Question:
Identify the product formed in the following reaction.
\[ \text{CH}_3 - \text{CH} = \text{CH} - \text{CH}_2 - \text{CHO} \xrightarrow[ii) \text{ H}_3\text{O}^+]{i) \text{ LiAlH}_4} \text{product} \]
Identify the product formed in the following reaction.
\[ \text{CH}_3 - \text{CH} = \text{CH} - \text{CH}_2 - \text{CHO} \xrightarrow[ii) \text{ H}_3\text{O}^+]{i) \text{ LiAlH}_4} \text{product} \]
\[ \text{CH}_3 - \text{CH} = \text{CH} - \text{CH}_2 - \text{CHO} \xrightarrow[ii) \text{ H}_3\text{O}^+]{i) \text{ LiAlH}_4} \text{product} \]
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LiAlH$_4$: Reduces C=O, not C=C.
Updated On: May 4, 2026
- \(\text{CH}_3 - (\text{CH}_2)_3 - \text{CH}_2\text{OH}\)
- \(\text{CH}_3 - \text{CH}_2 - \text{CH} = \text{CH} - \text{CH}_2 - \text{OH}\)
- \(\text{CH}_3 - \text{CH} = \text{CH} - \text{CH}_2 - \text{CH}_2 - \text{OH}\)
- \(\text{CH}_3 - \text{CH} = \text{CH} - \text{CH}_2 - \text{OH}\)
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The Correct Option is C
Solution and Explanation
Concept:
• LiAlH$_4$ reduces aldehydes to primary alcohols
• It does NOT affect C=C double bond
Step 1: Identify functional group. \[ -CHO \rightarrow -CH_2OH \]
Step 2: Apply reduction. \[ \text{CH}_3 - \text{CH} = \text{CH} - \text{CH}_2 - \text{CHO} \rightarrow \text{CH}_3 - \text{CH} = \text{CH} - \text{CH}_2 - \text{CH}_2OH \]
Step 3: Conclusion. \[ \text{Correct product = Option (C)} \]
• LiAlH$_4$ reduces aldehydes to primary alcohols
• It does NOT affect C=C double bond
Step 1: Identify functional group. \[ -CHO \rightarrow -CH_2OH \]
Step 2: Apply reduction. \[ \text{CH}_3 - \text{CH} = \text{CH} - \text{CH}_2 - \text{CHO} \rightarrow \text{CH}_3 - \text{CH} = \text{CH} - \text{CH}_2 - \text{CH}_2OH \]
Step 3: Conclusion. \[ \text{Correct product = Option (C)} \]
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