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9476 / 2027
Organic Chemistry

Lesson 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)

  • I can explain both nucleophilic-substitution mechanisms.Syllabus

    Syllabus 9476, 11.3(n). Explain both nucleophilic-substitution mechanisms.

  • I can explain substitution stereochemistry for optically active substrates.Syllabus

    Syllabus 9476, 11.5(b). Explain substitution stereochemistry for optically active substrates.

SN1 forms a carbocation before the nucleophile attacks

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

Make a guess. You are not marked.

2-bromo-2-methylpropane and 1-bromobutane are each warmed with water. Which is hydrolysed faster?

  1. 2-bromo-2-methylpropane, via a stable tertiary carbocation
  2. 1-bromobutane, because it is less crowded for the water
  3. 1-bromobutane, because its C-Br bond is weaker than the tertiary one
Show the answer

2-bromo-2-methylpropane, via a stable tertiary carbocation

It is tertiary, so it ionises to a tertiary carbocation stabilised by three alkyl groups. That SN1 route is fast.

First step: heterolytic C-Br cleavage

The curly arrow starts at the carbon-bromine bond and ends at bromine. The carbon has three carbon groups and becomes a positively charged carbocation after losing bromide.

Ionisation is the slow step in the simple SN1 model. A tertiary alkyl carbocation is stabilised by electron donation from its three alkyl groups.

The nucleophile then gives a lone pair to the positive carbon. With OH-, this gives the alcohol directly. With H2O, it first gives a protonated alcohol, which then loses H+. The rate equation is rate = k[halogenoalkane], because the slow step, ionisation, happens before the nucleophile attacks.

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

A lone pair on water oxygen points towards the positive carbon of the tert-butyl carbocation. Carbon has three methyl groups in a planar arrangement before attack. The resulting oxygen has three bonds and a positive charge in the protonated alcohol.

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
  1. Draw the protonated alcohol

    Oxygen is bonded to carbon and two H atoms, so place the positive charge on O.

  2. 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.

  3. 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 carbocations, tertiary is more stable than secondary, which is more stable than primary. Alkyl groups push electron density towards the positive carbon and spread out its charge. So a halogenoalkane that would form an unstable primary carbocation is unlikely to react by SN1.

The carbocation is flat (trigonal planar). If the starting compound is chiral, the nucleophile can attack from either face. In the simple model this gives equal amounts of both enantiomers, which is racemisation. In reality the leaving ion can partly block one face, so the mixture may not be exactly equal. The key contrast is with SN2, which gives inversion. The tert-butyl example shown is itself achiral, because its three methyl groups are identical.

Worked exampleUse 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.

Try it first. Then show the worked stepsHide the worked steps
  1. Its Br-bearing carbon is attached to Br, CH3, C2H5 and CH2CH2CH3: four different groups. It is both tertiary and chiral.
  2. 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.
  3. 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.
  4. 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

Each product carbon has methyl and ethyl attached by ordinary bonds in the plane of the page, OH on a solid wedge towards the viewer and n-propyl on a hashed wedge away. Reflecting every attached group exchanges left and right positions and gives non-superimposable mirror images.

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
FeatureSN1SN2
Elementary sequenceSlow ionisation, then fast attack.Concerted bond formation and breaking.
IntermediatePlanar carbocation.No carbocation; one transition state.
Rate law in the simple modelk[RX].k[RX][Nu].
Main substrate factorCarbocation stability.Steric access to the carbon.
Chiral substrate outcomeRacemisation through two-face attack.Inversion through backside attack.

Exam-style question

2-Bromo-2-methylpropane, (CH3)3CBr, is warmed with NaOH(aq). Name and describe the mechanism and explain why this halogenoalkane reacts by this route. [3]

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