Question:
Given the following sequence of reactions:
\[ \mathrm{CH_3CH_2I \xrightarrow{NaCN} A \xrightarrow{\text{Partial Hydrolysis}} B \xrightarrow{Br_2/NaOH} C} \]
The major product \(C\) is:
Given the following sequence of reactions:
\[ \mathrm{CH_3CH_2I \xrightarrow{NaCN} A \xrightarrow{\text{Partial Hydrolysis}} B \xrightarrow{Br_2/NaOH} C} \]
The major product \(C\) is:
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Hofmann bromamide reaction converts amide to amine with one carbon less.
Updated On: Apr 20, 2026
- \(\mathrm{CH_3CH_2NH_2}\)
- \(\mathrm{CH_3CH_2COONH_4}\)
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The Correct Option is A
Solution and Explanation
Step 1: Concept:
Identify each reaction type:
Step 2: Detailed Explanation:
First Reaction:
\[ \mathrm{CH_3CH_2I \xrightarrow{NaCN} CH_3CH_2CN} \] Product \(A\) is Propanenitrile.
Second Reaction (Partial Hydrolysis):
\[ \mathrm{CH_3CH_2CN \xrightarrow{Partial\ Hydrolysis} CH_3CH_2CONH_2} \] Product \(B\) is Propanamide.
Third Reaction (Hofmann Bromamide Reaction):
\[ \mathrm{CH_3CH_2CONH_2 \xrightarrow{Br_2/NaOH} CH_3CH_2NH_2} \] In Hofmann bromamide reaction: - Amide loses one carbon - Forms a primary amine
So, product \(C\) is Ethylamine.
Step 3: Final Answer:
\[ \boxed{\mathrm{CH_3CH_2NH_2}} \]
Identify each reaction type:
- \( \mathrm{NaCN} \) → Nucleophilic substitution (chain extension)
- Partial hydrolysis of nitrile → Amide
- \( \mathrm{Br_2/NaOH} \) → Hofmann bromamide reaction (gives amine with one less carbon)
Step 2: Detailed Explanation:
First Reaction:
\[ \mathrm{CH_3CH_2I \xrightarrow{NaCN} CH_3CH_2CN} \] Product \(A\) is Propanenitrile.
Second Reaction (Partial Hydrolysis):
\[ \mathrm{CH_3CH_2CN \xrightarrow{Partial\ Hydrolysis} CH_3CH_2CONH_2} \] Product \(B\) is Propanamide.
Third Reaction (Hofmann Bromamide Reaction):
\[ \mathrm{CH_3CH_2CONH_2 \xrightarrow{Br_2/NaOH} CH_3CH_2NH_2} \] In Hofmann bromamide reaction: - Amide loses one carbon - Forms a primary amine
So, product \(C\) is Ethylamine.
Step 3: Final Answer:
\[ \boxed{\mathrm{CH_3CH_2NH_2}} \]
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