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Organic Mechanisms

SN1 vs SN2: Choosing the Right Substitution Mechanism

One step or two? How to predict which mechanism a haloalkane follows, and why it matters.

Last reviewed 5 October 2026

The reaction

In a nucleophilic substitution, an electron-rich nucleophile (Nu⁻, such as OHX−\ce{OH-}, CNX−\ce{CN-} or water) replaces a leaving group (such as BrX−\ce{Br-}) on a carbon atom:

R−Br+OHX−→R−OH+BrX−\ce{R-Br + OH- -> R-OH + Br-}

The overall equation hides how it happens. There are two mechanisms, and students often mix them up.

Key idea

SN2 happens in one step: the nucleophile pushes in from the back as the leaving group leaves. SN1 happens in two steps: the leaving group leaves first to give a carbocation, which the nucleophile then attacks. The “1” and “2” refer to how many species are in the rate-determining step.

Read the curved arrows

A curved arrow shows a pair of electrons moving. It always starts at electrons (a lone pair or a bond) and points to where they go.

Two mechanisms. SN2, one step, rate = k[RX][Nu−]: a curved arrow goes from a lone pair on HO− to the carbon of H3C–Br while a second arrow moves the C–Br bonding pair onto bromine; through a transition state [HO···C···Br] the products HO–CH3 and Br− form. Backside attack inverts the configuration; best for methyl and primary carbons. SN1, two steps, rate = k[RX]: in a slow step the C–Br bond breaks to give the planar carbocation (CH3)3C+ and Br−; in a fast step water adds to give (CH3)3C–OH and H+. Attack from either face gives racemization at a stereocentre; best for tertiary carbons in polar protic solvents. Summary: methyl and primary go SN2, tertiary SN1, secondary either.
SN2 bonds and breaks at the same time; SN1 breaks first, then bonds.

Four factors decide the mechanism

FactorFavours SN2Favours SN1
Substratemethyl, 1° (little crowding)3° (stable carbocation)
Nucleophilestrong, negatively charged (OHX−\ce{OH-}, CNX−\ce{CN-})weak, neutral (HX2O\ce{H2O}, CHX3OH\ce{CH3OH})
Solventpolar aprotic (propanone, DMSO)polar protic (water, alcohols)
Leaving groupgood in both: I⁻ > Br⁻ > Cl⁻ ≫ F⁻good in both

Why the substrate matters most. In SN2 the nucleophile must reach the back of the carbon; three bulky groups on a 3° carbon block the way (steric hindrance). In SN1 the slow step makes a carbocation; alkyl groups push electron density towards the positive carbon and stabilize it, so 3° carbocations form far more easily than 1° ones.

Evidence from rates and shapes

  • Rate law. For SN2, rate = k[RX][Nu⁻]: doubling either concentration doubles the rate. For SN1, rate = k[RX]: the nucleophile is not in the slow step, so its concentration does not matter.
  • Stereochemistry. SN2 attack from the back turns the molecule inside out like an umbrella in the wind (inversion). The SN1 carbocation is planar, so the nucleophile can attack from either face, giving a mixture of both mirror-image products (racemization).

Worked example: Using the rate law

Question: For the SN2 reaction of bromomethane with hydroxide, the initial rate is 4.0×10−64.0 \times 10^{-6} M/s when [CHX3Br]=0.10[\ce{CH3Br}] = 0.10 M and [OHX−]=0.20[\ce{OH-}] = 0.20 M. Find k, and predict the rate if [OHX−][\ce{OH-}] is doubled.

  1. k=rate[CHX3Br][OHX−]=4.0×10−6 M/s(0.10 M)(0.20 M)=k = \dfrac{\text{rate}}{[\ce{CH3Br}][\ce{OH-}]} = \dfrac{4.0 \times 10^{-6}\ \text{M/s}}{(0.10\ \text{M})(0.20\ \text{M})} = 2.0×10−4 M−1 s−12.0 \times 10^{-4}\ \text{M}^{-1}\,\text{s}^{-1}
  2. The reaction is first order in OHX−\ce{OH-}, so doubling [OHX−][\ce{OH-}] doubles the rate: 8.0×10−68.0 \times 10^{-6} M/s.
  3. For an SN1 reaction, doubling the nucleophile concentration would leave the rate unchanged.

Worked example: Predicting the mechanism

Question: Predict SN1 or SN2: (a) 1-bromobutane + NaOH in propanone (b) 2-bromo-2-methylpropane in water (c) bromomethane + NaCN.

  1. (a) 1° substrate, strong nucleophile, aprotic solvent: SN2.
  2. (b) 3° substrate, weak nucleophile, protic solvent: SN1.
  3. (c) methyl substrate, strong nucleophile: SN2 (it also adds one carbon to the chain, a useful trick in synthesis).

Common mistakes

Common mistake: Arrows pointing the wrong way

Curved arrows show electrons, so they start at the nucleophile’s lone pair and end at the carbon, never from the carbon to the nucleophile.

Common mistake: Primary carbocations in SN1

1° carbocations are too unstable to form under normal conditions. A 1° haloalkane reacts by SN2, not SN1.

More precisely

Substitution competes with elimination (E1 and E2), which forms alkenes, especially with strong, bulky bases and at higher temperatures. Secondary substrates are the hardest to predict and often give mixtures. The labels SN1 and SN2 (“substitution, nucleophilic, unimolecular / bimolecular”) were introduced by Hughes and Ingold in 1935.

Timeline

  1. 1896Paul Walden discovers that some substitutions invert the configuration of a chiral carbon (the Walden inversion).
  2. 1935Edward Hughes and Christopher Ingold propose the SN1 and SN2 mechanisms and their labels.

Sources and further reading

Explore next

  • Electrophilic Addition to Alkenes and Markovnikov's Rule

    Why HBr adds to propene the way it does, explained by carbocation stability.

    The two-step mechanism of electrophilic addition, curved arrows, why the more stable carbocation wins (Markovnikov's rule), and the bromonium ion in bromine addition.

← All organic mechanisms

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