C H C H_{2}^{-}$	2-methylpropyl	1°
tert-butyl	$(C H_{3})_{3} C^{-}$	1,1-dimethylethyl	3°
neopentyl	$(C H_{3})_{3} C C H_{2}^{-}$	2,2-dimethylpropyl	1°

Naming a branched alkane. Apply these rules in order:

Find the longest continuous chain of carbons, however the drawing bends; it is the parent. If two chains are equally long, take the one with more substituents.
Number the chain from the end that gives the substituents the lowest locants. Compare locant sets term by term; the lower set at the first difference wins, so {2, 3, 5} beats {2, 4, 4}.
Write each substituent with its locant, in alphabetical order, before the parent name. Hyphens separate numbers from words, commas separate numbers, and the last prefix joins the parent with no space.
Combine identical substituents with di-, tri-, tetra- and give every locant (2,2-dimethyl). These multiplying prefixes are ignored when alphabetising.
If both ends give the same locant set, the lower number goes to the substituent that comes first alphabetically.
Number a branched substituent from its point of attachment (C1) and put its name in brackets: (2,2-dimethylpropyl). Its full name, including di- or tri-, is alphabetised under its first letter, here d.
Of the common prefixes, iso- and neo- count in alphabetising (isopropyl under i), but sec- and tert- do not (tert-butyl under b).

Fig. 8.7 applies the rules to a molecule whose drawing hides two different seven-carbon chains.

3/4Numbering from the left gives {2, 3, 5}, lower than {3, 5, 6}.

Fig. 8.7Longest chain, most substituents, lowest locants, alphabetical order.NCERT p. 264

A condensed structure: a seven-carbon horizontal chain with methyl on C2, ethyl on C3 and methyl on C5. Step 1 highlights two seven-carbon chains, one running up through the ethyl group. Step 2 keeps the horizontal chain, which has three branches. Step 3 numbers it from the left, locants 2, 3, 5, with the right-hand set 3, 5, 6 struck out. Step 4 shows the name 3-ethyl-2,5-dimethylheptane.

Cycloalkanes are named with cyclo before the alkane of the same carbon count: cyclopentane. One substituent needs no locant (methylcyclopentane). With more, use the lowest locant set, then give C1 to the group first in alphabetical order: 1-ethyl-2-methylcyclohexane; 1,1,3-trimethylcyclohexane, not 1,3,3-.

Example 8.7NCERT Problem 8.7 · p. 265

What is wrong with these names?

Each compound has been given the name in brackets. Explain the error and give the correct name.

$C_{2} H_{5} C (C H_{3})_{2} C H (C H_{3}) C_{2} H_{5}$ (3,4,4-trimethylhexane)
$(C_{2} H_{5})_{2} C H C H_{2} C (C H_{3})_{3}$ (2,2-dimethyl-4-ethylhexane)
$C H_{3} C H (C_{2} H_{5}) C H_{2} C H_{2} C H_{3}$ (2-ethylpentane)

Try it yourself first. Then reveal the solution one step at a time.

8.5.3 Nomenclature of Organic Compounds having Functional Group(s)

With a functional group present, that group takes charge of the name. Identify it first, because it fixes the suffix. Then choose the longest chain that contains it, and number that chain so that the functional group gets the lowest possible locant, even if a branch then gets a higher one.

Table 8.5Functional groups, prefixes and suffixesNCERT Table 8.4 · p. 267

Class	Group	Prefix	Suffix	Example
Alkene	>C=C<	–	-ene	propene, $C H_{2} = C H C H_{3}$
Alkyne	–C≡C–	–	-yne	propyne, $C H \equiv C C H_{3}$
Halide	–X	halo-	–	2-chloropropane, $C H_{3} C H C l C H_{3}$
Alcohol	–OH	hydroxy-	-ol	propan-1-ol, $C H_{3} C H_{2} C H_{2} O H$
Aldehyde	–CHO	formyl-, oxo-	-al	propanal, $C H_{3} C H_{2} C H O$
Ketone	>C=O	oxo-	-one	pentan-3-one, $(C_{2} H_{5})_{2} C O$
Carboxylic acid	–COOH	carboxy-	-oic acid	propanoic acid, $C_{2} H_{5} C O O H$
Carboxylate ion	$- C O O^{-}$	–	-oate	sodium ethanoate, $C H_{3} C O O^{-} N a^{+}$
Ester	–COOR	alkoxycarbonyl-	-oate	methyl ethanoate, $C H_{3} C O O C H_{3}$
Acyl halide	–COX	halocarbonyl-	-oyl halide	ethanoyl chloride, $C H_{3} C O C l$
Amide	$- C O N H_{2}$	carbamoyl-	-amide	propanamide, $C_{2} H_{5} C O N H_{2}$
Nitrile	–C≡N	cyano-	-nitrile	propanenitrile, $C_{2} H_{5} C N$
Amine	$- N H_{2}$	amino-	-amine	propan-2-amine, $(C H_{3})_{2} C H N H_{2}$
Ether	–OR	alkoxy-	–	methoxyethane, $C H_{3} O C_{2} H_{5}$
Nitro compound	$- N O_{2}$	nitro-	–	2-nitropropane, $(C H_{3})_{2} C H N O_{2}$
Sulphonic acid	$- S O_{3} H$	sulpho-	-sulphonic acid	benzenesulphonic acid, $C_{6} H_{5} S O_{3} H$

Choosing the principal group. When there are several functional groups, the one highest in this order becomes the suffix and the others become prefixes: –COOH, $- S O_{3} H$ , –COOR, –COCl, $- C O N H_{2}$ , –CN, –CHO, >C=O, –OH, $- N H_{2}$ , then C=C and C≡C. Some groups are always prefixes: alkyl (–R), phenyl ( $- C_{6} H_{5}$ ), halo (–F, –Cl, –Br, –I), nitro ( $- N O_{2}$ ) and alkoxy (–OR).

So $H_{2} N C H_{2} C H_{2} O H$ (ethanolamine) is 2-aminoethan-1-ol: the alcohol outranks the amine and takes the suffix. In $C H_{3} C H (N O_{2}) C H = C H_{2}$ the nitro group can only be a prefix, so the double bond sets the numbering: 3-nitrobut-1-ene.

Several identical groups take di-, tri- and so on; the final 'e' of the alkane stays before a consonant (propane-1,2,3-triol, butane-2,3-dione, the buttery smell of popcorn) and is dropped before a vowel (propan-1-ol). With two or more multiple bonds an 'a' is added: $C H_{2} = C H C H_{2} C H = C H_{2}$ is penta-1,4-diene.

Example 8.8NCERT Problem 8.8 · p. 266

Naming compounds with functional groups

Give the IUPAC name of each compound.

$(C H_{3})_{2} C H C H_{2} C H (O H) C H_{3}$
$O H C - C H_{2} C H_{2} - C O C H_{3}$
$C H_{3} C H C l C H_{2} C O O H$
$C H_{2} = C H C H_{2} C H_{2} C \equiv C H$

Try it yourself first. Then reveal the solution one step at a time.

Example 8.9NCERT Problem 8.9 · p. 268

Structures from names

Write the structure of (i) 3-bromo-2-methylpentane, (ii) 4-methylpent-3-en-2-one, (iii) 2-aminopropanoic acid and (iv) 2-chlorocyclopentan-1-ol.

Try it yourself first. Then reveal the solution one step at a time.

8.5.4 Nomenclature of Substituted Benzene Compounds

One substituent is named as a prefix to benzene: chlorobenzene $C_{6} H_{5} C l$ , ethylbenzene $C_{6} H_{5} C_{2} H_{5}$ , iodobenzene $C_{6} H_{5} I$ . Several common names are accepted by IUPAC and used as parents: phenol ( $C_{6} H_{5} O H$ ), toluene ( $C_{6} H_{5} C H_{3}$ ), benzaldehyde ( $C_{6} H_{5} C H O$ ) and benzoic acid ( $C_{6} H_{5} C O O H$ ).

Two substituents. Number the ring so that they get the lowest locants: 1,2, 1,3 or 1,4. Common names use the prefixes ortho (o-), meta (m-) and para (p-) for these three patterns, as Fig. 8.8 shows for the dimethylbenzenes (xylenes).

Fig. 8.8Locants 1,2, 1,3 and 1,4 correspond to ortho, meta and para.NCERT p. 269

Three benzene rings side by side, each with a methyl group at the top carbon, numbered 1. In the first ring the second methyl is on the next carbon, 2, labelled 1,2-dimethylbenzene, ortho-xylene. In the second it is on carbon 3, labelled 1,3, meta. In the third it is directly opposite on carbon 4, labelled 1,4, para.

Three or more substituents. o, m and p no longer work. Use the lowest set of locants and list the prefixes alphabetically; if the parent has a common name, its own group is C1. Examples: 1,2,4-trichlorobenzene (not 1,3,4- or 1,2,5-), 4-chloro-2-nitrophenol (OH on C1) and 2,4,6-trinitrotoluene, the explosive TNT ( $C H_{3}$ on C1).

When a benzene ring is attached to a chain that carries the functional group, the ring is named as the substituent phenyl, $C_{6} H_{5}^{-}$ (Ph). 2-Phenylethanol, $C_{6} H_{5} C H_{2} C H_{2} O H$ , gives roses much of their scent.

Example 8.10NCERT Problem 8.10 · p. 270

Structures of benzene derivatives

Write the structures of (i) m-chlorotoluene, (ii) p-aminophenol, (iii) 1-phenylpropan-2-one and (iv) 1,3-dichloro-5-nitrobenzene.

Try it yourself first. Then reveal the solution one step at a time.

Quick check

1 of 4

The IUPAC name of

C H_{3} C H (O H) C H_{2} C H O

4-oxobutan-2-ol3-hydroxybutanal2-hydroxybutan-4-al1-oxobutan-3-ol

Show answer

Answer (B) 3-hydroxybutanal

–CHO outranks –OH, so it is the suffix (-al) and its carbon is C1; OH becomes 3-hydroxy. Naming it as an alcohol ignores the priority order.

8.6 Isomerism

Ethanol, the alcohol in hand sanitisers, and methoxymethane, a gas used as an aerosol propellant, share the molecular formula $C_{2} H_{6} O$ . Yet ethanol is a liquid that boils at 78 °C, while methoxymethane boils at −24 °C.

Definition

Isomerism

The existence of two or more compounds with the same molecular formula but different structures, and therefore different properties. Such compounds are isomers.

Isomers come in two broad kinds. Structural isomers differ in which atoms are bonded to which; stereoisomers have the same bonds but a different arrangement of atoms in space.

Structural isomerism
- chain
- position
- functional group
- metamerism
Stereoisomerism
- geometrical
- optical

8.6.1 Structural Isomerism

Fig. 8.9 finds all four kinds of structural isomerism inside the single formula $C_{4} H_{10} O$ .

3/4Same formula, different functional group: alcohol vs ether.

Fig. 8.9One formula, C₄H₁₀O, four kinds of structural isomer.NCERT p. 270

Butan-1-ol on the left is compared with another C₄H₁₀O compound on the right. Step 1, chain isomer: 2-methylpropan-1-ol. Step 2, position isomer: butan-2-ol. Step 3, functional group isomer: ethoxyethane, an ether. Step 4, metamers: ethoxyethane on the left and 2-methoxypropane on the right, ethers with different groups on each side of oxygen.

(i) Chain isomers have different skeletons: $C_{6} H_{14}$ has five (hexane, 2-methylpentane, 3-methylpentane, 2,2-dimethylbutane, 2,3-dimethylbutane). (ii) Position isomers differ in where a group sits: 1-bromopropane and 2-bromopropane. (iii) Functional group isomers have different groups: ethanol and methoxymethane, or propanoic acid and methyl ethanoate. (iv) Metamers are in the same class but have different alkyl groups on either side of the functional group: 1-methoxybutane and ethoxypropane, both $C_{5} H_{12} O$ .

8.6.2 Stereoisomerism

Stereoisomers have the same atoms joined in the same order but arranged differently in space. Restricted rotation about C=C (Section 8.2.2) gives geometrical isomers: cis-but-2-ene has both $C H_{3}$ groups on the same side of the double bond, trans-but-2-ene on opposite sides. Optical isomers are non-superimposable mirror images, like left and right hands (Unit 9 and Class 12).

8.7 Fundamental Concepts in Organic Reaction Mechanism

In most organic reactions an organic molecule, the substrate, meets an attacking reagent. Bonds break and form, often through short-lived intermediates, and the products appear, sometimes with by-products.

An overall equation says where a reaction starts and ends. A reaction mechanism tells the story in between: the order in which bonds break and form, the path of every electron pair, and the energy and speed of each step. Once you know a few mechanisms, thousands of reactions stop being separate facts to memorise.

8.7.1 Fission of a Covalent Bond

A covalent bond is a shared electron pair, and it can break in two ways (Fig. 8.10). In heterolytic cleavage (heterolysis) one fragment keeps both electrons, so one becomes a cation and the other an anion. In homolytic cleavage (homolysis) each fragment keeps one electron.

2/4X takes the pair: a planar, sp² carbocation with an empty p orbital.

Fig. 8.10Who keeps the electrons decides the intermediate.NCERT Fig. 8.3 · p. 271

A carbon with three R groups bonded to X. Step 1: the bond with its shared pair drawn as two dots. Step 2: an arrow takes the pair to X, leaving a planar carbocation with an empty p orbital and X minus. Step 3: carbon keeps the pair as a lone pair, a pyramidal carbanion, beside Z plus. Step 4: two fish-hook arrows give one electron to each side: two free radicals.

Carbocations. If the departing group takes the pair, carbon is left with only six valence electrons and a positive charge: a carbocation (older name: carbonium ion). Heterolysis of 2-bromo-2-methylpropane, for example, gives the tert-butyl cation:

\begin{aligned} (C H_{3})_{3} C - B r \\ \overset{}{\to} (C H_{3})_{3} C^{+} \\ + B r^{-} \end{aligned}

(8.2)

With six valence electrons, the positive carbon is sp² hybridised: its three bonds spread flat at 120°, and the leftover p orbital, at right angles to that plane, is empty. How long a carbocation survives depends on what surrounds the charge. Count the carbons bonded to C⁺: one makes it primary ( $C H_{3} C H_{2} C H_{2}^{+}$ ), two secondary ( $C H_{3} C H^{+} C H_{2} C H_{3}$ ), three tertiary ( $(C H_{3})_{2} C^{+} C H_{2} C H_{3}$ ). Each alkyl group feeds electron density towards the charge, so 3° > 2° > 1° > $C H_{3}^{+}$ in stability; Section 8.7.5 and Section 8.7.9 explain how.

Carbanions. If carbon keeps the pair, it carries a negative charge and a lone pair: a carbanion. A simple carbanion such as $C H_{3}^{-}$ is sp³ and trigonal pyramidal. Alkyl groups push electrons onto the charge and destabilise it, so the order reverses: 3° < 2° < 1° < methyl. Reactions through heterolysis are called ionic, heteropolar or polar reactions.

Free radicals. Sunlight falling on chlorine splits some $C l_{2}$ molecules evenly, one electron to each atom ((8.3)). Bond-breaking of this kind, driven by heat or light and drawn with half-headed 'fish-hook' arrows, gives neutral fragments with an unpaired electron: free radicals. Carbon radicals such as $C H_{3}^{∙}$ are short-lived; like carbocations, they are more stable the more alkyl groups they carry (3° > 2° > 1° > methyl). Reactions that run through radicals are called free-radical, homopolar or nonpolar reactions.

C l_{2} \overset{light}{\to} 2 C l^{∙}

(8.3)

8.7.2 Substrate and Reagent

Who is the substrate and who the reagent? The usual rule: the substrate is the molecule whose carbon ends up in the new bond, and the other partner is the reagent. In (8.4) and (8.5), propene and bromoethane are the substrates. When a new C–C bond forms, both partners give carbon, and we call the molecule we are following the substrate. The partners are usually neutral molecules, not free ions.

\begin{aligned} C H_{2} = C H C H_{3} \\ + H B r \\ \overset{}{\to} C H_{3} C H B r C H_{3} \end{aligned}

(8.4)

\begin{aligned} C H_{3} C H_{2} B r \\ + C N^{-} \\ \overset{}{\to} C H_{3} C H_{2} C N \\ + B r^{-} \end{aligned}

(8.5)

Nucleophiles and electrophiles. Opposites attract here too. A reagent with electrons to spare (a lone pair or a negative charge) heads for the electron-poor part of the substrate: the positive end of a polar bond, or an atom short of an octet. A reagent short of electrons heads the other way, towards lone pairs or the exposed π electrons of a multiple bond.

Definition

Nucleophile (Nu:)

A reagent that brings an electron pair to the reaction site ('nucleus-loving'). The reaction is then called nucleophilic.

Example $H O^{-}$ , $C N^{-}$ , $C H_{3} O^{-}$ , carbanions; neutral $H_{2} O$ , $N H_{3}$ , $R N H_{2}$ through their lone pairs.

Definition

Electrophile (E⁺)

A reagent that takes an electron pair from the reaction site ('electron-loving'). The reaction is then called electrophilic.

Example $H^{+}$ , $N O_{2}^{+}$ , carbocations, $B F_{3}$ and $A l C l_{3}$ (incomplete octets); the carbon of C=O or of C–X.

Electrophiles need not carry a charge. Boron in $B F_{3}$ is two electrons short of an octet, and the carbon of a C=O or C–Cl bond is left partly positive by its electronegative partner; both accept electron pairs. Their meeting with a nucleophile is drawn as a curved arrow from the nucleophile's pair to the electrophilic centre.

Example 8.11NCERT Problem 8.11 · p. 273

Heterolysis with curved arrows

Show the heterolytic cleavage of the marked bond and name the intermediate formed: (i) $(C H_{3})_{3} C - C l$ , (ii) $C H_{3} - O C H_{3}$ , (iii) $C H_{3} C H_{2} - L i$ .

Try it yourself first. Then reveal the solution one step at a time.

Example 8.12NCERT Problems 8.12, 8.13 · p. 273

Nucleophiles, electrophiles and electrophilic centres

(i) Classify $N H_{3}$ , $A l C l_{3}$ , $C N^{-}$ , $N O_{2}^{+}$ , $C H_{3} O^{-}$ , $(C H_{3})_{3} C^{+}$ and $H_{2} O$ as nucleophiles or electrophiles. (ii) Identify the electrophilic centre in $C H_{3} C O C H_{3}$ , $C H_{3} C H_{2} C l$ and $C H_{3} C O C l$ .

Try it yourself first. Then reveal the solution one step at a time.

8.7.3 Electron Movement in Organic Reactions

Curved arrows are the shorthand of mechanisms. A full arrow moves an electron pair: its tail starts where the pair is now (a lone pair or a bond) and its head points to where the pair ends up (an atom, or between two atoms where a new bond forms). Fig. 8.11 shows the three moves you will draw most often.

Fig. 8.11Arrows start at electrons and end where the electrons go.NCERT p. 274

Three rows. Row 1: in CH₂=CH–CH₂⁺ an arrow moves the π pair from C1=C2 into the C2–C3 position, giving ⁺CH₂–CH=CH₂. Row 2: in H₂C=O an arrow moves the π pair onto oxygen, giving H₂C⁺–O⁻. Row 3: in H₂N–CH=CH₂ one arrow moves the nitrogen lone pair into the N–C bond position while a second moves the C=C π pair onto the end carbon, giving H₂N⁺=CH–CH₂⁻.

A half-headed fish-hook arrow moves a single electron, as in homolysis. In the substitution $H O^{-}$ + $C H_{3} C H_{2} I$ → $C H_{3} C H_{2} O H$ + $I^{-}$ , one arrow runs from a lone pair on oxygen to the carbon and a second from the C–I bond to iodine. In $C l_{2}$ → $2 C l^{∙}$ , two fish-hooks start at the middle of the Cl–Cl bond, one pointing to each atom.

8.7.4 Electron Displacement Effects in Covalent Bonds

A permanent displacement of bonding electrons exists in the molecule all the time: the inductive and resonance effects (and hyperconjugation). A temporary one appears only while a reagent attacks and vanishes when it leaves: the electromeric (polarisability) effect.

Compare

Permanent and temporary displacements

Permanent

Present in the molecule all the time
Inductive effect: along σ bonds
Resonance effect: through π systems
Show up in dipole moments and acid strength

Temporary

Only while a reagent attacks
Electromeric effect: a full shift of a π pair
Vanishes when the reagent is removed

8.7.5 Inductive Effect

Think of a tug-of-war passed down a line. Fluorine in 1-fluoropropane pulls the C–F electron pair towards itself (δ− on F, δ+ on C1). C1, now slightly electron-poor, tugs on the C1–C2 bond, leaving C2 a smaller positive charge (δδ+), and C3 less still (δδδ+). This step-by-step polarisation of σ bonds is the inductive effect. Each relay passes on only part of the pull, so beyond about three bonds it is negligible.

\overset{δ δ δ +}{C H_{3}} \to \overset{δ δ +}{C H_{2}} \to \overset{δ +}{C H_{2}} \to \overset{δ -}{F}

(8.6)

Groups are compared with hydrogen. Groups that draw electron density away from carbon show the −I effect: $- N O_{2}$ , –CN, –COOH, –COOR, the halogens (F > Cl > Br > I), –OR and $- O C_{6} H_{5}$ . Alkyl groups push electron density towards the carbon they are attached to and show the +I effect, which grows with branching: $- C H_{3}$ < $- C_{2} H_{5}$ < $- C H (C H_{3})_{2}$ < $- C (C H_{3})_{3}$ .

The effect shows clearly in acid strength. A −I group near –COOH spreads out the negative charge of the carboxylate ion and stabilises it, so the acid ionises more and its pKa is lower. Moving the chlorine away from –COOH weakens the effect almost to nothing (Table 8.6).

Table 8.6Chlorine's −I effect fades with distance (pKa at 25 °C)

Acid	Formula	pKa
butanoic acid	$C H_{3} C H_{2} C H_{2} C O O H$	4.8
4-chlorobutanoic acid	$C l C H_{2} C H_{2} C H_{2} C O O H$	4.5
3-chlorobutanoic acid	$C H_{3} C H C l C H_{2} C O O H$	4.1
2-chlorobutanoic acid	$C H_{3} C H_{2} C H C l C O O H$	2.9

Example 8.14NCERT Problems 8.14, 8.15 · p. 275

Bond polarity and the reach of induction

(i) Which bond is more polar: $H_{3} C - C l$ or $H_{3} C - I$ ; $H_{3} C - F$ or $H_{3} C - O H$ ; $H_{3} C - B r$ or $H_{3} C - L i$ ? (ii) In 1-nitropropane, $C H_{3} C H_{2} C H_{2} N O_{2}$ , which carbon carries the smallest induced positive charge?

Try it yourself first. Then reveal the solution one step at a time.

8.7.6 Resonance Structure

Some molecules cannot be described by any single Lewis structure. In the formate ion, $H C O O^{-}$ , a Lewis structure shows one C=O and one C–O⁻ bond, which should differ in length (about 120 pm and 134 pm, as in formic acid). Experiment finds the two C–O bonds identical, at about 125 pm. Two equally good Lewis structures can be drawn, and the real ion is a blend of both (Fig. 8.12).

3/3Real ion: both C–O bonds 125 pm, each O carries −½.

Fig. 8.12Two equivalent drawings; one real ion with equal bonds.NCERT p. 275

The formate ion. Step 1, structure I: C=O up and C–O minus down, with arrows moving the lower oxygen's lone pair into the bond and the C=O π pair onto the upper oxygen. Step 2, structure II: the double bond and the charge have swapped. Step 3, the hybrid: each C–O bond drawn solid plus dashed, each oxygen minus one half, both bonds 125 pm.

These drawings are resonance structures (canonical or contributing structures). They are imaginary: the ion does not flip between them. The real species, the resonance hybrid, has one structure that no single drawing captures, and contributors are joined by a double-headed arrow (↔).

Benzene is the classic case: its six C–C bonds are equal, and it is far more stable than a Kekulé structure suggests. Hydrogenating cyclohexene releases about 120 kJ/mol, so three such double bonds should release about 360 kJ/mol, but benzene releases only about 208 kJ/mol. The difference, about 150 kJ/mol, is its resonance energy: the energy by which the hybrid lies below the most stable contributor. It is largest when there are several important contributors of similar energy.

Writing resonance structures. Move only electrons (lone pairs and π electrons), never nuclei, and keep the number of unpaired electrons the same. Valid structures are not equally important. A structure counts for more when:

every atom has an octet (H a duplet);
it has the fewest separated charges;
any negative charge sits on the more electronegative atom, any positive charge on the more electropositive one;
it has the larger number of covalent bonds;
charge is spread over several atoms rather than piled on one.

Example 8.16NCERT Problems 8.16, 8.17 · p. 276

Drawing and ranking resonance structures

(i) Write the resonance structures of the allyl anion, $C H_{2} = C H - C H_{2}^{-}$ , with curved arrows. (ii) Write two structures for the enolate ion of propanone, $C H_{2} = C (O^{-}) C H_{3}$ , and say which contributes more.

Try it yourself first. Then reveal the solution one step at a time.

Example 8.18NCERT Problem 8.18 · p. 276

Why a structure is only a minor contributor

For methoxyethene, explain why $_{}^{}_{}^{-} C H_{2} - C H = O^{+} - C H_{3}$ (II) is only a minor contributor compared with $C H_{2} = C H - O - C H_{3}$ (I).

Try it yourself first. Then reveal the solution one step at a time.

8.7.7 Resonance Effect

Delocalisation does more than stabilise: it shifts electron density from one end of a conjugated system to the other. This shift, the resonance (mesomeric) effect, R or M, appears whenever a π bond sits next to another π bond or next to an atom with a lone pair.

+R effect. A group donates electron density into the conjugated system, usually from a lone pair. In methoxybenzene the oxygen lone pair is pushed into the ring, and the resonance structures put negative charge on the two ortho carbons and the para carbon (Fig. 8.13). +R groups: halogens, –OH, –OR, –OCOR, $- N H_{2}$ , –NHR, $- N R_{2}$ and –NHCOR.

−R effect. A group withdraws π electrons from the conjugated system. In benzaldehyde the ring's π electrons are drawn towards the carbonyl oxygen, leaving positive charge on the ortho and para carbons. −R groups: –COOH, –CHO, >C=O, –CN and $- N O_{2}$ .

Fig. 8.13+R makes ortho and para carbons electron-rich; −R makes them electron-poor.NCERT p. 276

Two benzene rings side by side. Left, methoxybenzene: a curved arrow from a lone pair on the OCH₃ oxygen into the ring, and δ− labels on the two ortho carbons and the para carbon. Right, benzaldehyde: a curved arrow from the ring towards the CHO oxygen, and δ+ labels on the ortho and para carbons.

Alternating single and double bonds, as in the long chain of β-carotene that colours carrots orange, or a π bond next to a lone pair, make a conjugated system. Its delocalised π electrons give it polarity and stability that no single Lewis structure predicts.

8.7.8 Electromeric Effect (E effect)

A double bond can also respond to a visitor. As a reagent closes in on a C=C or C=O bond, the whole π pair can shift onto one of the two atoms, and it slides back if the reagent leaves without reacting. This temporary, all-or-nothing shift, drawn as a curved arrow from the π bond, is the electromeric effect (E). Without a multiple bond and an attacking reagent there is no E effect.

In the +E effect the pair moves to the atom the reagent becomes attached to. When $H^{+}$ attacks propene, the π pair bonds C1 to the proton and leaves C2 positive:

\begin{aligned} C H_{2} = C H C H_{3} \\ + H^{+} \\ \overset{}{\to} C H_{3} C H^{+} C H_{3} \end{aligned}

(8.7)

In the −E effect the pair moves away from the atom the reagent attaches to. When cyanide attacks the carbonyl carbon of propanone, the π pair moves onto oxygen:

\begin{aligned} (C H_{3})_{2} C = O \\ + C N^{-} \\ \overset{}{\to} (C H_{3})_{2} C \\ (C N) O^{-} \end{aligned}

(8.8)

When the inductive and electromeric effects act in opposite directions, the electromeric effect wins: at the instant of attack it is the larger shift.

8.7.9 Hyperconjugation

Alkyl groups have one more trick. A C–H bond on the carbon next to an empty p orbital, or next to a π bond, can lean its σ electrons sideways into that orbital, provided it lines up parallel to it. This leakage of σ density, hyperconjugation, is a permanent, stabilising feature of the molecule. Fig. 8.14 shows it in the isopropyl cation, $(C H_{3})_{2} C H^{+}$ .

3/3Partial overlap spreads the positive charge onto the C–H hydrogen.

Fig. 8.14An aligned C–H bond leaks σ electron density into the empty p orbital.NCERT Fig. 8.4 · p. 277

The isopropyl cation: a central carbon with an empty p orbital, bonded to H and two methyl carbons. Step 1: one methyl C–H bond points sideways. Step 2: that C–H bond has turned parallel to the empty p orbital. Step 3: a shaded region joins the C–H bond and the p orbital, with δ+ on the hydrogen and on the central carbon.

It is often written as 'no-bond resonance', with structures in which an α hydrogen appears as $H^{+}$ with no bond to carbon. For the ethyl cation:

\begin{aligned} H - C H_{2} - C H_{2}^{+} \\ ⟷ H^{+} C H_{2} = C H_{2} \end{aligned}

(8.9)

The more α hydrogens, the greater the stabilisation: $(C H_{3})_{3} C^{+}$ (9 α-H) > $(C H_{3})_{2} C H^{+}$ (6) > $C H_{3} C H_{2}^{+}$ (3) > $C H_{3}^{+}$ (0). Alkenes benefit too: but-2-ene (6 α-H) is more stable than but-1-ene (2). This is why an alkyl group attached to a π system acts as an electron donor.

Example 8.19NCERT Problem 8.19 · p. 278

Ordering carbocations by hyperconjugation

Arrange $C H_{3} C H_{2}^{+}$ , $(C H_{3})_{2} C^{+} C H_{2} C H_{3}$ and $(C H_{3})_{2} C H^{+}$ in order of stability, and explain.

Try it yourself first. Then reveal the solution one step at a time.

8.7.10 Types of Organic Reactions and Mechanisms

Most organic reactions fall into four types, which you will study in detail in Unit 9 and in Class 12.

Table 8.7Four types of organic reactionNCERT p. 278

Type	What happens	Example
Substitution	one atom or group replaces another	$(C H_{3})_{3} C B r$ + $H_{2} O$ → $(C H_{3})_{3} C O H$ + $H B r$
Addition	two parts add across a multiple bond	$C H_{3} C H = C H_{2}$ + $B r_{2}$ → $C H_{3} C H B r C H_{2} B r$
Elimination	atoms or groups leave from neighbouring carbons, forming a multiple bond	$C H_{3} C H_{2} C H_{2} B r$ + $K O H$ → $C H_{3} C H = C H_{2}$ + $K B r$ + $H_{2} O$ (in ethanol)
Rearrangement	atoms of one molecule reorganise into an isomer	butane → 2-methylpropane ( $A l C l_{3}$ , HCl, heat)

8.8 Methods of Purification of Organic Compounds

Whether a chemist isolates a compound from neem leaves or makes it in a flask, the first product is a mixture. Choosing how to clean it up means asking: is the compound solid or liquid? Does it vaporise easily? Does it dissolve in water? The answers point to one of five methods: sublimation, crystallisation, distillation, differential extraction or chromatography.

How do we know a compound is pure? A pure solid melts sharply at a fixed temperature and a pure liquid boils at a fixed temperature, while impurities lower and widen the melting range. Chemists now also check purity by chromatography and spectroscopy.

8.8.1 Sublimation

Some solids turn straight into vapour on heating, without melting, and the vapour turns straight back into solid on a cool surface. Gentle heating under an inverted funnel or a cold dish therefore separates a solid that sublimes, such as camphor (kapoor) or naphthalene, from impurities that do not.

8.8.2 Crystallisation

Crystallisation uses a solvent in which the compound dissolves poorly when cold but well when hot. Dissolve the impure solid in the minimum of hot solvent, filter hot, and cool: the pure compound crystallises and is filtered off, while the mother liquor keeps the soluble impurities.

Water suits benzoic acid: about 5.6 g dissolves in 100 mL at 100 °C but only about 0.34 g at 25 °C. Coloured impurities are removed by boiling with a little activated charcoal before filtering. If no single solvent works, a pair is used, one that dissolves the compound easily and one that barely does. Impurities with solubilities close to the compound's need repeated crystallisation.

8.8.3 Distillation

Simple distillation separates (i) a volatile liquid from non-volatile impurities and (ii) liquids whose boiling points are far apart. The lower-boiling liquid vaporises first; its vapour is cooled in a water condenser and the liquid collected. When the thermometer reading jumps, the receiver is changed. Hexane (342 K) and nitrobenzene (484 K) separate this way (Fig. 8.15).

3/3A fractionating column restores two clean plateaus.

Fig. 8.15Flat steps mean a pure liquid is distilling.NCERT Figs. 8.5, 8.6 · p. 279

Temperature at the still head against volume collected. Step 1: two flat steps at two far-apart boiling points, with a sharp rise between, each fraction labelled. Step 2: a dashed curve for two close-boiling liquids rises gradually with no flat step: mixed fractions. Step 3: with a fractionating column the close-boiling pair gives two clean flat steps again.

Fractional distillation. With close boiling points both liquids vaporise together. A fractionating column (packed with glass beads or fitted with plates) between flask and condenser makes rising vapour meet falling condensate over a large surface, so condensation and re-vaporisation repeat many times; each stage is a theoretical plate. The vapour leaving the top is almost pure lower-boiling liquid. Refineries such as Jamnagar in Gujarat separate crude oil into LPG, petrol, kerosene, diesel and heavier fractions with columns of hundreds of plates.

Distillation under reduced pressure. A liquid boils when its vapour pressure equals the pressure above it, so a water or vacuum pump lets it boil well below its normal boiling point. This suits liquids that boil very high or decompose before boiling; refineries, for example, distil the heavy residue of crude oil under vacuum to obtain lubricating oils without cracking them.

Steam distillation. How can you distil a liquid that boils at 405 K, or one that would char at its boiling point, without heating it that far? If it does not mix with water and is volatile in steam, blow steam through it. Two immiscible liquids each keep their own vapour pressure, so the mixture boils as soon as the organic liquid's vapour pressure $p_{1}$ and water's $p_{2}$ add up to the pressure above it, $p$ :

p = p_{1} + p_{2}

(8.10)

Since $p_{1}$ adds to water's vapour pressure, the mixture boils below 373 K, far below the compound's own boiling point (Fig. 8.16). The condensate separates into two layers in a separating funnel. Essential oils such as eucalyptus and lemongrass oil are produced this way, and in the laboratory o-nitrophenol is separated from p-nitrophenol, which is not steam volatile.

Fig. 8.16Chlorobenzene (b.p. 405 K) steam-distils at about 364 K.NCERT p. 281

A graph of vapour pressure in mm Hg against temperature from 323 K to 378 K. Water's curve reaches the 760 mm line at 373 K. Chlorobenzene's lower curve rises slowly. The sum of the two curves crosses 760 mm at about 364 K, marked with a dot: the boiling point of the mixture, far below chlorobenzene's own boiling point of 405 K.

8.8.4 Differential Extraction

Tea leaves contain caffeine, which dissolves better in dichloromethane than in water. Shake a cooled tea infusion with dichloromethane in a separating funnel, let the two immiscible layers settle, run off the lower organic layer and evaporate it: caffeine is left behind. This is differential extraction. If the compound prefers the organic solvent only slightly, one shake removes little, so in continuous extraction the same solvent is recycled through the solution again and again.

8.8.5 Chromatography

In 1906 the botanist Mikhail Tsvet washed a leaf extract through a column of chalk powder with solvent; the pigments came apart as coloured bands, and he named the method chromatography ('colour writing'). Every version since has a stationary phase (a solid, or a liquid held on a solid) that holds the components back and a mobile phase (a liquid or a gas) that carries them along. Components held more strongly move more slowly, so they separate; the method separates, purifies and checks the purity of compounds.

Adsorption chromatography uses the different extents to which components are adsorbed on silica gel or alumina. In column chromatography the adsorbent is packed in a glass tube with a stopcock. The mixture is loaded at the top and the eluant flows down; the most strongly adsorbed component stays near the top, the others move down at different rates and are collected one by one.

Thin layer chromatography (TLC) uses an adsorbent layer about 0.2 mm thick on a glass plate (a chromaplate). A spot of the mixture is placed on a base line about 2 cm from the bottom, and the plate stands in a closed jar with a little eluant. As the eluant climbs, each component moves a characteristic distance, recorded as its retardation factor:

R_{f} = \frac{x}{y}

(8.11)

where

$x$: distance moved by the substance from the base line
$y$: distance moved by the solvent front from the base line

Coloured spots are seen directly. Colourless ones can be revealed by spraying a reagent (ninhydrin turns amino acids purple), by leaving the plate in a closed jar with a few iodine crystals (brown spots) or, for compounds that fluoresce, under ultraviolet light. Follow a run in Fig. 8.17, then try the TLC lab.

4/4Rf = x / y, both measured from the base line.

Fig. 8.17Each component stops at its own fraction of the solvent's climb.NCERT Fig. 8.12 · p. 283

A TLC plate in a jar with shallow solvent. Step 1: one spot on the pencil base line. Step 2: the solvent front has climbed part way and the spot starts to split. Step 3: three spots at different heights near the end of the run. Step 4: the front is marked and arrows show x to the middle spot and y to the front, with Rf = x/y.

Fig. 8.18Choose a mixture and the eluant's polarity, press Develop, then measure each spot's Rf.

Partition chromatography depends on components distributing between a stationary liquid and the mobile phase. In paper chromatography the stationary phase is water held in the cellulose of special paper; the solvent rises up the spotted strip by capillary action, and the developed strip (a chromatogram) is read like a TLC plate.

Try this

Unmix a black sketch pen

You’ll need

a strip of filter paper or coffee filter
a black water-based sketch pen
a tall glass and water
a pencil and a ruler

Predict first

What will the black dot do as water climbs the paper?

stay as one black spotsplit into coloured bandsdisappear completely

Draw a pencil line 2 cm from the bottom of a filter-paper strip and put a small dot of black sketch-pen ink on it.
Hang the strip in a tall glass with a little water, so that the water touches the paper below the line.
When the water has nearly reached the top, take the strip out and mark the front. Measure x for each colour and y for the water, and find each Rf.

Table 8.8

Colour	x (cm)	Rf

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1 of 4

Which method best separates naphthalene from common salt?

sublimationsteam distillationfractional distillationdifferential extraction

Show answer

Answer (A) sublimation

Naphthalene sublimes and salt does not. The distillations need liquids.

8.9 Qualitative Analysis of Organic Compounds

Every organic compound contains carbon and nearly all contain hydrogen; many also contain oxygen, nitrogen, sulphur, halogens or phosphorus. Qualitative analysis tells us which of these are present.

8.9.1 Detection of Carbon and Hydrogen

Heat the compound with dry copper(II) oxide. Carbon is oxidised to carbon dioxide, which turns lime water milky ( $C O_{2}$ + $C a (O H)_{2}$ → $C a C O_{3} ↓$ + $H_{2} O$ ), and hydrogen to water, which turns white anhydrous copper(II) sulphate blue ( $C u S O_{4}$ + $5 H_{2} O$ → $C u S O_{4} \cdot 5 H_{2} O$ ):

\begin{aligned} C + 2 C u O \\ \overset{Δ}{\to} 2 C u + C O_{2} \\ 2 H + C u O \\ \overset{Δ}{\to} C u + H_{2} O \end{aligned}

(8.12)

8.9.2 Detection of Other Elements

Nitrogen, sulphur, halogens and phosphorus in organic compounds are covalently bonded, so they do not give the usual ionic tests. In Lassaigne's test a little of the compound is fused with a small piece of sodium in an ignition tube, which turns them into ionic sodium salts (X = Cl, Br or I):

\begin{aligned} N a + C + N \\ \overset{Δ}{\to} N a C N \\ 2 N a + S \\ \overset{Δ}{\to} N a_{2} S \\ N a + X \overset{Δ}{\to} N a X \end{aligned}

(8.13)

The red-hot tube is plunged into distilled water, boiled and filtered. The filtrate, the sodium fusion extract, is tested for each element (Fig. 8.19).

4/4Boil with HNO₃, add AgNO₃: AgCl white, AgBr pale yellow, AgI yellow.

Fig. 8.19One fusion, then a separate test for each element.NCERT p. 284

A flow diagram. Step 1: compound and sodium are fused in an ignition tube, plunged into water and filtered to give the sodium fusion extract. Step 2: the nitrogen test, iron(II) sulphate then sulphuric acid, gives Prussian blue. Step 3: the sulphur tests give black lead sulphide or a violet colour. Step 4: after nitric acid, silver nitrate gives white, pale yellow or yellow halide precipitates.

(A) Nitrogen. Boil the extract with iron(II) sulphate, then acidify with concentrated sulphuric acid: a Prussian blue colour or precipitate shows nitrogen. Behind the colour are two steps: cyanide ions surround $F e^{2 +}$ to form the complex ion $[F e (C N)_{6}]^{4 -}$ , and the hot acid oxidises part of the iron(II) to iron(III), which joins these complex ions in the deep-blue solid iron(III) hexacyanidoferrate(II):

\begin{aligned} 6 C N^{-} + F e^{2 +} \\ \overset{}{\to} [F e (C N)_{6}]^{4 -} \\ 3 [F e (C N)_{6}]^{4 -} \\ + 4 F e^{3 +} \\ \overset{}{\to} F e_{4} [F e (C N)_{6}]_{3} \end{aligned}

(8.14)

(B) Sulphur. Acidified with acetic acid, the extract gives a black precipitate with lead acetate, $S^{2 -}$ + $P b^{2 +}$ → $P b S ↓$ , and a violet colour with sodium nitroprusside, from $[F e (C N)_{5} N O S]^{4 -}$ . A compound with both N and S may give sodium thiocyanate: the extract then turns blood red with iron(III), forming $[F e (S C N)]^{2 +}$ , and gives no Prussian blue. Excess sodium breaks it down, $N a S C N$ + $2 N a$ → $N a C N$ + $N a_{2} S$ .

(C) Halogens. Acidify the extract with nitric acid, boil, then add silver nitrate, $X^{-}$ + $A g^{+}$ → $A g X ↓$ . A white precipitate soluble in aqueous ammonia shows chlorine; pale yellow and sparingly soluble, bromine; yellow and insoluble, iodine. Boiling with nitric acid first destroys any cyanide and sulphide, which would otherwise precipitate as AgCN and $A g_{2} S$ and mislead.

(D) Phosphorus. Heat the compound with an oxidising agent such as sodium peroxide to convert phosphorus to phosphate. Boil with nitric acid, which turns phosphate into phosphoric acid ( $N a_{3} P O_{4}$ + $3 H N O_{3}$ → $H_{3} P O_{4}$ + $3 N a N O_{3}$ ), and add ammonium molybdate: a yellow colour or precipitate of ammonium phosphomolybdate shows phosphorus, $H_{3} P O_{4}$ + $12 (N H_{4})_{2} M o O_{4}$ + $21 H N O_{3}$ → $(N H_{4})_{3} P O_{4} \cdot 12 M o O_{3}$ + $21 N H_{4} N O_{3}$ + $12 H_{2} O$ .

Table 8.9Lassaigne's test at a glance

Element	Test on the extract	Positive result
N	$F e S O_{4}$ , boil; conc. $H_{2} S O_{4}$	Prussian blue
S	acetic acid + lead acetate	black PbS
S	sodium nitroprusside	violet
N and S	$F e^{3 +}$	blood red
Cl, Br, I	$H N O_{3}$ , boil; $A g N O_{3}$	white, pale yellow, yellow
P	after $N a_{2} O_{2}$ : $H N O_{3}$ + ammonium molybdate	yellow

8.10 Quantitative Analysis

Quantitative analysis finds the mass per cent of each element in a compound. As you saw in Unit 1, these percentages give the empirical formula, and with the molar mass, the molecular formula. Each classical method turns one element into something that can be weighed or measured.

8.10.1 Carbon and Hydrogen

A weighed sample is burnt completely in dry oxygen over heated copper(II) oxide, so that all its carbon becomes carbon dioxide and all its hydrogen becomes water:

\begin{aligned} C_{x} H_{y} + (x + \frac{y}{4}) O_{2} \\ ⟶ x C O_{2} + \frac{y}{2} H_{2} O \end{aligned}

(8.15)

The gases pass first through a weighed U-tube of anhydrous calcium chloride, which absorbs water, then through one of concentrated potassium hydroxide solution, which absorbs carbon dioxide (Fig. 8.20). The order matters, because the KOH solution would also take up water. The gain in mass of each tube gives the masses of water and carbon dioxide.

3/3KOH solution gains the mass of carbon dioxide, m₂.

Fig. 8.20Weigh the water and the carbon dioxide, in that order.NCERT Fig. 8.14 · p. 286

A combustion tube with oxygen entering from the left, a sample boat and copper(II) oxide, heated from below, followed by two U-tubes. Step 1: the sample burns and labelled arrows carry CO₂ and H₂O to the right. Step 2: the first U-tube, anhydrous calcium chloride, is highlighted as gaining the water. Step 3: the second, potassium hydroxide solution, is highlighted as gaining the carbon dioxide.

DerivationPer cent carbon and hydrogen

Let a sample of mass $m$ give $m_{1}$ g of water and $m_{2}$ g of carbon dioxide. Each mole of $C O_{2}$ (44 g) holds 12 g of carbon, and each mole of $H_{2} O$ (18 g) holds 2 g of hydrogen.

% C = \frac{12 m_{2} \times 100}{44 m}

(8.16)

% H = \frac{2 m_{1} \times 100}{18 m}

(8.17)

Example 8.20NCERT Problem 8.20 · p. 286

Percentage of carbon and hydrogen

On complete combustion, 0.290 g of an organic compound gave 0.660 g of carbon dioxide and 0.270 g of water. Find the percentages of carbon and hydrogen, the percentage of oxygen (the only other element) and the empirical formula.

Try it yourself first. Then reveal the solution one step at a time.

8.10.2 Nitrogen

(i) Dumas method. The compound is heated with copper(II) oxide in an atmosphere of carbon dioxide. Carbon and hydrogen become $C O_{2}$ and water, and nitrogen is set free as $N_{2}$ :

\begin{aligned} C_{x} H_{y} N_{z} \\ + (2 x + \frac{y}{2}) C u O \\ \to x C O_{2} + \frac{y}{2} H_{2} O \\ + \frac{z}{2} N_{2} \\ + (2 x + \frac{y}{2}) C u \end{aligned}

(8.18)

Nitrogen oxides are reduced to $N_{2}$ over hot copper gauze, and the gases are collected over KOH solution in a nitrometer, which absorbs $C O_{2}$ and leaves nitrogen to be measured. The gas is saturated with water vapour, so its own pressure is atmospheric pressure minus the aqueous tension.

DerivationPer cent nitrogen by the Dumas method

A mass $m$ of compound gives $V_{1}$ mL of nitrogen at $T_{1}$ and atmospheric pressure $p$ (mm Hg); the aqueous tension at $T_{1}$ is $f$ . Convert the volume to STP (273 K, 760 mm Hg), where 22 400 mL of $N_{2}$ has a mass of 28 g:

\begin{aligned} p_{1} & = p - f \\ V & = \frac{p_{1} V_{1}}{760} \times \frac{273}{T_{1}} \end{aligned}

(8.19)

% N = \frac{28 V \times 100}{22 400 m}

(8.20)

Using $p V = n R T$ directly gives the same result.

Example 8.21NCERT Problem 8.21 · p. 287

Dumas method

0.200 g of a compound gave 42.2 mL of nitrogen, collected over water at 290 K and 740 mm Hg. The aqueous tension at 290 K is 14.5 mm Hg. Find the percentage of nitrogen.

Try it yourself first. Then reveal the solution one step at a time.

(ii) Kjeldahl's method. Nitrogen is converted to ammonium sulphate by digestion with concentrated sulphuric acid, released as ammonia with sodium hydroxide, and absorbed in a measured excess of standard acid; back-titrating the leftover acid gives the acid used by the ammonia.

Step by step

Kjeldahl's method in four stages

Step 1 of 4

Digest

Heat with conc. $H_{2} S O_{4}$ (+ $C u S O_{4}$ , $K_{2} S O_{4}$ ): N → $(N H_{4})_{2} S O_{4}$ .

1 / 4

DerivationPer cent nitrogen by Kjeldahl's method

Let ammonia from $m$ g of compound be absorbed in $V_{a}$ mL of $H_{2} S O_{4}$ of molarity $M_{a}$ , and let the unused acid need $V_{b}$ mL of NaOH of molarity $M_{b}$ . In millimoles:

\begin{aligned} n (N H_{3}) & = 2 M_{a} V_{a} \\ - M_{b} V_{b} \end{aligned}

(8.21)

Each millimole of $N H_{3}$ contains 14 mg of nitrogen, so with $n$ in mmol and $m$ in g:

% N = \frac{1.4 n (N H_{3})}{m}

(8.22)

With acid and alkali of the same molarity $M$ , $n = M (2 V_{a} - V_{b})$ . With a monobasic acid such as HCl, $n = M_{a} V_{a} - M_{b} V_{b}$ .

Kjeldahl's method fails for nitrogen in nitro and azo groups and in rings such as pyridine, because that nitrogen is not converted to ammonium sulphate under these conditions.

Example 8.22NCERT Problem 8.22 · p. 288

Kjeldahl's method

The ammonia from 0.60 g of a compound exactly neutralised 15.0 mL of 0.50 M $H_{2} S O_{4}$ . Find the percentage of nitrogen.

Try it yourself first. Then reveal the solution one step at a time.

8.10.3 Halogens

Carius method. A weighed sample is heated with fuming nitric acid and silver nitrate in a sealed hard-glass Carius tube inside a furnace. Carbon and hydrogen are oxidised, and the halogen forms silver halide, AgX, which is filtered, washed, dried and weighed. One mole of AgX contains one mole of X:

\begin{aligned} % X & = \frac{A_{X}}{M_{A g X}} \\ \times \frac{m_{1}}{m} \times 100 \end{aligned}

(8.23)

where

$A_{X}$: atomic mass of the halogen
$M_{A g X}$: molar mass of AgX
$m_{1}$: mass of AgX formed

8.10.4 Sulphur

A weighed sample is heated in a Carius tube with fuming nitric acid or sodium peroxide, which oxidises its sulphur to sulphuric acid. Excess barium chloride precipitates barium sulphate, which is filtered, washed, dried and weighed. One mole of $B a S O_{4}$ (233 g) contains 32 g of sulphur:

% S = \frac{32 m_{1}}{233 m} \times 100

(8.24)

Example 8.23NCERT Problems 8.23, 8.24 · p. 289

Carius estimations of iodine and sulphur

(i) 0.156 g of an organic iodo compound gave 0.235 g of AgI (Ag = 108, I = 127). (ii) 0.180 g of another compound gave 0.350 g of $B a S O_{4}$ . Find the percentage of iodine in (i) and of sulphur in (ii).

Try it yourself first. Then reveal the solution one step at a time.

8.10.5 Phosphorus

The sample is heated with fuming nitric acid, which oxidises phosphorus to phosphoric acid. This is precipitated with ammonia and ammonium molybdate as ammonium phosphomolybdate, $(N H_{4})_{3} P O_{4} \cdot 12 M o O_{3}$ (1877 g/mol, one P), or with magnesia mixture as $M g N H_{4} P O_{4}$ , which on ignition gives $M g_{2} P_{2} O_{7}$ (222 g/mol, two P = 62 g):

\begin{aligned} % P & = \frac{31 m_{1}}{1877 m} \times 100 \\ or & \frac{62 m_{1}}{222 m} \times 100 \end{aligned}

(8.25)

8.10.6 Oxygen

Oxygen is usually found by difference: %O = 100 − (sum of all other percentages). It can also be measured directly. The compound is decomposed by heating in nitrogen, and the products pass over red-hot coke (1373 K), which turns all the oxygen into carbon monoxide. Warm iodine pentoxide then oxidises the CO to $C O_{2}$ and releases iodine:

\begin{aligned} 2 C + O_{2} \\ \overset{1373 K}{\to} 2 C O \\ I_{2} O_{5} + 5 C O \\ \overset{}{\to} I_{2} + 5 C O_{2} \end{aligned}

(8.26)

Multiplying the first equation by 5 and the second by 2 shows that each mole of $O_{2}$ (32 g) released ends up as two moles of $C O_{2}$ (88 g). So if $m$ g of compound gives $m_{1}$ g of $C O_{2}$ , %O = $32 m_{1} \times 100 / (88 m)$ ; the iodine can be titrated instead.

Modern laboratories use automatic CHN elemental analysers, which need only 1–3 mg of sample and report carbon, hydrogen and nitrogen within minutes. Practise the classical calculations in the estimation lab.

InteractiveEstimation lab

Fig. 8.21Pick a method, enter the data, and follow the calculation step by step; New problem sets a practice question to solve yourself.

Quick check

1 of 3

0.20 g of a compound gave 0.44 g of

C O_{2}

on complete combustion. What is the percentage of carbon?

Show answer

Answer 60 %

% C = 12 \times 0.44 \times 100 / (44 \times 0.20) = 60 %

Recap

Organic Chemistry – Some Basic Principles and Techniques: key points and summary

Carbon is sp³ (tetrahedral), sp² (trigonal planar) or sp (linear); more s character gives shorter, stronger bonds (Section 8.2.1).
A π bond needs parallel p orbitals: it fixes the geometry, blocks rotation and is a reactive site (Section 8.2.2).
Structures are drawn as complete, condensed or bond-line formulas, and in 3-D with wedges and dashes (Section 8.3).
Compounds are acyclic or cyclic (alicyclic, aromatic) and fall into homologous series by functional group (Section 8.4).
IUPAC names: longest chain, lowest locants, alphabetical prefixes; the principal functional group gives the suffix (Section 8.5).
Isomers share a molecular formula: structural (chain, position, functional group, metamers) or stereoisomers (Section 8.6).
Heterolysis gives carbocations (planar) or carbanions (pyramidal); homolysis gives free radicals (Section 8.7.1).
Nucleophiles donate an electron pair, electrophiles accept one; curved arrows track the pairs (Section 8.7.2, Section 8.7.3).
The inductive effect polarises σ bonds and fades within about three bonds (Section 8.7.5).
Resonance and the R effect delocalise π and lone-pair electrons; the hybrid is more stable than any contributor (Section 8.7.6, Section 8.7.7).
The electromeric effect is a temporary π shift; hyperconjugation by α C–H bonds stabilises carbocations and alkenes (Section 8.7.8, Section 8.7.9).
Organic reactions are substitutions, additions, eliminations or rearrangements (Section 8.7.10).
Purification uses sublimation, crystallisation, distillation, extraction and chromatography, with $R_{f} = x / y$ (Section 8.8).
Lassaigne's test detects N, S and halogens; combustion, Dumas, Kjeldahl and Carius methods give mass percentages (Section 8.9, Section 8.10).

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Quick check: MCQs on Organic Chemistry – Some Basic Principles and Techniques

8.2.2 Some Characteristic Features of π Bonds4 questions
8.3.2 Three-Dimensional Representation of Organic Molecules4 questions
8.4.2 Homologous Series4 questions
8.5.4 Nomenclature of Substituted Benzene Compounds4 questions
8.6.2 Stereoisomerism4 questions
8.7.2 Substrate and Reagent4 questions
8.7.6 Resonance Structure3 questions
8.7.10 Types of Organic Reactions and Mechanisms4 questions
8.8.5 Chromatography4 questions
8.9.2 Detection of Other Elements3 questions
8.10.6 Oxygen3 questions

Glossary · 15

Key terms (English–Hindi)

carbanionकार्बऋणायनkarbrinayan: A species with a negatively charged carbon that carries a lone pair; usually sp³ hybridised and pyramidal.First seen in the chapter →
carbocationकार्बधनायनkarbdhanayan: A species with a positively charged carbon that has only six valence electrons; it is sp² hybridised and planar.First seen in the chapter →
catenationश्रृंखलनshrinkhalan: The ability of atoms of an element, above all carbon, to bond to one another in chains and rings.First seen in the chapter →
chromatography: Separation of a mixture by the different rates at which its components move over a stationary phase with a mobile phase.First seen in the chapter →

electromeric effect: A temporary, complete shift of a π electron pair to one atom of a multiple bond while a reagent attacks.First seen in the chapter →
electrophileइलेक्ट्रॉनरागीelectronragi: A reagent that accepts an electron pair from an electron-rich centre.

free radicalमुक्त मूलकmukt moolak: An atom or group with an unpaired electron, formed by homolytic cleavage of a bond.First seen in the chapter →
functional groupप्रकार्यात्मक समूहprakaryatmak samooh: An atom or group of atoms that gives a compound its characteristic reactions, such as –OH or –COOH.

homologous seriesसमजातीय श्रेणीsamjatiya shreni: A family of compounds with the same functional group whose successive members differ by a $C H_{2}$ unit.
hyperconjugationअतिसंयुग्मनatisanyugman: Delocalisation of the σ electrons of a C–H bond into a neighbouring π system or empty p orbital.First seen in the chapter →

inductive effectप्रेरणिक प्रभावpreranik prabhav: Permanent polarisation passed along σ bonds from a polar bond; it fades within about three bonds.First seen in the chapter →
isomerismसमावयवताsamavayavata: The existence of different compounds with the same molecular formula; such compounds are isomers.

nucleophileनाभिकरागीnabhikragi: A reagent that donates an electron pair to an electron-poor centre.

resonanceअनुनादanunaad: Description of a molecule as a hybrid of Lewis structures that differ only in the positions of electrons.
retardation factor: $R_{f}$ : the distance moved by a substance divided by the distance moved by the solvent front, both from the base line.First seen in the chapter →

8.1 General Introduction

8.2 Tetravalence of Carbon: Shapes of Organic Compounds

8.2.1 The Shapes of Carbon Compounds

8.2.2 Some Characteristic Features of π Bonds

8.3 Structural Representations of Organic Compounds

8.3.1 Complete, Condensed and Bond-line Structural Formulas

8.3.2 Three-Dimensional Representation of Organic Molecules

8.4 Classification of Organic Compounds

8.4.1 Functional Group

8.4.2 Homologous Series

8.5 Nomenclature of Organic Compounds

8.5.1 The IUPAC System of Nomenclature

8.5.2 IUPAC Nomenclature of Alkanes

8.5.3 Nomenclature of Organic Compounds having Functional Group(s)

8.5.4 Nomenclature of Substituted Benzene Compounds

8.6 Isomerism

8.6.1 Structural Isomerism

8.6.2 Stereoisomerism

8.7 Fundamental Concepts in Organic Reaction Mechanism

8.7.1 Fission of a Covalent Bond

8.7.2 Substrate and Reagent

8.7.3 Electron Movement in Organic Reactions

8.7.4 Electron Displacement Effects in Covalent Bonds

8.7.5 Inductive Effect

8.7.6 Resonance Structure

8.7.7 Resonance Effect

8.7.8 Electromeric Effect (E effect)

8.7.9 Hyperconjugation

8.7.10 Types of Organic Reactions and Mechanisms

8.8 Methods of Purification of Organic Compounds

8.8.1 Sublimation

8.8.2 Crystallisation

8.8.3 Distillation

8.8.4 Differential Extraction

8.8.5 Chromatography

8.9 Qualitative Analysis of Organic Compounds

8.9.1 Detection of Carbon and Hydrogen

8.9.2 Detection of Other Elements

8.10 Quantitative Analysis

8.10.1 Carbon and Hydrogen

8.10.2 Nitrogen

8.10.3 Halogens

8.10.4 Sulphur

8.10.5 Phosphorus

8.10.6 Oxygen

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