E2 — Substitution & Elimination

E2 removes two groups from adjacent carbons and creates a pi bond between them, all in a single concerted step. It is SN2's direct competitor — same substrates, same reagents, same conditions — and the thing that most distinguishes it is a strict geometric requirement that no other mechanism in this chapter imposes.

Three arrows, one step

E2 in a single motion. Follow the electrons: the base takes the β-hydrogen, the pair that used to be the C–H bond swings down to become the π bond, and the leaving group departs with its own pair. Three arrows, one transition state, no intermediate — which is why the rate depends on both the base and the substrate, and why the reaction is called bimolecular.Nothing here is an intermediate. The base pulls the hydrogen, that C–H pair swings down into a π bond, and the bromide leaves with its pair — all at the same instant, in one concerted step.

E2 is concerted like SN2, but it takes three arrows rather than two, all firing at the same instant.

Arrow 1: a base removes a hydrogen from the carbon adjacent to the leaving group — the beta carbon. Tail on the base's lone pair, head on that beta hydrogen.

Arrow 2: the electrons from that C–H bond become a new pi bond between the beta carbon and the alpha carbon. Tail on the C–H bond, head between the two carbons.

Arrow 3: the C–leaving-group bond breaks and the leaving group departs with both electrons. Tail on that bond, head onto the leaving group.

None of these happens before the others. All three are one motion through one transition state, and there is no intermediate — which is why E2, unlike E1, never rearranges.

The vocabulary is worth fixing: the alpha carbon bears the leaving group; a beta carbon is any carbon adjacent to it. E2 is a beta elimination because the hydrogen comes from a beta carbon, and a substrate with no beta hydrogen simply cannot undergo it.

Why geometry matters: anti-periplanar

The geometric requirement, seen the only way it makes sense — down the C–C bond. The hydrogen being removed and the leaving group have to sit at 180° from each other, because the two electron pairs they leave behind are about to become a single pi bond, and a pi bond needs its two p orbitals parallel. On a cyclohexane ring this becomes an absolute rule: both groups must be axial, and a substrate whose favoured chair puts the leaving group equatorial has to flip into its minority chair before it can react at all.The two electron pairs that are about to become one π bond have to be lined up before they can merge. That single geometric demand is why a molecule may have to rotate — or ring-flip — into an unfavourable conformation before it can eliminate at all, and why some substrates simply cannot.

E2 only works when the beta hydrogen being removed and the leaving group are anti-periplanar — in the same plane, 180° apart, pointing in opposite directions.

The reason is orbital alignment. As the C–H bond breaks, its electrons must flow into the developing pi bond, and simultaneously into the antibonding orbital of the departing C–LG bond. That continuous overlap is only available when the two bonds are parallel and opposed. At any other dihedral angle the orbitals do not line up and the concerted process cannot proceed.

In an open chain this is rarely a problem — free rotation about the C–C bond means the required conformation is always accessible. On a ring it is a severe constraint, because the ring cannot rotate. On a cyclohexane, anti-periplanar means both groups axial, which is exactly the point the Conformational Analysis section of Module 4 built toward.

Because the geometry is fixed, E2 is stereospecific: which alkene geometry you get is determined by which diastereomer you start from. Eliminating HBr from the two diastereomers of 1-bromo-1,2-diphenylpropane gives, respectively, the E alkene and the Z alkene — different starting stereochemistry, different product geometry, from the same reaction. That is powerful evidence that the mechanism really is concerted with a defined geometry.

Worked example — the anti-periplanar constraint on a ring

Neomenthyl chloride eliminates rapidly with ethoxide and gives the Zaitsev product. Menthyl chloride, its diastereomer, eliminates about 200 times more slowly and gives only the less substituted alkene.

In neomenthyl chloride the favoured chair already has chlorine axial, with axial hydrogens on both neighbouring carbons — fast, and free to choose the more substituted product.

In menthyl chloride the favoured chair has everything equatorial, including chlorine. To eliminate at all the ring must flip into a strained triaxial chair, which is why it is slow; and in that chair only one neighbouring carbon carries an axial hydrogen, the one giving the less substituted alkene. Geometry overrides the usual product preference completely.

Zaitsev versus Hofmann: which hydrogen leaves

When several beta hydrogens are available, which one is removed decides the product.

A small base such as ethoxide or hydroxide can reach a crowded, more-substituted beta hydrogen, and does so because the resulting alkene is the more stable one. This gives the Zaitsev product: the more substituted alkene, favoured because alkene stability increases with substitution, for the same hyperconjugation reason that stabilizes carbocations.

A bulky base such as tert-butoxide, or the even bulkier LDA, cannot easily reach a crowded hydrogen and takes a more accessible one instead. This gives the Hofmann product: the less substituted alkene. The switch is dramatic — 2-bromo-2-methylbutane with ethoxide gives about 70% Zaitsev, and with tert-butoxide about 73% Hofmann.

So the base is not a spectator. Choosing between a small base and a bulky one is choosing which alkene you want, and that is a standard synthetic decision.

Zaitsev is a preference, not a rule, and it is outranked by geometry. Anti-periplanar availability comes first: if only one beta hydrogen can achieve the required alignment, that is the product regardless of which alkene would be more stable. The menthyl chloride case above is the standard demonstration, and it is a favourite exam question precisely because it punishes applying Zaitsev mechanically.

Rate law and conditions

E2 is bimolecular: rate = k[base][substrate], the same form as SN2, because both species participate in the single rate-determining step.

It is favoured by a strong base — and note that what matters here is basicity, not nucleophilicity, which is exactly what separates E2 from SN2. A reagent that is a strong base and a poor nucleophile (bulky alkoxides) drives elimination; one that is a strong nucleophile and a weak base (iodide, thiolate, azide, cyanide) drives substitution. E2 works on all substrate classes, and is the only bimolecular option for tertiary substrates, where SN2 is sterically impossible. Heat favours it further, since elimination makes two molecules from one and so gains entropy.

	SN2	E2
Reagent	strong nucleophile	strong base
3° substrate	impossible	preferred
Geometry	backside, 180°	anti-periplanar, 180°
Stereochem	inversion	stereospecific alkene
Heat	no effect	favoured

What carries forward

E2 is how alkenes are made from alkyl halides, which makes it the entry point to all of Module 7. Its anti-periplanar requirement is the clearest case in the course of conformation controlling reactivity, and the same geometric logic governs anti additions to alkenes and the trans-diaxial opening of epoxides in Module 8. The Zaitsev/Hofmann choice is the first time in the course that changing a reagent changes which product you get, rather than how fast you get it.