Carbocation Rearrangements
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Direct answer
Treat 3-methylbutan-2-ol with HBr and the bromine lands on the "wrong" carbon, C2 of 2-bromo-2-methylbutane. The reason: the first-formed secondary carbocation grabbed a hydride from the neighbouring carbon, becoming tertiary before bromide could attack. Carbocations rearrange by 1,2-shifts — hydride or alkyl migrating with its bonding pair from the adjacent carbon — whenever the destination is a more stable cation, or when a strained ring can expand (cyclobutylmethyl to cyclopentyl). Any reaction with a free cation intermediate (SN1, E1, acid-catalysed dehydration, pinacol rearrangement) is rearrangement territory; SN2 and E2, being concerted, never rearrange.
What you must remember
- Stability ladder: benzylic ≈ allylic ≈ 3° > 2° > 1° > methyl; shifts run strictly uphill on this ladder, never down.
- 1,2-hydride shift: H migrates with its electron pair from the adjacent carbon — the workhorse move, faster than nucleophile capture in most solvents.
- 1,2-alkyl shift: methyl moves when no hydride is available; a quaternary neighbour, having no hydrogens, forces the alkyl shift.
- Ring expansion: cyclobutylmethyl cations expand to cyclopentyl — relief of ring strain supplies the driving force; JEE Advanced keeps this in the options.
- Pinacol-pinacolone: vicinal diol + acid; after water leaves, a 1,2-methyl shift delivers the ketone — the named rearrangement inside the alcohols chapter.
- No-cation, no-shift rule: radical additions (HBr with peroxide) and SN2/E2 give unrearranged products — assertion-reason standard.
- Product consequence: after rearrangement, E1 follows Saytzeff, so dehydration of 3,3-dimethylbutan-2-ol finally gives 2,3-dimethylbut-2-ene, the tetrasubstituted alkene.
Following one cation through its journey
Protonate 3-methylbutan-2-ol; water departs, leaving a 2° cation at C2 with a 3° candidate next door at C3. A hydride shifts from C3 to C2: the positive charge transfers to C3, now tertiary, stabilised by hyperconjugation from three methyl groups. Two fates follow. At low temperature with a good nucleophile, Br^- captures it — 2-bromo-2-methylbutane. With heat and poor nucleophilia, E1 wins and removes a beta-hydrogen to give 2-methylbut-2-ene (Saytzeff). Both products are rearranged; neither matches the skeleton of the starting alcohol, and a JEE option list always includes the unrearranged red herrings.
Now the neopentyl horror story: neopentyl halide ionising to (CH3)3C-CH2+ creates a primary cation flanked by a quaternary carbon — no hydride exists there, so a methyl shift does the work, delivering the tertiary 2-methylbutan-2-yl cation in a heartbeat. This is why neopentyl derivatives betray their structure in every SN1 experiment.
Where students slip
The first error is writing products straight from Markovnikov logic without checking for a shift — always scan the adjacent carbons for a 3° upgrade before finalising any SN1, E1 or dehydration product. The second is inventing uphill shifts: a 3° cation does not rearrange to 2°, and 1,2-shifts never travel more than one carbon in one step. Third, students forget the elimination follow-through: after the shift, the Saytzeff alkene from the rearranged cation is usually the answer, not the alcohol replacement. And when a question mixes HBr with peroxides into an SN1-style setting, remember radicals do not rearrange — the anti-Markovnikov product comes out unrearranged, a contrast the exam has used more than once.
Frequently asked questions
What triggers a 1,2-hydride shift?
A carbocation with a more substituted (or resonance-richer) adjacent carbon; the hydride migrates with its bonding pair, relocating the charge to the stabler site.
Why can SN2 reactions never show rearrangement?
The nucleophile attacks as the leaving group departs in one concerted step, so no free carbocation ever exists to rearrange.
What is the pinacol-pinacolone rearrangement?
Acid protonates one OH of a vicinal diol; water leaves, a 1,2-alkyl shift migrates, and a ketone forms — the classic named carbocation rearrangement.
Why does 3,3-dimethylbutan-2-ol dehydrate to 2,3-dimethylbut-2-ene?
The first secondary cation methyl-shifts to tertiary, and subsequent Saytzeff elimination gives the tetrasubstituted alkene as the major product.
Do free-radical additions rearrange?
No — radicals redistribute far less readily than cations, so peroxide-HBr additions give unrearranged anti-Markovnikov products.