JEE Main 2022 · 29 Jul · Shift 1 · Chemistry Q76

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ChemistrySingle correct+4 / -1Question 76 of 90

'A' ( $\mathrm{C_8H_6Cl_2O}$ ) reacts with NH $_3$ to give $\mathrm{C_8H_8ClNO}$ , which with Br $_2$ /NaOH gives 4-chloro-3-methylaniline (shown).

Consider the above reaction, the compound 'A' is :

Scheme: 'A' (C8H6Cl2O) → (NH3) C8H8ClNO → (Br2, NaOH) 4-chloro-3-methylaniline.

4-Chloro-2-methylbenzoyl chloride

3-Chloro-4-methylbenzoyl chloride

4-Chloro-3-methylbenzoyl chloride

4-Chloro-3-(chloromethyl)benzaldehyde

Answer and worked solution

Official answer: C

Work back. Hofmann bromamide removes the carbonyl carbon, so the amide is 4-chloro-3-methylbenzamide and A is the acid chloride, 4-chloro-3-methylbenzoyl chloride ( $\mathrm{C_8H_6Cl_2O}$ ).

(C)

Topics:Amines Acid derivatives

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JEE Main 2022 · 29 Jul · Shift 1, Chemistry question 76