Question

Which of the following is the most basic compound?

• A. Aniline
• B. p-Nitro aniline
• C. Benzyl amine
• D. Acetanilide

Hint:

In benzyl amine the \( CH_2 \) separates \( NH_2 \) from the ring, so no resonance withdrawal; its lone pair stays fully available.

Answer 1

The Correct Option is C

Step 1: Basic strength of an amine depends on the availability of the lone pair on nitrogen. The more freely available the lone pair, the more basic the compound.
Step 2: Compare the four compounds:
Benzyl amine \( (C_6H_5CH_2NH_2) \): the \( NH_2 \) group is attached to an \( sp^3 \) \( CH_2 \) carbon, not directly to the ring. Its lone pair is not delocalised into the benzene ring, so it behaves like an aliphatic amine and the lone pair is fully available.
Aniline \( (C_6H_5NH_2) \): the \( NH_2 \) is directly on the ring, so the lone pair is delocalised into the ring by resonance, reducing basicity.
p-Nitroaniline: the electron withdrawing \( -NO_2 \) group at the para position pulls the lone pair even more, so it is less basic than aniline.
Acetanilide \( (C_6H_5NHCOCH_3) \): the lone pair is delocalised onto the carbonyl of the amide group, making it the least basic (almost neutral).
Step 3: Since only benzyl amine keeps its nitrogen lone pair free, it is the strongest base.
\[\boxed{\text{Benzyl amine}}\]