Topic 12 of 34

SN1 forms a carbocation before the nucleophile attacks

Carbocation stability controls ionisation; a planar intermediate loses the original stereochemical information.

A-Level 9476 (2026-2027)

SN1 forms a carbocation before the nucleophile attacks

Carbocation stability controls ionisation; a planar intermediate loses the original stereochemical information.

First step: heterolytic C-Br cleavage

Ionisation is the slow step in the simple S_N1 model. A tertiary alkyl carbocation is stabilised by electron donation from its three alkyl groups.

The nucleophile then donates a lone pair to the positive carbon. With OH^-, this gives the alcohol directly; with H₂O, it first gives a protonated alcohol that must lose H⁺. The simple rate law is rate = k[halogenoalkane], because the slow ionisation precedes nucleophile attack.

Second step with water: make the new C-O bond

The arrow starts at an oxygen lone pair, not at the positive carbon. Water is neutral, so the immediate product is (CH₃)₃C-OH₂⁺; it is not yet the neutral alcohol.

Finish water substitution by removing one proton

Draw the protonated alcohol
Oxygen is bonded to carbon and two H atoms, so place the positive charge on O.
Use another water molecule as a base
Draw an arrow from its oxygen lone pair to an H on the protonated alcohol. At the same time, draw an arrow from that O-H bond back to the alcohol oxygen.
Check atoms and charge
(CH₃)₃C-OH₂⁺ + H₂O → (CH₃)₃COH + H₃O⁺. Oxygen in the neutral alcohol now has two bonds; the overall charge remains +1.

For simple alkyl carbocations, tertiary is generally more stable than secondary, which is more stable than primary. Alkyl groups donate electron density towards the electron-deficient centre. A substrate that would form an unstable primary carbocation is therefore unlikely to hydrolyse by a simple S_N1 route.

For an initially chiral substrate, the carbocation is approximately trigonal planar. Attack can occur from either face, producing both enantiomers and racemisation in the ideal model. Real ion pairs can partly shield a face, so exact equality is an idealisation; the key contrast is loss of stereochemical specificity versus S_N2 inversion. The illustrated tert-butyl example itself is achiral because its three methyl groups are identical.

Worked example

Use a chiral substrate to deduce the SN1 products

Predict the stereochemical result when a single enantiomer of 3-bromo-3-methylhexane undergoes hydrolysis by the ideal SN1 model.

Its Br-bearing carbon is attached to Br, CH₃, C₂H₅ and CH₂CH₂CH₃: four different groups. It is both tertiary and chiral.
Draw C-Br cleavage to Br^-, then redraw the carbocation carbon and its three carbon groups as trigonal planar. The original wedge arrangement at this centre is lost.
A water oxygen lone pair can attack either face of this plane. After proton loss, each product carbon has OH, methyl, ethyl and propyl groups.
Draw one product in three dimensions and reflect the entire drawing to obtain the other. Because those four groups differ, the mirror images cannot be superimposed.

Answer

The ideal model gives equal amounts of the two 3-methylhexan-3-ol enantiomers, a racemic mixture. The reaction pathway, rather than a mere count of product functional groups, explains loss of the starting optical activity.

The two alcohols produced by opposite-face attack

Here C₃H₇ means CH₂CH₂CH₃. The two drawings have identical connectivity and opposite handedness. Equal quantities cancel their optical rotations in the ideal racemic product.

Use rate and stereochemistry as independent evidence
Feature	S_N1	S_N2
Elementary sequence	Slow ionisation, then fast attack.	Concerted bond formation and breaking.
Intermediate	Planar carbocation.	No carbocation; one transition state.
Rate law in the simple model	k[RX].	k[RX][Nu].
Main substrate factor	Carbocation stability.	Steric access to the carbon.
Chiral substrate outcome	Racemisation through two-face attack.	Inversion through backside attack.