Exams › JEE Advanced › Chemistry › General Organic Chemistry
3 questions with worked solutions.
Q1. Which of the following carbanions is the most stable?
Answer: HC≡C⁻
Carbanion stability is governed by the s-character of the carbon bearing the negative charge. HC≡C⁻ is sp hybridized (50% s-character), making it the most stable. C6H5⁻ (sp², 33%) is next, then vinyl-type (CH3)2C=CH⁻ (sp²), and the neopentyl-type carbanion (CH3)3C-CH2⁻ (sp³, 25%) is least stable.
Answer: A -> Q,S; B -> R; C -> P,S; D -> Q,S
-N=O (nitroso) is conjugated through N=O so it withdraws by resonance (-M) and is electronegative (-I). -CH3 donates by hyperconjugation (+H). -NH-C(=O)CH3 is bonded through N which has a lone pair to donate (+M) while the electronegative N gives -I. -C(=O)OCH3 is bonded through a carbonyl carbon so it withdraws by resonance (-M) and inductively (-I).
Answer: I < II < III
The nitrogen lone pair conjugates into the adjacent pi system. In (I) the lone pair is delocalized all the way onto the electron-withdrawing carbonyl, giving the C-N bond the most double-bond character and the shortest length. In (II) it delocalizes into an extended diene but with no strong acceptor, giving intermediate shortening. In (III) beyond the single C=C there is only a saturated chain, so the least delocalization and the longest C-N bond. Increasing bond length: I < II < III.