Learn · Organic Chemistry

The SN2 Mechanism

One concerted step, backside attack, and complete inversion of configuration

Quick answer One concerted step: the nucleophile attacks the carbon from the back as the leaving group leaves. That inverts the stereocenter, makes the rate depend on both partners (rate = k[substrate][Nu]), and only works on unhindered carbons — methyl > 1° > 2° ≫ 3°.

Hydroxide displaces bromide in one step → ethanol. Bond-making and bond-breaking happen together.

1. Backside attack inverts the stereocenter

The nucleophile comes in 180° opposite the leaving group and turns the carbon inside-out — the Walden inversion.

At a stereocenter, R → S every time — SN2 is stereospecific.

2. Sterics set the rate: methyl > 1° > 2° ≫ 3°

A crowded carbon walls off the backside, so more substitution means a slower reaction. Tertiary halides don’t do SN2 at all.

Methyl — fastest
1° — fast
2° — sluggish
3° — no SN2

3. Rate = k[substrate][nucleophile]

Both partners are in the single rate-determining step, so the reaction is second order — the “2” in SN2.

4. Strong nucleophile, aprotic solvent, good leaving group

Charged nucleophiles, polar aprotic solvents (DMSO, acetone), and I−/Br− leaving groups all speed it up.

Cyanide — strong Nu
Hydroxide — strong Nu
Iodide — great leaving group

Cyanide + iodomethane → a new C–C bond, one concerted step.

5. Summary

One step · backside attack · inversion · second-order · needs an open carbon.

Quiz yourself

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Because the mechanism has only one step, and both the substrate and the nucleophile are present in that single rate-determining step. So rate = k[substrate][nucleophile] — first order in each, second order overall.

Complete inversion of configuration at the stereocenter (Walden inversion). Because the nucleophile attacks from the backside, an R center becomes S (or vice versa). A racemic product would instead point to SN1.

Its central carbon carries three bulky methyl groups that block the required 180° backside approach. The nucleophile can’t reach the carbon, so the transition state is far too high in energy — tertiary substrates show essentially no SN2 (they go SN1/E1 instead).

An unhindered methyl or primary substrate, a strong negatively charged nucleophile (e.g. CN⁻, N₃⁻, OH⁻), a polar aprotic solvent (acetone, DMSO, DMF) that leaves the nucleophile unencumbered, and a good leaving group (I > Br > Cl ≫ F).

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