JEE Main 2020 · 9 January · morning · Single correct · +4 / −1
The increasing order of basicity for the following intermediates is (from weak to strong)
H₂C=CH-C(⊖)H₂ (ii) HC≡C⊖ (iii) C(⊖)H₃ (iv) C(⊖)N (v)
- (a) (iii) < (i) < (ii) < (iv) < (v)
- (b) (iii) < (iv) < (ii) < (i) < (v)
- (c) (v) < (i) < (iv) < (ii) < (iii)
- (d) (v) < (iii) < (ii) < (iv) < (i)
Answer and solution
Answer (d) (v) < (iii) < (ii) < (iv) < (i)
Solution
Chemistry ma’am · 3 steps
A carbanion is a weaker base the more its negative charge is spread out or held tightly. Rank stability first.
- Stable means weak. A base works by donating its lone pair. If the negative charge is stabilised, the anion is less eager to donate it. So the most stable carbanion is the weakest base.
- What stabilises charge. Three things stabilise a carbanion: more s-character at the carbon (sp holds electrons closest to the nucleus), resonance that spreads the charge, and an electronegative atom nearby. Alkyl groups do the opposite. Their +I effect pushes extra electron density onto the carbon, so the anion becomes less stable.
Carbanion Key effect Stability Basicity (v) ⁻C≡N sp carbon and an electronegative N pulling charge highest weakest (iii) HC≡C⁻ sp carbon, 50% s-character high weak (ii) H₂C=CH-CH₂⁻ resonance spreads charge over two carbons moderate moderate (iv) CH₃⁻ sp3 carbon, no stabilisation low strong (i) (CH₃)₃C⁻ three +I methyl groups destabilise it lowest strongest - So. From weak to strong base the order is (v) < (iii) < (ii) < (iv) < (i), which is option (d).
The idea to keep. For carbanions, more stable means less basic. sp carbon, resonance and electronegative neighbours lower basicity. Alkyl groups raise it, so a tertiary carbanion is the strongest base.
Answer from the NTA final answer key. Solution by PrepWiser.