Question

The major product P formed in the following transformation is

• A. A
• B. B
• C. C
• D. D

Hint:

MnO$_2$ selectively oxidizes benzylic alcohols, PBr$_3$ converts alcohols into bromides, and base then gives elimination to form the conjugated enone

Answer 1

The Correct Option is A

Step 1: Identify the functional groups in the starting compound.
The given compound contains a benzylic secondary alcohol and a terminal primary alcoholThe aromatic ring also contains a methoxy group, but it does not participate directly in the given sequence

Step 2: Action of MnO$_2$.
MnO$_2$ is a selective oxidizing agentIt oxidizes benzylic and allylic alcohols efficiently under mild conditions

Step 3: Formation of ketone.
The benzylic secondary alcohol is oxidized to a ketone, while the terminal primary alcohol remains unchanged under these conditions
\[ \text{Ar-CH(OH)-CH}_2\text{CH}_2\text{OH} \xrightarrow{\text{MnO}_2} \text{Ar-CO-CH}_2\text{CH}_2\text{OH} \]

Step 4: Action of PBr$_3$.
PBr$_3$ converts alcohols into alkyl bromidesTherefore, the terminal primary alcohol is converted into the corresponding bromide
\[ \text{Ar-CO-CH}_2\text{CH}_2\text{OH} \xrightarrow{\text{PBr}_3} \text{Ar-CO-CH}_2\text{CH}_2\text{Br} \]

Step 5: Action of NaOEt in ethanol.
NaOEt acts as a strong baseIt abstracts an acidic $\alpha$-hydrogen from the carbon next to the carbonyl group

Step 6: Elimination reaction.
Removal of the $\alpha$-hydrogen followed by elimination of bromide gives an $\alpha,\beta$-unsaturated ketoneThis product is stabilized by conjugation with the carbonyl group
\[ \text{Ar-CO-CH}_2\text{CH}_2\text{Br} \xrightarrow{\text{NaOEt}} \text{Ar-CO-CH=CH}_2 \]

Step 7: Conclusion.
Thus, the final product is the conjugated enone shown in option (A)
\[ \boxed{\text{A}} \]