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: Read the reactant.
The starting molecule is $\text{CH}_3-\text{CH}=\text{CH}-\text{CH}_2-\text{CHO}$. It has a double bond ($C=C$) and an aldehyde group ($-CHO$).

Step 2: Know the reagent.
$\text{LiAlH}_4$ is a strong reducing agent. It adds hydrogen to certain groups.

Step 3: See what it attacks.
$\text{LiAlH}_4$ reduces the polar carbonyl group, turning $-CHO$ into the alcohol $-CH_2OH$.

Step 4: See what it leaves alone.
A plain carbon-carbon double bond is not polar, so $\text{LiAlH}_4$ does not touch it. The double bond here is separated from the $-CHO$ by a $-CH_2-$ group, so it stays.

Step 5: Write the product.
\[ \text{CH}_3-\text{CH}=\text{CH}-\text{CH}_2-\text{CHO} \rightarrow \text{CH}_3-\text{CH}=\text{CH}-\text{CH}_2-\text{CH}_2-\text{OH} \]
Step 6: Conclusion.
Only the aldehyde becomes an alcohol; the double bond remains. \[ \boxed{\text{CH}_3-\text{CH}=\text{CH}-\text{CH}_2-\text{CH}_2-\text{OH (Option 4)}} \]