SN1 vs SN2 in five decisions

Every nucleophilic substitution at a saturated carbon is a competition between two mechanisms — a concerted one-step displacement (SN2) and a stepwise ionization-then-attack (SN1). Which one wins is not a mystery: it is decided by five features of the reaction, and once you have looked at each of them the answer is usually obvious. This guide walks the decision tree the way an experienced organic chemist actually runs it, so you can predict the product without back-of-the-book help.

The two mechanisms in one sentence

SN2 is a concerted, one-step process. The nucleophile attacks the electrophilic carbon from the side opposite to the leaving group, the C–Nu bond forms and the C–LG bond breaks in the same transition state, and the stereochemistry at that carbon inverts (a “Walden inversion”). Rate law: $\text{rate} = k[\text{substrate}][\text{Nu}]$ — second-order overall.

SN1 is stepwise. The C–LG bond breaks first to give a carbocation intermediate, and then the nucleophile attacks the (roughly) planar cation from either face. Rate law: $\text{rate} = k[\text{substrate}]$ — first-order in substrate only, because the slow step does not involve the nucleophile.

The five decisions below are exactly the five knobs that shift the balance between “concerted attack looks easy” and “ionization looks easy.”

Decision 1 — substrate class

Look at the carbon carrying the leaving group. How many alkyl groups are attached to it?

• Methyl and primary (CH₃– or R–CH₂–) → strongly SN2. No SN1 pathway to speak of: primary carbocations are too unstable to form under normal conditions.
• Tertiary (R₃C–) → strongly SN1. SN2 is dead here — the transition state is too crowded for a nucleophile to reach the backside. Meanwhile the tertiary cation is stable enough to form readily.
• Secondary (R₂CH–) → the ambiguous middle. Either can win, and the next four decisions are what breaks the tie.
• Neopentyl and other β-branched primaries → SN2 in principle, but painfully slow. The β-methyl groups sterically block the SN2 backside attack, and yet the primary cation is still too poor for SN1. Practically, you may need to switch to E2 conditions and accept an alkene.
• Vinyl and aryl (C=C–X, Ar–X) → neither. The C–X bond has partial double-bond character; nucleophilic substitution needs a different mechanism (addition-elimination or benzyne).

For a substrate that resembles CH₃Br or PhCH₂Cl you already know the answer. The five-decision tree is mostly for that awkward secondary substrate.

Decision 2 — nucleophile strength

A “strong” nucleophile favours SN2 because it can push into the backside without waiting for the leaving group to ionize; a “weak” nucleophile favours SN1 because the mechanism does not need it to participate in the slow step.

Rough scale of nucleophilicity in protic solvents (Anslyn & Dougherty §5.3, Clayden 2e §15.5):

• Very strong: RS⁻, I⁻, CN⁻, HO⁻ in polar aprotic, R₂N⁻
• Strong: RO⁻, HO⁻, N₃⁻, Br⁻ (in polar aprotic)
• Moderate: RNH₂, RCOO⁻, F⁻, Cl⁻
• Weak: H₂O, ROH, RCOOH

“Weak nucleophile” is roughly synonymous with “the solvent itself is the nucleophile” — a scenario called solvolysis. Solvolysis is SN1’s home ground. If your reaction mixture is a tertiary halide dissolved in methanol with no added nucleophile, expect SN1.

Decision 3 — leaving group

Both mechanisms require a good leaving group, so this decision affects how fast rather than which mechanism. Good leaving groups are weak bases — stable in their negative form. Order for common ones (Clayden 2e §15.6):

$\mathrm{TfO^- > TsO^- > I^- > Br^- > Cl^- \gg F^- \gg OH^- \approx NH_2^-}$

Two practical consequences:

• If the leaving group is very poor (–OH or –NH₂), neither substitution mechanism works on it directly; you have to activate it first (protonate –OH, tosylate it, etc.).
• Between iodide and tosylate, tosylate slightly favours SN1 because its ionization step is more thermodynamically favourable — but the effect is modest and rarely swings a mechanism on its own.

Decision 4 — solvent

This is where the “polar aprotic vs polar protic” distinction earns its keep.

• Polar aprotic — DMSO, DMF, acetonitrile, acetone, HMPA. These solvate cations well but leave anions relatively free (“naked” nucleophiles). Anionic nucleophiles are massively enhanced. SN2 wins.
• Polar protic — water, methanol, ethanol, acetic acid. These stabilise both cations and anions through hydrogen bonding, so the transition state that stabilises charge (the SN1 cation) is preferred and nucleophiles are attenuated. SN1 wins.
• Non-polar — hexane, benzene. Neither mechanism runs well; ionization is disfavoured (no charge stabilisation for SN1) and polar transition states are disfavoured (no dipolar stabilisation for SN2). Substitution reactions typically require at least a modestly polar medium.

The rule of thumb: if the solvent can hydrogen-bond to your nucleophile, expect SN1; if it cannot, expect SN2.

Decision 5 — temperature

SN1 has a positive entropy of activation (one molecule dissociates into two ions in the rate-limiting step). SN2 has a negative entropy of activation (two molecules come together into one transition state). By the Eyring equation, the ratio $k_{SN1}/k_{SN2}$ increases with temperature — hotter conditions favour SN1.

The temperature effect is modest — maybe a factor of 2–5 between 0 °C and 80 °C for a borderline substrate — but combined with the other four decisions it can be the deciding vote when everything else looks balanced.

Running the tree — worked example

Suppose you dissolve 2-bromobutane in dry acetone with NaN₃ at 25 °C. Which mechanism runs, and what is the product?

• Decision 1 (substrate): secondary — ambiguous.
• Decision 2 (nucleophile): N₃⁻ is a very good nucleophile. → SN2.
• Decision 3 (leaving group): Br⁻ is good. No preference.
• Decision 4 (solvent): acetone is polar aprotic. → SN2.
• Decision 5 (temperature): 25 °C is unremarkable. No preference.

Two strong SN2 votes and no strong SN1 votes. Result: SN2, with inversion of configuration at C2.

Now change one condition. Replace acetone with methanol and drop the NaN₃; the substrate ionizes in the protic solvent and methanol itself becomes the nucleophile:

• Decision 1 (substrate): secondary — ambiguous.
• Decision 2 (nucleophile): MeOH is weak. → SN1.
• Decision 3 (leaving group): Br⁻ still good. No preference.
• Decision 4 (solvent): MeOH is polar protic. → SN1.
• Decision 5 (temperature): still unremarkable.

Same substrate, same leaving group, but two strong SN1 votes. Product: racemic 2-methoxybutane — both faces of the intermediate cation are attacked, so any starting stereochemistry is lost.

What the tree does not tell you

• Elimination. SN2 competes with E2 (also concerted, also bimolecular); SN1 competes with E1 (also stepwise via a carbocation). A strong bulky base like KOtBu, or a heated tertiary substrate, will run mostly E2 or E1 instead of substitution. The five decisions still apply, but you also need to ask “is the nucleophile so basic and bulky that it deprotonates instead of attacking C?”
• Rearrangements. SN1 intermediates are carbocations, and carbocations rearrange (1,2-hydride and 1,2-methyl shifts) to more stable cations before the nucleophile arrives. If your product looks scrambled, the mechanism is SN1 and you missed the rearrangement.
• Neighbouring-group participation. A ≪β-heteroatom≫ (S, N, O) can attack intramolecularly before the external nucleophile does, giving a bicyclic intermediate and net retention of configuration. This looks like SN1 kinetically (unimolecular) but SN2 stereochemically (well-defined transition state). Anslyn & Dougherty §11.5 has the detailed treatment.

Practice problems

Where to go from here

Once you can predict which mechanism runs, the next step is predicting when substitution loses to elimination — E1 competes with SN1, E2 competes with SN2, and the branch point is usually the basicity and bulk of the nucleophile. Clayden 2e Ch. 17 lays out the full four-mechanism decision (SN1/SN2/E1/E2) as a two-dimensional grid. It looks intimidating; it is not. The same five decisions you learned here still do most of the work — you are just adding one more axis about the nucleophile/base’s ability to abstract a β-hydrogen.

For product prediction on real substrates, having the molar mass and degrees-of-unsaturation tools ready helps confirm that whatever you drew is chemically plausible.