Question

The major product of the following reaction is:
$Ph-C(=O)-CH_2CH_2-CN$
(i) $CH_{3}MgBr (excess) (ii) H_{+}H_{2}O$

• A. \(Ph-CH(CH_3)-CH_2-COCH_3\)
• B. $Ph-C(CH_3)(OH)-CH(CH_3)_2$
• C. $Ph-CO-CH_2CH_2CH_3$
• D. $Ph-CH(CH_3)-CH_2-CN$

Hint:

Grignard reagents attack both carbonyl and nitrile groups; nitriles give ketones, while aldehydes/ketones give alcohols after hydrolysis.

Answer 1

The Correct Option is A

The given compound is a substituted aryl ketone with a cyano group at the terminal position. Reagent: excess CH₃MgBr (a Grignard reagent) + acidic hydrolysis.

Step 1: Grignard Reaction with Ketone
- CH₃MgBr adds a methyl group to the carbonyl carbon of the ketone (nucleophilic attack), forming a tertiary alcohol after hydrolysis.

Step 2: Reaction with –CN group
- Since excess Grignard reagent is used, it also reacts with the nitrile group (–CN) to give an imine intermediate, which on hydrolysis gives a ketone.

Net Result:
- The carbonyl group (initial ketone) gives a tertiary alcohol: Ph-C(OH)(CH₃)-CH₂-CH₂-CN
- The –CN gets converted into a ketone group: -CH₂-COCH₃

After tautomerization or rearrangement, we obtain the product:
Ph-CH(CH₃)-CH₂-COCH₃ — this matches option (1).