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?

• A. \includegraphics[]{156a.png}
• B. \includegraphics[]{156b.png}
• C. \includegraphics[]{156c.png}
• D. \includegraphics[]{156d.png}

Hint:

Grignard reagents react with D\(_2\)O to replace MgBr with D, forming a deuterated hydrocarbon.

Answer 1

The Correct Option is D

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).