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Almost every organic mechanism in JEE and NEET passes through a carbocation, a carbanion or a free radical. If you can say which one forms and how stable it is, you can predict products instead of memorising them. This guide covers how each forms, what makes each stable, two worked predictions with numbers, and the mistakes that cost marks.
How do bonds break to give these intermediates?
A covalent bond can break in two ways. The OpenStax Organic Chemistry text describes them: in heterolytic cleavage, both bonding electrons go to one fragment; in homolytic cleavage, each fragment keeps one electron.
Heterolytic cleavage gives ions. If carbon loses the pair, you get a carbocation (positive carbon). If carbon keeps the pair, you get a carbanion (negative carbon).
Homolytic cleavage gives free radicals: neutral species with an unpaired electron.
An intermediate is not a transition state. OpenStax explains that an intermediate sits in an energy minimum between two steps. It usually cannot be isolated, but it is more stable than the transition states on either side.
What makes a carbocation stable?
OpenStax section 7.9 gives the structure: the positive carbon is sp² hybridised and planar, with an empty p orbital. Stability rises with the number of alkyl groups: tertiary > secondary > primary > methyl.
Two effects explain it:
Inductive effect. Alkyl groups push electron density towards the positive carbon.
Hyperconjugation. C–H bonds on the neighbouring carbon, lined up with the empty p orbital, share some electron density with it. More alkyl groups give more such bonds.
Resonance helps even more where it is possible, as in allyl and benzyl cations, because the charge spreads over several carbons.
The 1994 Nobel Prize in Chemistry went to George Olah "for his contribution to carbocation chemistry", work that let chemists study these short-lived ions directly.
Worked example 1: predicting a rearrangement
Question. What forms when HCl adds to 3-methyl-1-butene, CH2=CH–CH(CH3)2?
H⁺ adds to the end carbon, giving a secondary carbocation next to a carbon carrying one H and two methyls.
A hydride (H with its electron pair) shifts from that neighbour, giving a tertiary carbocation, which is more stable.
Cl⁻ attacks either cation.
OpenStax reports that the reaction gives a substantial amount of rearranged 2-chloro-2-methylbutane alongside the expected 2-chloro-3-methylbutane. The rearranged product is evidence that a carbocation formed.
What makes a free radical stable?
The order is the same as for carbocations: tertiary > secondary > primary > methyl. The evidence is bond strength. OpenStax section 10.2 gives C–H bond dissociation energies of 421 kJ/mol (primary), 410 kJ/mol (secondary) and 400 kJ/mol (tertiary). Less energy is needed to make the tertiary radical, so it is the most stable. Allyl radicals are stabilised further by resonance, as OpenStax shows.
Worked example 2: product ratios from radical chlorination
OpenStax uses chlorination of butane, which gives 1-chlorobutane and 2-chlorobutane in a 30 : 70 ratio. Butane has 6 primary and 4 secondary hydrogens.
Per primary H: 30 ÷ 6 = 5.
Per secondary H: 70 ÷ 4 = 17.5.
Relative reactivity: 17.5 ÷ 5 = 3.5. A secondary H reacts 3.5 times as often as a primary H.
Now predict propane (6 primary H, 2 secondary H) using that factor:
Primary share: 6 × 1 = 6.
Secondary share: 2 × 3.5 = 7.
Predicted 1-chloropropane: 6 ÷ 13 = 46%. 2-chloropropane: 7 ÷ 13 = 54%.
This is an estimate from one set of conditions, but it shows the method: number of hydrogens times reactivity per hydrogen.
What makes a carbanion stable?
A carbanion carries a lone pair, so the trends reverse. Alkyl groups, which push electrons in, destabilise it. Three things stabilise it:
s character. NCERT's Hydrocarbons chapter explains that ethyne's hydrogens are acidic because sp-hybridised carbon (50% s character) holds electron pairs more tightly than sp² or sp³ carbon.
Resonance with a C=O or NO2 group next door.
Aromaticity. The cyclopentadienyl anion has 6 π electrons, which is why OpenStax notes cyclopentadiene has pKa 16, very low for a hydrocarbon.
What are the common mistakes?
Using the carbocation order for carbanions. Alkyl groups stabilise cations and radicals but destabilise anions.
Calling a radical charged. A free radical is neutral.
Ignoring rearrangement. If a more stable cation is one shift away, expect some rearranged product.
Judging species by charge alone. Anzovino and Bretz (2015) found organic chemistry students relied mainly on charges to identify nucleophiles and electrophiles. Look at structure and orbitals too.
Syllabus and practice
The JEE (Main) 2026 syllabus lists "free radicals, carbocations and carbanions, stability of carbocations and free radicals". The JEE (Advanced) 2026 brochure lists their formation, structure and stability, and the NEET (UG) 2026 syllabus repeats the JEE Main wording. Check the current bulletin. Our post on inductive, resonance and hyperconjugation effects goes deeper into the effects used here.
A three-day routine:
Day 1: Draw ten intermediates and label each 1°, 2°, 3°, allylic or benzylic. Rank them.
Day 2: Five alkene + HX problems. For each, ask whether a shift is possible.
Day 3: Predict chlorination ratios for two alkanes using the 3.5 factor.
When a mechanism step does not make sense, the Fermi Problem Library has worked problems with a step-by-step solver that explains the reasoning rather than printing an answer.
Pick one alkene from your textbook today and write its carbocation, then ask whether it can rearrange.
