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A question gives you an alkene and HCl, or an alcohol and HBr, and the "obvious" product is wrong because the carbon skeleton has changed. That change is a carbocation rearrangement. It catches students who apply rules without drawing the intermediate. This post gives a three-question check, works three examples that use different reactions and lists the mistakes to avoid.
Where is this in the syllabus?
The JEE (Advanced) 2026 brochure lists "Rearrangement reactions of alkyl carbocation" and the formation, structure and stability of carbocations. The JEE (Main) 2026 syllabus and the NEET (UG) 2026 syllabus list the stability of carbocations and include rearrangement among the common types of organic reactions. The NCERT Class 12 chapter Alcohols, Phenols and Ethers (2026-27 reprint) has an exercise on it, which we work below. Check the current year's documents.
What is a 1,2-shift?
In a 1,2-shift, a group on the carbon next to the positive carbon moves over, taking its bonding electron pair with it. A hydrogen moves as a hydride (H with its pair); a methyl group moves the same way. The positive charge ends up on the carbon the group left. OpenStax Organic Chemistry describes both kinds and notes that in each case a less stable carbocation rearranges to a more stable one. OpenStax credits F. C. Whitmore with finding, in the 1930s, rearranged products in HX additions, which became key evidence that a carbocation forms.
Which three questions tell you to expect a shift?
Does the reaction go through a carbocation? HX addition to alkenes, acid-catalysed hydration, SN1 and E1 reactions, and Friedel–Crafts alkylation do. SN2 reactions, hydroboration and Friedel–Crafts acylation do not.
What is on the neighbouring carbons? Look for an H or a CH₃ on a carbon directly bonded to the positive carbon.
Would moving it make a more stable cation? Secondary to tertiary is the common case. If the new cation is no more stable, do not shift.
Worked example 1: a methyl shift
3,3-Dimethylbut-1-ene, CH₂=CH–C(CH₃)₃, reacts with HCl.
H⁺ adds to the CH₂ end, giving a secondary cation: CH₃–C⁺H–C(CH₃)₃.
The neighbouring carbon has no hydrogen, but it has three methyl groups. A methyl shift gives a tertiary cation: (CH₃)₂C⁺–CH(CH₃)₂.
Cl⁻ can attack either cation.
OpenStax reports an equal mixture of unrearranged 3-chloro-2,2-dimethylbutane and rearranged 2-chloro-2,3-dimethylbutane. Both are C₆H₁₃Cl (M = 120.6 g/mol). So if 0.30 mol of alkene reacted completely, you would expect about 0.15 mol, or 18.1 g, of each. The masses cannot tell the products apart; only the structures can.
Worked example 2: Friedel–Crafts alkylation
OpenStax's Friedel–Crafts section gives two clear cases:
Benzene with 1-chlorobutane and AlCl₃ at 0 °C gives about 2 : 1 rearranged sec-butylbenzene to unrearranged butylbenzene, through a hydride shift. From 0.30 mol of product, that is about 0.20 mol of sec-butylbenzene.
Benzene with 1-chloro-2,2-dimethylpropane gives (1,1-dimethylpropyl)benzene, through a methyl shift.
OpenStax also notes that Friedel–Crafts acylation does not rearrange, because the acyl cation is stabilised by resonance. So to make butylbenzene cleanly, acylate first and then reduce the C=O group to CH₂; the JEE (Main) 2026 syllabus lists the Clemmensen and Wolff–Kishner reductions that do this. Our post on electrophilic substitution in benzene covers the rest of the mechanism.
Worked example 3: NCERT's own question
NCERT Exercise 7.33 asks for the mechanism when 3-methylbutan-2-ol is treated with HBr. Its hint: the secondary carbocation formed rearranges to a more stable tertiary carbocation by a hydride shift from the third carbon.
Protonate the OH and lose water: a secondary cation at C2.
C3 carries one H and two methyls. A hydride shift puts the charge on C3, now tertiary.
Br⁻ attacks it, giving 2-bromo-2-methylbutane.
Which mistakes cost marks?
Shifting from two carbons away. Only groups on the adjacent carbon move.
Shifting with no gain. A secondary cation that would become another secondary cation stays put.
Moving the hydrogen as H⁺. It moves as a hydride, with its electron pair.
Rearranging where there is no carbocation. SN2 and hydroboration keep the skeleton; see our post on hydroboration-oxidation.
Ignoring the unrearranged product. Example 1 gives both in equal amounts.
For carbocation stability itself, see our guide to reaction intermediates.
A practice routine
Apply the three questions to OpenStax Problem 7-21: vinylcyclohexane with HBr gives 1-bromo-1-ethylcyclohexane. Draw the shift.
Predict the Friedel–Crafts product of benzene with 1-chloro-2-methylpropane.
Redo NCERT 7.33 with HCl instead of HBr.
For each, write one sentence on why the new cation is more stable.
When you cannot tell whether a shift happens, the Fermi Problem Library has worked problems with a step-by-step solver that explains the reasoning rather than printing an answer.
Today, take three carbocation reactions from your notes and run each through the three questions.
