JEE Mains Chemistry · Organic Reaction Mechanisms
Carbocations, Rearrangements and Addition Regiochemistry
A carbocation forms faster the more stable it is, moves a hydrogen or a methyl group from the next carbon when that makes it more stable, and decides where H and X or OH land when a reagent adds across a C=C or C≡C bond.
Why this matters
Eleven PYQs, one of them asking for a number, and two from 2026. Five turn on a carbocation: which one forms faster, which reactions rearrange by a hydride or methyl shift, and when a ring next to the cation grows by one carbon. Six are addition steps inside a longer scheme: Markovnikov with HX or Hg²⁺ and H₂SO₄, anti-Markovnikov with B₂H₆ then alkaline H₂O₂, and a bromonium ion opened by the molecule's own carboxylate.
Concept 1 of 2: Carbocation stability and 1,2-shifts
A carbocation is a carbon with only six electrons around it. Alkyl groups next to it push electron density towards it, by the inductive effect and by hyperconjugation, so the more alkyl groups it carries, the more stable it is. A cation next to a benzene ring or a C=C spreads its charge by resonance. If a hydrogen or a methyl group on the NEXT carbon can move over, with its bond pair, and leave behind a more stable cation, it does. The product then comes from the new cation, not the first one.
Definition
- Stability: . A benzylic or allylic cation is resonance-stabilised and ranks with the secondary and tertiary ones.
- The more stable cation also forms FASTER: the transition state leading to a cation looks like the cation. This is why HX adds by Markovnikov's rule.
- 1,2-hydride shift: an H on the next carbon moves with its bond pair to the cation carbon; the charge moves to the carbon the H left.
- 1,2-methyl shift: when the next carbon has no H but carries groups (a quaternary carbon), a methyl group moves instead.
- A shift happens only if it gives a MORE stable cation. A cation that is already tertiary does not rearrange.
- Where shifts show up: Friedel-Crafts alkylation with a primary halide, acid dehydration and reactions of alcohols, HX addition to an alkene, and isomerisation of an n-alkane by anhydrous and HCl.
- Ring expansion: a cation on a carbon attached to a cyclobutane or cyclopentane ring moves one ring C–C bond; the ring grows from 4 to 5 or from 5 to 6 carbons and loses strain.
Carbocation stability order
Worked example
3,3-Dimethylbutan-2-ol, , is heated with concentrated . Find the major alkene.
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The same idea in a real exam question:
JEE Mains · 2026 · 8 Apr 2026 Shift 2 · Q38Moderate
Example 1 · Organic Reaction Mechanisms · Carbocations, Rearrangements and Addition Regiochemistry
The major product of which of the following reaction is not obtained by rearrangement reaction?
A primary halide in Friedel-Crafts alkylation gives the branched product
The primary cation (or its complex with ) rearranges before it attacks the ring. 1-Chloropropane gives isopropylbenzene, not n-propylbenzene. To put a straight chain on a ring, acylate and then reduce the C=O.
No shift without a better cation at the end
A 1,2-shift happens only when the new cation is more stable. A tertiary cation, or a benzylic one next to a ring, has nothing better to reach, so its product is not rearranged.
Stability and rate go together
The more stable carbocation is also the one formed faster, because the transition state resembles the cation. An option that pairs 'more stable' with 'formed more slowly' is wrong.
Concept 2 of 2: Addition regiochemistry by reagent
When a reagent adds across a C=C or C≡C bond, two questions decide the product: which carbon the new group lands on, and whether a carbocation forms on the way. Reagents that go through a cation (HX, acid and water, Hg²⁺ with an alkyne) put H on the carbon that already has more H, so the cation sits on the more substituted carbon. Hydroboration goes through no cation, so boron, and later OH, lands on the LESS substituted carbon and nothing rearranges.
Definition
- Markovnikov: in HX or addition, H goes to the carbon with more H; the more stable cation carries the X or OH.
- Anti-Markovnikov, HBr with a peroxide: a free-radical chain; only HBr does this, not HCl or HI.
- Hydroboration-oxidation (, then ): boron adds to the less hindered carbon, H and OH add to the same face (syn), and the OH replaces B. No cation, so no rearrangement.
- Alkyne hydration (): Markovnikov; the enol changes to the ketone, so a terminal alkyne gives a methyl ketone. Ethyne alone gives ethanal.
- Hydroboration of a terminal alkyne puts OH on the end carbon; the enol changes to an ALDEHYDE.
- adds through a cyclic bromonium ion, opened from the back: anti addition. If the molecule carries its own carboxylate (acid plus ), the carboxylate opens the ion and closes a lactone ring, a five-membered ring when it can.
| Reagent | Where the new group goes | Why | Example |
|---|---|---|---|
| HBr (no peroxide) | Br on the more substituted carbon | The more stable cation forms; it can rearrange | |
| HBr with a peroxide | Br on the less substituted carbon | Free-radical chain; works for HBr only | |
| , dilute | OH on the more substituted carbon | Through a cation; it can rearrange | |
| , then | OH on the less substituted carbon | No cation: syn addition, no rearrangement | |
| , on an alkyne | O on the inner carbon; a ketone | Markovnikov enol changes to the keto form | Only ethyne gives an aldehyde (ethanal) this way. |
| , then on a terminal alkyne | O on the end carbon; an aldehyde | Anti-Markovnikov enol changes to the aldehyde | |
| in | One Br on each carbon, on opposite faces | Cyclic bromonium ion opened from the back | Cyclohexene gives trans-1,2-dibromocyclohexane |
| , on an unsaturated acid | Ring O on the inner carbon, Br outside the ring | The molecule's carboxylate opens the bromonium ion | An alkenoic acid gives a bromo-lactone, a five-membered ring when possible |
Through a cation: Markovnikov, and a shift is possible. Through boron or a radical: the other carbon, and no shift.
Practice this conceptself-check · 4 quick reps
The same idea in a real exam question:
JEE Mains · 2023 · 15 Apr 2023 · Q32Moderate
Example 2 · Organic Reaction Mechanisms · Carbocations, Rearrangements and Addition Regiochemistry
The product formed in the following multistep reaction is:
Hydroboration puts OH on the less substituted carbon
followed by alkaline is anti-Markovnikov: a terminal alkene gives a primary alcohol, which PCC can then take to an aldehyde. Acid hydration of the same alkene gives the secondary or tertiary alcohol.
The peroxide effect works for HBr only
With a peroxide, HBr adds anti-Markovnikov by a radical chain. HCl and HI still add by Markovnikov's rule with or without a peroxide.
Alkyne hydration ends at a carbonyl
The first product is an enol, which changes to its keto form at once. With a terminal alkyne gives a methyl ketone; by hydroboration it gives an aldehyde. Neither route stops at an alcohol.
Summary — formulas & gotchas at a glance
A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.
Formulas (1)
- Carbocation stability and 1,2-shifts
Carbocation stability order
Reference tables (1)
Addition regiochemistry by reagent8 rows
| Reagent | Where the new group goes | Why | Example |
|---|---|---|---|
| HBr (no peroxide) | Br on the more substituted carbon | The more stable cation forms; it can rearrange | |
| HBr with a peroxide | Br on the less substituted carbon | Free-radical chain; works for HBr only | |
| , dilute | OH on the more substituted carbon | Through a cation; it can rearrange | |
| , then | OH on the less substituted carbon | No cation: syn addition, no rearrangement | |
| , on an alkyne | O on the inner carbon; a ketone | Markovnikov enol changes to the keto form | Only ethyne gives an aldehyde (ethanal) this way. |
| , then on a terminal alkyne | O on the end carbon; an aldehyde | Anti-Markovnikov enol changes to the aldehyde | |
| in | One Br on each carbon, on opposite faces | Cyclic bromonium ion opened from the back | Cyclohexene gives trans-1,2-dibromocyclohexane |
| , on an unsaturated acid | Ring O on the inner carbon, Br outside the ring | The molecule's carboxylate opens the bromonium ion | An alkenoic acid gives a bromo-lactone, a five-membered ring when possible |
Through a cation: Markovnikov, and a shift is possible. Through boron or a radical: the other carbon, and no shift.
Watch out for (6)
- A primary halide in Friedel-Crafts alkylation gives the branched product→ Carbocation stability and 1,2-shifts
- No shift without a better cation at the end→ Carbocation stability and 1,2-shifts
- Stability and rate go together→ Carbocation stability and 1,2-shifts
- Hydroboration puts OH on the less substituted carbon→ Addition regiochemistry by reagent
- The peroxide effect works for HBr only→ Addition regiochemistry by reagent
- Alkyne hydration ends at a carbonyl→ Addition regiochemistry by reagent
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