JEE Main 2022 · 27 July · morning · Single correct · +4 / −1
Match List I with List II.
Choose the correct answer from the options given below:
| List I | List II | ||
|---|---|---|---|
| (A) | Benzenesulphonyl chloride | (I) | Test for primary amines |
| (B) | Hoffmann bromamide reaction | (II) | Anti Saytzeff |
| (C) | Carbylamine reaction | (III) | Hinsberg reagent |
| (D) | Hoffmann orientation | (IV) | Known reaction of Isocyanates. |
- (a) A-IV, B-III, C-II, D-I
- (b) A-IV, B-II, C-I, D-III
- (c) A-III, B-IV, C-I, D-II
- (d) A-IV, B-III, C-I, D-II
Answer and solution
Answer (c) A-III, B-IV, C-I, D-II
Solution
Chemistry ma’am · 2 steps
Pin each name to one fact: Hinsberg is a reagent, and the isocyanate sits in the middle of bromamide.
- Match by identity. Each item in List I has one defining fact. Write that fact down first, and the matching List II entry follows from it.
List I What it really is List II (A) Benzenesulphonyl chloride C₆H₅SO₂Cl is the reagent that tells 1°, 2° and 3° amines apart (III) Hinsberg reagent (B) Hoffmann bromamide reaction RCONH₂ with Br₂ and NaOH rearranges to the isocyanate R-N=C=O, which is hydrolysed to RNH₂ (IV) Known reaction of isocyanates (C) Carbylamine reaction RNH₂ + CHCl₃ + 3KOH gives the foul-smelling isocyanide RNC, only with primary amines (I) Test for primary amines (D) Hoffmann orientation Hofmann elimination of quaternary ammonium hydroxides gives the less substituted alkene (II) Anti Saytzeff - So. The match is A-III, B-IV, C-I, D-II, which is option (c).
The idea to keep. Tie each named reaction to one signature: Hinsberg is C₆H₅SO₂Cl, bromamide runs through an isocyanate, carbylamine tests 1° amines, Hofmann elimination is anti-Saytzeff.
Answer from the NTA final answer key. Solution by PrepWiser.