SN2

Hydroxide ion and bromomethane react in a single step. A lone pair on the oxygen forms a new C–O bond while the C–Br bond breaks, and out come methanol and bromide ion. You met this reaction, and its name, SN2, in Nucleophiles and Leaving groups. This page takes it apart: where the nucleophile has to come in, what that does to a stereocenter, which carbons can react at all, and what sets the speed.

SN2 is the first mechanism the course studies in full. Its rate law can be measured, its stereochemistry can be predicted, and its sensitivity to crowding follows from its shape. Most later mechanisms are described by how they differ from this one.

One step, two arrows

Start by finding the carbon under attack. Bromine is more electronegative than carbon, so the C–Br bond is polarized: the carbon is δ+ and the bromine δ−. That electron-poor carbon is the electrophile, and in CH₃Br it is the only candidate. The molecule that contains it, here the alkyl halide, is called the substrate.

Two curved arrows describe the change. Arrow 1 starts at a lone pair on the oxygen and ends on the carbon; it makes the new C–O bond. Arrow 2 starts on the C–Br bond and ends on the bromine, which leaves as bromide ion and takes both electrons of the old bond with it. A bond that breaks this way, with one atom keeping both electrons, breaks heterolytically (see Bonding).

Both arrows act at once: the new bond forms as the old one breaks. A step in which bonds form and break together is called concerted. That is the mechanism's central claim, and the rest of this page follows from it.

In general terms, an SN2 reaction is one concerted step in which a nucleophile, Nu⁻, bonds to a carbon and pushes out a leaving group, which takes the old bond's electrons with it.

The transition state: one hill, no valley

Halfway through the step, the carbon is bonded to five atoms at once, a pentacoordinate carbon. Three of them are the hydrogens, still fully bonded. The other two are the oxygen and the bromine, each held by a partial bond: the C–O bond half formed, the C–Br bond half broken. The negative charge that hydroxide brought in is shared between those two ends, so each carries a partial negative charge, δ−.

The three hydrogens lie flat, in one plane with the carbon, 120° apart. The oxygen and the bromine sit on a line through the carbon at right angles to that plane, 180° from each other. The carbon is close to sp² at this instant. Its three sp² orbitals hold the hydrogens, and its leftover p orbital, which runs along the O–C–Br line, holds both partial bonds.

This structure is a transition state, not an intermediate. Energy diagrams set out the difference: a transition state sits at the top of a hill and has no lifetime, while an intermediate sits in a valley and does. The SN2 diagram has one hill and no valley, so nothing forms along the way. The charges in the transition state are partial ones, and no free carbocation ever exists, because the carbon never drops to three bonds.

Because nothing sits in a valley, nothing has time to rearrange the way a carbocation can, and the carbon never loses track of the side the nucleophile came from. The next two sections show that this side is always the back. SN1, the next mechanism, has two hills instead, with a carbocation in the valley between them.

Why the nucleophile attacks from the back

The simple reason is crowding. The bromine and its three lone pairs fill the front of the carbon, so a nucleophile arriving from that side would run into them.

The deeper reason is about where the incoming electrons go. Every bond has an empty antibonding orbital to go with it (Orbitals introduced these). For the C–Br bond it is written σ*, read "sigma star". Electrons that flow into σ* weaken the C–Br bond and break it, which is exactly what has to happen here. So the nucleophile must push its pair into the C–Br σ* orbital.

That orbital is lopsided. Its large lobe sits on the carbon and points directly away from the bromine; a node separates it from a small lobe beyond the bromine. The nucleophile has to aim at the large lobe, so the back side is more than the least crowded approach. It is the only approach that leads to reaction.

Inversion of configuration

Look again at the hydrogens in the first figure. Before the reaction they lean toward the incoming hydroxide. At the transition state, drawn in the second figure, they lie flat. Afterward they lean toward the side the bromine left. The three groups that stay on the carbon flip over, the way an umbrella turns inside out in a strong wind.

On bromomethane the flip changes nothing you could detect, because the three hydrogens are identical. On a stereocenter it reverses the arrangement of the three groups that stay, so SN2 inverts the configuration of every molecule. This is called Walden inversion. Every molecule inverts the same way, so no racemic mixture forms.

Inversion describes the arrangement in space, and it does not automatically change the R/S letter. The letter depends on CIP priorities, and the incoming group need not take the rank the leaving group held. So work out the geometry first, then assign R or S to the product from scratch. The two worked examples below show one case of each.

On a ring, inversion shows up as a switch between cis and trans rather than as a change of letter. Treat cis-1-bromo-4-methylcyclohexane with ⁻SH. The bromine and the methyl start on the same face of the ring. The nucleophile has to arrive on the face opposite the bromine, so the sulfur ends up on the other face from the methyl, and the product is trans-4-methylcyclohexane-1-thiol.

Steric effects: which carbons can react

Compare alkyl bromides reacting with the same nucleophile under the same conditions. Bromomethane, whose carbon carries three hydrogens, reacts about 30 times faster than bromoethane, which has one methyl group in place of a hydrogen. 2-Bromopropane, with two methyl groups, is about 30 times slower again. The table adds two more crowded cases, taken up below.

The reason is the backside attack itself. The nucleophile has to reach the back of the carbon, and every group attached to that carbon stands in its way. A hydrogen is small; a methyl group is much larger. The labels 1°, 2° and 3° (primary, secondary and tertiary) count the carbons attached to the carbon that carries the leaving group: one, two or three (see Carbocations).

The tertiary entry is different in kind. Its three methyl groups close off the back of the carbon entirely, so SN2 does not happen there. Proposing an SN2 at a tertiary carbon is a mechanism error, not a slow reaction.

Crowding one carbon further out counts too. Neopentyl bromide, (CH₃)₃C–CH₂Br, is primary, because the carbon that carries the bromine has only one carbon attached. That neighbor, though, is a quaternary carbon, bonded to four other carbons, and its methyl groups hang over the back of the reacting carbon. Neopentyl bromide reacts about 10⁵ times more slowly than bromoethane.

A bridgehead carbon is the extreme case. It is a carbon where the rings of a bicyclic (two-ring) cage meet, like C1 of the cage in the figure, where three two-carbon bridges join C1 to C4 (its name is 1-bromobicyclo[2.2.2]octane). Its back side points into the middle of the cage, where no nucleophile can get, and the cage could not turn inside out even if one did. Bridgehead halides do not undergo SN2 at any rate worth measuring.

Two kinds of substrate react faster than their place on the ladder suggests. An allylic halide has its leaving group on the carbon next to a C=C. A benzylic halide has it on the carbon next to a benzene ring. Both react by SN2 roughly ten to a hundred times faster than a comparable ordinary substrate.

No cation forms; the reaction is still one concerted step. The difference lies in the transition state. There, the attacked carbon uses a p orbital to hold its two partial bonds, and that p orbital lies side by side with the p orbitals of the neighboring π bond or ring. They overlap, so some of the charge that builds up at the transition state spreads onto the π system, and the hill is lower. Allylic and benzylic substrates are fast by other pathways as well, which is why they come up again in SN1.

Rate law

Start with hydroxide and bromomethane reacting at a measured rate. Double the hydroxide concentration, and the rate doubles. Double the bromomethane instead, and the rate also doubles. Double both, and it goes up four times.

That behavior is the rate law rate = k[Nu][substrate]. Square brackets mean concentration, and k, the rate constant, is a number fixed for a given reaction at a given temperature. Both concentrations appear because both species take part in the one step, which is also the rate-determining step. A reaction whose rate-determining step involves two species is called bimolecular, and that is the 2 in SN2. The measured rate law shows that the nucleophile is present in the slow step. That is what a concerted mechanism predicts, though the rate law alone cannot prove the step is concerted; the clean inversion is the other half of the evidence.

Solvent

Polar aprotic solvents, such as acetone, DMSO (dimethyl sulfoxide), DMF (dimethylformamide) and acetonitrile, strongly favor SN2. They have no O–H or N–H bond, so they cannot donate hydrogen bonds. An anionic nucleophile comes as a salt, such as NaOH or NaCN. These solvents surround the Na⁺ well but leave the anion bare, with its lone pairs free to attack.

Polar protic solvents, such as water and alcohols, hydrogen-bond to an anionic nucleophile and cage it. The nucleophile has to shed part of that shell before it can attack, and that can cost several orders of magnitude in rate. The same reaction can take minutes in DMSO and days in methanol. Nucleophiles draws the cage.

The nucleophile

The rate law puts the nucleophile in the slow step, so its identity changes the rate. Nucleophiles ranked nucleophiles by charge, electronegativity, polarizability, crowding and solvent, and all of it applies here. Three of those results come up again and again in SN2 problems.

• Charge, then position in the row. An anion beats its own neutral partner, usually by several orders of magnitude: HO⁻ beats H₂O, H₂N⁻ beats NH₃, and RS⁻ beats RSH. Across a row, strength follows basicity: H₃C⁻ > H₂N⁻ > HO⁻ > F⁻.
• The halide order depends on the solvent. In water or an alcohol it runs I⁻ > Br⁻ > Cl⁻ > F⁻, because hydrogen bonds cage large, polarizable iodide least of the four. In a polar aprotic solvent such as DMSO it reverses to F⁻ > Cl⁻ > Br⁻ > I⁻. Check the solvent before you rank halides.
• Bulk cuts nucleophilicity but not basicity. tert-Butoxide is a stronger base than ethoxide and a far poorer nucleophile. It can reach a hydrogen on the outside of a molecule, but not the back of a carbon. That difference is what lets a chemist steer between SN2 and elimination, which E2 takes up.

The four variables, summarized

SN2 is favored by an unhindered substrate (methyl > 1° > 2°, never 3°), a strong nucleophile, usually an anion, a good leaving group (I⁻ > Br⁻ > Cl⁻ ≫ F⁻, or a sulfonate such as tosylate), and a polar aprotic solvent. All four come back in Substrate & solvent effects at the end of this chapter, where they decide among SN2, SN1, E2 and E1. SN2 is the mechanism that wants each of them in its strongest form.

What carries forward

SN2 is the model for concerted mechanisms in general, and its inversion is a basic tool of stereochemical planning in synthesis. It competes directly with E2 for the same substrates and reagents, and judging that competition is the central skill of this chapter. Backside attack into a σ* orbital also returns later in the course, whenever a nucleophile opens a three-membered ring, as with the bromine-bridged ion in alkene bromination and with epoxides in Alcohols, Ethers & Related Chemistry.