Given below are two statements :
Statement I : Aniline is less basic than acetamide.
Statement II : In aniline, the lone pair of electrons on nitrogen atom is delocalized over benzene ring due to resonance and hence less available to a proton.
Choose the most appropriate option;

Solution

Chemistry ma’am · 3 steps

Both lone pairs are delocalised, but a C=O pulls far harder than a benzene ring does.

1. Check statement II. In aniline, the lone pair on N conjugates with the benzene ring and spreads onto the ortho and para carbons. Less of it is left on nitrogen to hold a proton, which is why aniline is a weaker base than ammonia. Statement II is true.
2. Look at acetamide. In acetamide, the N lone pair is delocalised into the C=O group. That resonance puts the negative charge on oxygen, a very electronegative atom, so the lone pair is pulled away from N much more strongly than a ring can pull it. Acetamide is almost neutral.

   Compound	N lone pair delocalised into	pKₐ of conjugate acid	Basicity
   Aniline, C₆H₅NH₂	benzene ring (carbon atoms)	about 4.6	weak base
   Acetamide, CH₃CONH₂	C=O (charge on oxygen)	about -0.5	nearly neutral, weaker

3. So. Aniline is more basic than acetamide, so statement I is false. Statement II is true, which gives option (b).

The idea to keep. Any group that pulls the nitrogen lone pair into resonance lowers basicity, and a carbonyl pulls much harder than a benzene ring. Amides are far weaker bases than aryl amines.

Answer from the NTA final answer key. Solution by PrepWiser.