D is N-methyl aniline in the given below sequence of reaction: $A \xrightarrow{\text{reduction}} B \xrightarrow{CHCl_3/KOH} C \xrightarrow{\text{reduction}} D$. Structure of A can be:

Let's deduce the reaction sequence in reverse:

1. Compound D: Given as N-methyl aniline ($C_6H_5-NH-CH_3$), which is a secondary amine.
2. Compound C: Formed by the reaction of B with $CHCl_3$ and KOH, and upon reduction gives D. The reaction of a primary amine with $CHCl_3/KOH$ is the Carbylamine reaction, which yields an isocyanide ($R-NC$). Reduction of an isocyanide ($R-NC + 4[H] \rightarrow R-NH-CH_3$) always produces a secondary amine with one methyl group. Since D is N-methyl aniline, C must be phenyl isocyanide ($C_6H_5-NC$).
3. Compound B: Since B undergoes the carbylamine reaction to form phenyl isocyanide, B must be the corresponding primary aromatic amine, aniline ($C_6H_5-NH_2$).
4. Compound A: A is reduced to form aniline (B). The most common precursor that yields aniline upon reduction is nitrobenzene ($C_6H_5-NO_2$).

Therefore, the structure of A is nitrobenzene.