An SN2 reaction is a one-step substitution where a nucleophile replaces a leaving group on carbon. “One step” means bond making and bond breaking happen together in a single transition state. The rate is second order overall, so rate = k[substrate][nucleophile]; doubling either concentration doubles the speed. That rate law is the original clue that both partners are present in the slow step.

The nucleophile approaches from the backside of the carbon–leaving group bond and overlaps with the bond’s antibonding orbital, written σ*.

Think of the backside as a mail slot and the nucleophile as a paper plane. You must fly into the slot. Anything that narrows the opening cuts down the success rate. Sterics control access to that slot. Methyl halides react fastest because three small hydrogens give a wide opening. Primary halides are still good because only one alkyl group crowds the carbon. Secondary halides slow down more as two alkyl groups start to become cumbersome to get past. Tertiary halides essentially shut down the possibility of SN2 because three alkyl groups block the path too much.

If the reacting carbon is chiral, backside attack causes the product to show Walden inversion, where the three groups that stay on carbon rotate to the opposite handedness. It’s like an umbrella flipping in the wind. This inversion is experimental evidence that SN2 passes through one aligned transition state rather than through a free carbocation.

Leaving-group (LG) quality controls how easily the C-LG bond can stretch and break in the transition state. Good leaving groups form stable anions and bind carbon weakly. Iodide is larger and less basic than bromide or chloride since its charge is spread over a bigger volume. Iodine’s large size makes the C–I bond long, and its diffuse orbitals overlap poorly with carbon, which makes the bond weaker. Both points explain why it’s easier for I to leave. The typical order is I⁻ > Br⁻ > Cl⁻, while F⁻ is the worst because it is the least polarizable and holds onto carbon the most tightly.

Nucleophile strength matters because the same lone pair must both donate electron density and reach the proper alignment with the electrophilic carbon. A high-energy, readily available lone pair reacts faster. In polar protic solvents such as water or alcohols, anions are caged by hydrogen bonding and must shed that shell before attack, which slows them. In polar aprotic solvents such as acetone, acetonitrile, DMF, or DMSO, anions are not strongly caged, so desolvation costs less and nucleophilicity rises. As a result, iodide and bromide feel sluggish in ethanol but much more powerful in DMSO.

Electronics can also influence how well the carbon accepts electron density into σ*. Groups that pull electron density away from the reacting carbon lower the energy of σ* and make overlap easier, which speeds attack. Allylic and benzylic positions are especially reactive because the transition state can spread partial charge into a nearby π system.

Intramolecular SN2 often beats intermolecular attack because tethering keeps the nucleophile and electrophilic carbon in the same neighbourhood, which creates a high effective molarity and cuts the entropic cost of bringing two molecules together. The tether also preorganises geometry so the lone pair can find the C–LG line more easily. Ring-closing SN2 is fastest when the tether length allows a near-linear approach, with five- and six-membered closures.

Let’s look at a concrete example to tie these ideas together: In DMSO, hydroxide substitutes on 1-bromoethane with a clean SN2 because the anion is not tightly caged by hydrogen bonds and the primary carbon leaves a wide backside path. Change only the substrate to tert-butyl bromide and the rate drops sharply since three alkyl groups hinder the carbon from nucleophilic attack. Keep the primary substrate but switch the solvent to ethanol and the reaction slows again because hydrogen bonding cages hydroxide so it takes more energy for it to desolvate and attack. Go back to DMSO and swap Br for I and the rate rises because the C–I bond is weaker and I- holds the negative charge more happily. Now make the attack intramolecular: deprotonate 5-bromopentanol and the alkoxide cyclises to tetrahydrofuran. Under the same conditions, this ring closure typically is much faster than, for example if a hydroxide were to attack to form the diol.