Question:

Hydrolysis of an alkyl bromide X (C\(_4\)H\(_9\)Br) follows first-order kinetics. Reaction of X with Mg in dry ether followed by treatment of D\(_2\)O gave Y. What is Y?

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Grignard reagents react with D\(_2\)O to replace MgBr with D, forming a deuterated hydrocarbon.
Updated On: Mar 24, 2025

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The Correct Option is D

Solution and Explanation

Step 1: Identifying the Type of Alkyl Bromide
- The hydrolysis follows first-order kinetics, which suggests an S\(_N1\) mechanism.
- S\(_N1\) reactions occur faster for tertiary > secondary > primary alkyl halides.
- Hence, X must be a tertiary alkyl bromide. Step 2: Formation of Grignard Reagent
- When X (tert-butyl bromide) is reacted with Mg in dry ether, a Grignard reagent is formed: \[ (CH_3)_3CBr + Mg \rightarrow (CH_3)_3CMgBr \] Step 3: Reaction with D\(_2\)O
- Grignard reagents act as strong nucleophiles and react with D\(_2\)O to give a deuterated product: \[ (CH_3)_3CMgBr + D_2O \rightarrow (CH_3)_3CD + Mg(OD)Br \] Step 4: Identifying the Product (Y)
- The final product (Y) is tert-butyl deuteride.
- The correct option corresponds to the tert-butyl group with deuterium (D) replacing hydrogen.
Thus, the correct answer is option (4). \bigskip
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