Question

Identify product 'B' in following reaction.
$\text{CH}_3\text{Br} + \text{AgNO}_2 \longrightarrow \text{A} \xrightarrow{\text{Sn / HCl}} \text{B}$

• A. $\text{CH}_3\text{NO}_2$
• B. $\text{CH}_3\text{NH}_2$
• C. $\text{CH}_3\text{Cl}$
• D. $\text{CH}_3\text{OH}$

Hint:

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 ($\text{R-NH}_2$).

Answer 1

The Correct Option is B

Step 1: Read the reaction chain.
We start with $\text{CH}_3\text{Br}$, react it with silver nitrite $\text{AgNO}_2$ to make an intermediate A, then reduce A with $\text{Sn / HCl}$ to make the final product B. Our job is to name B.
Step 2: Identify what AgNO2 does.
Silver nitrite is the classic reagent that introduces a nitro group. Because nitrogen carries the active lone pair here, the methyl group bonds to nitrogen, giving nitromethane.
Step 3: Write intermediate A.
So $A = \text{CH}_3\text{NO}_2$ (nitromethane), with $\text{AgBr}$ leaving as a side product.
Step 4: Recognise the reducing agent.
The combination $\text{Sn / HCl}$ supplies nascent hydrogen and is a standard tool for reducing a nitro group $(-\text{NO}_2)$ all the way down to an amine group $(-\text{NH}_2)$.
Step 5: Apply the reduction.
Reducing $\text{CH}_3\text{NO}_2$ converts $-\text{NO}_2$ into $-\text{NH}_2$, giving $\text{CH}_3\text{NH}_2$ (methylamine), with water released.
Step 6: Match with the options.
The product B is $\text{CH}_3\text{NH}_2$, which is option 2 and agrees with the key.
\[ \boxed{B = \text{CH}_3\text{NH}_2} \]