Question

The major product formed in the following reaction: 

 is: 

• A. A
• B. B
• C. C
• D. D

Hint:

LDA is used for deprotonating the alpha position of carbonyl compounds to form enolates. This can then undergo nucleophilic attack to form new C-C bonds, followed by reduction to the desired product.

Answer 1

The Correct Option is A

The given reaction involves three steps: 1. Step 1: Deprotonation by LDA. LDA (Lithium Diisopropylamide) is a strong base and deprotonates the amide at the alpha position to form an enolate. This enolate is highly reactive and will attack electrophilic species. 2. Step 2: Nucleophilic substitution with PhCH\(_2\)Br. The enolate formed in the first step attacks the benzyl bromide (PhCH\(_2\)Br), leading to the formation of a new C-C bond between the enolate carbon and the benzyl group. The resulting product is an intermediate compound where the newly formed bond is between the alpha-carbon and the phenyl group attached to a methyl group. 3. Step 3: Reduction by LiAlH\(_4\). LiAlH\(_4\) is a strong reducing agent that reduces the carbonyl group (C=O) to an alcohol (-OH). This results in the final product, which is a hydroxymethylated phenyl compound, i.e., \(\text{Me}-\text{CH}_2\text{OH}-\text{Ph}\). Thus, the major product of this reaction is \(\text{Me}-\text{CH}_2\text{OH}-\text{Ph}\). Final Answer: (A) \(\text{Me}-\text{CH}_2\text{OH}-\text{Ph}\)