Chemistry · General Principles Of Organic Chemistry · NEET
A free radical is a neutral species (an atom or group) that has one unpaired (odd) electron. It forms when a covalent bond breaks by homolysis, where each atom keeps one electron of the shared pair. We show this single-electron movement with a half-headed (fish-hook) curved arrow. Example: CH3-Cl breaking under UV light gives a methyl radical CH3• and a chlorine radical Cl•.
Neither. A free radical is neutral. This is the key difference from a carbocation (positive, no unpaired electron, 6 electrons on carbon) and a carbanion (negative, has a lone pair). A radical simply has one unpaired electron and no overall charge. This is a very common NEET mix-up, so remember: radical = neutral + one odd electron.
A tertiary radical has three alkyl groups attached to the radical carbon, while a primary has only one. Each alkyl group provides C-H bonds that can share electron density with the odd electron. This sharing is called hyperconjugation. More alkyl groups means more hyperconjugation, so the odd electron is spread out more and the radical is more stable. Order: 3° > 2° > 1° > CH3•.
For carbocations and free radicals, the stability order is the SAME: 3° > 2° > 1° > methyl, because both are electron-poor at that carbon and both are helped by hyperconjugation and +I effect of alkyl groups. Carbanions are the OPPOSITE: methyl > 1° > 2° > 3°, because carbanions are electron-rich and extra alkyl groups push more electrons onto an already negative carbon, which destabilises it.
Bond dissociation energy (BDE) is the energy needed to break a C-H bond by homolysis to make a radical. If the radical formed is very stable, less energy is needed to make it, so the BDE is LOW. If the radical is unstable, more energy is needed, so the BDE is HIGH. So BDE is inversely related to radical stability: more stable radical = lower BDE. NEET 2025 tested exactly this idea.
In an allyl or benzyl radical, the carbon with the odd electron sits next to a double bond or a benzene ring. This lets the unpaired electron spread over more than one atom through resonance (delocalisation). Spreading the odd electron over several atoms lowers the energy a lot. That is why allyl and benzyl radicals are even more stable than a simple tertiary alkyl radical.
Among the given compounds I-III, the correct order of bond dissociation energy of the C-H bond marked with an asterisk (*) is:
A tertiary butyl carbocation is more stable than the secondary butyl carbocation because of which one of the following?
Try the real previous-year questions from this chapter — each with the answer and a full solution.
For alkyl radicals: 3° > 2° > 1° > methyl. When resonance is possible, benzyl and allyl radicals are even more stable than a simple tertiary radical because the odd electron is delocalised.
Yes, both follow 3° > 2° > 1° > CH3. Both are stabilised by hyperconjugation and the +I effect of alkyl groups. Only carbanions have the reversed order.
By homolytic fission (homolysis) of a covalent bond, usually with heat or UV light. Each atom keeps one electron of the shared pair, giving neutral species with one unpaired electron each.
Low. A more stable radical needs less energy to form, so its C-H bond dissociation energy is lower. BDE is inversely proportional to radical stability.
