Identify the compounds

• $A$ = Phenyl methanamine = benzylamine, $C_6H_5CH_2NH_2$
• $B$ = $N,N$-Dimethylaniline, $C_6H_5N(CH_3)_2$
• $C$ = $N$-Methylaniline, $C_6H_5NHCH_3$
• $D$ = Benzenamine = aniline, $C_6H_5NH_2$

Basicity depends on availability of lone pair on nitrogen

Greater the availability of the lone pair for protonation, greater is the basic strength.

Compare benzylamine ($A$) with aromatic amines

In benzylamine, the nitrogen is not directly attached to the benzene ring: $C_6H_5-CH_2-NH_2$ So the lone pair on nitrogen is not involved in resonance with the aromatic ring.

Hence, benzylamine behaves like an aliphatic amine and is more basic than aniline and its $N$-methyl derivatives.

Therefore, $A$ is the most basic.

Compare aniline derivatives $D$, $C$, and $B$

In all these compounds, nitrogen is directly attached to the benzene ring, so the lone pair is delocalized into the ring by resonance, reducing basicity.

Now consider the effect of methyl groups on nitrogen:

• Methyl groups show $+I$ effect, which tends to increase electron density on nitrogen.
• But in aromatic amines, alkyl substitution on nitrogen also affects solvation of the conjugate acid and can increase resonance donation of the lone pair into the ring.

In aqueous medium, the general order is: $\text{aniline} > N\text{-methylaniline} > N,N\text{-dimethylaniline}$

because increased alkyl substitution decreases solvation of the protonated species, making the base effectively weaker.

Thus: $D > C > B$

Combine both comparisons

Therefore, the overall basicity order is: $A > D > C > B$

Match with given options

Options given are:

• A: $D > C > B > A$
• B: $A > C > B > D$
• C: $D > B > C > A$
• D: $A > B > C > D$

Our derived order $A > D > C > B$ is not present among the options.

Closest chemical conclusion

From standard basicity trends:

• benzylamine is strongest
• among aniline, $N$-methylaniline, and $N,N$-dimethylaniline, the usual aqueous order is $D > C > B$

So the correct order should be: $A > D > C > B$

Hence the stored answer $D$ is not correct.