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: Understanding the Concept
The aldehyde group has to become CH2OH while the C=C double bond stays unchanged.

Step 2: Why not H2/Ni or H2/Pd
Catalytic hydrogenation also saturates the C=C bond and would give pentan-1-ol.

Step 3: Right reagent
\(\text{LiAlH}_{4}\) gives hydride to the polar carbonyl carbon only. Acidic work-up with \(\text{H}_{3}\text{O}^{+}\) gives the alcohol. The non-polar C=C is not attacked.

Final Answer:
The best reagent is \(\text{LiAlH}_{4}\) followed by \(\text{H}_{3}\text{O}^{+}\). \[ \boxed{\text{(B) }\text{LiAlH}_{4}\text{, H}_{3}\text{O}^{+}} \]