Question

Identify the product formed in the following reaction,
$\text{CH}_3-\text{CH}=\text{CH}-\text{CH}_2-\text{CHO} \xrightarrow{\text{(i) LiAlH}_4 \text{ (ii) H}_2\text{O}} \text{Products}$

• A. $\text{CH}_3-\text{CH}=\text{CH}-\text{CH}_2-\text{OH}$
• B. $\text{CH}_3-(\text{CH}_2)_3-\text{CH}_2-\text{OH}$
• C. $\text{CH}_3-\text{CH}_2-\text{CH}=\text{CH}-\text{CH}_2-\text{OH}$
• D. $\text{CH}_3-\text{CH}=\text{CH}-\text{CH}_2-\text{CH}_2-\text{OH}$

Hint:

$\text{LiAlH}_4$ acts specifically on polar bonds (like C=O or C-N). It leaves non-polar carbon-carbon double and triple bonds completely untouched unless they are part of a specific conjugated system.

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
We need to identify the final organic product when an unsaturated aldehyde is reduced using lithium aluminium hydride ($\text{LiAlH}_4$).

Step 2: Detailed Explanation:
Lithium aluminium hydride ($\text{LiAlH}_4$) is a strong reducing agent. It effectively reduces polar carbonyl groups, converting aldehydes (-CHO) into primary alcohols ($\text{-CH}_2\text{OH}$).
However, $\text{LiAlH}_4$ generally does not reduce isolated or non-conjugated carbon-carbon double bonds (C=C) because the hydride ion ($\text{H}^-$) is a nucleophile and repels the electron-rich pi bond.
The reactant given is pent-3-enal: $\text{CH}_3-\text{CH}=\text{CH}-\text{CH}_2-\text{CHO}$.
The double bond is isolated from the carbonyl group by a $-\text{CH}_2-$ spacer.
Therefore, only the aldehyde group is reduced, while the double bond remains intact:
$$\text{CH}_3-\text{CH}=\text{CH}-\text{CH}_2-\text{CHO} \rightarrow \text{CH}_3-\text{CH}=\text{CH}-\text{CH}_2-\text{CH}_2-\text{OH}$$

Step 3: Final Answer:
The correct product is $\text{CH}_3-\text{CH}=\text{CH}-\text{CH}_2-\text{CH}_2-\text{OH}$, matching option (D).