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Question #43993

Difficulty: 8/10 | 🟢 Free

The Question

Silver 3-methyl-2-(trifluoromethyl)butanoate, $\ce{(CH_3)_2CH-CH(CF_3)-COOAg}$ , is treated with $\ce{Br_2}$ in refluxing $\ce{CCl_4}$ . Instead of the expected Hunsdiecker alkyl halide, the reaction yields a different major organic product. Identify the major product formed.

Options

A.
$\ce{(CH_3)_2CH-CH(Br)-CF_3}$
B.
$\ce{(CH_3)_2C=CH-CF_3}$
C.
$\ce{(CH_3)_2C(Br)-CH_2-CF_3}$
D.
$\ce{(CH_3)_2CH-CH(CF_3)-COO-CH(CF_3)-CH(CH_3)_2}$

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Exception_Exposure

Learn to identify edge cases and boundary exceptions that JEE examiners hide in math problems.

Concepts Behind This Question

Hunsdiecker/Hell-Volhard-Zelinsky|Decarboxylation Mechanisms|Inductive Effect Trends

Why This Question Is Tricky

Common Mistakes

The question tests the 'Chemical Exception' in the Hunsdiecker reaction where the inclusion of a strong electron-withdrawing group (EWG) at the alpha-carbon destabilizes the radical intermediate.

Students typically memorize the Hunsdiecker reaction as a 'black box' decarboxylative bromination.

The trap is failing to recognize that the $-I$ effect of the $-\ce{CF_3}$ group makes the alpha-radical thermodynamically inaccessible, shifting the mechanism from radical substitution to an E2-like elimination of the acyl hypobromite intermediate.

Option C serves as a secondary trap for students who incorrectly assume an ionic carbocation rearrangement.

Quick Facts About This Question

Question Information

Subject	Chemistry
Chapter	Aldehydes, Ketones, and Carboxylic Acids - Class 12
Concept	Hunsdiecker/Hell-Volhard-Zelinsky
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free