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JEE Main 4 August 2021, Shift 1, Chemistry Q44

Question 44 of 90 in this shift, Chemistry question 14 of 30, Section A.

Identify the reagent in the following reaction.

\mathrm{CH_3-CH_2-C(=O)-OH \xrightarrow{Reagent} CH_3-CH(X)-C(=O)-OH}

(X = Cl, – Br)

The figure, in words

Reaction scheme: CH3-CH2-C(=O)-OH (C=O drawn with O above the carbonyl carbon) on the left, arrow labelled 'Reagent' above it, and on the right CH3-CH-C(=O)-OH with X drawn on a vertical bond below the CH carbon; '(X = Cl, – Br)' printed after the product.

(1) $\mathrm{SOCl_2}$
(2)NBS/h $\nu$ or heat
(3) $\mathrm{X_2}$ /Red Phosphorus, $\mathrm{H_2O}$ Official answer
(4)Halogen at high temperature (400-500°C)

Official answer

Option 3

NTA final key.

Chapter: Aldehydes, Ketones and Carboxylic Acids
Topic: Carboxylic Acids and Derivatives
Idea tested: Hell-Volhard-Zelinsky Reaction

JEE Main 4 August 2021, Shift 1, Chemistry Q44

Same idea in other shifts