JEE Main 2020 · 6 September · morning · Single correct · +4 / −1
The increasing order of pK b values of the following compounds is :
- (a) I < II < IV < III
- (b) I < II < III < IV
- (c) II < I < III < IV
- (d) II < IV < III < I
Answer and solution
Answer (a) I < II < IV < III
Solution
Chemistry ma’am · 3 steps
Lower pK_b means a stronger base, so rank how freely each nitrogen can give its lone pair.
- Flip the question. Increasing pK_b is the same as decreasing basicity. So the strongest base comes first. In every compound the lone pair on N is partly delocalised into the ring, and the ring substituent makes it more or less available.
- Para versus meta. A group at the para position can push or pull electrons by resonance and by induction. A group at the meta position cannot reach the nitrogen-bearing carbon by resonance, so only its -I effect counts.
Compound Substituent and position Effect on N lone pair Basicity I -OCH₃ at para +R beats -I: more electron density on N highest II none reference C₆H₅N(CH₃)₂ second IV -OH at meta only -I, a mild pull third III -CN at meta strong -I, a large pull lowest - So. Basicity runs I > II > IV > III, so pK_b runs I < II < IV < III, which is option (a).
The idea to keep. For substituted anilines, electron donors at ortho or para raise basicity. At meta, only the inductive effect works, so every electronegative group there lowers basicity, and -CN lowers it the most.
Answer from the NTA final answer key. Solution by PrepWiser.