Question:
Identify product 'B' in following reaction.
$\text{CH}_3\text{Br} + \text{AgNO}_2 \longrightarrow \text{A} \xrightarrow{\text{Sn / HCl}} \text{B}$
Identify product 'B' in following reaction.
$\text{CH}_3\text{Br} + \text{AgNO}_2 \longrightarrow \text{A} \xrightarrow{\text{Sn / HCl}} \text{B}$
$\text{CH}_3\text{Br} + \text{AgNO}_2 \longrightarrow \text{A} \xrightarrow{\text{Sn / HCl}} \text{B}$
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Be very careful with nitrite reagents! Alkyl halide + $\text{KNO}_2$ (ionic) gives an alkyl nitrite (R-ONO), whereas Alkyl halide + $\text{AgNO}_2$ (covalent) gives a nitroalkane ($\text{R-NO}_2$). Once you have a nitroalkane, adding Sn/HCl or Fe/HCl always reduces it directly to a primary amine ($\text{R-NH}_2$).
Updated On: Jun 12, 2026
- $\text{CH}_3\text{NO}_2$
- $\text{CH}_3\text{NH}_2$
- $\text{CH}_3\text{Cl}$
- $\text{CH}_3\text{OH}$
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The Correct Option is B
Solution and Explanation
Step 1: Understanding the Question:
The question presents a two-step organic transformation starting with an alkyl halide (bromomethane) and asks us to identify the final product 'B'.
Step 2: Key Formula or Approach:
1. The reaction of an alkyl halide with silver nitrite ($\text{AgNO}_2$) is a nucleophilic substitution reaction where the nitrogen atom acts as the nucleophile due to the covalent nature of the Ag-O bond, primarily yielding a nitroalkane.
2. The subsequent treatment of a nitroalkane with a metal in a mineral acid (Sn / HCl) reduces the nitro group ($\text{-NO}_2$) completely to a primary amino group ($\text{-NH}_2$).
Step 3: Detailed Explanation:
3.
First Step (Formation of A): Bromomethane ($\text{CH}_3\text{Br}$) reacts with silver nitrite ($\text{AgNO}_2$). Because the bond between silver and oxygen is highly covalent, the lone pair on the nitrogen atom attacks the methyl group, leading to the formation of nitromethane ($\text{CH}_3\text{NO}_2$) as the major product 'A': $$\text{CH}_3\text{Br} + \text{AgNO}_2 \rightarrow \text{CH}_3\text{NO}_2\text{ (A)} + \text{AgBr}$$ 4.
Second Step (Formation of B): Nitromethane (A) is treated with tin and concentrated hydrochloric acid (Sn / HCl). This is a powerful reducing combination that supplies nascent hydrogen to completely reduce the nitro group into a primary amine group: $$\text{CH}_3\text{NO}_2 \xrightarrow{\text{Sn / HCl}} \text{CH}_3\text{NH}_2\text{ (B)} + 2\text{H}_2\text{O}$$ The resulting molecule 'B' is methanamine (methylamine).
Step 4: Final Answer:
The final product 'B' is $\text{CH}_3\text{NH}_2$, which corresponds to option (B).
The question presents a two-step organic transformation starting with an alkyl halide (bromomethane) and asks us to identify the final product 'B'.
Step 2: Key Formula or Approach:
1. The reaction of an alkyl halide with silver nitrite ($\text{AgNO}_2$) is a nucleophilic substitution reaction where the nitrogen atom acts as the nucleophile due to the covalent nature of the Ag-O bond, primarily yielding a nitroalkane.
2. The subsequent treatment of a nitroalkane with a metal in a mineral acid (Sn / HCl) reduces the nitro group ($\text{-NO}_2$) completely to a primary amino group ($\text{-NH}_2$).
Step 3: Detailed Explanation:
3.
First Step (Formation of A): Bromomethane ($\text{CH}_3\text{Br}$) reacts with silver nitrite ($\text{AgNO}_2$). Because the bond between silver and oxygen is highly covalent, the lone pair on the nitrogen atom attacks the methyl group, leading to the formation of nitromethane ($\text{CH}_3\text{NO}_2$) as the major product 'A': $$\text{CH}_3\text{Br} + \text{AgNO}_2 \rightarrow \text{CH}_3\text{NO}_2\text{ (A)} + \text{AgBr}$$ 4.
Second Step (Formation of B): Nitromethane (A) is treated with tin and concentrated hydrochloric acid (Sn / HCl). This is a powerful reducing combination that supplies nascent hydrogen to completely reduce the nitro group into a primary amine group: $$\text{CH}_3\text{NO}_2 \xrightarrow{\text{Sn / HCl}} \text{CH}_3\text{NH}_2\text{ (B)} + 2\text{H}_2\text{O}$$ The resulting molecule 'B' is methanamine (methylamine).
Step 4: Final Answer:
The final product 'B' is $\text{CH}_3\text{NH}_2$, which corresponds to option (B).
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