The SN2 reaction — substitution, nucleophilic, bimolecular — is the cleanest mechanism in introductory organic chemistry. A nucleophile displaces a leaving group at a saturated carbon in a single concerted step: no intermediate, no carbocation, no waiting. Everything happens at once, and that one fact explains the kinetics, the stereochemistry, and every trend you are asked to predict.
The canonical SN2: hydroxide displaces bromide from bromoethane in one step to give ethanol.
In the scheme above, the C–Br bond breaks at the same moment the new C–O bond forms. Bond-making and bond-breaking are perfectly synchronized. The rest of this tutorial unpacks what that concerted step forces to be true.
1. The reaction happens in one concerted step
There is no carbocation intermediate and no discrete stopping point along the way. The nucleophile begins bonding to carbon before the leaving group has fully departed, so the molecule passes smoothly through a single high-energy transition state and drops into product. One barrier, one event.
Cyanide displaces iodide from iodomethane, forging a new C–C bond in the same concerted step.
Because the whole thing is one step, that step is the rate-determining step, and both partners are present in it. That is the seed of everything below.
2. The rate is second order in both partners
Since the substrate and the nucleophile both appear in the single rate-determining step, the rate law contains both:
rate = k[substrate][nucleophile]
The reaction is second order overall and first order in each species — this is exactly the "2" in SN2 (bimolecular). Double the nucleophile concentration and the rate doubles; double the substrate and it doubles again. Contrast this with SN1, whose slow step involves only the substrate.
Azide is a superb SN2 nucleophile; here it converts bromoethane to ethyl azide. Its rate responds to both [azide] and [substrate].
3. The nucleophile attacks from the backside
The nucleophile cannot approach from the same face as the leaving group — that space is occupied, and the electrons of the C–LG bond repel it. Instead it approaches 180° opposite the leaving group, straight through the back lobe of the carbon’s orbital. At the top of the barrier the central carbon is trigonal-bipyramidal: the three unchanging groups splay out into a flat plane, with the incoming nucleophile and the departing leaving group on the axis, each partially bonded.
This geometric requirement is the hinge of the whole mechanism: it dictates the stereochemistry and it explains why crowding is fatal.
4. Backside attack inverts the stereocenter
As the nucleophile pushes in from behind, the three other groups on carbon are turned inside-out, like an umbrella flipping in the wind. The result is inversion of configuration, known historically as the Walden inversion. If the starting carbon is a stereocenter, an R center becomes S (or vice versa) essentially every time — SN2 is stereospecific.
Hydroxide displaces bromide from 2-bromopropane. At a true stereocenter the same backside attack flips the configuration cleanly.
This is a signature you can test for in the lab: run SN2 on an enantiopure substrate and you get the inverted product, not a racemic mixture (that would signal SN1 and a planar carbocation instead).
5. Substrate sterics set the rate: methyl > 1° > 2° ≫ 3°
Because the nucleophile must squeeze into the backside of the carbon, anything bulky bolted to that carbon blocks the approach and raises the barrier. More alkyl groups mean more crowding at the transition state, so SN2 reactivity falls steeply as substitution rises:
Tertiary substrates do not undergo SN2 at any useful rate — the three surrounding groups wall off the backside completely. (Tertiary carbons instead favor SN1/E1 via a stable carbocation.) A methyl or 1° substrate is the ideal SN2 partner; a 1-bromobutane like reacts readily because its reacting carbon is still primary.
6. Strong nucleophiles, aprotic solvents, and good leaving groups accelerate it
Three external factors push the rate up. First, a strong, negatively charged nucleophile attacks faster: hydroxide, cyanide, azide, and iodide all outrun their neutral conjugate acids. Second, a polar aprotic solvent (acetone, DMSO, DMF) dissolves the salt but leaves the anion "naked" and reactive, whereas a protic solvent cages the nucleophile in hydrogen bonds and slows it down.
Third, the reaction requires a good leaving group — one that departs happily as a stable anion. Among the halides the order is I > Br > Cl ≫ F: iodide is large and polarizable and holds its charge well, while fluoride is a terrible leaving group and effectively shuts SN2 down. This is why iodides and bromides, not fluorides, are the usual SN2 substrates.
A primary iodide with a strong nucleophile is close to an ideal SN2: good leaving group, unhindered carbon.
7. Summary
The SN2 mechanism is a single concerted step, and that one property predicts everything else. Both the substrate and nucleophile enter the rate-determining step, giving the second-order rate law rate = k[substrate][nucleophile]. The nucleophile attacks 180° backside from the leaving group, threads a trigonal-bipyramidal transition state, and inverts the stereocenter (Walden inversion). Sterics rule the substrate, so reactivity is methyl > 1° > 2° ≫ 3° with tertiary essentially inert. And the reaction is fastest with a strong anionic nucleophile, a polar aprotic solvent, and a good leaving group (I > Br > Cl ≫ F). Memorize the concerted step and you can re-derive the rest on the spot.
Worked example
- Hydroxide is a strong, small nucleophile and acetone is a polar aprotic solvent — classic SN2 conditions.
- The nucleophile attacks the carbon bearing Br from the backside, 180° opposite the leaving group.
- That backside attack inverts the stereocentre (Walden inversion): the three other groups flip through like an umbrella.
- So an (R) starting material gives the (S) product (the priorities don't change here, so the descriptor flips).
Answer. (S)-butan-2-ol, formed with complete inversion of configuration at the reacting 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).
Draw this on the whiteboard
Open the OChem Board whiteboard — benzene rings, wedge/dash bonds, and a clickable periodic table built in. No account needed.