Chemistry · General Principles Of Organic Chemistry · NEET
Two reasons work together. First, alkyl groups push electron density toward the electron-poor positive carbon (+I effect), and a 3-degree carbon has three alkyl groups doing this while a 2-degree has only two. Second, hyperconjugation: each alpha C-H bond next to the positive carbon can overlap with the empty p orbital and share its electrons. More attached alkyl groups means more alpha C-H bonds and more hyperconjugation. So the order is 3-degree > 2-degree > 1-degree > CH3 positive. This exact idea was asked in NEET 2020.
All three are carbon centres from a broken bond, but they differ in electrons and charge. A carbocation has 6 electrons on carbon, is positive, sp2 hybridised and flat (trigonal planar). A carbanion has 8 electrons including a lone pair, is negative, sp3 hybridised and pyramidal. A free radical has 7 electrons including one unpaired electron, is neutral, roughly sp2 and nearly planar. Simple check: count electrons on the carbon - 6 means cation, 7 means radical, 8 means anion.
A simple carbanion (like CH3 negative) is sp3 hybridised and pyramidal, because the lone pair sits in one sp3 orbital just like ammonia. This is different from a carbocation, which is sp2 and flat. Careful: NEET can show a carbanion that is part of a triple bond (like an acetylide), and that carbon is sp because of the two pi bonds - the hybridisation depends on the whole structure, not just the negative charge.
A carbocation is short of electrons, so anything that pushes electrons in (alkyl +I, hyperconjugation, electron-donating resonance) makes it happier. A carbanion already has extra electrons, so electron-pushing groups make it worse; instead it needs electron-withdrawing groups (-I, -R) to pull the extra charge away and spread it out. That is why carbanion stability order is the reverse: 1-degree > 2-degree > 3-degree. Do not blindly write '3-degree most stable' - that is only for carbocations and radicals.
By breaking the covalent bond in two ways. Heterolytic fission: both bonding electrons go to one atom, giving one positive and one negative fragment - this makes carbocations and carbanions, and is shown with a full curved arrow. Homolytic fission: each atom keeps one electron, giving two neutral radicals - shown with a half-headed (fish-hook) arrow. Homolysis usually needs heat or UV light; heterolysis is favoured by polar solvents.
A tertiary butyl carbocation is more stable than the secondary butyl carbocation because of which one of the following?
The most stable carbocation among the following is:
Among compounds I-III, the correct order of bond dissociation energy (BDE) of the marked C-H bond is (radical stability III > I > II):
Try the real previous-year questions from this chapter — each with the answer and a full solution.
3-degree (tertiary) > 2-degree (secondary) > 1-degree (primary) > CH3 positive. Resonance-stabilised cations like allyl and benzyl can beat even a tertiary cation because the charge is spread over several atoms.
It is the reverse of carbocations: CH3 negative > 1-degree > 2-degree > 3-degree, because a carbanion is electron-rich and prefers fewer electron-donating alkyl groups. Electron-withdrawing groups (like -NO2, -CN) stabilise carbanions strongly.
Nearly. A simple carbon free radical is roughly sp2 hybridised and close to planar, with the lone unpaired electron in a p orbital. This lets hyperconjugation and resonance stabilise it, which is why radical stability follows the same 3-degree > 2-degree > 1-degree order as carbocations.
A half-headed 'fish-hook' arrow shows the movement of a single electron in homolytic fission (radical formation). A full double-headed curved arrow shows a pair of electrons moving in heterolytic fission (carbocation/carbanion formation).
Almost every organic mechanism in Classes 11 and 12 (SN1, SN2, addition, elimination, aromatic substitution) passes through one of these intermediates, and NEET repeatedly asks which cation/radical is most stable. Getting the stability order and hybridisation right lets you predict the product and the reaction rate quickly.
