Amines: Class 12 Chemistry · Chapter 9

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

Published 12 Oct 2026 · Updated 12 Oct 2026

90 min
27 practice questions
39 MCQs

What you will learn in Amines

Describe amines as derivatives of ammonia with a pyramidal sp3 nitrogen that carries a lone pair
Classify amines as primary, secondary or tertiary and name them by common and IUPAC rules
Choose a preparation route (reduction of nitro compounds, nitriles or amides, ammonolysis, Gabriel synthesis or Hoffmann degradation) and predict the carbon count of the product
Explain the boiling points and solubility of amines through hydrogen bonding
Rank amines by basic strength in the gas phase and in water using inductive, solvation, steric and resonance effects
Use the carbylamine, nitrous acid and Hinsberg tests to tell primary, secondary and tertiary amines apart, and write their acylation and ring-substitution reactions
Prepare arenediazonium salts and use their replacement and coupling reactions to make aromatic compounds, including azo dyes

Original companion text aligned to NCERT Chemistry Class 12, Chapter 9, pp. 259–280. Not affiliated with or endorsed by NCERT.

Walk through a fish market on a warm afternoon and the sharp smell that reaches you comes mostly from one small molecule, trimethylamine, $(C H_{3})_{3} N$ . Decaying meat releases two more, with the blunt names putrescine and cadaverine. All three are amines (ऐमीन): compounds built from ammonia by putting carbon groups where its hydrogen atoms were.

Amines run all through biology and medicine. Nerve cells signal to each other with dopamine and serotonin; tobacco owes its hold on smokers to nicotine; local anaesthetics such as lidocaine and many antihistamines carry an amino group. In industry, hexane-1,6-diamine is one of the two building blocks of nylon-6,6, and long-chain ammonium salts are the active part of fabric softeners and many disinfectants.

The chapter has two halves. Sections 9.1 to 9.6 deal with amines themselves: their shape, classes, names, preparation, properties and reactions. Sections 9.7 to 9.10 follow the arenediazonium salts made from aromatic amines. These ions survive only in ice-cold water, yet they let chemists place F, I, CN, OH and other groups on a benzene ring where direct methods fail, and they join two rings into the bright azo dyes.

9.1 Structure of Amines

Begin with ammonia. Nitrogen has five valence electrons: three pair up with electrons from hydrogen atoms in σ bonds, and two stay together as a lone pair. The four electron pairs occupy four $s p^{3}$ hybrid orbitals aimed roughly at the corners of a tetrahedron. The three bonded atoms therefore sit at the base of a low pyramid, with nitrogen at its apex.

An amine keeps exactly this arrangement. Replace one, two or all three hydrogens with alkyl or aryl groups and the nitrogen still forms three σ bonds and keeps one lone pair in the fourth $s p^{3}$ orbital (Fig. 9.1).

Definition

Amine

An organic compound in which one, two or three hydrogen atoms of ammonia, $N H_{3}$ , are replaced by alkyl or aryl groups. The nitrogen keeps its lone pair.

ExampleEthanamine, $C H_{3} C H_{2} N H_{2}$ , and aniline, $C_{6} H_{5} N H_{2}$ .

3/4Two replaced: a secondary amine.

Fig. 9.1The pyramid and its lone pair stay; only the groups at the base change.NCERT Fig. 9.1 · p. 260

A nitrogen atom sits at the apex of a pyramid, with a shaded lone-pair lobe above it and three bonds below: one to the left, one dashed behind to the right and one wedged towards the viewer. Counters on the left and right give the number of N–H bonds and of carbon groups. Step 1 shows ammonia with three hydrogens: 3 N–H bonds and 0 carbon groups. Step 2 swaps the left hydrogen for a methyl group, giving methanamine, a primary amine. Step 3 swaps the right hydrogen too, giving dimethylamine, a secondary amine. Step 4 has three methyl groups, trimethylamine, a tertiary amine with no N–H bond left.

A lone pair is held by only one nucleus, so it spreads wider than a bonding pair and pushes the bonds a little closer together. The angles round nitrogen therefore fall slightly short of the tetrahedral 109.5°; in ammonia the H–N–H angle is about 107°. Bulky groups push back against this squeeze, so in trimethylamine the C–N–C angle stays close to the tetrahedral value.NCERT quotes 108° for trimethylamine; gas-phase measurements give about 111°. Either way the molecule is a pyramid, not flat.

Keep the lone pair in view through the whole chapter. It is why amines take up protons (they are bases), why they attack electron-poor carbon atoms (they are nucleophiles), and why an amino group makes a benzene ring so reactive.

Let's explore

Build a nitrogen pyramid

You’ll need

a small potato or a ball of clay
four toothpicks
three peas or small beads
three larger beads or grapes
a cotton ball

Predict first

The toothpicks point to the corners of a tetrahedron. If the model rests on three of its ends, what shape do you see?

a low pyramid with the potato on topa flat triangle with the potato in the middlea straight line of three balls

Push four toothpicks into the potato so that they point as far from each other as possible, about 109° apart.
Put a pea on three of the sticks: these are hydrogen atoms. Fix the cotton ball on the fourth: that is the lone pair.
Stand the model on its three peas and look at it from the side.
Swap the peas one at a time for the larger beads (methyl groups) and notice how the beads start to crowd each other.

Quick check

1 of 3

What is the shape of trimethylamine around nitrogen?

trigonal planartetrahedral, with four atoms round Npyramidallinear

Show answer

Answer (C) pyramidal

The four electron pairs point to the corners of a tetrahedron, but one is a lone pair, so the three carbon atoms and nitrogen form a pyramid. Trigonal planar would need

s p^{2}

nitrogen with no lone pair in the plane.

9.2 Classification

Amines are sorted by counting the carbon groups on nitrogen. One group gives a primary (1°) amine, $R N H_{2}$ ; two give a secondary (2°) amine, $R_{2} N H$ ; three give a tertiary (3°) amine, $R_{3} N$ . The groups may be alkyl (R) or aryl (Ar) and need not be alike. A secondary or tertiary amine whose groups are all the same is called simple; one with different groups is mixed.

Compare

Three classes of amine

Primary (1°)

general formula $R N H_{2}$
two N–H bonds
e.g. $C H_{3} C H_{2} C H_{2} N H_{2}$

Secondary (2°)

$R_{2} N H$ (groups may differ)
one N–H bond
e.g. $C_{6} H_{5} N H C_{2} H_{5}$

Tertiary (3°)

$R_{3} N$
no N–H bond
e.g. $(C H_{3} C H_{2})_{3} N$

A fourth group can attach only if nitrogen uses its lone pair to form a fourth bond. The result is a quaternary ammonium ion, $R_{4} N^{+}$ , found in salts such as tetramethylammonium chloride, $(C H_{3})_{4} N^{+} C l^{-}$ . With no lone pair left it is not an amine, but it is where repeated alkylation of an amine ends (Ammonolysis of alkyl halides).

Watch one trap. For alcohols and haloalkanes, 1°, 2° and 3° describe the carbon atom that carries the group. For amines they describe the nitrogen. So 2-methylbutan-2-amine, $C H_{3} C H_{2} C (C H_{3})_{2} N H_{2}$ , is a primary amine even though its $N H_{2}$ sits on a tertiary carbon, while the matching alcohol, 2-methylbutan-2-ol, is a tertiary alcohol.

2/2For the amine, count carbons on nitrogen: one, so 1°.

Fig. 9.2Same skeleton, different counting rule: 2-methylbutan-2-ol is 3°, 2-methylbutan-2-amine is 1°.NCERT p. 260

A central carbon carries a methyl group on the left, an ethyl group on the right, a methyl group below and a functional group on top. In step 1 the top group is OH; the central carbon is ringed and its three carbon neighbours are numbered 1, 2 and 3, so the label reads 3° alcohol, because for alcohols you count the carbons on the carbon. In step 2 the top group is NH2; the nitrogen is ringed and only one carbon, the one below it, is numbered, so the label reads 1° amine, because for amines you count the carbons on the nitrogen.

Definition

Primary, secondary and tertiary amines

Amines with one ( $R N H_{2}$ ), two ( $R_{2} N H$ ) or three ( $R_{3} N$ ) carbon groups on nitrogen. The label counts groups on N, not on the carbon next to it.

Quick check

1 of 3

Which of these is a secondary amine?

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

(C H_{3} C H_{2})_{3} N

C_{6} H_{5} N H C H_{2} C H_{3}

(C H_{3})_{4} N^{+} C l^{-}

Show answer

Answer (C) $C_{6} H_{5} N H C H_{2} C H_{3}$

N-Ethylaniline has a phenyl and an ethyl group on N, so it is 2°. Option A has its

N H_{2}

on a secondary carbon, but only one group on N, so it is a 1° amine; option B is 3° and D is a quaternary salt.

9.3 Nomenclature

Every amine name has to say two things: which carbon framework carries the nitrogen, and what else is attached to that nitrogen. Two systems do this. The older common names simply list the groups on nitrogen, alphabetically, and finish with amine, all in one word: methylamine, ethylmethylamine, triethylamine (with di- and tri- for repeated groups).

IUPAC names build on the alkane instead, giving an alkanamine. For a primary amine the alkane loses its final e, amine takes its place, and a locant marks the carbon that carries $N H_{2}$ : butan-1-amine is $C H_{3} C H_{2} C H_{2} C H_{2} N H_{2}$ and butan-2-amine is $C H_{3} C H (N H_{2}) C H_{2} C H_{3}$ . When there are two amino groups the e stays, diamine follows and both locants are given: butane-1,4-diamine is $H_{2} N (C H_{2})_{4} N H_{2}$ .

For secondary and tertiary amines, the largest carbon group (the one with the longest chain) becomes the parent alkanamine. Every other group on nitrogen becomes a prefix carrying the locant N, and the prefixes go in alphabetical order. Fig. 9.3 builds one such name step by step.

4/4N-ethyl-N-methylbutan-1-amine.

Fig. 9.3Largest group → parent alkanamine; the others → N-prefixes.NCERT p. 261

A nitrogen atom carries a four-carbon chain on the left, an ethyl group on the right and a methyl group below. Step 1 boxes the three groups and labels them butyl 4 C, ethyl 2 C and methyl 1 C. Step 2 numbers the butyl chain 1 to 4, starting at the carbon bonded to nitrogen, and names the parent butan-1-amine. Step 3 labels the other two groups N-ethyl and N-methyl and notes that prefixes go in alphabetical order. Step 4 shows the full name, N-ethyl-N-methylbutan-1-amine.

In arylamines the nitrogen is bonded directly to a benzene ring. The simplest, $C_{6} H_{5} N H_{2}$ , keeps its traditional name aniline, which IUPAC also accepts; the systematic form is benzenamine. Ring substituents are numbered from the carbon that holds $N H_{2}$ (C-1), and groups on nitrogen again take N: $C_{6} H_{5} N H C H_{2} C H_{3}$ is N-ethylaniline (N-ethylbenzenamine). Benzylamine, $C_{6} H_{5} C H_{2} N H_{2}$ , is not an arylamine. Its nitrogen sits on a $C H_{2}$ group, so IUPAC calls it phenylmethanamine, and it behaves like an alkylamine.

Table 9.1IUPAC and (common) names of some amines

Amine	Name
$C H_{3} (C H_{2})_{3}$ – $N H_{2}$	butan-1-amine (n-butylamine)
$C H_{3} C H (N H_{2})$ – $C H_{2} C H_{3}$	butan-2-amine (sec-butylamine)
$(C H_{3})_{2} N H$	N-methylmethanamine (dimethylamine)
$C H_{3} C H_{2} C H_{2}$ – $N H C H_{3}$	N-methylpropan-1-amine (methylpropylamine)
$(C_{2} H_{5})_{2} N$ – $C_{3} H_{7}$	N,N-diethylpropan-1-amine (diethylpropylamine)
$H_{2} N (C H_{2})_{4}$ – $N H_{2}$	butane-1,4-diamine (putrescine)
$C H_{2} = C H$ – $C H_{2} C H_{2} N H_{2}$	but-3-en-1-amine
$C_{6} H_{11} N H_{2}$	cyclohexanamine (cyclohexylamine)
$C_{6} H_{5} C H_{2} N H_{2}$	phenylmethanamine (benzylamine)
$C_{6} H_{5} N H C_{2} H_{5}$	N-ethylaniline or N-ethylbenzenamine
m- $C H_{3} C_{6} H_{4} N H_{2}$	3-methylaniline (m-toluidine)
p- $O_{2} N C_{6} H_{4} N H_{2}$	4-nitroaniline (p-nitroaniline)

Example

Naming amines both ways

Give the IUPAC name and the class of each amine: (a) $(C H_{3})_{2} C H N H C H_{2} C H_{3}$ ; (b) $C H_{3} C H_{2} C H (C H_{3})$ – $C H_{2} N H_{2}$ ; (c) $C_{6} H_{5} N (C H_{3}) C H_{2} C H_{3}$ .

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

Quick check

1 of 3

What is the IUPAC name of

C H_{3} C H_{2} C H_{2} N (C H_{3})_{2}

dimethylpropylamineN,N-dimethylpropan-1-amine1-(dimethylamino)propaneN-propyl-N-methylmethanamine

Show answer

Answer (B) N,N-dimethylpropan-1-amine

Propyl is the largest group, so the parent is propan-1-amine and the two methyl groups are N-prefixes. Option A is the common name, and option D wrongly picks methyl as the parent.

Practice · 2 questions

Practice: classifying and naming: questions with answers

1
Quick answerEasyNCERT Intext Question 9.1
Match each species with its class.
Tap an item on the left, then its match on the right.
Show answer
Answer
$(C H_{3})_{3} C C H_{2} N H_{2}$ : primary amine
$C_{6} H_{5} C H_{2} N H C H_{3}$ : secondary amine
$C_{6} H_{5} N (C_{2} H_{5})_{2}$ : tertiary amine
$C_{6} H_{5} C H_{2} N (C H_{3})_{3}^{+} B r^{-}$ : quaternary salt
2,2-Dimethylpropan-1-amine has one group on N (the crowded carbon next door does not matter); N-methyl-1-phenylmethanamine has two; N,N-diethylaniline has three; benzyltrimethylammonium bromide has four and a positive charge.
2
Short answerHardNCERT Intext Question 9.2
The formula $C_{5} H_{13} N$ has 17 amine isomers: eight primary, six secondary and three tertiary.
1. (a)
  Write condensed formulas and IUPAC names for the six secondary amines.
  Show solution
  $C H_{3} N H$ – $(C H_{2})_{3}$ – $C H_{3}$ : N-methylbutan-1-amine
  $C H_{3} N H$ – $C H (C H_{3})$ – $C_{2} H_{5}$ : N-methylbutan-2-amine
  $C H_{3} N H$ – $C H_{2}$ – $C H (C H_{3})_{2}$ : N,2-dimethylpropan-1-amine
  $C H_{3} N H$ – $C (C H_{3})_{3}$ : N,2-dimethylpropan-2-amine
  $C_{2} H_{5} N H$ – $C H_{2}$ – $C H_{2} C H_{3}$ : N-ethylpropan-1-amine
  $C_{2} H_{5} N H$ – $C H (C H_{3})_{2}$ : N-ethylpropan-2-amine
2. (b)
  Do the same for the three tertiary amines.
  Show solution
  $(C H_{3})_{2} N$ – $C H_{2}$ – $C H_{2} C H_{3}$ : N,N-dimethylpropan-1-amine
  $(C H_{3})_{2} N$ – $C H (C H_{3})_{2}$ : N,N-dimethylpropan-2-amine
  $C H_{3}$ – $N (C_{2} H_{5})_{2}$ : N-ethyl-N-methylethanamine
3. Name the kind of structural isomerism shown by each pair: (i) N-methylbutan-1-amine and N-ethylpropan-1-amine; (ii) N-methylbutan-1-amine and N-methylbutan-2-amine; (iii) N-ethylpropan-1-amine and N,N-dimethylpropan-1-amine.
  Show solution
  (i) Metamerism: the same functional group, but different groups on either side of nitrogen. (ii) Position isomerism: the nitrogen is on C-1 in one and C-2 in the other. (iii) Functional isomerism: one is a secondary amine, the other tertiary.

9.4 Preparation of Amines

Every route to an amine either forms a new C–N bond or reduces a nitrogen group that is already bonded to carbon. To choose one, ask two questions: which class of amine do you need, and should the carbon count stay the same, rise by one or fall by one? Table 9.2 at the end of this section puts the six methods side by side.

Reduction of nitro compounds

Most arylamines start life as nitro compounds, because nitration is the easiest way to attach nitrogen to a benzene ring. Turning $- N O_{2}$ into $- N H_{2}$ is a six-electron reduction, and two kinds of reagent supply it: a metal with acid (tin or iron with hydrochloric acid), or hydrogen gas on a finely powdered catalyst such as Ni, Pd or Pt ((9.1)). The same reagents turn nitroalkanes into alkylamines; nitroethane, for example, gives ethanamine.

4-nitrotoluene to 4-methylaniline

\begin{aligned} C H_{3} C_{6} H_{4} N O_{2} + 3 H_{2} \\ \to_{e t h a n o l}^{P d} C H_{3} C_{6} H_{4} N H_{2} \\ + 2 H_{2} O \end{aligned}

(9.1)

For large batches, iron filings and hydrochloric acid are the cheapest pair, and they need surprisingly little acid. As the reduction runs, the iron(II) chloride it produces reacts with water and regenerates HCl, so the same acid is used again and again. Groups that are harder to reduce survive: 3-chloronitrobenzene gives 3-chloroaniline with its chlorine untouched.

with iron and acid

\begin{aligned} C l C_{6} H_{4} N O_{2} + 6 [H] \\ \overset{F e, H C l}{\to} C l C_{6} H_{4} N H_{2} \\ + 2 H_{2} O \end{aligned}

Ammonolysis of alkyl halides

Ammonia is a nucleophile, so it can displace a halide ion from an alkyl or benzyl halide by the SN2 route of Unit 6. Heating the halide with ammonia in ethanol in a sealed tube (about 373 K, so that the ammonia cannot boil away) gives the alkylammonium salt, and a strong base such as NaOH then releases the free amine ((9.2)). Splitting a C–X bond with ammonia like this is called ammonolysis.

ammonolysis, then freeing the amine

\begin{aligned} N H_{3} + R - X \overset{}{\to} R - N H_{3}^{+} X^{-} \\ R - N H_{3}^{+} X^{-} + N a O H \\ \overset{}{\to} R - N H_{2} + H_{2} O + N a X \end{aligned}

(9.2)

The catch is that the product is itself a nucleophile, and a slightly better one than ammonia, because its alkyl group feeds electron density to nitrogen. It competes for the remaining halide and climbs the staircase in Fig. 9.4: primary, then secondary, then tertiary amine, and finally a quaternary ammonium salt. Ammonolysis therefore gives a mixture.

A large excess of ammonia makes it far more likely that each halide molecule meets an ammonia molecule rather than an amine, so the primary amine becomes the main product. Iodides react fastest, then bromides, then chlorides (RI > RBr > RCl), because the C–I bond is the weakest of the three.

5/5Swamping the halide with ammonia keeps most of it on the first step.

Fig. 9.4Each amine formed can attack the halide again.NCERT p. 263

A staircase rises from ammonia at the bottom left. Each step adds an arrow labelled plus RX. Step 1 shows ammonia turning into the primary amine RNH2. Step 2 adds the secondary amine R2NH, step 3 the tertiary amine R3N and step 4 the quaternary salt R4N+ X− at the top right, labelled 1°, 2°, 3° and 4°. Step 5 keeps the whole staircase and adds a box saying that a large excess of ammonia gives mostly RNH2.

Example 9.1NCERT Example 9.1 · p. 263

Writing ammonolysis equations

Write equations, naming each organic product, for: (a) 1-bromopropane heated with ethanolic ammonia, followed step by step to the quaternary salt; (b) butan-1-amine treated with two moles of iodomethane.

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

Reduction of nitriles

A nitrile, $R - C \equiv N$ , adds four hydrogen atoms across its triple bond to give a primary amine in which nitrogen sits on a new $C H_{2}$ group. Lithium aluminium hydride, $L i A l H_{4}$ , does this, and so do hydrogen over nickel and sodium in ethanol. A nitrile is easily made from an alkyl halide and cyanide ion, so the sequence halide → nitrile → amine adds one carbon to the chain. This is the classic way to ascend the amine series.

nitrile to primary amine

\begin{aligned} R - C \equiv N + 4 [H] \\ \overset{L i A l H_{4}}{\to} R - C H_{2} - N H_{2} \end{aligned}

(9.3)

Reduction of amides

Amides are reduced by $L i A l H_{4}$ , followed by water, in a similar way: the C=O group becomes $C H_{2}$ , so the amine has the same number of carbons as the amide. An amide that already carries a group on nitrogen gives a secondary or tertiary amine; N-methylethanamide, $C H_{3} C O N H C H_{3}$ , for instance, gives N-methylethanamine, $C H_{3} C H_{2} N H C H_{3}$ .

amide to amine

\begin{aligned} R - C O - N H_{2} \to_{(i i) H_{2} O}^{(i) L i A l H_{4}} \\ R - C H_{2} - N H_{2} \end{aligned}

(9.4)

Example 9.2NCERT Example 9.2 · p. 264

Adding one carbon

Show how to make (a) pentan-1-amine from 1-bromobutane and (b) 3-methylbutan-1-amine from 1-chloro-2-methylpropane.

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

Gabriel phthalimide synthesis

Gabriel synthesis makes primary amines without the mixtures that spoil ammonolysis. Its nitrogen source is phthalimide, in which an N–H group sits between two C=O groups. Both carbonyl groups pull electron density from nitrogen, so the hydrogen is acidic enough to be removed by ethanolic potassium hydroxide. The resulting phthalimide anion is a good nucleophile and displaces halide from a primary alkyl halide (Fig. 9.5).

The N-alkylphthalimide formed cannot react again: its nitrogen has no hydrogen left, and its lone pair is shared with two carbonyl groups. Heating with aqueous sodium hydroxide then hydrolyses both C–N bonds, releasing the primary amine and the phthalate ion. The method fails for arylamines: a halogen on a benzene ring cannot be displaced by the phthalimide anion, for the reasons met in Unit 6.

3/4The anion displaces $X^{-}$ from a primary halide (SN2).

Fig. 9.5Nitrogen is delivered only once, so the product is a pure primary amine.NCERT p. 264

A benzene ring is fused to a five-membered ring containing two C=O groups and a nitrogen. Step 1, phthalimide, highlights the hydrogen on nitrogen between the two carbonyl groups. Step 2 shows that hydrogen removed by KOH: a negative charge on nitrogen and a K+ ion beside it. Step 3 shows an R group now bonded to the nitrogen, N-alkylphthalimide, after heating with R–X. Step 4 shows the ring opened by NaOH: both carbonyl carbons now carry O−, and the primary amine R–NH2 is released on the right.

Hoffmann bromamide degradation reaction

In 1881 August Wilhelm von Hofmann found that bromine and sodium hydroxide convert a primary amide into a primary amine with one carbon fewer.The chemist spelled his name Hofmann; NCERT names the reaction the Hoffmann bromamide degradation. This Hoffmann bromamide degradation works because the group R attached to the carbonyl carbon migrates onto the nitrogen, and the carbonyl carbon leaves as carbonate (Fig. 9.6). Chemists use it to step down a homologous series.

Hoffmann bromamide degradation

\begin{aligned} R - C O - N H_{2} + B r_{2} \\ + 4 N a O H \overset{}{\to} R - N H_{2} \\ + N a_{2} C O_{3} + 2 N a B r \\ + 2 H_{2} O \end{aligned}

(9.5)

The reaction works for aromatic amides too, and the migrating group keeps its structure. An aryl amide therefore gives an arylamine: 4-chlorobenzamide gives 4-chloroaniline.

3/3The carbonyl carbon leaves as carbonate: one carbon fewer.

Fig. 9.6Hexanamide (6 C) gives pentan-1-amine (5 C).NCERT p. 264

Five carbon circles in a row form the group R, joined to a sixth, dashed carbon that carries C=O and an NH2 group on the right. Step 1 is headed hexanamide, 6 C, and notes that the C=O carbon will leave. Step 2, headed Br2 + NaOH, adds a curved arrow from the chain to the nitrogen labelled R moves to N. Step 3 slides the five-carbon chain onto the nitrogen while the carbonyl carbon drifts away and fades, labelled as carbonate; the heading reads pentan-1-amine, 5 C, and the note says one carbon fewer.

Example 9.3NCERT Example 9.3 · p. 265

Working back from a Hoffmann product

(a) Which amide gives cyclohexanamine, $C_{6} H_{11} N H_{2}$ , in a Hoffmann degradation? (b) What amine does 4-methylbenzamide give with bromine and NaOH, and what does it give with $L i A l H_{4}$ ?

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

Table 9.2Choosing a route to an amine

Method	Gives	Carbons
Reduce a nitro compound	1° (main route to arylamines)	same
Ammonolysis of $R - X$	mixture of 1°, 2°, 3° and $R_{4} N^{+}$	same
$R - X$ → nitrile, then reduce	1°	one more than $R - X$
Reduce an amide	1° (2° or 3° from N-substituted amides)	same
Gabriel synthesis	pure 1° alkylamine	same
Hoffmann degradation	1° (alkyl or aryl)	one fewer than the amide

Quick check

1 of 4

You need butan-1-amine from 1-bromobutane, free of secondary and tertiary amines. Which route is best?

Heat with ethanolic ammoniaNaCN, then

L i A l H_{4}

Hoffmann degradation of the bromideGabriel synthesis

Show answer

Answer (D) Gabriel synthesis

Gabriel synthesis gives only the primary amine. Ammonolysis gives a mixture, the nitrile route gives pentan-1-amine (one carbon too many), and Hoffmann degradation needs an amide, not a halide.

Practice · 1 question

Practice: making amines: questions with answers

1
Short answerMediumNCERT Intext Question 9.3
How would you make each compound? Give reagents for every step.
1. 4-Chloroaniline from chlorobenzene
  Show solution
  Nitrate with concentrated $H N O_{3}$ and $H_{2} S O_{4}$ . Chlorine directs ortho/para, and the main product, 1-chloro-4-nitrobenzene, is separated. Reduce it with Fe and HCl (or Sn and HCl) and add NaOH to free 4-chloroaniline.
2. N,N-Diethylaniline from benzene
  Show solution
  Nitrate benzene ( $H N O_{3}$ and $H_{2} S O_{4}$ , about 330 K), reduce the nitrobenzene with Sn and HCl and free the aniline with NaOH. Then heat the aniline with two moles of bromoethane and a base such as $N a_{2} C O_{3}$ to absorb the HBr. More bromoethane would go on to the quaternary salt.
  $\begin{aligned} C_{6} H_{5} N H_{2} + 2 C_{2} H_{5} B r \\ \overset{}{\to} C_{6} H_{5} N (C_{2} H_{5})_{2} \\ + 2 H B r \end{aligned}$
3. Pentane-1,5-diamine from 1,3-dibromopropane
  Show solution
  The product has five carbons and the dibromide three, so add one carbon at each end through nitriles. Heat with ethanolic KCN to get pentanedinitrile, $N C (C H_{2})_{3} C N$ , then reduce both C≡N groups with $L i A l H_{4}$ (or $H_{2}$ over Ni) to $H_{2} N (C H_{2})_{5}$ – $N H_{2}$ .

9.5 Physical Properties

You have already met the smell. The lightest amines (methanamine, N-methylmethanamine, trimethylamine and ethanamine) boil below room temperature and escape as gases that smell of ammonia and stale fish. Lengthen the chain and the boiling point climbs, so propan-1-amine and its larger relatives are liquids and long-chain amines are waxy solids. Freshly distilled aniline is colourless; left in a part-filled bottle, it is slowly oxidised by air and darkens through yellow to brown.

In water the amino group mixes well. Its lone pair accepts a hydrogen bond from a water molecule's O–H, and in primary and secondary amines the N–H hydrogens can return the favour by bonding to water's oxygen. So methanamine, ethanamine and their small cousins dissolve freely. Every extra $C H_{2}$ adds hydrocarbon that water cannot grip, and beyond about six carbons an amine barely dissolves. Organic solvents such as ethanol, diethyl ether and even hexane take up amines of every size.

Primary and secondary amines also hydrogen-bond to each other: the N–H of one molecule bonds to the lone pair of a neighbour. A primary amine has two N–H hydrogens to offer, a secondary amine one and a tertiary amine none (Fig. 9.7). The more the molecules associate, the more energy it takes to pull them apart, so among isomeric amines the boiling points fall in the order 1° > 2° > 3°.

1/3Propan-1-amine: each molecule has two N–H hydrogens to donate; b.p. 321 K.

Fig. 9.7Fewer N–H hydrogens, fewer hydrogen bonds, lower boiling point (the three

C_{3} H_{9} N

isomers).NCERT Fig. 9.2 · p. 266

Six amine molecules are drawn in two rows, each a nitrogen with an R group above it. Dotted lines mark hydrogen bonds from an N–H hydrogen to a neighbouring nitrogen. In step 1 (primary) each molecule has two N–H hydrogens, one bonding to the next molecule in its row and one to a molecule in the row below, making a network; a bar gauge on the right reads 321 K. In step 2 (secondary) each molecule has one N–H, so only chains along each row remain, and the gauge falls to 310 K. In step 3 (tertiary) no N–H is left, no dotted lines remain, and the gauge falls to 276 K.

Table 9.3Boiling points of compounds with molar mass M close to 59 g mol⁻¹

Compound	M	b.p./K
$C H_{3} C H_{2} C H_{2}$ – $N H_{2}$ (1°)	59	321
$C H_{3} C H_{2}$ – $N H C H_{3}$ (2°)	59	310
$(C H_{3})_{3} N$ (3°)	59	276
$C H_{3} C H_{2}$ – $C H_{2} C H_{3}$ (alkane)	58	273
$C H_{3} C H_{2} C H_{2}$ – $O H$ (alcohol)	60	370

The table also places amines among neighbours of similar mass. They boil well above the alkane, which has only weak dispersion forces, but well below the alcohol. Oxygen (electronegativity 3.5) is more electronegative than nitrogen (3.0), so the O–H bond is more polar and O–H⋯O hydrogen bonds are stronger than N–H⋯N bonds.

Quick check

1 of 3

Arrange in order of boiling point, highest first.

propan-1-amine
trimethylamine
propan-1-ol
N-methylethanamine

Show answer

Answer

propan-1-ol
propan-1-amine
N-methylethanamine
trimethylamine

370 K, 321 K, 310 K and 276 K: O–H⋯O bonds beat N–H⋯N bonds, and among the amines more N–H hydrogens mean more hydrogen bonding.

9.6 Chemical Reactions

Two features of the amino group explain nearly all of its reactions. The lone pair on nitrogen makes amines bases (they take up protons) and nucleophiles (they attack electron-poor carbon). The number of N–H hydrogens decides what happens next: a primary amine has two to lose, a secondary amine one and a tertiary amine none, so the three classes often give different products. Chemists use those differences to tell them apart.

Basic character of amines

Amines react with acids the way ammonia does, by using the lone pair to bond to a proton. The products are substituted ammonium salts: propan-1-amine and hydrogen chloride give propylammonium chloride, and aniline gives anilinium chloride.

salt formation

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

These salts are ionic, so they dissolve in water but not in non-polar solvents such as ether, and a stronger base such as NaOH turns them back into the free amine. Together these facts give a simple separation. Shake a mixture of organic compounds in ether with dilute hydrochloric acid, and only the amine moves into the water layer, as its salt; adding NaOH to that layer releases the amine again.

Try this

Lemon after fish

You’ll need

a small piece of raw fish or a few prawns from the kitchen
half a lemon
two bowls of tap water
soap

Predict first

After handling the fish you rinse one hand in plain water and the other in lemon water. Which hand will smell less fishy?

both will smell the samethe hand rinsed in plain waterthe hand rinsed in lemon water

Rub both palms on the fish for a few seconds.
Rinse your left hand in plain water and your right hand in water with the juice of half a lemon.
Dry both hands and compare the smell of each.
Finish by washing both hands with soap.

Table 9.4What did you notice?

Rinse	Smell afterwards
Plain water
Lemon water

Your entries are saved automatically.

In water an amine takes a proton from a water molecule and leaves hydroxide ions behind, so its solution is alkaline. The equilibrium constant of that reaction, the base dissociation constant $K_{b}$ , measures the base's strength ((9.6)).

an amine in water

\begin{aligned} R N H_{2} + H_{2} O \\ \overset{}{⇌} R N H_{3}^{+} + O H^{-} \end{aligned}

base dissociation constant

K_{b} = \frac{[R N H_{3}^{+}] [O H^{-}]}{[R N H_{2}]}

(9.6)

where

$K_{b}$: base dissociation constant (water's concentration is folded into it)
$[X]$: equilibrium concentration in $m o l L^{- 1}$

Because $K_{b}$ values span many powers of ten, chemists usually quote p $K_{b}$ $= - \log K_{b}$ . A larger $K_{b}$ , or a smaller p $K_{b}$ , means a stronger base. Ammonia has p $K_{b} = 4.75$ . Simple alkylamines come in between about 3.0 and 4.2, so they are stronger bases than ammonia, while aniline, at 9.38, is weaker by almost five powers of ten.

Example

From pKb to Kb and pH

Methanamine has p $K_{b} = 3.38$ and aniline 9.38. (a) Find $K_{b}$ for each and compare them. (b) Find the pH of a $0.10 m o l L^{- 1}$ solution of methanamine at 298 K.

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

Structure and basicity: alkanamines versus ammonia

Start with the cleanest experiment: take the solvent away. In the gas phase a base can be ranked by its proton affinity, the energy released when the isolated molecule picks up a proton. Table 9.5 shows that each methyl group added to ammonia raises it by a few tens of kilojoules per mole, and ethyl groups raise it a little more: in the gas, 3° > 2° > 1° > $N H_{3}$ without exception.

Inductive effect. That steady climb is the +I effect at work. Alkyl groups release electron density, so the more of them nitrogen carries, the more readily its lone pair is shared with a proton, and the better the positive charge of the new cation is spread out and stabilised. A base is strong when its conjugate cation is comfortable.

Table 9.5Gas-phase proton affinity, PA, in kJ mol⁻¹ (energy released when the gaseous base takes up a proton; larger = more basic) and p

K_{b}

in water at 298 K

Base	PA	p $K_{b}$
$N H_{3}$	854	4.75
$C H_{3} N H_{2}$	899	3.38
$(C H_{3})_{2} N H$	930	3.27
$(C H_{3})_{3} N$	949	4.22
$C_{2} H_{5} N H_{2}$	912	3.29
$(C_{2} H_{5})_{2} N H$	952	3.00
$(C_{2} H_{5})_{3} N$	982	3.25
$C_{6} H_{5} C H_{2}$ – $N H_{2}$	–	4.70
$C_{6} H_{5} N H_{2}$	–	9.38

Water changes the picture, because it can stabilise the cations too. Two further effects come into play.

Solvation. In water a cation is also stabilised by hydrogen bonds from its N–H hydrogens to water molecules; this is solvation. $R N H_{3}^{+}$ has three such hydrogens, $R_{2} N H_{2}^{+}$ two and $R_{3} N H^{+}$ only one (Fig. 9.8). On this count alone the order would be reversed: 1° > 2° > 3°.

Crowding. Bulky groups round the nitrogen get in the way of the water molecules that solvate the cation. The bigger the groups, the more this matters; this steric hindrance hurts tertiary amines most.

2/4 $R N H_{3}^{+}$ is held by three hydrogen bonds to water.

Fig. 9.8+I effect alone in the gas; +I, solvation and crowding together in water.NCERT p. 268

Step 1, gas phase: a tertiary ammonium ion, N+ with three R groups and one H, has arrows on its R–N bonds showing the alkyl groups pushing electron density onto N+; the heading gives the gas-phase order 3° > 2° > 1° > NH3. Steps 2 to 4 show single ions in water, with water molecules (OH2) hydrogen-bonded to each N–H by dotted lines. Step 2, RNH3+, has three waters attached. Step 3, R2NH2+, has two. Step 4, R3NH+, has one water and three larger R groups crowding the nitrogen.

In water all three effects act together, and the measured orders are a compromise. For the methyl series the order is $(C H_{3})_{2} N H$ > $C H_{3} N H_{2}$ > $(C H_{3})_{3} N$ > $N H_{3}$ ; for the ethyl series it is $(C_{2} H_{5})_{2} N H$ > $(C_{2} H_{5})_{3} N$ > $C_{2} H_{5} N H_{2}$ > $N H_{3}$ . In both, the secondary amine wins: it gains from two alkyl groups and still keeps two N–H hydrogens for solvation. The tertiary amine is the one that moves, because for it the gain from a third alkyl group and the loss of solvation almost cancel, so a small change in the groups tips the balance.

Arylamines versus ammonia

Cyclohexanamine and aniline both carry an $N H_{2}$ group on a six-carbon ring, yet cyclohexanamine (p $K_{b}$ 3.36) is about a million times stronger as a base than aniline (9.38), which is weaker even than ammonia. The difference is the ring's π system. In aniline the nitrogen's lone pair lies beside the ring's p orbitals and spreads into them, placing negative charge on the ortho and para carbons (Fig. 9.9). Written as resonance, aniline has five contributing structures, three of them with the pair pushed into the ring.

Protonation locks that pair into a new N–H bond. The anilinium ion, $C_{6} H_{5} N H_{3}^{+}$ , is left with only the two Kekulé structures of an ordinary benzene ring, so aniline gives up its extra stabilisation when it accepts a proton, and the equilibrium lies far to the left. The $s p^{2}$ ring carbon also tugs on the nitrogen electrons inductively.

4/5The real molecule: δ− at C-2, C-4 and C-6.

Fig. 9.9Shared with the ring, the lone pair is less available for a proton.NCERT p. 269

A benzene ring drawn with alternating double bonds carries an NH2 group at the top. Step 1 shows the lone pair as two dots on nitrogen. Step 2 shows a curved arrow moving the pair into the N–C bond, now a double bond, with a positive charge on nitrogen and a negative charge on the upper-right ortho carbon. Step 3 shows the same with the negative charge moved to the para carbon at the bottom. Step 4 summarises the real molecule: delta minus at both ortho carbons and the para carbon, delta plus on nitrogen. Step 5 shows the anilinium ion, NH3+, whose pair is used for the extra N–H bond, so no charge spreads into the ring.

A group on the ring tunes the effect. At the para position a methyl group (4-methylaniline, p $K_{b}$ 8.92) or a methoxy group (4-methoxyaniline, 8.66) donates electron density back towards nitrogen and raises the basicity a little. A chlorine atom (4-chloroaniline, about 10.0) or a nitro group (4-nitroaniline, about 13.0) withdraws density and lowers it, and so do $- C O O H$ and $- S O_{3} H$ . Benzylamine (4.70) shows that the effect needs nitrogen on the ring itself: with one $C H_{2}$ in between, the amine behaves as an ordinary alkylamine.

InteractiveBasicity explorer: gas or water?

Buttons switch between the gas phase and water and between three series: ammonia with the methylamines, ammonia with the ethylamines, and ring-substituted anilines with benzylamine and cyclohexanamine (water only). The learner first taps the bases in the order they think, strongest first, then presses Reveal. Horizontal bars then show the measured values, proton affinity in kilojoules per mole for the gas or pKb for water, sorted strongest first, and the learner's guesses are marked right or wrong. Tapping any base fills a factor panel listing its alkyl groups, the N–H hydrogens its cation can offer to water, how crowded the nitrogen is and whether the lone pair is shared with a ring, with one sentence explaining its place in the order.

Fig. 9.10Pick a phase and a series, rank the bases yourself, then reveal the measured order and the reasons.

Example 9.4NCERT Example 9.4 · p. 270

Ranking bases in water

Rank these four bases from strongest to weakest in water: 4-nitroaniline, ammonia, N-methylmethanamine and 4-methylaniline.

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

Alkylation

Because amines are nucleophiles, they react with alkyl halides just as ammonia does (Unit 6), each step swapping an N–H hydrogen for an alkyl group. With enough halide the reaction runs all the way to the quaternary ammonium salt. Heating propan-1-amine with excess iodomethane and a base to remove the HI, for example, gives N,N,N-trimethylpropan-1-aminium iodide, $C_{3} H_{7} N (C H_{3})_{3}^{+} I^{-}$ .

Acylation

Mix a primary or secondary amine with an acid chloride and the nitrogen lone pair attacks the carbonyl carbon; chloride leaves, and an N–H hydrogen is lost as HCl. The product is an amide, and the overall swap of an N–H hydrogen for an acyl group, $R C O^{-}$ , is called acylation. Acid anhydrides do the same job more gently (a carboxylate leaves instead of chloride), and esters react only slowly.

N-propylethanamide

\begin{aligned} C_{3} H_{7} N H_{2} + C H_{3} C O C l \\ \overset{p y r i d i n e}{\to} C_{3} H_{7} N H C O C H_{3} \\ + H C l \end{aligned}

(9.7)

Each mole of acid chloride releases a mole of HCl, and that acid would protonate the remaining amine, turning it into an ammonium ion with no lone pair left to attack with. The reaction is therefore run in pyridine, or with another base stronger than the amine, which mops up the HCl and keeps the reaction moving towards the amide. Tertiary amines have no hydrogen on nitrogen, so they cannot end up as amides.

Two acylations matter later. Ethanoic anhydride converts aniline into N-phenylethanamide (acetanilide), the protected form used in Electrophilic substitution. Benzoyl chloride, $C_{6} H_{5} C O C l$ , gives benzamides, a reaction called benzoylation; N-methylmethanamine, for example, becomes N,N-dimethylbenzamide. A carboxylic acid itself is too weak an acylating agent at room temperature: with an amine it simply forms a salt, and the amide appears only on strong heating.

acetanilide

\begin{aligned} C_{6} H_{5} N H_{2} + (C H_{3} C O)_{2} O \\ \overset{}{\to} C_{6} H_{5} N H C O C H_{3} \\ + C H_{3} C O O H \end{aligned}

Carbylamine reaction

One of the most memorable tests in organic chemistry relies on a smell. Warm a little of an amine with chloroform and ethanolic KOH: if a powerful, sickening odour appears, the amine is primary. The smell comes from an isocyanide (or carbylamine), $R - N \equiv C$ , in which the carbon from chloroform ends up on nitrogen after both N–H hydrogens have been removed. Secondary and tertiary amines lack those two hydrogens and give no smell. Aliphatic and aromatic primary amines both respond, and the test goes by two names: the carbylamine reaction and the isocyanide test.

carbylamine reaction

\begin{aligned} R - N H_{2} + C H C l_{3} \\ + 3 K O H \overset{h e a t}{\to} R - N C \\ + 3 K C l + 3 H_{2} O \end{aligned}

(9.8)

Reaction with nitrous acid

Nitrous acid, $H N O_{2}$ , is unstable, so it is made in the flask as it is needed, from sodium nitrite and a mineral acid such as HCl. The three classes of amine react with it differently.

Primary aliphatic amines form alkyldiazonium salts that fall apart at once. Nitrogen gas bubbles out and the alkyl part ends up mainly as an alcohol, along with alkenes and other by-products. Each $N H_{2}$ group gives one $N_{2}$ molecule, so measuring the gas measures the amino groups, a method once used to estimate amino acids and proteins.
Primary aromatic amines at 273–278 K give arenediazonium salts that stay in solution as long as they are kept cold (Section 9.7).
Secondary amines give N-nitrosamines, $R_{2} N - N = O$ , usually yellow oils, while tertiary aliphatic amines simply dissolve in the acid as their salts.

primary aliphatic amine and nitrous acid

\begin{aligned} R - N H_{2} \overset{N a N O_{2} + H C l}{\to} [R - N_{2}^{+} C l^{-}] \\ \overset{H_{2} O}{\to} R - O H + N_{2} ↑ + H C l \end{aligned}

(9.9)

Reaction with arylsulphonyl chloride

Primary and secondary amines can also be sulphonylated. The reagent is benzenesulphonyl chloride, $C_{6} H_{5} S O_{2} C l$ , called Hinsberg's reagent; it behaves like an acid chloride and turns these amines into sulphonamides. The value of the test lies in what those sulphonamides then do in alkali (Fig. 9.11).

Primary amine: the product, such as $C_{6} H_{5} S O_{2} N H C_{4} H_{9}$ from butan-1-amine, still has one H on nitrogen. The strongly electron-withdrawing $S O_{2}$ group makes that H acidic, so the sulphonamide dissolves in aqueous KOH as its potassium salt. Acidifying brings it back out as a solid.
Secondary amine: the product, such as $C_{6} H_{5} S O_{2} N (C H_{3}) C_{4} H_{9}$ from N-methylbutan-1-amine, has no H on nitrogen. With nothing to lose to the alkali, it stays undissolved in KOH.
Tertiary amine: no N–H, no reaction. The amine stays unchanged and dissolves only when acid is added, as its salt.

a primary amine and Hinsberg's reagent

\begin{aligned} C_{6} H_{5} S O_{2} C l + H_{2} N C_{4} H_{9} \\ \overset{}{\to} C_{6} H_{5} S O_{2} N H C_{4} H_{9} \\ + H C l \end{aligned}

2/31°: the sulphonamide dissolves. 2°: a solid. 3°: no reaction.

Fig. 9.11Solubility in KOH, before and after adding acid, sorts the three classes.NCERT p. 272

Three test tubes are labelled 1° amine, 2° amine and 3° amine. Step 1: benzenesulphonyl chloride is added to each and the tubes are shaken; each liquid amine floats as an oily layer. Step 2: after shaking with aqueous KOH, the 1° tube is clear, the 2° tube has solid at the bottom and the 3° tube still has an oily layer. Step 3: after adding dilute hydrochloric acid, solid forms in the 1° tube, the solid stays in the 2° tube and the 3° tube becomes clear as the amine dissolves as its salt.

Run on a mixture, the same reactions also separate the three amines: the unreacted tertiary amine can be distilled or extracted off, and the two sulphonamides are parted by their solubility in alkali, then hydrolysed back to the amines. Modern laboratories prefer p-toluenesulphonyl chloride, $C H_{3} C_{6} H_{4} S O_{2} C l$ ; it reacts in the same way and its sulphonamides crystallise better.

Example

Which amine is which?

Bottles X, Y and Z contain pentan-1-amine, N-ethylpropan-1-amine and N,N-dimethylpropan-1-amine (all $C_{5} H_{13} N$ ), in some order. X gives a foul smell when warmed with chloroform and ethanolic KOH; Y and Z do not. With benzenesulphonyl chloride and KOH, Y gives a solid that does not dissolve, and Z shows no change. Identify X, Y and Z.

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

Electrophilic substitution

Look again at Fig. 9.9: the lone pair of $- N H_{2}$ feeds electron density into the ring, mostly at the ortho and para carbons. The amino group is therefore a powerful activating group and an ortho/para director. Aniline reacts with electrophiles so much faster than benzene does that the real problem is stopping after one substitution.

Bromination. Add bromine water to aniline at room temperature and a white solid appears at once: 2,4,6-tribromoaniline. All three activated positions take a bromine, and no catalyst is needed.

2,4,6-tribromoaniline

\begin{aligned} C_{6} H_{5} N H_{2} + 3 B r_{2} \\ \overset{}{\to} C_{6} H_{2} B r_{3} N H_{2} + 3 H B r \end{aligned}

To stop at one bromine, chemists protect the amino group by acetylating it with ethanoic anhydride (acetanilide). In acetanilide the nitrogen lone pair is also shared with the neighbouring C=O group (the pair can be drawn as an N=C double bond with a negative charge on oxygen), so it donates less to the ring. The ring is still activated, but gently. Bromine in ethanoic acid now adds a single bromine, mostly para, because the bulky $- N H C O C H_{3}$ group shields the ortho positions. Hydrolysis with acid or alkali then removes the acetyl group to give 4-bromoaniline (Fig. 9.12).

3/4One Br, para to the bulky amide group.

Fig. 9.12Cap, substitute once, uncap: the standard way to tame the amino group.NCERT p. 272

One benzene ring stays in the centre throughout. Step 1: with bromine water, aniline picks up three bromine atoms, at both ortho positions and the para position, labelled too reactive. Step 2: the NH2 group is capped as NHCOCH3 by ethanoic anhydride, labelled tamer ring. Step 3: with bromine in ethanoic acid, only one bromine adds, at the para position. Step 4: hydrolysis removes the cap, giving 4-bromoaniline.

Nitration. Mixed acid attacks aniline in two unwanted ways. Nitric acid oxidises much of it to tarry by-products, and in the strongly acidic mixture much of the aniline is protonated to the anilinium ion. The $- N H_{3}^{+}$ group has no lone pair and withdraws electrons, so it is deactivating and meta-directing. The nitro products therefore contain roughly as much meta isomer as para, with very little ortho. Protecting the amino group as acetanilide first solves both problems: nitration with $H N O_{3}$ and $H_{2} S O_{4}$ near 288 K then gives mainly 4-nitroacetanilide, which is hydrolysed to 4-nitroaniline.

Sulphonation. Concentrated sulphuric acid first forms the salt anilinium hydrogensulphate. Heated to 453–473 K this gives 4-aminobenzenesulphonic acid, called sulphanilic acid. The molecule carries both an acidic $- S O_{3} H$ group and a basic $- N H_{2}$ group, so it exists mainly as a zwitterion, $H_{3} N^{+} C_{6} H_{4} S O_{3}^{-}$ .

sulphanilic acid

\begin{aligned} C_{6} H_{5} N H_{2} + H_{2} S O_{4} \\ \overset{}{\to} C_{6} H_{5} N H_{3}^{+} H S O_{4}^{-} \\ \overset{453-473 K}{\to} H_{2} N C_{6} H_{4} S O_{3} H \\ + H_{2} O \end{aligned}

No Friedel–Crafts reactions. Ring alkylation or acylation with aluminium chloride fails for aniline. The Lewis acid $A l C l_{3}$ grabs the nitrogen lone pair, which leaves nitrogen positively charged, and a strongly activating group becomes a strongly deactivating one.

Practice · 5 questions

Practice: reactions of amines: questions with answers

1
Quick answerMediumNCERT Intext Question 9.4
Arrange each set as asked.
1. (a)
  Increasing basic strength in water
  aniline
  4-methylaniline
  4-nitroaniline
  cyclohexanamine
  Show answer
  Answer
  4-nitroaniline
  aniline
  4-methylaniline
  cyclohexanamine
  p $K_{b}$ about 13.0, 9.38, 8.92 and 3.36. Ring delocalisation weakens the anilines, a nitro group weakens further, a methyl group strengthens slightly, and cyclohexanamine has no ring on nitrogen at all.
2. (b)
  Increasing basic strength in water
  $(C H_{3})_{3} N$
  $N H_{3}$
  $C H_{3} N H_{2}$
  $(C H_{3})_{2} N H$
  Show answer
  Answer
  $N H_{3}$
  $(C H_{3})_{3} N$
  $C H_{3} N H_{2}$
  $(C H_{3})_{2} N H$
  p $K_{b}$ 4.75, 4.22, 3.38 and 3.27: in water the secondary amine is strongest and the tertiary amine falls below the primary because its cation is poorly solvated.
2
Short answerEasyNCERT Intext Question 9.5
Complete each acid–base reaction and name the product.
1. $(C H_{3})_{2} N H$ with $H B r$
  Show solution
  $(C H_{3})_{2} N H + H B r \overset{}{\to} (C H_{3})_{2} N H_{2}^{+} B r^{-}$
  Dimethylammonium bromide (N-methylmethanaminium bromide).
2. $C_{6} H_{5} N H_{2}$ with $H_{2} S O_{4}$
  Show solution
  $\begin{aligned} C_{6} H_{5} N H_{2} + H_{2} S O_{4} \\ \overset{}{\to} C_{6} H_{5} N H_{3}^{+} H S O_{4}^{-} \end{aligned}$
  Anilinium hydrogensulphate.
Short answerMediumNCERT Intext Question 9.6
Benzylamine is heated with excess iodoethane, with sodium carbonate present to neutralise the acid. Write the overall equation and name the final product.
Show solution
$\begin{aligned} C_{6} H_{5} C H_{2} N H_{2} + 3 C_{2} H_{5} I \\ + N a_{2} C O_{3} \\ \overset{}{\to} C_{6} H_{5} C H_{2} N (C_{2} H_{5})_{3}^{+} I^{-} \\ + 2 N a I + C O_{2} + H_{2} O \end{aligned}$
Two ethyl groups replace the two N–H hydrogens, and the tertiary amine then takes a third ethyl group. The product is N-benzyl-N,N-diethylethanaminium iodide (benzyltriethylammonium iodide). Sodium carbonate neutralises the two moles of HI formed in the first two steps.
Short answerEasyNCERT Intext Question 9.7
Write the equation for the reaction of 4-methylaniline with ethanoyl chloride in pyridine, and name the product.
Show solution
$\begin{aligned} C H_{3} C_{6} H_{4} N H_{2} + C H_{3} C O C l \\ \overset{}{\to} C H_{3} C_{6} H_{4} N H C O C H_{3} \\ + H C l \end{aligned}$
The product is N-(4-methylphenyl)ethanamide (4-methylacetanilide); pyridine takes up the HCl.
5
Short answerMediumNCERT Intext Question 9.8
Five aromatic amines have the formula $C_{7} H_{9} N$ .
1. (a)
  Write their structures and names.
  Show solution
  benzylamine (phenylmethanamine), $C_{6} H_{5}$ – $C H_{2} N H_{2}$
  N-methylaniline, $C_{6} H_{5}$ – $N H C H_{3}$
  2-, 3- and 4-methylaniline (o-, m- and p-toluidine), $C H_{3} C_{6} H_{4}$ – $N H_{2}$
2. Which one gives off nitrogen gas at once with $N a N O_{2}$ and HCl at 273 K, and why?
  Show solution
  Benzylamine. Its $N H_{2}$ is on a $C H_{2}$ group, not on the ring, so it behaves as a primary aliphatic amine: its diazonium ion has no ring to stabilise it and loses $N_{2}$ immediately. The three methylanilines form diazonium salts that stay in the cold solution, and N-methylaniline, a secondary amine, gives an N-nitrosamine.

Diazonium Salts

A diazonium salt pairs an aryl group carrying $- N_{2}^{+}$ with an anion: $A r N_{2}^{+} X^{-}$ , where $X^{-}$ may be $C l^{-}$ , $B r^{-}$ , $H S O_{4}^{-}$ , $B F_{4}^{-}$ and so on. The $- N_{2}^{+}$ group, that is $- \overset{+}{N} \equiv N$ , is the diazonium group. Names are built from the arene, the word diazonium and the anion: $C_{6} H_{5} N_{2}^{+} C l^{-}$ is benzenediazonium chloride, and the salt from 4-nitroaniline and fluoroboric acid is 4-nitrobenzenediazonium tetrafluoroborate.

Primary aliphatic amines also form diazonium ions, but these lose nitrogen the moment they form (Reaction with nitrous acid). Arenediazonium ions last much longer, provided they are kept cold. The next four sections show how they are made, what they are like, how they react and why they are so useful.

9.7 Method of Preparation of Diazonium Salts

Making a diazonium salt is quick, but it must be done cold. The amine is dissolved in an excess of dilute hydrochloric acid and chilled in ice to 273–278 K (0–5 °C); sodium nitrite solution is then run in, and the nitrous acid it forms in the acid converts $- N H_{2}$ into $- N_{2}^{+}$ . This step is diazotisation. Nobody bottles the product: the cold solution goes straight into the next reaction (lab steps below).

diazotisation of aniline

\begin{aligned} C_{6} H_{5} N H_{2} + N a N O_{2} \\ + 2 H C l \overset{273-278 K}{\to} \\ C_{6} H_{5} N_{2}^{+} C l^{-} + N a C l \\ + 2 H_{2} O \end{aligned}

(9.10)

Why so cold? The arenediazonium ion is stabilised because its positive charge is shared with the ring by resonance, but only just (Fig. 9.13). Above about 278 K water attacks it, nitrogen gas is lost and a phenol forms. An alkyldiazonium ion has no ring to share its charge with, so it loses $N_{2}$ instantly, even in the cold.

2/3The + charge spreads onto the ring; the ion lasts at 273–278 K.

Fig. 9.13A shared charge buys the arenediazonium ion time in the cold, but not at room temperature.NCERT p. 274

Step 1 shows an alkyldiazonium ion, R–N2+, breaking at once into R+ and nitrogen gas, with bubbles, even at 273 K. Step 2 shows a benzene ring carrying N+≡N, with small plus signs on the two ortho carbons and the para carbon to show the charge shared with the ring, and a cold label reading 273–278 K. Step 3 shows the same ring after warming above 278 K: the group is now OH, bubbles of nitrogen rise, and the label reads phenol plus N2 plus HCl.

Step by step

Diazotising aniline in the lab

Step 1 of 5

Dissolve

Stir the aniline into about three equivalents of dilute hydrochloric acid; it dissolves as anilinium chloride.

1 / 5

Quick check

1 of 3

Why is diazotisation carried out at 273–278 K?

so that sodium nitrite dissolvesto speed the reaction upto keep the aniline solidso that the diazonium salt does not lose

N_{2}

and form a phenol

Show answer

Answer (D) so that the diazonium salt does not lose $N_{2}$ and form a phenol

The arenediazonium ion is only slightly stabilised; warm water attacks it, nitrogen is lost and a phenol forms. Cooling slows the reaction rather than speeding it up.

9.8 Physical Properties

You will rarely see solid benzenediazonium chloride. It does form colourless crystals, but chemists almost never isolate them, because the dry salt can decompose violently. In water it dissolves easily and survives as long as the solution stays cold; warm it and the salt starts reacting with the water. Changing the anion changes everything: benzenediazonium tetrafluoroborate, $C_{6} H_{5} N_{2}^{+} B F_{4}^{-}$ , does not dissolve in water and is stable enough to be filtered off and kept at room temperature.

9.9 Chemical Reactions

Diazonium salts react in two broad ways. In most reactions the $- N_{2}^{+}$ group leaves as nitrogen gas and another group takes its place on the ring. In coupling reactions the two nitrogen atoms stay, as a bridge between two aromatic rings.

Reactions in which nitrogen is replaced

The diazonium group is an outstanding leaving group. It departs as $N_{2}$ , one of the most stable molecules there is, and the gas bubbles out of the solution so the reaction cannot reverse. That is why so many different groups can take its place (Table 9.6).

Table 9.6Replacing the diazonium group of

A r N_{2}^{+}

Reagent	Product (note)
$C u C l$ / $H C l$	$A r C l$ (Sandmeyer)
$C u B r$ / $H B r$	$A r B r$ (Sandmeyer)
$C u C N$ / $K C N$	$A r C N$ (Sandmeyer)
Cu powder with HCl or HBr	$A r C l$ or $A r B r$ (Gattermann, lower yield)
$K I$ solution	$A r I$ (no catalyst needed)
$H B F_{4}$ , then heat	$A r F$ (via $A r N_{2}^{+} B F_{4}^{-}$ )
$H_{3} P O_{2}$ and water, or ethanol	$A r H$ (group removed)
water, warm	$A r O H$ (a phenol)
$H B F_{4}$ , then $N a N O_{2}$ and Cu, heat	$A r N O_{2}$ (a nitro compound)

Chloride, bromide and cyanide. These three come in with the matching copper(I) salt: CuCl, CuBr or CuCN. Copper(I) is what makes these Sandmeyer reactions work; chloride or cyanide ions on their own do little. An older variant, the Gattermann reaction, uses copper powder with HCl or HBr instead, but its yields are usually poorer.

Sandmeyer reaction with cyanide

\begin{aligned} A r N_{2}^{+} C l^{-} + C u C N \\ \overset{}{\to} A r C N + N_{2} ↑ + C u C l \end{aligned}

Iodide. Iodide needs no copper at all. Stirring potassium iodide solution into the diazonium salt is enough to give the aryl iodide, a compound that is awkward to make by direct iodination.

aryl iodide

\begin{aligned} A r N_{2}^{+} C l^{-} + K I \\ \overset{}{\to} A r I + K C l + N_{2} ↑ \end{aligned}

(9.11)

Fluoride. Fluorine goes on in two stages. First, fluoroboric acid swaps the anion, and the insoluble arenediazonium tetrafluoroborate comes out as a solid. Then the dried solid is heated, and it breaks down into the aryl fluoride, boron trifluoride and nitrogen.

aryl fluoride

\begin{aligned} A r N_{2}^{+} C l^{-} + H B F_{4} \\ \overset{}{\to} A r N_{2}^{+} B F_{4}^{-} + H C l \\ A r N_{2}^{+} B F_{4}^{-} \overset{h e a t}{\to} \\ A r F + B F_{3} + N_{2} ↑ \end{aligned}

Hydrogen. To delete the group altogether, use a mild reducing agent. Phosphinic (hypophosphorous) acid, $H_{3} P O_{2}$ , ends up as phosphorous acid and ethanol ends up as ethanal; either way the ring gains a hydrogen where the nitrogen was. Removing an amino group from a ring sounds pointless, but it is one of the most useful tricks in synthesis (Section 9.10).

removing the group

\begin{aligned} A r N_{2}^{+} C l^{-} + H_{3} P O_{2} \\ + H_{2} O \overset{}{\to} A r H + N_{2} \\ + H_{3} P O_{3} + H C l \end{aligned}

Hydroxyl. No extra reagent is needed for a phenol: water itself replaces the diazonium group once the solution warms past about 283 K.

phenol

\begin{aligned} A r N_{2}^{+} C l^{-} + H_{2} O \\ \overset{}{\to} A r O H + N_{2} ↑ + H C l \end{aligned}

Nitro. A nitro group can even be put back where the amine came from: warming the tetrafluoroborate salt with aqueous sodium nitrite and copper powder gives the nitroarene.

Coupling reactions

In a coupling reaction the diazonium ion keeps its nitrogen and acts as an electrophile. It is a weak one, so it attacks only strongly activated rings, those of phenols and arylamines, almost always at the para position (or ortho if para is taken). The two rings end up joined by an azo group, –N=N–. The long conjugated system that results absorbs visible light, so azo compounds are brightly coloured (Fig. 9.14).

3/3Loss of $H^{+}$ restores the ring: orange 4-hydroxyazobenzene.

Fig. 9.14The diazonium ion is the electrophile; the activated ring supplies the electrons.NCERT p. 276

Step 1: a benzene ring carrying N+≡N on its right faces a phenoxide ring, with O− on its far right, a little way off; the phenoxide's para carbon carries an H, and a curved arrow runs from that ring to the end nitrogen. Step 2: the phenoxide ring has moved in and a new bond joins the end nitrogen to its para carbon, which now carries both the nitrogen and an H. Step 3: the H has gone as H+, the nitrogen–nitrogen bond is now a double bond, O− has become OH, and the product is labelled 4-hydroxyazobenzene, an orange azo dye whose –N=N– group links the two rings.

With phenol in mildly alkaline solution, where it is present as the very reactive phenoxide ion, benzenediazonium chloride gives orange 4-hydroxyazobenzene. With aniline in mildly acidic solution it gives yellow 4-aminoazobenzene. Both are electrophilic substitutions on the phenol or aniline ring. The classic laboratory test for a primary aromatic amine uses 2-naphthol: diazotise the amine at 273–278 K, add the cold solution to 2-naphthol in NaOH, and an orange-red dye precipitates at once ((9.12)).

the 2-naphthol dye test

\begin{aligned} C_{6} H_{5} N_{2}^{+} + C_{10} H_{7} O H \\ + O H^{-} \overset{}{\to} \\ C_{6} H_{5} N = N C_{10} H_{6} O H + H_{2} O \end{aligned}

(9.12)

Let's explore

An azo dye from your chemistry lab

You’ll need

methyl orange indicator solution
dilute hydrochloric acid
dilute sodium hydroxide solution
distilled water
three test tubes
safety goggles

Predict first

Methyl orange is an azo dye. What do you expect when a few drops go into acid, water and alkali?

the same orange colour in all threered in acid, yellow-orange in water and yellow in alkalicolourless in acid and orange in alkali

Half-fill three test tubes with dilute HCl, distilled water and dilute NaOH.
Add two drops of methyl orange to each and shake.
Record each colour, then add acid drop by drop to the alkali tube and watch.

Table 9.7Colours of methyl orange

Tube	Colour
Acid
Water
Alkali

Your entries are saved automatically.

Methyl orange is red in acid and yellow in alkali; it changes between about pH 3.1 and 4.4, so distilled water (pH about 5 to 7) shows yellow-orange. The dye is made by coupling diazotised sulphanilic acid with N,N-dimethylaniline, so two reactions from this chapter meet in one molecule. Protonating the azo group alters the conjugated system, which changes the light absorbed and the colour you see.

InteractiveDiazonium reaction map

A card at the top shows benzenediazonium chloride kept at 273 to 278 K. Below it, eleven reagent buttons cover the Sandmeyer reactions with copper(I) chloride, bromide and cyanide, the Gattermann reaction, potassium iodide, fluoroboric acid then heat, phosphinic acid, warm water, fluoroborate then sodium nitrite with copper, and coupling with phenol in alkali or aniline in acid. Choosing a reagent draws the product on a benzene ring and shows its name, the balanced equation, the name of the reaction, whether nitrogen gas is lost, and why the route is useful. Pressing Play then shows nitrogen bubbles leaving or the azo link forming. A Target practice mode names a product, such as fluorobenzene or 4-hydroxyazobenzene, and asks the learner to pick the right reagent, with feedback after each choice.

Fig. 9.15Choose a reagent to see what replaces or keeps the diazonium group, or switch to target practice.

Timeline

Amines, dyes and named reactions

1856
Perkin's mauve
Eighteen-year-old William Perkin oxidised impure aniline while trying to make quinine and obtained mauveine, the first synthetic dye.
1858
Diazonium salts
Peter Griess treated aromatic amines with nitrous acid and opened the way to azo dyes.
1881
Hofmann's degradation
Amides were found to lose a carbon atom and give amines.
1884
Sandmeyer reaction
Traugott Sandmeyer, trying to make phenylacetylene, found that copper(I) chloride gave chlorobenzene from a diazonium salt.
1887
Gabriel synthesis
Siegmund Gabriel used phthalimide to make pure primary amines.
1890
Hinsberg test
Oscar Hinsberg used benzenesulphonyl chloride to tell the classes of amine apart.
1932
Prontosil
An azo dye became the first sulpha drug.

9.10 Importance of Diazonium Salts in Synthesis of Aromatic Compounds

Put the replacement reactions together and a diazonium salt becomes a hub. From one aromatic amine you can reach the aryl fluoride, chloride, bromide, iodide, nitrile, phenol, nitro compound or the bare arene. Several of these are hard or impossible to make directly. Fluorine reacts explosively with benzene, so aryl fluorides cannot be made by direct fluorination. Iodination with iodine alone is slow and reversible and needs an oxidising agent. And chlorobenzene shrugs off cyanide ion, so the only easy road to benzonitrile runs through the diazonium salt.

The second great use is steering. An amino group, or its tamer acetyl form, is a strong ortho/para director. You can put one on a ring, use it to guide a new substituent into place, and then delete it through the diazonium salt and $H_{3} P O_{2}$ . The result is a substitution pattern that no direct reaction would give, as Example 9.5 shows.

Example 9.5NCERT Example 9.5 · p. 277

Making 3-bromotoluene

Bromination of toluene gives 2- and 4-bromotoluene, never the 3-isomer, because $- C H_{3}$ directs ortho/para. Show how to make 3-bromotoluene from 4-methylaniline.

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

6/6 $H_{3} P O_{2}$ swaps $- N_{2}^{+}$ for H: 3-bromotoluene.

Fig. 9.16Use the amino group to place the bromine, then delete it.NCERT Example 9.5 · p. 277

One benzene ring carries a methyl group at the bottom throughout. Step 1: NH2 at the top, 4-methylaniline. Step 2: the NH2 is capped as NHCOCH3 by ethanoic anhydride. Step 3: bromine adds at the upper-right carbon, next to the cap. Step 4: hydrolysis turns the cap back into NH2. Step 5: the NH2 becomes N2+ with sodium nitrite and HCl at 273 to 278 K. Step 6: phosphinic acid replaces N2+ by H, leaving 3-bromotoluene with Br meta to CH3.

Quick check

1 of 3

Which compound cannot be made well by direct substitution on benzene, but is easy to make from benzenediazonium chloride?

chlorobenzenenitrobenzenefluorobenzenebromobenzene

Show answer

Answer (C) fluorobenzene

Direct fluorination is explosive and uncontrollable, but heating benzenediazonium tetrafluoroborate gives fluorobenzene cleanly. Chlorobenzene, nitrobenzene and bromobenzene are all made directly.

Practice · 1 question

Practice: using diazonium salts: questions with answers

1
Short answerHardNCERT Intext Question 9.9
Show how to carry out each conversion.
1. 4-Chloroaniline into 1-chloro-4-fluorobenzene
  Show solution
  Diazotise ( $N a N O_{2}$ and HCl, 273–278 K), then add $H B F_{4}$ to precipitate 4-chlorobenzenediazonium tetrafluoroborate. Filter, dry carefully and heat it: $N_{2}$ and $B F_{3}$ leave and fluorine takes the place of the diazonium group.
2. 4-Nitroaniline into 1,3-dibromo-5-nitrobenzene
  Show solution
  Shake with bromine water. The strongly activating $- N H_{2}$ puts Br on both free ortho positions (its para position already holds $N O_{2}$ ), giving 2,6-dibromo-4-nitroaniline. Diazotise in strong acid at 273–278 K and treat with $H_{3} P O_{2}$ to replace $- N_{2}^{+}$ by H. The two bromine atoms and the nitro group are now all meta to one another.

Recap

Amines: key points and summary

Amines replace one, two or three hydrogens of ammonia by alkyl or aryl groups; nitrogen is $s p^{3}$ , pyramidal and keeps a lone pair (Section 9.1).
Primary, secondary and tertiary amines carry one, two or three carbon groups on N; four groups give a quaternary ammonium ion (Section 9.2).
IUPAC names are alkanamines with N-prefixes for the smaller groups on nitrogen; aniline (benzenamine) is the parent arylamine (Section 9.3).
Amines come from reducing nitro compounds, nitriles (one carbon more) or amides, from ammonolysis (a mixture), from Gabriel synthesis (pure 1°) and from Hoffmann degradation (one carbon fewer) (Section 9.4).
Hydrogen bonding through N–H makes boiling points fall 1° > 2° > 3° and lets small amines dissolve in water; alcohols boil higher than amines (Section 9.5).
Amines are bases that form salts with acids; a smaller p $K_{b}$ means a stronger base (Basic character of amines).
In the gas phase basicity follows the +I effect, 3° > 2° > 1° > $N H_{3}$ ; in water solvation and crowding reorder it, and the secondary amine is strongest in both the methyl and ethyl series (Structure and basicity: alkanamines versus ammonia).
Arylamines are much weaker bases because the lone pair is delocalised into the ring; donor groups raise and acceptor groups lower their basicity (Arylamines versus ammonia).
Amines are alkylated up to quaternary salts, and primary and secondary amines are acylated; the carbylamine test detects 1° amines and the Hinsberg test sorts 1°, 2° and 3° amines (Section 9.6).
$- N H_{2}$ strongly activates the ring towards ortho/para attack; acetylation tames it, and in strong acid the anilinium ion directs meta (Electrophilic substitution).
Diazotisation at 273–278 K turns $A r N H_{2}$ into $A r N_{2}^{+} X^{-}$ , stabilised by resonance but unstable when warm or dry (Section 9.7).
Replacing $- N_{2}^{+}$ gives aryl chlorides, bromides and nitriles (Sandmeyer), iodides, fluorides, arenes, phenols and nitro compounds; coupling with phenols or arylamines gives azo dyes (Section 9.9, Section 9.10).

Practice · 18 questions

Practice questions with answers (NCERT “Exercises”)

1
Short answerEasyNCERT Exercises Q9.1
Give the IUPAC name of each amine and say whether it is primary, secondary or tertiary.
1. $(C H_{3})_{2} C H$ – $C H_{2} C H_{2} N H_{2}$
  Show solution
  3-Methylbutan-1-amine; primary.
2. $C H_{3} C H_{2} C H_{2}$ – $N H C H_{2} C H_{3}$
  Show solution
  N-Ethylpropan-1-amine; secondary.
3. $(C H_{3})_{2} N$ – $(C H_{2})_{3} C H_{3}$
  Show solution
  N,N-Dimethylbutan-1-amine; tertiary.
4. $C_{6} H_{5}$ – $C H_{2} C H_{2} N H_{2}$
  Show solution
  2-Phenylethan-1-amine (2-phenylethanamine); primary.
5. p- $C H_{3} O C_{6} H_{4} N H_{2}$
  Show solution
  4-Methoxyaniline (4-methoxybenzenamine); primary.
6. $C_{6} H_{5} N H$ – $C H (C H_{3})_{2}$
  Show solution
  N-(Propan-2-yl)aniline, also called N-isopropylaniline; secondary.
7. o- $C l C_{6} H_{4} N H_{2}$
  Show solution
  2-Chloroaniline (2-chlorobenzenamine); primary.
2
Short answerMediumNCERT Exercises Q9.2
For each pair, name a chemical test whose result differs for the two compounds, and say what you would observe.
1. Propan-1-amine and N-methylethanamine
  Show solution
  Carbylamine test: warm each with $C H C l_{3}$ and ethanolic KOH. Propan-1-amine (1°) gives the foul smell of propyl isocyanide; N-methylethanamine (2°) does not.
2. N-Methylaniline and N,N-dimethylaniline
  Show solution
  Hinsberg test: shake each with $C_{6} H_{5} S O_{2} C l$ and KOH. N-Methylaniline (2°) gives a sulphonamide that stays undissolved as a solid; N,N-dimethylaniline (3°) does not react.
3. Cyclohexanamine and aniline
  Show solution
  Azo-dye test: treat each with $N a N O_{2}$ and HCl at 273–278 K, then add the solution to 2-naphthol in NaOH. Aniline gives an orange-red precipitate; cyclohexanamine gives off $N_{2}$ and no dye.
4. 4-Methylaniline and benzylamine
  Show solution
  The same azo-dye test: only 4-methylaniline, an arylamine, gives a coloured dye. Bromine water also works: it gives a white precipitate of 2,6-dibromo-4-methylaniline only with 4-methylaniline.
5. Aniline and N-ethylaniline
  Show solution
  Carbylamine test: only aniline (1°) gives the isocyanide smell; N-ethylaniline (2°) does not.
3
Long answerMediumNCERT Exercises Q9.3
Account for each observation.
1. 4-Methylaniline is a slightly stronger base than aniline, but 4-nitroaniline is a much weaker one.
  Show solution
  In all three the lone pair is partly delocalised into the ring. A para methyl group releases electrons, pushing density back towards nitrogen and stabilising the cation, so 4-methylaniline is a little stronger (p $K_{b}$ 8.92 against 9.38). A para nitro group withdraws electrons strongly by resonance and induction, pulling the lone pair further into the ring, so 4-nitroaniline is weaker by about 3.6 p $K_{b}$ units (p $K_{b}$ about 13.0).
2. Propan-1-amine mixes with water in all proportions, but N,N-dimethylaniline hardly dissolves.
  Show solution
  Propan-1-amine is small: its lone pair accepts hydrogen bonds from water, its two N–H hydrogens donate them, and its hydrocarbon part is short. N,N-Dimethylaniline has no N–H hydrogen, its lone pair is partly tied up with the ring, and its eight-carbon hydrocarbon part repels water.
3. Adding ethanamine solution to iron(II) sulphate solution gives a dirty green precipitate.
  Show solution
  Ethanamine is a base and makes hydroxide ions in water:
  $\begin{aligned} C_{2} H_{5} N H_{2} + H_{2} O \\ \overset{}{⇌} C_{2} H_{5} N H_{3}^{+} + O H^{-} \end{aligned}$
  The hydroxide ions precipitate dirty green iron(II) hydroxide:
  $F e^{2 +} + 2 O H^{-} \overset{}{\to} F e (O H)_{2} ↓$
4. Nitration of N,N-dimethylaniline in concentrated acid gives mainly the meta nitro compound, although $- N (C H_{3})_{2}$ directs ortho/para.
  Show solution
  In concentrated $H_{2} S O_{4}$ most of the amine is protonated to $C_{6} H_{5}$ – $N H (C H_{3})_{2}^{+}$ . The positively charged group has no free lone pair and withdraws electrons, so it deactivates the ring and directs meta; nitration happens on this cation.
5. Anisole, $C_{6} H_{5} O C H_{3}$ , undergoes Friedel–Crafts methylation, but aniline does not.
  Show solution
  Aniline's nitrogen is a much stronger Lewis base than anisole's oxygen. It bonds to the catalyst $A l C l_{3}$ , putting a positive charge on nitrogen that strongly deactivates the ring, and it uses up the catalyst.
6. Benzenediazonium chloride survives for a while in ice-cold water, but ethanediazonium chloride cannot be kept even at 273 K.
  Show solution
  In the arenediazonium ion the positive charge is delocalised over the ring, and losing $N_{2}$ would leave a very unstable aryl cation. The ethanediazonium ion has no such stabilisation and loses $N_{2}$ at once.
7. Gabriel synthesis gives a purer primary amine than ammonolysis.
  Show solution
  In ammonolysis the primary amine formed is itself a nucleophile and goes on to give secondary, tertiary and quaternary products. In Gabriel synthesis the nitrogen of the N-alkylphthalimide has no H left and its lone pair is shared with two C=O groups, so it cannot be alkylated again; hydrolysis then releases only the primary amine.
4
Quick answerMediumNCERT Exercises Q9.4
Arrange each set as asked.
1. (a)
  Decreasing p $K_{b}$ (weakest base first)
  $(C H_{3})_{2} N H$
  $C H_{3} N H_{2}$
  $C_{6} H_{5} N H_{2}$
  $C_{6} H_{5} N (C H_{3})_{2}$
  Show answer
  Answer
  $C_{6} H_{5} N H_{2}$
  $C_{6} H_{5} N (C H_{3})_{2}$
  $C H_{3} N H_{2}$
  $(C H_{3})_{2} N H$
  p $K_{b}$ 9.38 > 8.92 > 3.38 > 3.27. The two anilines are weak because of delocalisation; N-methyl groups help a little.
2. (b)
  Increasing basic strength in water
  4-nitroaniline
  4-methoxyaniline
  aniline
  Show answer
  Answer
  4-nitroaniline
  aniline
  4-methoxyaniline
  p $K_{b}$ about 13.0, 9.38 and 8.66: a withdrawing nitro group lowers basicity and a donating methoxy group raises it.
3. (c)
  Increasing basic strength in water
  $C_{6} H_{11} N H_{2}$
  $C_{6} H_{5} C H_{2} N H_{2}$
  $C_{6} H_{5} N H_{2}$
  Show answer
  Answer
  $C_{6} H_{5} N H_{2}$
  $C_{6} H_{5} C H_{2} N H_{2}$
  $C_{6} H_{11} N H_{2}$
  Aniline 9.38 (lone pair delocalised), benzylamine 4.70 (the $C H_{2}$ insulates nitrogen, but phenyl still withdraws a little), cyclohexanamine 3.36 (an alkyl group releases electrons).
4. (d)
  Decreasing basic strength in the gas phase
  $N H_{3}$
  $C_{2} H_{5} N H_{2}$
  $(C H_{3})_{2} N H$
  $(C_{2} H_{5})_{3} N$
  Show answer
  Answer
  $(C_{2} H_{5})_{3} N$
  $(C H_{3})_{2} N H$
  $C_{2} H_{5} N H_{2}$
  $N H_{3}$
  Proton affinities 982, 930, 912 and 854 kJ mol⁻¹. In the gas only electron release by the alkyl groups counts: three groups beat two, and two beat one.
5. (e)
  Increasing boiling point
  $(C H_{3})_{2} N (C H_{2})_{2} C H_{3}$
  $C H_{3} (C H_{2})_{4} N H_{2}$
  $C H_{3} (C H_{2})_{4} O H$
  $C H_{3} N H (C H_{2})_{3} C H_{3}$
  Show answer
  Answer
  $(C H_{3})_{2} N (C H_{2})_{2} C H_{3}$
  $C H_{3} N H (C H_{2})_{3} C H_{3}$
  $C H_{3} (C H_{2})_{4} N H_{2}$
  $C H_{3} (C H_{2})_{4} O H$
  N,N-Dimethylpropan-1-amine 338 K (no N–H), N-methylbutan-1-amine 364 K (one N–H), pentan-1-amine 377 K (two N–H), pentan-1-ol 411 K (stronger O–H⋯O bonds).
6. (f)
  Increasing solubility in water
  $C H_{3} C H_{2} C H_{2} N H_{2}$
  $C_{6} H_{5} N H_{2}$
  $C_{6} H_{5} N (C H_{3})_{2}$
  Show answer
  Answer
  $C_{6} H_{5} N (C H_{3})_{2}$
  $C_{6} H_{5} N H_{2}$
  $C H_{3} C H_{2} C H_{2} N H_{2}$
  N,N-Dimethylaniline (no N–H and a large hydrocarbon part) < aniline (about 3.6 g per 100 g of water) < propan-1-amine (mixes in all proportions).
5
Short answerHardNCERT Exercises Q9.5
Show how to carry out each conversion, with reagents and conditions.
1. Propanoic acid into ethanamine
  Show solution
  Convert the acid to propanamide ( $S O C l_{2}$ , then $N H_{3}$ ), then use $B r_{2}$ and NaOH: the Hoffmann degradation removes the carbonyl carbon and gives $C H_{3} C H_{2} N H_{2}$ .
2. Pentanenitrile into butan-1-amine
  Show solution
  Partly hydrolyse the nitrile with alkaline $H_{2} O_{2}$ to pentanamide, $C_{4} H_{9} C O N H_{2}$ ; then $B r_{2}$ and NaOH give $C_{4} H_{9} N H_{2}$ .
3. Ethanol into propanoic acid
  Show solution
  $P B r_{3}$ turns ethanol into bromoethane; KCN in ethanol gives propanenitrile, $C_{2} H_{5} C N$ ; hydrolysis with hot dilute acid gives $C_{2} H_{5} C O O H$ .
4. Propan-1-amine into ethanamine
  Show solution
  Nitrous acid ( $N a N O_{2}$ and HCl) gives propan-1-ol and $N_{2}$ ; acidified $K M n O_{4}$ oxidises it to propanoic acid; $S O C l_{2}$ then $N H_{3}$ give propanamide; $B r_{2}$ and NaOH give ethanamine.
5. Ethanamine into propan-1-amine
  Show solution
  Nitrous acid gives ethanol; $P B r_{3}$ gives bromoethane; KCN gives propanenitrile; $L i A l H_{4}$ reduces it to propan-1-amine.
6. Nitroethane into N-methylethanamine
  Show solution
  Reduce with Sn and HCl (then NaOH) to ethanamine; heat with $C H C l_{3}$ and ethanolic KOH to get ethyl isocyanide, $C_{2} H_{5} N C$ ; reduce it with $H_{2}$ over Pt (or $L i A l H_{4}$ ) to $C_{2} H_{5} N H C H_{3}$ .
7. Butanoic acid into propanoic acid
  Show solution
  $S O C l_{2}$ then $N H_{3}$ give butanamide; $B r_{2}$ and NaOH give propan-1-amine; nitrous acid gives propan-1-ol; acidified $K M n O_{4}$ gives propanoic acid.
8. Benzoic acid into phenylmethanamine
  Show solution
  $S O C l_{2}$ gives benzoyl chloride; $N H_{3}$ gives benzamide; $L i A l H_{4}$ followed by water gives $C_{6} H_{5} C H_{2} N H_{2}$ . (A Hoffmann degradation would give aniline instead.)
Long answerMediumNCERT Exercises Q9.6
Bottles hold butan-1-amine, N-methylpropan-1-amine and triethylamine, unlabelled. Describe how p-toluenesulphonyl chloride (TsCl, $C H_{3} C_{6} H_{4} S O_{2} C l$ ) and aqueous KOH let you identify each one, with equations.
Show solution
Shake a little of each amine with TsCl and excess aqueous KOH, then acidify any clear solution with dilute HCl.
Butan-1-amine (1°) forms a sulphonamide that still has one N–H. The $S O_{2}$ group makes that H acidic, so the product dissolves in KOH as a salt and gives a clear solution; acid brings it out again as a solid.
$\begin{aligned} T s C l + H_{2} N C_{4} H_{9} \\ \overset{}{\to} T s N H C_{4} H_{9} + H C l \\ T s N H C_{4} H_{9} + K O H \\ \overset{}{\to} [T s N C_{4} H_{9}]^{-} K^{+} + H_{2} O \end{aligned}$
N-Methylpropan-1-amine (2°) forms a sulphonamide with no N–H, which does not dissolve in KOH and separates as a solid or oil.
$\begin{aligned} T s C l + H N (C H_{3}) C_{3} H_{7} \\ \overset{}{\to} T s N (C H_{3}) C_{3} H_{7} + H C l \end{aligned}$
Triethylamine (3°) has no N–H and does not react. It stays as a separate layer in the alkali and dissolves when HCl is added:
$\begin{aligned} (C_{2} H_{5})_{3} N + H C l \\ \overset{}{\to} (C_{2} H_{5})_{3} N H^{+} C l^{-} \end{aligned}$
7
Short answerMediumNCERT Exercises Q9.7
For each named reaction, explain in a sentence what it does and give one balanced equation for the compound stated.
1. Carbylamine reaction, with 4-methylaniline
  Show solution
  Primary amines give foul-smelling isocyanides with chloroform and ethanolic KOH; a test for 1° amines.
  $\begin{aligned} C H_{3} C_{6} H_{4} N H_{2} + C H C l_{3} \\ + 3 K O H \overset{}{\to} C H_{3} C_{6} H_{4} N C \\ + 3 K C l + 3 H_{2} O \end{aligned}$
2. Diazotisation, with 4-chloroaniline
  Show solution
  A primary arylamine and nitrous acid at 273–278 K give a diazonium salt, used at once.
  $\begin{aligned} C l C_{6} H_{4} N H_{2} + N a N O_{2} \\ + 2 H C l \overset{}{\to} C l C_{6} H_{4} N_{2}^{+} C l^{-} \\ + N a C l + 2 H_{2} O \end{aligned}$
3. Hoffmann bromamide reaction, with propanamide
  Show solution
  A primary amide loses its carbonyl carbon and gives an amine with one carbon fewer.
  $\begin{aligned} C H_{3} C H_{2} C O N H_{2} + B r_{2} \\ + 4 N a O H \overset{}{\to} C H_{3} C H_{2} N H_{2} \\ + N a_{2} C O_{3} + 2 N a B r \\ + 2 H_{2} O \end{aligned}$
4. Coupling reaction, of 4-methylbenzenediazonium chloride with phenol
  Show solution
  The diazonium ion attacks the para position of phenoxide, giving an orange azo dye.
  $\begin{aligned} C H_{3} C_{6} H_{4} N_{2}^{+} C l^{-} \\ + C_{6} H_{5} O H + N a O H \\ \overset{}{\to} C H_{3} C_{6} H_{4} N = N C_{6} H_{4} O H \\ + N a C l + H_{2} O \end{aligned}$
5. Ammonolysis, of bromoethane
  Show solution
  Ammonia displaces bromide (SN2); NaOH frees the amine, and excess ammonia limits further alkylation.
  $\begin{aligned} C_{2} H_{5} B r + N H_{3} \\ \overset{}{\to} C_{2} H_{5} N H_{3}^{+} B r^{-} \end{aligned}$
6. Acetylation, of N-methylaniline with ethanoic anhydride
  Show solution
  An acetyl group replaces the N–H hydrogen, giving an amide.
  $\begin{aligned} C_{6} H_{5} N H C H_{3} + (C H_{3} C O)_{2} O \\ \overset{}{\to} C_{6} H_{5} N (C H_{3}) C O C H_{3} \\ + C H_{3} C O O H \end{aligned}$
7. Gabriel phthalimide synthesis, making 2-methylpropan-1-amine
  Show solution
  Heat potassium phthalimide with 1-bromo-2-methylpropane to get N-(2-methylpropyl)phthalimide, then hydrolyse with hot NaOH(aq): $(C H_{3})_{2} C H$ – $C H_{2} N H_{2}$ and sodium phthalate form. Only the primary amine is produced.
8
Short answerHardNCERT Exercises Q9.8
Plan each synthesis: give the reagents and the intermediate compounds.
1. 4-Nitrotoluene into 4-methylbenzoic acid
  Show solution
  Fe and HCl give 4-methylaniline; $N a N O_{2}$ and HCl at 273–278 K give the diazonium salt; CuCN and KCN give 4-methylbenzonitrile; hot dilute acid hydrolyses it to 4-methylbenzoic acid.
2. Nitrobenzene into 3-chlorophenol
  Show solution
  $C l_{2}$ with $F e C l_{3}$ chlorinates meta to the nitro group, giving 1-chloro-3-nitrobenzene; Sn and HCl reduce it to 3-chloroaniline; diazotise at 273–278 K; warm with water to get 3-chlorophenol.
3. Phenylethanoic acid into phenylmethanamine
  Show solution
  $S O C l_{2}$ then $N H_{3}$ give 2-phenylethanamide, $C_{6} H_{5} C H_{2}$ – $C O N H_{2}$ ; $B r_{2}$ and NaOH give $C_{6} H_{5} C H_{2} N H_{2}$ , one carbon fewer.
4. Aniline into 2,4,6-tribromobenzonitrile
  Show solution
  Bromine water gives 2,4,6-tribromoaniline; diazotise it in strong acid at 273–278 K; CuCN and KCN replace $- N_{2}^{+}$ by CN.
5. 1-Bromo-3-phenylpropane into 4-phenylbutan-1-amine
  Show solution
  KCN in ethanol gives 4-phenylbutanenitrile, $C_{6} H_{5} (C H_{2})_{3}$ – $C N$ ; $L i A l H_{4}$ reduces it to $C_{6} H_{5} (C H_{2})_{4}$ – $N H_{2}$ .
6. Toluene into 4-methylaniline
  Show solution
  Nitrate with concentrated $H N O_{3}$ and $H_{2} S O_{4}$ ( $C H_{3}$ directs ortho/para), separate 4-nitrotoluene, and reduce it with Sn and HCl or $H_{2}$ over Pd.
7. Aniline into 4-nitroaniline
  Show solution
  Acetylate ( $(C H_{3} C O)_{2} O$ , pyridine) to acetanilide; nitrate ( $H N O_{3}$ and $H_{2} S O_{4}$ , about 288 K) to 4-nitroacetanilide; hydrolyse with acid or alkali. Direct nitration would oxidise the ring and give much meta product.
8. 4-Methylbenzamide into toluene
  Show solution
  $B r_{2}$ and NaOH give 4-methylaniline; $N a N O_{2}$ and HCl at 273–278 K give the diazonium salt; $H_{3} P O_{2}$ (or ethanol) replaces $- N_{2}^{+}$ by H, giving toluene.
9
Short answerMediumNCERT Exercises Q9.9
Identify A, B and C in each sequence.
1. 1-Bromobutane is heated with KCN to give A; alkaline $H_{2} O_{2}$ partly hydrolyses A to B; $B r_{2}$ and NaOH turn B into C.
  Show solution
  A is pentanenitrile, $C_{4} H_{9} C N$ ; B is pentanamide, $C_{4} H_{9} C O N H_{2}$ ; C is butan-1-amine, $C_{4} H_{9} N H_{2}$ .
2. Aniline with $N a N O_{2}$ and HCl at 273 K gives A; A with CuBr and HBr gives B; B with Mg in dry ether gives C.
  Show solution
  A is benzenediazonium chloride; B is bromobenzene; C is phenylmagnesium bromide, $C_{6} H_{5} M g B r$ , a Grignard reagent (Unit 6).
3. Benzyl chloride with KCN gives A; $L i A l H_{4}$ reduces A to B; B heated with $C H C l_{3}$ and ethanolic KOH gives C.
  Show solution
  A is phenylethanenitrile, $C_{6} H_{5} C H_{2} C N$ ; B is 2-phenylethanamine, $C_{6} H_{5}$ – $C H_{2} C H_{2} N H_{2}$ ; C is the isocyanide $C_{6} H_{5}$ – $C H_{2} C H_{2} N C$ .
4. 3-Nitrotoluene with Sn and HCl gives A; A with $N a N O_{2}$ and HCl at 273 K gives B; B warmed with ethanol gives C.
  Show solution
  A is 3-methylaniline; B is 3-methylbenzenediazonium chloride; C is toluene (the ethanol is oxidised to ethanal).
5. Benzoyl chloride with $N H_{3}$ gives A; A with $B r_{2}$ and NaOH gives B; B with ethanoic anhydride gives C.
  Show solution
  A is benzamide; B is aniline; C is acetanilide (N-phenylethanamide).
6. Nitration of toluene gives A (the para isomer); Fe and HCl reduce A to B; B is diazotised at 273 K and coupled with phenol in NaOH to give C.
  Show solution
  A is 4-nitrotoluene; B is 4-methylaniline; C is the orange azo dye $C H_{3} C_{6} H_{4}$ – $N = N$ – $C_{6} H_{4} O H$ , 4-[(4-methylphenyl)diazenyl]phenol.
Short answerHardNCERT Exercises Q9.10
An aromatic compound P, $C_{8} H_{8} O_{2}$ , is heated with aqueous ammonia to give Q. Q reacts with bromine and NaOH to give R, $C_{7} H_{9} N$ . R, diazotised at 273 K and added to phenol in NaOH, gives an orange dye. Alkaline $K M n O_{4}$ oxidises P to benzene-1,4-dicarboxylic acid. Identify P, Q and R and name them.
Show solution
R has one carbon fewer than Q and is a primary amine, so Q is an amide and the step is a Hoffmann degradation. Because R forms an azo dye, its NH₂ is on the ring.
P oxidises to a 1,4-dicarboxylic acid, so P carries two carbon groups para to each other: a $C H_{3}$ and a $C O O H$ .
P is 4-methylbenzoic acid, $C H_{3} C_{6} H_{4} C O O H$ ( $C_{8} H_{8} O_{2}$ ); Q is 4-methylbenzamide, $C H_{3} C_{6} H_{4} C O N H_{2}$ (the ammonium salt loses water on heating); R is 4-methylaniline, $C H_{3} C_{6} H_{4} N H_{2}$ ( $C_{7} H_{9} N$ ).
11
Short answerMediumNCERT Exercises Q9.11
Complete each reaction and name the organic product.
1. Propan-1-amine heated with $C H C l_{3}$ and ethanolic KOH
  Show solution
  $\begin{aligned} C_{3} H_{7} N H_{2} + C H C l_{3} \\ + 3 K O H \overset{}{\to} C_{3} H_{7} N C \\ + 3 K C l + 3 H_{2} O \end{aligned}$
  Propyl isocyanide.
2. 4-Chlorobenzenediazonium chloride + KI
  Show solution
  $\begin{aligned} C l C_{6} H_{4} N_{2}^{+} C l^{-} + K I \\ \overset{}{\to} C l C_{6} H_{4} I + K C l + N_{2} \end{aligned}$
  1-Chloro-4-iodobenzene.
3. 4-Methylaniline + HBr
  Show solution
  $\begin{aligned} C H_{3} C_{6} H_{4} N H_{2} + H B r \\ \overset{}{\to} C H_{3} C_{6} H_{4} N H_{3}^{+} B r^{-} \end{aligned}$
  4-Methylanilinium bromide.
4. 3-Nitrobenzenediazonium chloride + water, warm
  Show solution
  $\begin{aligned} O_{2} N C_{6} H_{4} N_{2}^{+} C l^{-} \\ + H_{2} O \overset{}{\to} O_{2} N C_{6} H_{4} O H \\ + N_{2} + H C l \end{aligned}$
  3-Nitrophenol.
5. 4-Methylaniline + excess bromine water
  Show solution
  $\begin{aligned} C H_{3} C_{6} H_{4} N H_{2} + 2 B r_{2} \\ \overset{}{\to} C H_{3} C_{6} H_{2} B r_{2} N H_{2} \\ + 2 H B r \end{aligned}$
  2,6-Dibromo-4-methylaniline; the para position is already taken by the methyl group.
6. N-Methylaniline + benzoyl chloride
  Show solution
  $\begin{aligned} C_{6} H_{5} N H C H_{3} + C_{6} H_{5} C O C l \\ \overset{}{\to} C_{6} H_{5} N (C H_{3}) C O C_{6} H_{5} \\ + H C l \end{aligned}$
  N-Methyl-N-phenylbenzamide.
7. 4-Methylbenzenediazonium chloride: (i) $H B F_{4}$ , (ii) heat
  Show solution
  $\begin{aligned} C H_{3} C_{6} H_{4} N_{2}^{+} C l^{-} + H B F_{4} \\ \overset{}{\to} C H_{3} C_{6} H_{4} N_{2}^{+} B F_{4}^{-} + H C l \\ \overset{h e a t}{\to} C H_{3} C_{6} H_{4} F \\ + B F_{3} + N_{2} \end{aligned}$
  1-Fluoro-4-methylbenzene (4-fluorotoluene).
Short answerMediumNCERT Exercises Q9.12
A student plans to make 4-methylaniline by heating 4-bromotoluene with potassium phthalimide and then hydrolysing. Explain why this fails, and suggest a route that works.
Show solution
Gabriel synthesis needs the phthalimide anion to displace bromide in an SN2 reaction. In 4-bromotoluene the C–Br carbon is an $s p^{2}$ ring carbon, the bond has partial double-bond character, and the ring blocks attack from the back, so no substitution occurs.
A route that works: nitrate toluene, separate 4-nitrotoluene, and reduce it with Fe and HCl (or Sn and HCl) to 4-methylaniline.
Short answerMediumNCERT Exercises Q9.13
Butan-1-amine and 4-chloroaniline are each treated with $N a N O_{2}$ and dilute HCl at 273–278 K, and both solutions are later allowed to warm. Write equations and compare what you would see.
Show solution
Butan-1-amine: nitrogen bubbles off at once, even in the cold, and butan-1-ol forms (with other products).
$\begin{aligned} C_{4} H_{9} N H_{2} + H N O_{2} \\ \overset{}{\to} C_{4} H_{9} O H + N_{2} ↑ + H_{2} O \end{aligned}$
4-Chloroaniline: no gas in the cold; a clear solution of 4-chlorobenzenediazonium chloride forms.
$\begin{aligned} C l C_{6} H_{4} N H_{2} + N a N O_{2} \\ + 2 H C l \overset{}{\to} C l C_{6} H_{4} N_{2}^{+} C l^{-} \\ + N a C l + 2 H_{2} O \end{aligned}$
On warming, nitrogen is now given off and 4-chlorophenol forms.
$\begin{aligned} C l C_{6} H_{4} N_{2}^{+} C l^{-} + H_{2} O \\ \overset{}{\to} C l C_{6} H_{4} O H + N_{2} ↑ + H C l \end{aligned}$
14
Short answerMediumNCERT Exercises Q9.14
Give a reason for each.
1. Ethanol and ethanamine have almost the same molar mass, yet ethanol gives up a proton from O–H far more easily than ethanamine gives one up from N–H.
  Show solution
  Oxygen is more electronegative than nitrogen, so the O–H bond is more polar and the ethoxide ion holds its negative charge better than the $C_{2} H_{5} N H^{-}$ ion does. A more stable conjugate base means a stronger acid, so amines are much weaker acids than alcohols.
2. Propan-1-amine boils about 45 K higher than its isomer trimethylamine.
  Show solution
  Propan-1-amine has two N–H hydrogens and its molecules hydrogen-bond to each other. Trimethylamine has none, so only weaker dipole and dispersion forces hold its molecules together (321 K against 276 K).
3. Cyclohexanamine is about a million times stronger as a base than aniline.
  Show solution
  Their p $K_{b}$ values are 3.36 and 9.38, six units apart. In aniline the lone pair is delocalised into the ring, and protonation would destroy that resonance; in cyclohexanamine the lone pair is localised and the alkyl group releases electrons.
Quick answerEasy
Which reagent converts benzenediazonium chloride into iodobenzene?
$K I$ solution $I_{2}$ with $F e C l_{3}$ $H I$ with copper powder $N a I$ in acetone
Show answer
Answer (A) $K I$ solution
Iodide ion replaces the diazonium group directly when potassium iodide solution is added; nitrogen gas is released.
Quick answerEasy
The conversion of a primary aromatic amine into a diazonium salt is called .
Show answer
Answer diazotisation
In diazotisation, $A r N H_{2}$ reacts with $N a N O_{2}$ and two moles of HCl at 273–278 K to give $A r N_{2}^{+} C l^{-}$ , NaCl and water.
Quick answerMedium
Hoffmann degradation of 3-phenylpropanamide, $C_{6} H_{5} C H_{2} C H_{2} C O N H_{2}$ , gives an amine. How many carbon atoms does the amine contain?
Show answer
Answer 8
The amide has 9 carbons; the carbonyl carbon is lost, leaving 2-phenylethanamine, $C_{6} H_{5} C H_{2} C H_{2} N H_{2}$ , with 8.
18
Quick answerMedium
Which of these give a positive carbylamine test? Choose all that apply.
anilineN-methylanilinebenzylaminetriethylamine
Select all that apply
Show answer
Answer
(A) aniline
(C) benzylamine
Aniline and benzylamine are both primary amines (aromatic and aliphatic), so both give isocyanides. N-Methylaniline is secondary and triethylamine tertiary.

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Our practice questions are written for ncert.ai and follow the NCERT exercises. Find the one that matches the question in your textbook; its answer is with it.

In your NCERT textbook	Practice question here
NCERT Intext Question 9.1	Practice: classifying and naming · Q1
NCERT Intext Question 9.2	Practice: classifying and naming · Q2
NCERT Intext Question 9.3	Practice: making amines · Q1
NCERT Intext Question 9.4	Practice: reactions of amines · Q1
NCERT Intext Question 9.5	Practice: reactions of amines · Q2
NCERT Intext Question 9.6	Practice: reactions of amines · Q3
NCERT Intext Question 9.7	Practice: reactions of amines · Q4
NCERT Intext Question 9.8	Practice: reactions of amines · Q5
NCERT Intext Question 9.9	Practice: using diazonium salts · Q1
NCERT Exercises Q9.1	Exercises · Q1
NCERT Exercises Q9.2	Exercises · Q2
NCERT Exercises Q9.3	Exercises · Q3
NCERT Exercises Q9.4	Exercises · Q4
NCERT Exercises Q9.5	Exercises · Q5
NCERT Exercises Q9.6	Exercises · Q6
NCERT Exercises Q9.7	Exercises · Q7
NCERT Exercises Q9.8	Exercises · Q8
NCERT Exercises Q9.9	Exercises · Q9
NCERT Exercises Q9.10	Exercises · Q10
NCERT Exercises Q9.11	Exercises · Q11
NCERT Exercises Q9.12	Exercises · Q12
NCERT Exercises Q9.13	Exercises · Q13
NCERT Exercises Q9.14	Exercises · Q14

Quick check: MCQs on Amines

9.1 Structure of Amines3 questions
9.2 Classification3 questions
9.3 Nomenclature3 questions
Hoffmann bromamide degradation reaction4 questions
9.5 Physical Properties3 questions
Arylamines versus ammonia4 questions
Reaction with arylsulphonyl chloride4 questions
Electrophilic substitution4 questions
9.7 Method of Preparation of Diazonium Salts3 questions
Coupling reactions5 questions
9.10 Importance of Diazonium Salts in Synthesis of Aromatic Compounds3 questions

Glossary · 15

Key terms (English–Hindi)

acylationऐसिलनaisilan: Replacing a hydrogen on nitrogen (or oxygen) by an acyl group, $R C O^{-}$ , using an acid chloride or anhydride.First seen in the chapter →
amineऐमीनaimeen: A compound made by replacing one, two or three hydrogens of ammonia with alkyl or aryl groups; the nitrogen keeps a lone pair.First seen in the chapter →
ammonolysisअमोनी-अपघटनamoni-apaghatan: Breaking a C–X bond with ammonia, which replaces the halogen by an amino group.First seen in the chapter →

carbylamine reaction: Formation of a foul-smelling isocyanide when a primary amine is heated with $C H C l_{3}$ and ethanolic KOH; a test for 1° amines.First seen in the chapter →
coupling reactionयुग्मन अभिक्रियाyugman abhikriya: Attack of a diazonium ion on a phenol or arylamine ring, joining the rings by –N=N– to give a coloured azo compound.First seen in the chapter →

diazonium saltडाइऐज़ोनियम लवणdaaiazoniyam lavan: A salt $A r N_{2}^{+} X^{-}$ made from a primary aromatic amine; stable only in cold solution.First seen in the chapter →
diazotisationडाइऐज़ोटीकरणdaaiazoteekaran: Conversion of a primary aromatic amine into a diazonium salt with $N a N O_{2}$ and acid at 273–278 K.First seen in the chapter →

Gabriel phthalimide synthesis: A route to pure primary amines: potassium phthalimide is alkylated by $R - X$ and the product is hydrolysed.First seen in the chapter →

Hinsberg's reagent: Benzenesulphonyl chloride, $C_{6} H_{5} S O_{2} C l$ , used to distinguish and separate 1°, 2° and 3° amines.First seen in the chapter →
Hoffmann bromamide degradation: Conversion of a primary amide by $B r_{2}$ and NaOH into a primary amine with one carbon fewer.First seen in the chapter →

pKb: $- \log K_{b}$ , a measure of base strength: the smaller the pKb, the stronger the base.First seen in the chapter →
primary, secondary and tertiary aminesप्राथमिक, द्वितीयक और तृतीयक ऐमीनpraathmik, dviteeyak aur triteeyak aimeen: Amines with one, two or three carbon groups on nitrogen: $R N H_{2}$ , $R_{2} N H$ and $R_{3} N$ .First seen in the chapter →

quaternary ammonium saltचतुष्क अमोनियम लवणchatushk amoniyam lavan: A salt of the ion $R_{4} N^{+}$ , in which nitrogen carries four carbon groups and no lone pair.First seen in the chapter →

Sandmeyer reaction: Replacement of $- N_{2}^{+}$ by Cl, Br or CN using the matching copper(I) salt.First seen in the chapter →
solvationविलायकनvilaayakan: Stabilisation of a dissolved ion by solvent molecules clustering round it; in water mainly through hydrogen bonds.First seen in the chapter →

9.1 Structure of Amines

9.2 Classification

9.3 Nomenclature

9.4 Preparation of Amines

Reduction of nitro compounds

Ammonolysis of alkyl halides

Reduction of nitriles

Reduction of amides

Gabriel phthalimide synthesis

Hoffmann bromamide degradation reaction

9.5 Physical Properties

9.6 Chemical Reactions

Basic character of amines

Structure and basicity: alkanamines versus ammonia

Arylamines versus ammonia

Alkylation

Acylation

Carbylamine reaction

Reaction with nitrous acid

Reaction with arylsulphonyl chloride

Electrophilic substitution

Diazonium Salts

9.7 Method of Preparation of Diazonium Salts

9.8 Physical Properties

9.9 Chemical Reactions

Reactions in which nitrogen is replaced

Coupling reactions

9.10 Importance of Diazonium Salts in Synthesis of Aromatic Compounds

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