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CBSE · Class 12 · Chemistry

Amines

Standard difficulty · 20 questions · 40 marks

Name: Date: Score: ____ / 40

Show your working for Sections B and C. Answers are on the last page — try every question first!

Section A — Choose the correct answer8 × 1 = 8 marks
1.
Amines are basic because the nitrogen atom has:
  1. no lone pair
  2. a double bond
  3. a positive charge
  4. a lone pair of electrons
2.
The IUPAC name of C2H5NH2 (ethylamine) is:
  1. ethanamine
  2. propan-1-amine
  3. N-methylethanamine
  4. N,N-dimethylaniline
3.
N,N-dimethylaniline (C6H5N(CH3)2) is a:
  1. primary (1°) amine
  2. secondary (2°) amine
  3. tertiary (3°) amine
  4. quaternary ammonium salt
4.
With Hinsberg's reagent (benzenesulphonyl chloride), N,N-dimethylaniline (C6H5N(CH3)2) gives:
  1. a sulphonamide that does not dissolve in alkali
  2. a sulphonamide that dissolves in alkali
  3. a yellow oily nitrosoamine only
  4. no reaction
5.
The correct order of increasing basic strength in water is:
  1. phenylmethanamine < methanamine < N-methylmethanamine
  2. N-methylmethanamine < phenylmethanamine < methanamine
  3. N-methylmethanamine < methanamine < phenylmethanamine
  4. phenylmethanamine < N-methylmethanamine < methanamine
6.
Benzenediazonium chloride reacts with Cu powder with HCl. The product is:
  1. 4-aminoazobenzene (a yellow dye)
  2. chlorobenzene
  3. 4-hydroxyazobenzene (an orange dye)
  4. iodobenzene
7.
2-phenylethanamide is warmed with Br2 and aqueous NaOH. The product is:
  1. butan-1-amine
  2. ethanamine
  3. phenylmethanamine
  4. propan-1-amine
8.
Assertion (A): Methanamine (CH3NH2) gives the carbylamine (isocyanide) test.
Reason (R): Only tertiary amines give an isocyanide with chloroform and alcoholic KOH.
  1. Both A and R are true, and R is the correct explanation of A.
  2. Both A and R are true, but R is not the correct explanation of A.
  3. A is true, but R is false.
  4. A is false, but R is true.
Section B — Short answer8 × 2 = 16 marks
9.
Why are aliphatic amines stronger bases than ammonia?
10.
Give the IUPAC name of (C2H5)3N (triethylamine), classify it, and say what it gives with Hinsberg's reagent.
11.
Arrange in increasing order of basic strength in aqueous solution: phenylmethanamine, benzenamine, N-methylmethanamine.
12.
What is formed when benzenediazonium chloride reacts with H3PO2 (hypophosphorous acid)? Name the reaction.
13.
116.1 g of heptanamide undergoes Hoffmann bromamide degradation with 100% yield. Name the amine and find its mass. (C = 12, H = 1, N = 14, O = 16)
14.
How would you tell phenylmethanamine from N-methylmethanamine?
15.
How would you convert aniline to benzene?
16.
What happens when N-methylaniline (C6H5NHCH3) is treated with nitrous acid (NaNO2 and HCl) in the cold?
Section C — Word problems4 × 4 = 16 marks
17.
Describe three methods of preparing amines, with equations.
18.
Identify the compounds in: nitrobenzene —Fe/HCl→ A —Br2 water→ B.
19.
Arrange in increasing order of basic strength in water, and explain using the inductive effect, solvation and resonance: phenylmethanamine, benzenamine, N-methylaniline, N-ethylethanamine.
20.
57.5 g of hexanamide (CH3(CH2)4CONH2) is treated with Br2 and NaOH. Write the equation, name the product and find its mass for a 80% yield. Why can this method not make secondary amines? (C = 12, H = 1, N = 14, O = 16)
AmbifulWorksheets
Practice that never repeats
CBSE · Class 12 · Chemistry

Answer key

Amines · AW-12C-AMN-2026-V1 · For parents — fold back or tear off before handing the sheet to your child.

QAnswerWorking
1(d) a lone pair of electrons—
2(a) ethanamineN-substituents are named with the prefix N-
3(c) tertiary (3°) aminecount the carbon groups on nitrogen
4(d) no reactionN,N-dimethylaniline is tertiary (3°)
5(a) phenylmethanamine < methanamine < N-methylmethanaminepKb: N-methylmethanamine 3.27, phenylmethanamine 4.7, methanamine 3.38 (lower pKb, stronger base)
6(b) chlorobenzene—
7(c) phenylmethanamineHoffmann bromamide degradation: the amine has one carbon fewer than the amide
8(c) A is true, but R is false.A is true: it is a primary (1°) amine. R is false.
9Alkyl groups release electrons (+I effect), raising the electron density on nitrogen, so the lone pair is more available to accept a proton.—
10N,N-diethylethanamine; tertiary (3°); no reactionHinsberg: 1° soluble in alkali, 2° insoluble, 3° no reaction
11benzenamine < phenylmethanamine < N-methylmethanaminepKb: phenylmethanamine 4.7, benzenamine 9.38, N-methylmethanamine 3.27; aryl amines are weak because the lone pair is spread into the ring
12benzene; reductionthe N2+ group is replaced or the ion couples
13hexan-1-amine; 90.9 gRCONH2 + Br2 + 4NaOH → RNH2 + Na2CO3 + 2NaBr + 2H2O; M(amine) = 101
14Hinsberg's reagent: phenylmethanamine gives a sulphonamide that dissolves in alkali; N-methylmethanamine gives a sulphonamide that does not dissolve in alkali. Also, only the primary amine (phenylmethanamine) gives the carbylamine testHinsberg separates 1°, 2° and 3° amines
15diazotise, then H3PO2—
16a yellow oily N-nitrosoamine formsthe products depend on the class and on whether the amine is aromatic
17Reduction of nitro compounds: C6H5NO2 + 6[H] (Sn/HCl) → C6H5NH2 + 2H2O. Ammonolysis of alkyl halides gives a mixture of 1°, 2° and 3° amines and the quaternary salt. Reduction of nitriles (LiAlH4 or H2/Ni) and of amides (LiAlH4). Gabriel synthesis gives pure primary aliphatic amines. Hoffmann bromamide degradation gives a primary amine with one carbon fewer.—
18A = aniline, B = 2,4,6-tribromoanilinefollow each reagent in turn
19benzenamine < N-methylaniline < phenylmethanamine < N-ethylethanaminealkyl groups (+I) raise basicity, but in water solvation and crowding make the order of methyl amines 2° > 1° > 3°; in aryl amines the lone pair is delocalised into the ring, so they are much weaker. pKb: phenylmethanamine 4.7, benzenamine 9.38, N-methylaniline 9.3, N-ethylethanamine 3
20CH3(CH2)4CONH2 + Br2 + 4NaOH → CH3(CH2)4NH2 + Na2CO3 + 2NaBr + 2H2O; pentan-1-amine; 34.8 g; the reaction removes the carbonyl carbon and always leaves an –NH2 group, so only primary amines formmol = 0.5