Question

Which is the best reagent for the following conversion?
\(\text{CH}_{3}\text{-CH=CH-CH}_{2}\text{-CHO}\longrightarrow \text{CH}_{3}\text{-CH=CH-CH}_{2}\text{-CH}_{2}\text{-OH}\)

• A. \(\text{H}_{2}\) / Ni
• B. \(\text{LiAlH}_{4}\), \(\text{H}_{3}\text{O}^{+}\)
• C. \(\text{H}_{2}\) / Pd
• D. Anhydrous \(\text{ZnCl}_{2}\), conc. HCl

Hint:

Only the carbonyl group should be reduced, so use a hydride reagent, not catalytic hydrogenation.

Answer 1

The Correct Option is B

Step 1: Eliminate
$\text{H}_{2}$/Ni and $\text{H}_{2}$/Pd reduce C=C as well. Anhydrous $\text{ZnCl}_{2}$ with conc. HCl is Lucas reagent and is not a reducing agent.

Step 2: Remaining option
Only $\text{LiAlH}_{4}$ with acid work-up reduces just the carbonyl group.

Step 3: Final Check
Hydride reagents reduce only polar multiple bonds. This confirms option (B).

Final Answer:
Option (B) is correct. \[ \boxed{\text{(B) }\text{LiAlH}_{4}\text{, H}_{3}\text{O}^{+}} \]