HBr addition with carbocation rearrangement: practice and mechanism
Protonation gives a secondary cation. A methyl group from the neighbouring quaternary carbon moves over (1,2-shift), giving a tertiary cation, where Br⁻ attacks.
Example
Protonation first gives a secondary carbocation. Right next to it is a quaternary carbon: a methyl group moves over with its electron pair (1,2-methyl shift), giving the more stable tertiary cation. Only there does bromide attack.
Mechanism step by step
- 1
H⁺ adds to the terminal CH₂ (Markovnikov): a secondary carbocation forms next to the quaternary carbon.
- 2
A methyl group moves to the cation with its electron pair (1,2-methyl shift). The positive charge is now on a tertiary carbon.
- 3
Bromide attacks the tertiary cation: 2-bromo-2,3-dimethylbutane.
Practice tasks
Which product forms, which reagent is needed? After each answer you see the explanation.
Which main product forms?
Typical mistakes
- That would be the product without rearrangement. The secondary cation next to a quaternary carbon rearranges faster than bromide attacks: Br ends up on the tertiary carbon.
Often confused with: Radical addition of HBr (anti-Markovnikov) · Oxymercuration–demercuration
Frequently asked questions
When should I expect rearrangements?
Whenever a carbocation forms (SN1, E1, HX and H₂O/H⁺ addition) and a neighbouring carbon has an H or alkyl group whose shift gives a more stable cation.
How do I avoid the rearrangement?
With reactions without a free cation: oxymercuration (for OH), hydroboration or HBr with peroxide (anti-Markovnikov).