Question

An organic compound \(\text{CH}_3\text{-CH=CH-CH}_2\text{-CHO}\) is taken in two different containers A and B. Sample in A is treated with \(\text{H}_2\) / Ni forming new compound P. Sample in B is treated with \(\text{LiAlH}_4\) and hydrolyzed further forming new compound Q. Identify P and Q.

• A. \(\text{P} = \text{Q} = \text{CH}_3\text{-CH=CH-CH}_2\text{-CH}_2\text{OH}\)
• B. \(\text{P} = \text{Q} = \text{CH}_3(\text{CH}_2)_3\text{CH}_2\text{OH}\)
• C. \(\text{P} = \text{CH}_3(\text{CH}_2)_3\text{CHO}\) and \(\text{Q} = \text{CH}_3(\text{CH}_2)_3\text{CH}_2\text{OH}\)
• D. \(\text{P} = \text{CH}_3(\text{CH}_2)_3\text{CH}_2\text{OH}\) and \(\text{Q} = \text{CH}_3\text{-CH=CH-CH}_2\text{-CH}_2\text{OH}\)

Hint:

H2/Ni reduces C=C and CHO; LiAlH4 reduces only CHO.

Answer 1

The Correct Option is D

Step 1: Reagent selectivity:
$\text{H}_2/\text{Ni}$ is a non-selective hydrogenation catalyst. $\text{LiAlH}_4$ delivers hydride to polar $\text{C=O}$ groups.

Step 2: Sample A:
Both $\text{C=C}$ and $\text{CHO}$ take up hydrogen, giving the saturated alcohol $\text{CH}_3(\text{CH}_2)_3\text{CH}_2\text{OH}$.

Step 3: Sample B:
Only the aldehyde is reduced, so the alkene survives: $\text{CH}_3\text{CH=CHCH}_2\text{CH}_2\text{OH}$. This is option (D).

Final Answer:
P is saturated alcohol and Q retains the double bond. \[ \boxed{\text{(D) }\text{P = pentan-1-ol, Q = pent-3-en-1-ol}} \]