Question

Benzonitrile on reaction with a reagent in dry ether followed by hydrolysis forms benzophenone along with ammonia and hydroxymagnesium bromide. Identify the reagent used in above transformation.

• A. \(\text{C}_6\text{H}_5\text{MgCl}\)
• B. \(\text{C}_2\text{H}_5\text{MgBr}\)
• C. \(\text{C}_6\text{H}_5\text{MgI}\)
• D. \(\text{C}_6\text{H}_5\text{MgBr}\)

Hint:

The Grignard reagent must supply a phenyl group, and bromide shows in the by-product.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
A nitrile reacts with a Grignard reagent to form an imine salt, which hydrolyses to a ketone: \(\text{RCN} + \text{R}'\text{MgX} \to \text{RC(=NMgX)R}'\), then \(\to \text{RCOR}'\).

Step 2: Key Formula or Approach:
Benzophenone is \(\text{C}_6\text{H}_5\text{COC}_6\text{H}_5\). One phenyl comes from benzonitrile, so the Grignard must supply the second phenyl group.

Step 3: Detailed Explanation:
The by-product is hydroxymagnesium bromide, \(\text{Mg(OH)Br}\), so the halide in the reagent is bromide.
That fits \(\text{C}_6\text{H}_5\text{MgBr}\).
\[ \text{C}_6\text{H}_5\text{CN} + \text{C}_6\text{H}_5\text{MgBr} \to \text{C}_6\text{H}_5\text{C(=NMgBr)C}_6\text{H}_5 \xrightarrow{\text{H}_2\text{O}} \text{C}_6\text{H}_5\text{COC}_6\text{H}_5 + \text{NH}_3 + \text{Mg(OH)Br} \]
The ethyl reagent would give propiophenone. The chloride and iodide reagents would give \(\text{Mg(OH)Cl}\) and \(\text{Mg(OH)I}\) instead.

Final Answer:
The reagent is phenylmagnesium bromide, option (D). \[ \boxed{\text{C}_6\text{H}_5\text{MgBr}} \]