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: Read the reaction sequence from the text.
The starting material is $\beta$-phenethyl chloride $C_6H_5CH_2CH_2Cl$, which is first treated with magnesium in dry ether and then with carbon dioxide followed by acid. We trace each step to find the final product $Y$.
Step 2: Form the Grignard reagent.
Alkyl halides react with magnesium in dry ether to give a Grignard reagent: \[ C_6H_5CH_2CH_2Cl \xrightarrow[\text{dry ether}]{Mg} C_6H_5CH_2CH_2MgCl. \]
Step 3: React the Grignard reagent with carbon dioxide.
The carbon of the Grignard adds to $CO_2$, giving the magnesium salt of a carboxylic acid: $C_6H_5CH_2CH_2COOMgCl$.
Step 4: Hydrolyse with acid.
Treatment with dilute acid releases the free carboxylic acid: \[ C_6H_5CH_2CH_2COOMgCl \xrightarrow{H_3O^+} C_6H_5CH_2CH_2COOH. \]
Step 5: Note the chain growth.
The $CO_2$ step adds one carbon, so the two carbon chloride becomes a three carbon acid, namely 3-phenylpropanoic acid $C_6H_5CH_2CH_2COOH$.
Step 6: Match with the option.
The major product $Y$ is 3-phenylpropanoic acid, which corresponds to option 3. So the answer is
\[ \boxed{C_6H_5CH_2CH_2COOH} \]