The decreasing order of basicity of the following amines is
(I) (see structure) (II) (see structure) (III) (see structure) (IV) (see structure)IIIIIIIV
(A)(III) > (I) > (II) > (IV)
(B)(II) > (III) > (IV) > (I)
(C)(I) > (III) > (IV) > (II)
(D)(III) > (II) > (I) > (IV)✓ Correct
Answer: (D)
Solution
The basicity of each nitrogen is set by how freely its lone pair is offered to a proton. In cyclohexylamine the lone pair sits on an sp3 nitrogen with an electron-donating alkyl group and no delocalization, making it the most basic. In pyridine the lone pair lies in an sp2 orbital in the ring plane, outside the aromatic π system, so it is available but held more tightly than in an aliphatic amine. In aniline the lone pair is drawn into the benzene ring by resonance, lowering its availability. In pyrrole the lone pair is part of the aromatic sextet itself, so offering it would destroy aromaticity, leaving it least basic. The decreasing order is cyclohexylamine > pyridine > aniline > pyrrole.
JEE Main · 2020 · 09 Jan 2020 · Shift 2 · Single correct option