Nucleophilic substitution mechanisms

Pour hydroxide into a flask of a haloalkane and the halogen leaves, a nucleophile takes its place, and a new bond forms — but the story of how that swap happens splits into two completely different plots. In one, the nucleophile shoulders the leaving group out in a single concerted push, like someone squeezing through a revolving door from the opposite side. In the other, the leaving group departs first, leaving behind a flat, short-lived carbocation that a nucleophile then attacks from either face. These are the $S_N2$ and $S_N1$ mechanisms, and knowing which one a given haloalkane will follow — primary, secondary or tertiary, in a strong or weak nucleophile, in a protic or aprotic solvent — is what lets a chemist predict both the rate of a substitution and the stereochemistry of its product, down to whether an optically active starting material yields an inverted product or a lifeless racemic mixture.

The core insight

Primary haloalkanes react by a one-step, backside-attack $S_N2$ pathway that inverts stereochemistry, while tertiary haloalkanes react through a two-step $S_N1$ pathway via a planar carbocation that racemises the product, with secondary halides sitting on the fence between the two depending on the nucleophile and solvent.

The core idea

Two pathways, one net result. Both mechanisms convert $R\text{-}X$ into $R\text{-Nu}$ , but they differ in how many bonds are breaking and forming at the rate-determining step, and that difference cascades into everything else — kinetics, stereochemistry, and which class of haloalkane prefers which route.

$S_N2$ is concerted. The nucleophile attacks the carbon from the face directly opposite the C–X bond while that bond is still breaking, passing through a single pentacoordinate transition state.

\text{Rate} = k[R\text{-}X][\text{Nu}^-]

where: $k$ is the rate constant, $[R\text{-}X]$ and $[\text{Nu}^-]$ are the molar concentrations of the haloalkane and nucleophile. Both concentrations appear because both species are present in the single rate-determining step — the reaction is second order overall.

$S_N1$ is stepwise. The C–X bond ionises first, on its own, to give a carbocation; only afterwards does the nucleophile attack.

\text{Rate} = k[R\text{-}X]

where: the rate depends only on the haloalkane's concentration because the nucleophile plays no part in the slow, bond-breaking first step — it only shows up in the fast second step, which does not control the overall rate. The reaction is first order overall.

Why the halide class decides the winner. Backside attack in $S_N2$ needs open space behind the carbon bearing the leaving group; three bulky alkyl groups on a tertiary carbon block that approach almost completely, while a primary carbon (only one alkyl group) leaves the backside wide open. Carbocation formation in $S_N1$ needs the opposite: a cation stabilised by nearby alkyl groups through hyperconjugation and the positive inductive effect, so stability runs $3° > 2° > 1°$ and only tertiary (and some resonance-stabilised) cations form fast enough to matter. The two trends point in opposite directions across the same series, which is precisely why primary halides are $S_N2$ specialists, tertiary halides are $S_N1$ specialists, and secondary halides are genuinely ambidextrous.

Figure

Two reaction pathway diagrams side by side showing SN2 as one concerted step with a single transition state and SN1 as two steps through a planar carbocation intermediate

Stereochemistry: inversion versus racemisation

The mechanism's shape at the carbon centre decides what happens to a chiral haloalkane. In $S_N2$ , the nucleophile can only approach from the one open face — directly opposite the leaving group — so the three other groups flip through the carbon like an umbrella turning inside out. This is Walden inversion: a single enantiomer gives a single, inverted-configuration product, so the product stays optically active. In $S_N1$ , the carbocation is planar and $sp^2$ -hybridised, with the empty $p$ orbital exposed equally on both faces; the nucleophile attacks from either side with roughly equal probability, so a single enantiomer gives a near 50:50 mixture of both configurations — racemisation — and the product is close to optically inactive. Solvent reinforces the split: polar protic solvents (water, alcohols) hydrogen-bond to and stabilise both the developing carbocation and the departing halide ion, favouring $S_N1$ ; polar aprotic solvents (acetone, DMSO) leave the nucleophile "naked" and highly reactive without solvating it away, favouring $S_N2$ .

Derivation APredicting rate and stereochemical outcome from halide class, nucleophile and solvent

The goal is to build a three-question checklist — is the halide primary, secondary or tertiary; is the nucleophile strong or weak; is the solvent polar protic or polar aprotic — that lets you predict, before writing a single arrow, both the rate law a haloalkane will obey and whether its optically active form will invert or racemise, using only the steric and electronic reasoning already established for $S_N1$ and $S_N2$ .

Step 1. Classify the carbon bearing the leaving group by counting the alkyl groups attached to it: one alkyl group means primary, two means secondary, three means tertiary. This single count fixes both the steric bulk blocking backside attack and the number of hyperconjugating C–H/C–C bonds available to stabilise a cation, so it is the first and most decisive filter.

Step 2. Weigh the nucleophile. A strong, highly reactive, often negatively charged nucleophile (such as $OH^-$ , $CN^-$ , $RO^-$ ) present in reasonable concentration favours the bimolecular $S_N2$ pathway because it can force its way into the rate-determining step; a weak, often neutral nucleophile (such as $H_2O$ or an alcohol, used as solvent) cannot force that step and instead waits for the haloalkane to ionise on its own, favouring $S_N1$ .

Step 3. Weigh the solvent. A polar protic solvent stabilises the ionic transition state and the eventual carbocation and halide ion through hydrogen bonding, lowering the energy barrier to ionisation and favouring $S_N1$ ; a polar aprotic solvent solvates cations well but leaves anions (and hence the nucleophile) poorly shielded and highly reactive, favouring $S_N2$ .

Step 4. Combine the three answers. Primary halide plus strong nucleophile plus aprotic solvent locks in $S_N2$ with clean inversion; tertiary halide plus weak nucleophile plus protic solvent locks in $S_N1$ with racemisation; a secondary halide is decided by whichever of nucleophile strength and solvent polarity dominates, since its steric and electronic profile does not strongly favour either pathway on its own. Verified against the two worked examples below: (R)-2-bromobutane with strong $OH^-$ gives $S_N2$ /inversion to (S)-butan-2-ol; 2-bromo-2-methylbutane in aqueous ethanol with weak $H_2O$ gives $S_N1$ /racemisation to (±)-2-methylbutan-2-ol.

Worked example

Predict the mechanism, rate law and stereochemical outcome for two separate reactions: (a) (R)-2-bromobutane with $NaOH$ in a polar aprotic-like environment, and (b) optically active 2-bromo-2-methylbutane hydrolysed in aqueous ethanol.

Step 1. Classify each halide. In (a), Br sits on $C2$ of butane, attached to $CH_3$ and $CH_2CH_3$ — a secondary carbon. In (b), Br is attached to a carbon bonded to three other carbons — a tertiary carbon, so steric bulk rules out backside $S_N2$ attack almost entirely.

Step 2. Weigh the reagents. In (a), $OH^-$ is a strong nucleophile at meaningful concentration, and the medium behaves like a polar aprotic solvent — both push a secondary halide toward $S_N2$ . In (b), the nucleophile is $H_2O$ , weak, and the solvent is polar protic — both push the tertiary halide toward $S_N1$ .

Step 3. Write the mechanism and rate law for each. (a) is single-step: $OH^-$ attacks the carbon opposite Br, passing through a pentacoordinate transition state, so $\text{Rate} = k[R\text{-}X][OH^-]$ . (b) is two-step: the C–Br bond first ionises (slow) to a planar tertiary carbocation stabilised by hyperconjugation from three alkyl groups, then water attacks the cation from either face (fast), so $\text{Rate} = k[R\text{-}X]$ only.

Step 4. Read off the stereochemistry. (a) proceeds with Walden inversion, so the product is (S)-butan-2-ol, optically active because a single enantiomer reacted through a stereospecific backside attack. (b) proceeds through an achiral planar intermediate attacked from both faces with near-equal probability, giving (±)-2-methylbutan-2-ol as a racemic mixture — optically inactive even though the starting material was optically active, because the chiral information is destroyed at the flat carbocation stage.

Competitive stretch

If the same tertiary bromide in (b) were instead treated with a strong nucleophile like $CN^-$ in DMSO (a polar aprotic solvent), the steric bulk of the tertiary carbon would still block backside attack almost completely, so $S_N1$ would still dominate — the halide class can override an otherwise $S_N2$ -favouring reagent set, which is why classifying the carbon always comes first in the checklist above.

Common traps

Assuming $S_N2$ always gives an "optically pure" product without first checking that the starting haloalkane was itself a single enantiomer — inversion only matters if there was a defined configuration to invert.
Treating $S_N1$ as giving exactly 50:50 racemisation in every case — real carbocations often show a slight excess of inversion because the departing halide ion can briefly shield one face, but for exam purposes "racemisation" is the expected answer.
Confusing molecularity (how many species take part in the rate-determining step, giving the mechanism its "1" or "2" label) with the experimentally measured order of reaction — for $S_N1$ and $S_N2$ the two numbers happen to coincide, but they are conceptually different ideas.
Forgetting that solvent choice can flip the outcome for a secondary halide even when the nucleophile looks unambiguous, since polarity and protic/aprotic character act independently of nucleophile strength.

Ask the tutor: Want to try classifying a few more haloalkanes by mechanism, or see how changing just the solvent flips a secondary halide's pathway?

If a haloalkane could plausibly go by either mechanism, how do I decide which one an exam question wants

Work through the checklist in order: classify the carbon first (primary pushes hard toward $S_N2$ , tertiary pushes hard toward $S_N1$ ), then look at the nucleophile's strength and concentration, then the solvent's polarity and protic/aprotic character. Only secondary halides are genuinely ambiguous, and in that case the nucleophile and solvent clues in the question are there deliberately — a strong nucleophile in an aprotic solvent signals $S_N2$ , a weak nucleophile in a protic solvent signals $S_N1$ .

Does a racemic SN1 product mean literally zero net rotation of light

Not always exactly zero in real chemistry — the departing halide ion can linger near one face of the carbocation for a moment and shield it slightly, giving a small excess of inverted product and a faint residual rotation. For board-level purposes, however, treat $S_N1$ hydrolysis of a single enantiomer as producing a racemic (optically inactive) mixture; the idealised 50:50 answer is what is expected unless a question explicitly asks about the real-world partial-inversion effect.

Why does a polar protic solvent favour SN1 specifically, rather than just speeding up both mechanisms equally

A polar protic solvent has O–H or N–H bonds that hydrogen-bond directly to ions, so it stabilises the carbocation and the departing halide ion that $S_N1$ produces in its slow step, lowering that step's energy barrier. At the same time, hydrogen bonding to the nucleophile itself makes it less "naked" and less reactive, which specifically slows $S_N2$ down. The two effects push in opposite directions, so a protic solvent does not just speed everything up uniformly — it tilts the balance toward $S_N1$ .