Warm 2-bromopropane, CH3CHBrCH3, with sodium ethoxide in ethanol, and most of it turns into propene, CH2=CH–CH3. Ethoxide has pulled a hydrogen off one carbon, bromide has left the carbon next to it, and a C=C double bond now joins the two. Only about a fifth of the 2-bromopropane takes the SN2 route and becomes an ether.
A reaction that removes two groups from neighboring carbons and puts a π bond between them is an elimination. This one happens in a single concerted step, and it is called E2. It competes with SN2 for the same alkyl halides and the same reagents. What sets it apart is a demand on the substrate's conformation: before anything can happen, the hydrogen and the leaving group have to line up in one particular way. No other mechanism in this chapter makes that demand.
One step, three arrows
Two carbons in the substrate get special names. The alpha (α) carbon carries the leaving group. A beta (β) carbon is any carbon bonded to the α carbon, and a hydrogen on it is a β-hydrogen. In 2-bromopropane the α carbon is C2, and both CH3 groups are β carbons, with three β-hydrogens each.
E2 removes a β-hydrogen together with the leaving group, so it is a β-elimination. A substrate with no β-hydrogen cannot do it at all. Bromomethane, CH3Br, has no β carbon, so it can only substitute.
Three curved arrows describe the step, one more than SN2's two, and all three move at the same instant.
- Arrow 1. A lone pair on the base takes the β-hydrogen. The tail sits on the lone pair and the head on the hydrogen.
- Arrow 2. The pair that held that hydrogen to the β carbon swings over to become the new π bond. The tail sits on the C–H bond and the head on the C–C bond.
- Arrow 3. The C–Br bond breaks, and bromide leaves with both of its electrons. The tail sits on the C–Br bond and the head on the bromine.
Because the three arrows move together, the reaction passes through one transition state and forms no intermediate. In particular it never forms a carbocation. The hydride and methyl shifts of SN1 need a carbocation to move toward, so an E2 never rearranges the carbon skeleton.
Why geometry matters: anti-periplanar
Look at 2-bromobutane down the bond from C2, which carries the bromine, to C3, one of its β carbons. C3 holds two hydrogens and a CH3. Turning C3 changes which of them sits opposite the bromine, and the three conformations in the figure behave very differently.
When a hydrogen on C3 sits 180° from the bromine, E2 can happen. The two bonds that break, C–H and C–Br, then lie in one plane and point in opposite directions. That arrangement is called anti-periplanar. Anti means 180° apart, as in the anti conformation of butane in Newman projections, and periplanar means lying in one plane.
When both hydrogens are gauche to the bromine, 60° away, nothing happens. The molecule first has to turn C3 until a hydrogen reaches the anti position. In an open chain that turn is cheap and fast, so the conformation E2 needs is always within reach.
The reason lies in the orbitals. As the C–H bond breaks, its electron pair flows into the antibonding orbital of the C–Br bond, the σ* orbital met in SN2. That orbital's large lobe on the α carbon points straight away from the bromine. When the C–H bond is anti to the C–Br bond, the C–H pair and that lobe point the same way, side by side, and the pair can flow straight across. As the step goes on, the two become the two p orbitals of the new π bond, and a π bond needs its p orbitals parallel. At 60°, the gauche angle, the orbitals do not line up, and the concerted step cannot run.
One other angle also lines the orbitals up: 0°, with the hydrogen directly behind the bromine. This is syn-periplanar, where syn means on the same side. It works less well than anti, because the molecule must hold an eclipsed conformation, the high-energy one from Newman projections. When an anti hydrogen is available, syn elimination almost never happens.
Some rigid bicyclic molecules, two rings that share atoms and lock each other in place, have no hydrogen anti to the leaving group, and the locked rings cannot turn to bring one there. For them syn elimination is the only route, and it runs slowly. Treat anti-periplanar as the working rule. The syn exception shows that the rule is really about the orbitals being parallel, not about the two groups being on opposite sides.
A cyclohexane ring cannot turn about its ring bonds the way a chain can, so the rule bites harder there. Take chlorocyclohexane. In the chair with chlorine axial, the axial hydrogens on the two neighboring carbons point the other way, to the opposite face, and each is anti-periplanar to the C–Cl bond. In the chair with chlorine equatorial, no hydrogen is anti to the C–Cl bond; the bonds anti to it are ring C–C bonds.
Chlorocyclohexane prefers the chair with chlorine equatorial, as Axial/equatorial showed. It still eliminates, because it flips into the other chair, where chlorine is axial, and reacts from there. That chair is the minor one, so fewer molecules are ready to react at any moment and the reaction runs slower.
Zaitsev versus Hofmann: which hydrogen leaves
2-Bromo-2-methylbutane has β-hydrogens in two different places. Its α carbon, C2, is bonded to two CH3 groups (C1 and a methyl branch) and to a CH2 (C3). A base can take a hydrogen from either kind of β carbon, and each choice gives a different alkene.
Taking a hydrogen from C3 puts the double bond between C2 and C3, which gives 2-methylbut-2-ene. Taking one from either CH3 puts the double bond at the end of the chain, which gives 2-methylbut-1-ene.
To compare the two alkenes, count the carbons bonded to the two carbons of the C=C. That number says how substituted the alkene is. 2-Methylbut-2-ene has three such carbons, so it is trisubstituted. 2-Methylbut-1-ene has two, both on the same end, so it is disubstituted. The scale runs from monosubstituted, with one carbon, to tetrasubstituted, with four carbons and no hydrogen left on the C=C.
The more substituted an alkene is, the more stable it is. The reason is hyperconjugation, from Carbocations: a C–H bond on a carbon next to the C=C spreads a little of its electron pair into the empty antibonding orbital of the π bond, much as it spreads into the empty p orbital of a carbocation. Each alkyl group on the C=C brings more of those C–H bonds. The energy differences are small, but they are enough to decide which alkene forms faster.
A small base such as ethoxide or hydroxide gives mostly the more substituted alkene, the Zaitsev product. The preference is called Zaitsev's rule. It holds because the E2 transition state already has part of the new double bond, so whatever stabilizes the alkene also lowers the barrier on the way to it.
A bulky base such as tert-butoxide, (CH3)3CO−, gives mostly the less substituted alkene, the Hofmann product. Its oxygen hangs off a carbon that carries three CH3 groups, and it cannot easily reach the hydrogens on C3, which are tucked in between the crowded C2 and the CH3 of C4. The hydrogens on the two CH3 groups stick out from the edge of the molecule, so tert-butoxide takes one of those instead. The switch is large: ethoxide gives about 70% of the Zaitsev alkene, and tert-butoxide gives about 73% of the Hofmann alkene.
So the base is not a spectator. Choosing a small base or a bulky one is choosing which alkene you get, and chemists make that choice routinely when they plan a synthesis.
Stereospecific: the diastereomer decides E or Z
1-Bromo-1,2-diphenylpropane, PhCHBr–CH(Ph)CH3, has two stereocenters, C1 and C2, so it comes as two diastereomers. Here Ph is short for a phenyl group, the C6H5– ring from Functional groups. With a strong base, the (1S,2R) diastereomer gives only the E alkene, (E)-1,2-diphenylprop-1-ene. The (1S,2S) diastereomer gives only the Z alkene. The base and the mechanism are the same, and the alkene depends on which diastereomer went in.
Newman projections down the C1–C2 bond explain it. Turn each diastereomer until the bromine on C1 is anti to the hydrogen on C2, the only conformation that can react. When those two leave, the other four groups stay where they were. The two groups that sat on the same side of the Newman, one on the front carbon and one on the back, end up on the same side of the new double bond. In the (1S,2R) diastereomer that puts the two phenyls on opposite sides, so the alkene is E. In the (1S,2S) diastereomer the phenyl and the methyl on C2 have swapped places, the phenyls end up on the same side, and the alkene is Z.
A reaction in which the stereochemistry of the starting material fixes the stereochemistry of the product is stereospecific: change the starting stereoisomer, and the product stereoisomer changes with it. E2 is stereospecific because its geometry is fixed.
The result is also evidence for the mechanism. Suppose the C–H bond broke first, or the C–Br bond. The half-reacted molecule could then turn about the C1–C2 bond before the second bond broke, and both diastereomers would give the same mixture of E and Z. They do not, so the molecule never turns between one bond breaking and the other: both break in the same step.
Rate law and conditions
Both the base and the substrate take part in the one step, so the rate depends on both: rate = k[base][substrate]. A reaction with two species in its rate-determining step is bimolecular, and that is the 2 in E2. The rate law has the same form as SN2's, for the same reason.
E2 needs a strong base, such as hydroxide, ethoxide or tert-butoxide. What matters is basicity, how strongly the reagent pulls on a proton, and not nucleophilicity, how strongly it attacks a carbon. That difference is what separates E2 from SN2. A strong base that is a poor nucleophile, such as the bulky tert-butoxide from SN2, drives elimination. A good nucleophile that is a weak base, such as iodide, a thiolate (RS−), azide (N3−) or cyanide (CN−), drives substitution.
Every class of substrate with a β-hydrogen can do E2, but not at the same speed: 3° > 2° > 1°. That matches the SN1 order, yet no cation is involved. The reason is the same partial double bond in the transition state that explains Zaitsev's rule. A more substituted substrate forms a more substituted alkene, so the transition state on the way to it is lower. On a tertiary substrate SN2 is blocked, and E2 is the only bimolecular route left.
Heat favors elimination over substitution. E2 turns two particles, the base and the substrate, into three: the alkene, the protonated base and the leaving group. SN2 turns two particles into two. More particles means more entropy. Entropy measures how many ways the particles and their energy can be arranged, and the higher the temperature, the more weight it carries in deciding which reaction wins.
One more experiment identifies E2. Replace the β-hydrogens with deuterium and the reaction slows by a factor of six or seven. This is the kinetic isotope effect from Atomic structure: a C–D bond breaks more slowly than a C–H bond. When the isotope sits on a bond that breaks in the rate-determining step, the effect is this large, and it is called a primary isotope effect. So the C–H bond breaks in the rate-determining step, which is just what a one-step mechanism predicts.
Looking ahead. E1, the other elimination in this chapter, breaks its C–H bond in a fast step after the slow one. Deuterium on its β carbons leaves its rate almost unchanged, so the same measurement tells the two eliminations apart.
| SN2 | E2 | |
|---|---|---|
| Reagent | strong nucleophile | strong base |
| 3° substrate | impossible | preferred |
| No β-hydrogen (CH3Br) | fastest | impossible |
| Geometry | backside, 180° | anti-periplanar, 180° |
| Stereochem | inversion | stereospecific alkene |
| Heat | speeds it, but less than E2 | speeds it more; favored |
When geometry overrules Zaitsev: menthyl chloride
Menthyl chloride is menthol, the cooling compound in mint, with a chlorine in place of the OH. It is a cyclohexane with a chlorine on C1, an isopropyl group on C2 and a methyl on C5. Neomenthyl chloride is its diastereomer: the same three groups, with the chlorine on the other face of the ring. Both eliminate HCl with ethoxide. Which reacts faster, and which alkene does each give?
Neomenthyl chloride. Its favored chair has the isopropyl and the methyl equatorial, and that puts the chlorine axial. Both neighboring carbons, C2 and C6, then carry an axial hydrogen on the opposite face, anti-periplanar to the C–Cl bond. The molecule eliminates straight from its favored chair, so it reacts fast. It also has a choice of two hydrogens, and it follows Zaitsev's rule: it gives mostly the trisubstituted alkene, with the C=C between C1 and C2.
Menthyl chloride. Its favored chair has all three groups equatorial, the chlorine included, and nothing is anti to an equatorial chlorine. The ring has to flip first. In the flipped chair all three groups are axial, so that chair is strained and scarce, and menthyl chloride reacts about 200 times more slowly than neomenthyl chloride.
The flipped chair also decides the product. The chlorine is now axial, but so is the isopropyl on C2, which leaves C2 only an equatorial hydrogen, and that hydrogen is not anti to the chlorine. Only C6 has an axial hydrogen on the opposite face. So menthyl chloride gives only the disubstituted alkene, with the C=C between C1 and C6. Geometry has overruled Zaitsev's rule completely.
What carries forward
E2 is how alkenes are made from alkyl halides, which makes it the entry point to Alkenes & Alkynes. Its anti-periplanar requirement is the course’s clearest case of conformation controlling reactivity. The same geometric logic governs anti additions to alkenes and the trans-diaxial opening of epoxides in Alcohols, Ethers & Related Chemistry.