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: Understanding the Concept:
The compound \(\text{CH}_3\text{CH=CH-CH}_2\text{CHO}\) has two reducible groups: a \(\text{C=C}\) bond and an aldehyde. The two reagents differ in what they reduce.

Step 2: Detailed Explanation:
In A, catalytic hydrogenation with \(\text{H}_2/\text{Ni}\) adds hydrogen to the double bond and also reduces \(-\text{CHO}\) to \(-\text{CH}_2\text{OH}\).
\[ \text{P} = \text{CH}_3(\text{CH}_2)_3\text{CH}_2\text{OH} \]
In B, \(\text{LiAlH}_4\) reduces \(-\text{CHO}\) to \(-\text{CH}_2\text{OH}\) but does not touch an isolated \(\text{C=C}\).
\[ \text{Q} = \text{CH}_3\text{CH=CH-CH}_2\text{CH}_2\text{OH} \]
So P is pentan-1-ol and Q is pent-3-en-1-ol.

Step 3: Final Answer:
Option (D) matches both products.

Final Answer:
H2/Ni reduces both groups; LiAlH4 only the aldehyde. \[ \boxed{\text{(D) }\text{P = pentan-1-ol, Q = pent-3-en-1-ol}} \]