`A overset( "reduction ") to B overset(CHCl_3 //KOH ) to C overset("reduction ") to `N - methyl aniline then A is

Options

A.
B.
C. `CH_3NH_2`
D.

Answer

To solve the problem step by step, we need to identify compound A based on the given transformations leading to N-methyl aniline.

### Step 1: Identify the final product
The final product is N-methyl aniline. This compound has the structure where a methyl group (–CH₃) is attached to the nitrogen atom of an aniline (C₆H₅NH₂) structure.

### Step 2: Determine the precursor to N-methyl aniline
To form N-methyl aniline, we can deduce that it is formed from an aniline derivative. The methyl group on nitrogen suggests that the precursor must have a nitrogen atom that can be methylated.

### Step 3: Identify the reaction leading to aniline
The question states that there is a reduction step that leads to compound C. Aniline can be produced by the reduction of a nitro group (–NO₂) or a similar nitrogen-containing compound. The most likely precursor for aniline is nitrobenzene (C₆H₅NO₂), which can be reduced to aniline using reducing agents like Sn/HCl.

### Step 4: Identify the reaction with CHCl₃ and KOH
The reaction of aniline with CHCl₃ in the presence of KOH is known as the carbylamine reaction, which leads to the formation of N-methyl aniline. In this reaction, the amine group (–NH₂) reacts with chloroform (CHCl₃) and a strong base (KOH) to form a methylated product.

### Step 5: Conclusion about compound A
Since we have established that:
- A reduction of nitrobenzene (compound A) leads to aniline (compound B).
- The reaction of aniline with CHCl₃ and KOH leads to N-methyl aniline (compound C).

Thus, compound A is **nitrobenzene (C₆H₅NO₂)**.

### Final Answer:
**A is nitrobenzene (C₆H₅NO₂).**

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