CBSE Class 12 Chemistry 2026: question 29
Set in the CBSE Class 12 board exam of 2026, paper 56/1/1, section D, question 29, for 4 marks. Try it first; the hints and our step-by-step solution are below.
Q29 · 4 marks · Case-basedCompetency-basedAmines
The reaction of amines with mineral acids to form ammonium salts shows that these are basic in nature. Aliphatic amines are stronger bases than ammonia whereas aromatic amines are weaker bases than ammonia. Aliphatic and aromatic primary and secondary amines react with acid chlorides, anhydrides and esters by nucleophilic substitution reaction. The main problem encountered during electrophilic substitution reactions of aromatic amines is that of their high reactivity. Substitution tends to occur at ortho-and para-positions. Hinsberg reagent is used for the identification and distinction between primary, secondary and tertiary amines. Aryldiazonium salts, usually obtained from arylamines, undergo replacement of the diazonium group with a variety of nucleophiles to provide advantageous methods for producing aryl halides, cyanides, phenols and arenes.
Answer the following questions :
- (a)2 marks
(i) Why is a stronger base than in aqueous solution ?
(ii) Write structural formulae of the compound A and B : - (b)1 mark
A compound 'X' with molecular formula reacts with Hinsberg reagent to give a product insoluble in alkali. Identify 'X'.
OR
(b)1 markHow can you convert aniline to benzonitrile ?
- (c)1 mark
Why is – group of aniline acetylated before carrying out nitration ?
Source: official question paper from cbse.gov.in (PDF). See it in the full 2026 board paper, set 56/1/1.
Hints
Open one hint at a time and try again before the next.
Hint 1
In water, consider how well each protonated amine is stabilised by H-bonding with water.
Hint 2
NaOBr on an amide removes one carbon (Hoffmann bromamide reaction).
Hint 3
A Hinsberg product insoluble in alkali has no acidic N–H, which points to a secondary amine; for –CN, think diazonium salt and Sandmeyer.
Step-by-step solution
Each step shows the marks it earns, as an examiner would award them. Checked by an independent re-solve from the question paper.
- 1 mark
(a)(i) In water, basicity is decided by the combined effect of the +I effect, solvation (H-bonding) of the conjugate acid, and steric hindrance. has three N–H hydrogens to H-bond with water, so it is well stabilised by solvation. has only one, and its bulky methyl groups also hinder solvation. Hence is the stronger base in aqueous solution.
- ½ mark
(a)(ii) Hoffmann bromamide degradation gives **A = ** (methanamine).
- ½ mark
Benzoylation with in base gives **B = ** (N-methylbenzamide).
- 1 mark
(b) [first alternative] A product insoluble in alkali means the amine is secondary. **X = ** (N-methylethanamine).
(b) [second alternative] Aniline is diazotised with at 273–278 K to form benzenediazonium chloride, which is treated with CuCN/KCN (Sandmeyer reaction):
- 1 mark
(c) –NH₂ strongly activates the ring and gets protonated in the nitrating mixture to the meta-directing anilinium ion; it is also oxidised by . Acetylation to lowers the activating effect, prevents anilinium-ion formation and oxidation, and allows controlled nitration giving the p-nitro derivative as the major product.
Answer: (a)(i) Greater solvation (H-bonding) and less steric hindrance of CH₃NH₃⁺; (ii) A = CH₃NH₂, B = CH₃NHCOC₆H₅; (b) X = CH₃CH₂NHCH₃ / aniline → NaNO₂, HCl (0–5 °C) → CuCN/KCN → benzonitrile; (c) to control nitration and prevent anilinium ion formation, giving the p-nitro product.
Where the marks are
- Combined inductive, solvation and steric effects explained (1 mark)
- A = CH₃NH₂ (½ mark), B = CH₃NHCOC₆H₅ (½ mark)
- X = N-methylethanamine, or correct diazotisation + Sandmeyer conversion (1 mark)
- Controls nitration / prevents meta-directing anilinium ion (1 mark)
Revise the chapter
- Chapter: Amines (notes, formulas, where marks are lost and every past question)
- Topic: Basicity of Amines
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