SN2 Reaction Coordinate

Introduction

When hydroxide replaces bromide in bromoethane, the new C–O bond is not completed before the C–Br bond starts to break. The two changes occur together in a single elementary event called SN2. The term describes substitution by a nucleophile with a bimolecular rate-determining step. A reaction-coordinate diagram makes this concerted nature visible and helps distinguish SN2 from pathways that form a carbocation intermediate.

Core explanation

Consider Nu⁻ + R–Br → R–Nu + Br⁻, using a primary alkyl bromide. The nucleophile's electron pair approaches the carbon from the side opposite bromine. The carbon–nucleophile bond begins forming as the C–Br bonding pair moves onto bromine. At the highest-energy point, carbon is partially bonded to both incoming and departing groups. This arrangement is a transition state , conventionally placed in square brackets with a double dagger. It is not an isolable chemical species and cannot be assigned the normal tetrahedral valence of the starting carbon.

The approach is from the back because the C–Br sigma-star antibonding orbital has a favourable accepting lobe opposite the leaving group. The nucleophile donates electron density into this orbital as bromide departs. A front-side approach would give poorer orbital overlap and greater repulsion from the electron-rich leaving group. As the tetrahedral carbon is traversed, the three groups that remain attached move through an approximately planar arrangement. If that carbon is stereogenic, its configuration is inverted, a result developed on the stereochemistry page.

On a reaction-coordinate plot, put free energy on the vertical axis and progress along the elementary reaction on the horizontal axis. There is one peak, corresponding to the SN2 transition state, and no energy minimum between reactants and products. The vertical distance from reactants to peak is the activation free energy. The product energy may be higher or lower than reactants depending on the bonds and conditions. A favourable overall free-energy change does not remove the activation barrier.

Four major factors change the barrier. First, a strong nucleophile raises the likelihood of productive attack. Second, low crowding around the reacting carbon allows backside approach: methyl and primary substrates are usually best, secondary substrates slower, and ordinary tertiary substrates do not undergo normal SN2. Third, a good leaving group such as iodide, bromide or sulfonate accepts the bonding pair readily. Fourth, a polar aprotic solvent often keeps anionic nucleophiles reactive. Each factor affects a different part of the same transition-state comparison; no single rule replaces the whole analysis.

Two curved arrows depict the event. One arrow starts at the nucleophile's lone pair and ends at the electrophilic carbon. A second starts at the C–Br bond and ends at Br. Both arrows belong to the same step. Do not draw a free carbocation between them, and do not split the attack and departure into separate elementary steps.

Step-by-step reasoning

Locate the carbon bearing the leaving group, then identify the nucleophilic lone pair. Draw its arrow into carbon from the side opposite the leaving group. Draw the leaving-bond arrow onto the departing atom simultaneously. Check that carbon has only four full bonds in reactant and product. Finally draw a one-hump energy profile; the peak, not a valley, represents the shared transition-state arrangement.

Visual explanation

Sketch a three-dimensional carbon with three persistent substituents forming a shallow umbrella. The nucleophile enters under the umbrella while bromide exits above it. In the transition state, use dashed lines for the incoming and outgoing bonds. As the process finishes, the umbrella turns inside out. Underneath, draw one peak between two valleys labelled reactants and products, with no central intermediate valley.

Real-world analogy

Imagine passing a ball directly from one person to another in a narrow doorway. For a brief moment both hands touch the ball, but the ball is never left on the floor between them. That shared moment resembles the transition state: partially formed and partially broken connections, with no free intermediate waiting between steps.

Real-world example

Making an ether by the Williamson reaction commonly uses an alkoxide and a primary alkyl halide. For example, sodium ethoxide can displace bromide from bromoethane to form diethyl ether. Choosing a primary substrate leaves room for backside attack; a tertiary bromide under the same basic conditions would be far more likely to eliminate.

Why?

Why does steric crowding matter so strongly? In the SN2 transition state, the nucleophile must get close to carbon while the leaving group remains partly attached. Alkyl groups around that carbon obstruct the required trajectory and raise the activation barrier. Crowding at a remote atom may matter less, so the location of branching, not merely total molecular size, is important.

Common misconception

"The nucleophile bonds first and then the leaving group leaves." That sequence would give an overbonded carbon or a separate intermediate. In SN2, bond formation and bond cleavage overlap in a single elementary step. The diagram has one transition-state peak and no carbocation valley.

Worked example

Question: Draw the mechanism and reaction coordinate for CH₃CH₂Br + HO⁻ → CH₃CH₂OH + Br⁻.

Reasoning: The ethyl carbon is primary and accessible. Hydroxide donates a lone pair to the carbon as the C–Br pair moves onto bromine. Both arrows are drawn in one step. The transition state contains partial C–O and C–Br bonds.

Answer: This is a concerted SN2 substitution with one energy maximum, no carbocation intermediate and ethanol as the organic product.

Quick check

1. How many transition-state peaks and intermediates are on an ideal SN2 reaction-coordinate plot? Answer: One peak and no intermediate minimum, because the bond changes occur in one concerted step.

Exam focus

Draw both curved arrows at once, start them at electrons rather than atoms, and label partial bonds in the transition state. If asked for an energy profile, show one peak. Use the substrate's substitution level, nucleophile and leaving group to justify the mechanism instead of saying simply that substitution occurred.

Advanced insight

The transition state is a statistical ensemble of configurations rather than a single frozen picture; solvent reorganisation and bond vibrations contribute to the measured activation free energy. Computational orbitals provide a deeper explanation for backside attack: donation into the C–X sigma-star orbital weakens that bond while building the new C–Nu bond. This orbital account and the curved-arrow account describe the same electron movement.

Summary

SN2 is a one-step substitution in which an electron-pair donor attacks carbon from behind as the leaving group departs. Its transition state has partial bonds to both groups and produces one peak on the reaction-coordinate diagram, without an intermediate. Strong nucleophiles, accessible carbons, good leaving groups and suitable solvents lower the barrier. Backside geometry leads to inversion at a stereogenic carbon.

Practice questions

1. Why must an SN2 energy profile have no intermediate valley? Answer: Because attack and leaving-group departure occur in one elementary event; there is no separately stable species between reactants and products. 2. Which reacts faster by typical SN2, bromoethane or 2-bromo-2-methylpropane, and why? Answer: Bromoethane, because its primary carbon allows backside approach; the tertiary carbon is too crowded for normal SN2. 3. Where does the curved arrow for leaving-group departure begin? Answer: At the C–leaving-group bond, since that electron pair moves onto the leaving-group atom. 4. Does a lower product energy imply no activation barrier? Answer: No. Even an overall favourable reaction must pass through a higher-energy transition state.