The decreasing order of basicity of the following amines is :

Solution

Chemistry ma’am · 3 steps

Ask one question for each nitrogen: is its lone pair free, or is the ring using it?

1. What basicity needs. An amine is a base because its nitrogen lone pair picks up H⁺. The more freely available that lone pair is, the stronger the base.

   Amine	Nitrogen	Lone pair	pKₐ of conjugate acid
   (III) Cyclohexylamine	sp³, on an alkyl group	Fully free, and the alkyl group pushes electrons in	about 10.6
   (II) Pyridine	sp², in the ring	Free, not part of the aromatic sextet, but held closer by more s character	about 5.2
   (I) Aniline	attached to a benzene ring	Partly spread into the ring by resonance	about 4.6
   (IV) Pyrrole	sp², in the ring	Part of the aromatic sextet, so it is not available	about -3.8

2. Pyridine versus aniline. The sp² nitrogen of pyridine holds its lone pair a little tighter. Even so, that pair points away from the ring and is not delocalised, while the aniline lone pair is shared with the benzene ring. Pyridine therefore comes out slightly more basic.
3. So. The order is (III) > (II) > (I) > (IV), which is option (a). Protonating pyrrole would destroy its aromaticity, so it is the weakest base by far.

The idea to keep. Find the nitrogen lone pair first. A pair that is localised and on an alkyl carbon gives the strongest base. A pair shared with a ring by resonance is weaker, and a pair inside an aromatic sextet hardly acts as a base.

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