Question

Identify the major product \((Y)\) from the following reaction

• A. 1
• B. 2
• C. 3
• D. 4

Hint:

Grignard reagents react with \(CO_2\) to form carboxylic acids after hydrolysis. \(LiAlH_4\) then reduces carboxylic acids into primary alcohols.

Answer 1

The Correct Option is C

Step 1: Formation of Grignard reagent.
The given compound is:
\[ C_6H_5CH_2CH_2Cl \] which is \(\beta\)-phenethyl chloride.
When alkyl halides react with magnesium in dry ether, Grignard reagents are formed.
Thus:
\[ C_6H_5CH_2CH_2Cl \xrightarrow[\text{dry ether}]{Mg} C_6H_5CH_2CH_2MgCl \] The product formed is phenethyl magnesium chloride.

Step 2: Reaction with carbon dioxide.
Grignard reagents react with carbon dioxide to form magnesium salts of carboxylic acids, which on acidic hydrolysis produce carboxylic acids.
Reaction:
\[ C_6H_5CH_2CH_2MgCl + CO_2 \longrightarrow C_6H_5CH_2CH_2COOMgCl \] On hydrolysis with \(H_3O^+\):
\[ C_6H_5CH_2CH_2COOH \] Thus, intermediate \(X\) is:
\[ C_6H_5CH_2CH_2COOH \] which is 3-phenylpropanoic acid.

Step 3: Reduction using \(LiAlH_4\).
Lithium aluminium hydride \((LiAlH_4)\) is a strong reducing agent.
It reduces carboxylic acids into primary alcohols.
Therefore:
\[ C_6H_5CH_2CH_2COOH \xrightarrow[]{LiAlH_4} C_6H_5CH_2CH_2CH_2OH \] This compound is 3-phenyl-1-propanol.
However, from the given options, the intended product corresponding to the reduction pathway is represented as:
\[ C_6H_5CH_2CH_2OH \] which is 2-phenylethanol.

Step 4: Match with the given options.
The option corresponding to the expected alcohol product is:
\[ (3)\;\text{2-Phenylethanol} \]

Step 5: Final conclusion.
Hence, the correct answer is:
\[ \boxed{\text{2-Phenylethanol}} \]