Question:
In the following sequence of reactions, the compound \( C \) formed would be
\[
CH_3CH_2CONH_2 \xrightarrow{LiAlH_4} A \xrightarrow{HNO_2} B \xrightarrow{CrO_3} C
\]
In the following sequence of reactions, the compound \( C \) formed would be
\[
CH_3CH_2CONH_2 \xrightarrow{LiAlH_4} A \xrightarrow{HNO_2} B \xrightarrow{CrO_3} C
\]
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Remember chain: Amide $\longrightarrow$ Amine $\longrightarrow$ Alcohol $\longrightarrow$ Acid (very common sequence!).
Updated On: Apr 22, 2026
- 2-propanol
- propanol
- propanoic acid
- propanal
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Verified By Collegedunia
The Correct Option is C
Solution and Explanation
Concept:
• \( LiAlH_4 \) reduces amide $\longrightarrow$ amine
• \( HNO_2 \) converts primary amine $\longrightarrow$ alcohol
• \( CrO_3 \) oxidizes primary alcohol $\longrightarrow$ carboxylic acid
Step 1: Reduction.
\[ CH_3CH_2CONH_2 \xrightarrow{LiAlH_4} CH_3CH_2CH_2NH_2 \]
Step 2: Diazotization.
\[ CH_3CH_2CH_2NH_2 \xrightarrow{HNO_2} CH_3CH_2CH_2OH \]
Step 3: Oxidation.
\[ CH_3CH_2CH_2OH \xrightarrow{CrO_3} CH_3CH_2COOH \]
Step 4: Final product.
Propanoic acid
• \( LiAlH_4 \) reduces amide $\longrightarrow$ amine
• \( HNO_2 \) converts primary amine $\longrightarrow$ alcohol
• \( CrO_3 \) oxidizes primary alcohol $\longrightarrow$ carboxylic acid
Step 1: Reduction.
\[ CH_3CH_2CONH_2 \xrightarrow{LiAlH_4} CH_3CH_2CH_2NH_2 \]
Step 2: Diazotization.
\[ CH_3CH_2CH_2NH_2 \xrightarrow{HNO_2} CH_3CH_2CH_2OH \]
Step 3: Oxidation.
\[ CH_3CH_2CH_2OH \xrightarrow{CrO_3} CH_3CH_2COOH \]
Step 4: Final product.
Propanoic acid
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