Question

Consider the following reaction, the major product 'P' is: 

• A. A
• B. B
• C. C
• D. D

Hint:

Grignard addition to \(\alpha,\beta\)-unsaturated aldehydes/ketones follows 1,2-addition unless \(Cu(I)\) salts are added (which favor 1,4-addition). Always look for potential side reactions like \(HX\) addition if the workup uses a concentrated mineral acid.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
Grignard reagents (\(RMgX\)) undergo nucleophilic addition to carbonyl groups. With \(\alpha,\beta\)-unsaturated ketones, alkyl Grignard reagents typically prefer 1,2-addition to the carbonyl group over 1,4-addition. The second step involving aqueous acid workup with \(HCl\) not only protonates the alkoxide to alcohol but can also lead to the addition of \(HCl\) across the double bond if present in excess.
Step 2: Key Formula or Approach:
1. Nucleophilic addition of \(Et^-\) to the \(C=O\) carbon.
2. Protonation of \(O^-\) to \(OH\).
3. Electrophilic addition of \(HCl\) to the remaining alkene.
Step 3: Detailed Explanation:
1. Addition of Grignard: The starting material is but-3-en-2-one (\(CH_2=CH-COCH_3\)). Reaction with \(C_2H_5MgBr\) occurs at the carbonyl group (1,2-addition) to give the intermediate alkoxide: \(CH_2=CH-C(O^-MgBr)(CH_3)(C_2H_5)\).
2. Acidic Workup: Addition of \(H_2O/HCl\) results in the formation of the alcohol: 3-methyl-pent-1-en-3-ol, \(CH_2=CH-C(OH)(CH_3)(C_2H_5)\).
3. Reaction with HCl: The presence of \(HCl\) facilitates the addition across the double bond. Following Markovnikov's rule (and considering the stability of the carbocation intermediate), the chloride ion adds to the more substituted/stable position. In many competitive exam contexts for this specific reaction, the final product involves the addition of the halogen to the double bond.
Step 4: Final Answer:
The major product is the tertiary alcohol with the halogen added to the alkyl chain, matching Option (A).