Identify the compound 'X' in the following reaction:
`CH_3NO_2overset(Cl_2//NaOH)underset("Excess")rarrX`

Options

A. `ClCH_2NO_2`
B. `Cl_2CHNO_2`
C. `Cl_3CNO_2`
D. `CH_3Cl`

Explanation

To identify the compound 'X' in the reaction given, we will analyze the reaction step by step.

### Step 1: Identify the Reactant
The reactant provided is CH₃NO₂, which is methyl nitrite.

### Step 2: Understand the Reaction Conditions
The reaction is carried out in the presence of Cl₂ (chlorine gas) and NaOH (sodium hydroxide) in excess. The presence of NaOH indicates that it will act as a base in this reaction.

### Step 3: Deprotonation by NaOH
The NaOH will deprotonate the methyl nitrite (CH₃NO₂) at the methyl group. This means that the hydroxide ion (OH⁻) from NaOH will extract a hydrogen atom (H) from CH₃, forming water (H₂O) and generating a carbanion (CH₂NO₂):

\[ \text{CH}_3\text{NO}_2 + \text{NaOH} \rightarrow \text{CH}_2\text{NO}_2^{-} + \text{H}_2\text{O} + \text{Na}^+ \]

### Step 4: Chlorination Reaction
Now, the carbanion (CH₂NO₂⁻) can react with Cl₂. Chlorine will act as an electrophile and will substitute the hydrogen atom on the carbon with chlorine atoms. Since we have excess Cl₂, this substitution can happen multiple times:

1. The first Cl atom replaces one H atom:
\[ \text{CH}_2\text{NO}_2^{-} + \text{Cl}_2 \rightarrow \text{ClCH}_2\text{NO}_2 + \text{HCl} \]

2. The second Cl atom can replace another H atom:
\[ \text{ClCH}_2\text{NO}_2 + \text{Cl}_2 \rightarrow \text{Cl}_2\text{CH}\text{NO}_2 + \text{HCl} \]

3. Finally, the third Cl atom replaces the last H atom:
\[ \text{Cl}_2\text{CH}\text{NO}_2 + \text{Cl}_2 \rightarrow \text{CCl}_3\text{NO}_2 + \text{HCl} \]

### Step 5: Final Product
After all substitutions, the final product 'X' is identified as trichloroacetic acid (CCl₃NO₂).

### Conclusion
Thus, the compound 'X' formed in the reaction is **CCl₃NO₂**.

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