The SN2 reaction

At a glance

The name was given by Hughes and Ingold in the 1930s: "2" means two species, the substrate and the nucleophile, take part in the single, rate-determining step.

The mechanism

Take (R)-2-bromobutane reacting with hydroxide ion. Everything happens in one smooth step, but it is easiest to understand in three snapshots:

Curved arrows, the exam way: arrow 1 starts at a lone pair on O and ends at the carbon; arrow 2 starts in the middle of the C–Br bond and ends on Br. Both arrows are drawn in the same diagram, because both happen in the same step.

Transition state and energy profile

At the top of the energy hill, the carbon has five groups around it: Nu and LG on a straight axis, partly bonded, and the three other groups lying flat in one plane at 120°, like a trigonal bipyramid. The negative charge is shared between Nu and LG (δ− on each).

Kinetics: the rate law

Both species are in the only step, so the rate depends on both: first order in each, second order overall. The units of k are L mol⁻¹ s⁻¹ (M⁻¹ s⁻¹).

Stereochemistry: Walden inversion

Because the nucleophile enters from the back, the three other groups are pushed through a flat arrangement to the other side, like an umbrella turned inside out by wind. The carbon's 3D arrangement is inverted. This is called Walden inversion, after Paul Walden, who first observed that one enantiomer could be converted into the other (1896).

• Every molecule reacts with inversion, so an enantiomerically pure substrate gives an enantiomerically pure product. The product is optically active, not racemic.
• Proof that every step inverts (1935): Hughes and co-workers treated optically active 2-iodooctane with radioactive iodide. The optical activity was lost exactly twice as fast as radioactive iodine was taken up: each exchange inverts one molecule, which cancels the rotation of one unreacted molecule.

Four factors that control the rate

a. The substrate: steric hindrance decides

The nucleophile must reach the back of the carbon. Every extra alkyl group around that carbon blocks the path and crowds the five-groups transition state.

• Branching next door matters too: neopentyl bromide, (CH₃)₃C–CH₂Br, is primary, yet its S_N2 is extremely slow because the bulky tert-butyl group shields the back of the CH₂ carbon.
• Allylic and benzylic halides (CH₂=CH–CH₂–X, C₆H₅–CH₂–X) react fast: the π system overlaps with and stabilises the transition state.
• Vinyl and aryl halides do not undergo S_N2: the back side is blocked by the molecule itself, and the C–X bond is stronger.

b. The nucleophile: stronger reacts faster

• An anion is a better nucleophile than its neutral conjugate acid: HO⁻ > H₂O, CH₃O⁻ > CH₃OH, NH₂⁻ > NH₃.
• In polar protic solvents, nucleophilicity increases down a group: I⁻ > Br⁻ > Cl⁻ > F⁻ and HS⁻ > HO⁻. Larger ions are more polarisable and less tightly held by hydrogen bonds.

Data: OpenStax Organic Chemistry, section 11.3 (via LibreTexts). The bar is on a log scale.

c. The leaving group: weak bases leave best

A good leaving group is stable on its own after it takes the bonding pair, which means it is a weak base (the conjugate base of a strong acid).

HO⁻, RO⁻ and NH₂⁻ are strong bases and do not leave. To substitute an alcohol, first turn OH into a good leaving group: protonate it with HBr (water leaves) or convert it to a tosylate (TsO⁻ leaves).

d. The solvent: polar aprotic is best

• Polar aprotic solvents such as DMSO, DMF, acetone, acetonitrile and HMPA dissolve the salt but cannot hydrogen-bond to the anion. The nucleophile is left "naked" and very reactive.
• Polar protic solvents such as water and methanol wrap the anion in a cage of hydrogen bonds, which must be broken before it can attack. The reaction is slower.
• Example: azide ion with 1-bromobutane is about 200,000 times faster in HMPA than in methanol (OpenStax, section 11.3).

S_N2 versus S_N1