The mechanism
Step 1 is slow: the C-X bond breaks by itself, helped by a polar solvent, to give a carbocation and a halide ion. Step 2 is fast: any nucleophile present adds to the cation.
hydrolysis of tert-butyl bromide
The C-Br bond breaks by itself, and both electrons go to bromine. This slow step sets the rate, so rate .
The empty p orbital accepts a lone pair from water. Any nucleophile present can do this, because the cation is so reactive.
A neutral water molecule removes a proton. Overall, .
Watch the animation: SN1: two steps and a flat cation
The nucleophile is not involved until after the slow step. So its strength and concentration do not change the rate. In many reactions the solvent itself is the nucleophile, and this is called solvolysis.
Carbocation stability
- Hyperconjugation and induction. Each alkyl group donates electron density into the empty p orbital. More alkyl groups give a more stable cation.
- Resonance. Allyl and benzyl cations spread the charge. A donor such as at the para position helps more, and hurts.
- Vinyl and phenyl cations are very unstable, because the charge sits in an orbital with high s-character. So vinylic and aryl halides do not ionise.
Watch the animation: Why a tertiary cation is stable
Rearrangement
A carbocation shifts a neighbouring hydrogen or methyl group, with its bonding pair, whenever that gives a more stable cation.
Example from April 2023: isobutyl bromide with ethoxide stays isobutyl, because the reaction is . With ethanol alone, the skeleton changes and tert-butyl ethyl ether forms.
3-Methylbutan-2-ol with HBr
Acid turns OH, a poor leaving group, into water, a good one.
Water leaves. The cation is secondary, and that is not the best available.
A 1,2-hydride shift takes a hydrogen with its pair to the cationic carbon. The charge moves to a tertiary carbon.
Bromide adds where the charge now is. The main product is 2-bromo-2-methylbutane, not the 2-bromo-3-methylbutane that a simple swap would give.
Watch the animation: Hydride and methyl shifts
A real primary cation is too unstable to form freely, so textbooks often describe the shift as happening while the leaving group departs. In exam answers, treat this as the rearranged cation path.
The syllabus for JEE Advanced names rearrangement of alkyl carbocations directly.
and side by side
| Steps | Two, through a carbocation | One, through a transition state |
| Rate law | ||
| Best substrate | , benzylic, allylic | Methyl, , allylic, benzylic |
| Nucleophile | Weak is fine | Strong helps |
| Solvent | Polar protic | Polar aprotic |
| Stereochemistry | Racemisation | Inversion |
| Rearrangement | Possible | Never |
Worked example: rank for
Rank by rate: (a) , (b) , (c) 4-methoxybenzyl chloride, (d) 4-nitrobenzyl chloride.
Show the solution
Draw each cation and ask who holds the charge. Two rings spread it more than one, so (b) beats (a). The methoxy group gives electrons by resonance, so (c) also beats (a). Nitro pulls electrons, so (d) is last. The firm order is (b), (c) > (a) > (d). Between (b) and (c), a question would need to give data.
JEE Advanced: when does not give a racemate
Racemic means equal amounts of two enantiomers. If the molecule has a second stereocentre that the reaction does not touch, the two faces of the cation are no longer equal. The products are then diastereomers, and they form in unequal amounts.