Predict the product of the following reaction:

Question

Accepted Answer

To understand why Option 2 is the correct answer for the reaction $C_6H_5NHCH_3 + NaNO_2 + HCl \rightarrow$ Product, let's analyze the reaction step by step.

### Explanation for Option 2:

1. **Reaction Type**: The reaction involves an aromatic amine ($C_6H_5NHCH_3$) with sodium nitrite ($NaNO_2$) and hydrochloric acid ($HCl$). This is a classic setup for a diazotization reaction, which typically occurs with primary aromatic amines. However, in this case, we have a secondary amine ($N-methylaniline$).

2. **Mechanism**: 
   - The sodium nitrite ($NaNO_2$) in the presence of hydrochloric acid ($HCl$) generates nitrous acid ($HNO_2$) in situ.
   - The nitrous acid then reacts with the secondary amine ($C_6H_5NHCH_3$) to form an N-nitroso compound.
   - Specifically, the secondary amine undergoes nitrosation to form $C_6H_5N(CH_3)-N=O$, which is an N-nitrosoamine.

3. **Product Formation**: The product of this reaction is an N-nitroso compound, which matches Option 2: $C_6H_5N(CH_3)-N=O$.

### Why Other Options are Incorrect:

1. **Option 1: $C_6H_5N(CH_3)OH$**
   - This option suggests the formation of a hydroxylamine derivative. However, the reaction conditions (nitrous acid with a secondary amine) do not favor the formation of hydroxylamines. Instead, nitrosation occurs, leading to the N-nitroso compound.

2. **Option 3: $C_6H_5N(CH_3)-NO_2$**
   - This option implies the formation of a nitro compound. Nitro compounds are typically formed through nitration reactions involving nitric acid, not through nitrosation with nitrous acid. Therefore, this product is not feasible under the given reaction conditions.

3. **Option 4: o-nitro-N-methylaniline + p-nitro-N-methylaniline**
   - This option suggests nitration at the ortho and para positions of the aromatic ring. However, the reaction conditions (nitrous acid with a secondary amine) do not support electrophilic aromatic substitution leading to nitration. Instead, nitrosation of the amine group occurs.

In conclusion, the reaction of $C_6H_5NHCH_3$ with $NaNO_2$ and $HCl$ leads to the formation of an N-nitroso compound, specifically $C_6H_5N(CH_3)-N=O$, which is why Option 2 is the correct answer.