The increasing order of basicity of the following compounds is :

Solution

Chemistry ma’am · 4 steps

Ask one thing of each nitrogen: is its lone pair free to take a proton, and how tightly is it held?

1. Name the four. (A) is pyridine, (B) is pyrrole, (C) is pyrrolidine and (D) is imidazole. Each base uses a nitrogen lone pair to take up H⁺.
2. Two deciding factors. A lone pair that is part of the aromatic sextet is not available, so that base is very weak. A free lone pair in an sp³ orbital is held more loosely than one in an sp² orbital, which has more s character. Free sp³ lone pairs are therefore more basic.

   Compound	Basic nitrogen lone pair	pKₐ of conjugate acid (approx.)	Basicity
   (B) pyrrole	in the aromatic sextet, not available	-3.8	weakest
   (A) pyridine	free, sp²	5.2	weak
   (D) imidazole	free, sp² on the =N⁻ nitrogen, and the cation is resonance-stabilised	7.0	moderate
   (C) pyrrolidine	free, sp³, with alkyl groups donating electrons	11.3	strongest

3. Why D beats A. Both use an sp² lone pair. When imidazole is protonated, the positive charge is shared equally over both nitrogens by resonance. That stabilises the conjugate acid, so imidazole is more basic than pyridine.
4. So. The increasing order of basicity is (B) < (A) < (D) < (C), which is option (d).

The idea to keep. For nitrogen heterocycles, first check whether the lone pair is part of the aromatic ring. If it is not, an sp³ lone pair beats sp², and resonance that spreads the charge in the protonated form adds basicity.

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