JEE Main 2026 · 23 January · evening · Single correct · +4 / −1
Given below are two statements:
In the light of the above statements, choose the correct answer from the options given below
- (a) Both Statement I and Statement II are true
- (b) Statement I is false but Statement II is true
- (c) Statement I is true but Statement II is false
- (d) Both Statement I and Statement II are false
Answer and solution
Answer (a) Both Statement I and Statement II are true
Solution
Chemistry ma’am · 5 steps
In I each amide loses a carbon; in II the NH₂ places both Br atoms, then leaves.
- Statement I, oxidation. Acidic KMnO₄ oxidises any alkyl chain with a benzylic H right down to -COOH, whatever its length. So 1,2-dipropylbenzene becomes phthalic acid, C₆H₄(COOH)₂.
- Amide, then Hofmann. Ammonia converts both -COOH groups into -CONH₂, giving phthalamide. Bromine and alkali then run the Hofmann bromamide degradation on each amide, so every -CONH₂ loses its carbon and becomes -NH₂. You get o-phenylenediamine. Statement I is true.
- Statement II, bromination. In p-toluidine, -NH₂ is a strong ortho/para director, and its para position is blocked by CH₃. Bromine water therefore puts Br on both positions ortho to NH₂, giving 2,6-dibromo-4-methylaniline.
- Remove the NH2. NaNO₂/HCl at 0 to 5 °C turns -NH₂ into -N₂⁺, and H₃PO₂ replaces it with H. The two Br atoms stay put and are now meta to each other and meta to CH₃: 3,5-dibromotoluene. Statement II is true.
- So. Both statements are true, which is option (a).
The idea to keep. Use NH₂ as a temporary director: let it place substituents, then remove it by diazotisation and H₃PO₂. And Hofmann on an amide always gives an amine with one carbon fewer.
Answer from the NTA final answer key. Solution by PrepWiser.