Warm 2-bromo-2-methylbutane in ethanol, and part of it turns into alkenes. No strong base is present, yet the molecule still loses H and Br. The route is E1, short for elimination, unimolecular. It builds the alkene in two steps, through the same carbocation as SN1.
E1 is the last of the chapter's four mechanisms. It is unimolecular like SN1 and forms an alkene like E2. Because its first step is exactly SN1's first step, the two run together. That makes E1 a poor way to make a compound. It is still important to recognize, because it turns up whenever SN1 does.
E1 starts exactly like SN1
Step 1 is the slow step of SN1, unchanged. In 2-bromo-2-methylbutane, the C–Br bond breaks on its own, and bromine leaves with both electrons as Br⁻. What remains is a tertiary carbocation at C2: a flat, sp² carbon with a positive charge and an empty p orbital. Nothing else takes part in this step.
Everything the Carbocations section says about cation stability applies here. Tertiary beats secondary, secondary beats primary, and allylic and benzylic cations gain extra stability from resonance.
Step 2 is where E1 parts from SN1. In SN1, a nucleophile bonds to the positive carbon. In E1, a weak base removes a β-hydrogen, one on a carbon next to the positive carbon. The pair of electrons from that C–H bond becomes a new π bond between the two carbons.
So step 2 needs two curved arrows. The first runs from a lone pair on the base to the β-hydrogen. The second runs from the C–H bond into the space between the β carbon and the positive carbon, where the pair becomes the π bond. Here the base is an ethanol molecule, which becomes EtOH₂⁺ once it holds the proton.
The base is usually a weak one that is already in the flask: the solvent, or the leaving group that has just departed. E1 needs no strong base. The hydrogen it removes sits next to a full positive charge, which makes it acidic enough for even water to take.
It follows that E1 is not stereospecific: the starting stereoisomer does not decide the product's stereochemistry. Two diastereomers that differ only at the carbon bearing the leaving group both lose it and give the same flat cation, so they give the same mixture of alkenes. When the alkene can be cis or trans, the more stable isomer, usually trans, is the major product, whichever diastereomer you start from.
Zaitsev's rule
Go back to the cation from 2-bromo-2-methylbutane. It has β-hydrogens in two kinds of position. C3, the CH₂ of the ethyl group, carries two of them. The two methyls on the positive carbon carry three each, and they are equivalent, since losing a hydrogen from either one gives the same alkene.
Losing a hydrogen from C3 gives 2-methylbut-2-ene, the major product. Losing one from a methyl gives 2-methylbut-1-ene, the minor product. Ethanol can also bond to the positive carbon, as it would in SN1, and give the ether 2-ethoxy-2-methylbutane. All three products come out of the same flask.
The major alkene is the more substituted one, as Zaitsev's rule predicts. Counting hydrogens alone would predict the opposite, since the two methyls offer six and C3 offers only two. As in E2, count the carbon groups on the two alkene carbons: 2-methylbut-2-ene has three and 2-methylbut-1-ene has two.
The same count sorts every alkene. But-1-ene has a single ethyl on its C=C, so it is monosubstituted. But-2-ene has one methyl on each alkene carbon and is disubstituted. 2,3-Dimethylbut-2-ene has two methyls on each and is tetrasubstituted. Four is the most there can be, because each alkene carbon has only two bonds left after the double bond. A group counts once however long it is, so the ethyl of but-1-ene counts exactly as a methyl would.
The reason is alkene stability, as in E2. 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. More groups on the C=C means more of this sharing, and so a more stable alkene. Step 2 is where the π bond forms, so the more stable alkene also has the lower barrier and forms faster.
The differences are small. They are measured by the heat of hydrogenation, the heat released when H₂ adds across the C=C; the more stable the alkene, the less heat it releases. By that measure, a tetrasubstituted alkene is about 4 kcal/mol more stable than a monosubstituted one. The steps are uneven: about 2 kcal/mol from mono to di, but only a fraction of a kcal/mol from tri to tetra. Differences that small are still enough to set a product ratio.
In E2 you can steer the product. A bulky base such as tert-butoxide takes the more exposed hydrogen and gives the less substituted alkene. E1 has no such option, because its base is always a weak one. A strong, bulky base would not wait for a cation to form, and would bring on E2 instead. So E1 gives the Zaitsev product whatever weak base is present.
Rate law and the SN1 connection
E1 is unimolecular: rate = k[substrate]. Only the substrate takes part in the slow step. The base acts afterwards, so its concentration does not appear in the rate law.
The kinetic isotope effect confirms it. In E2, putting deuterium in place of the β-hydrogens slows the reaction six- or sevenfold, because the C–H bond breaks in the slow step. In E1 that bond breaks in the fast second step, so the same swap leaves the rate almost unchanged.
E1 and SN1 share one cation, so they also share conditions: a weak nucleophile that is also a weak base, a polar protic solvent, and a substrate that ionizes easily. Under those conditions both always run. Any reaction you would call SN1 makes some E1 product, and the reverse is true too. That is why neither is chosen for a synthesis where yield matters.
Heat shifts the balance toward elimination, for the reason given in E2: elimination makes more particles, and so more entropy. The choice is made at the cation, which meets one ethanol, so two particles go in. E1 keeps two, the alkene and EtOH₂⁺. SN1 joins them into one, the protonated ether.
Entropy enters free energy through ΔG = ΔH − TΔS, where ΔS is the change in entropy and T is the temperature in kelvin. The −TΔS term helps whichever reaction has the larger ΔS, and it counts for more the hotter the flask. Here that is E1. The barriers follow the same pattern, because the elimination transition state is heading toward two particles and the substitution one toward one. So heating the flask lowers the elimination barrier relative to the substitution one, and elimination takes a larger share.
More substitution around the cation also favors elimination. A crowded positive carbon is hard for a nucleophile to reach, while the base only needs a hydrogen on the outside. A more substituted cation also leads to a more substituted, more stable alkene.
Warm 3-bromo-2,2-dimethylbutane, (CH₃)₃C–CHBr–CH₃, in aqueous ethanol.
Step 1. Bromide leaves, giving a secondary cation at C3.
Step 2. Check for a shift before anything else. The neighbor, C2, is quaternary: it carries three methyls and no hydrogen, so no hydride can move. A whole methyl group can move instead, taking its bonding pair with it, just as a hydride would. The arrow starts on the C2–CH₃ bond and ends on C3. The charge moves to C2, which now carries two methyls and the rest of the chain, so the new cation is tertiary. The shift is faster than the weak base can find a hydrogen, so it happens first.
Step 3. Now eliminate from the rearranged cation, (CH₃)₂C⁺–CH(CH₃)₂. Its β-hydrogens sit in two places. Losing one from the two methyls on C2 gives the disubstituted 2,3-dimethylbut-1-ene. Losing the single hydrogen on C3 gives the tetrasubstituted 2,3-dimethylbut-2-ene, and Zaitsev's rule picks this one.
The point. Without the shift, the only β-hydrogens on the secondary cation are on C4, and losing one gives the monosubstituted 3,3-dimethylbut-1-ene. The real product has its double bond between C2 and C3 instead, a place the unshifted cation could never reach, because C2 started with no hydrogen. Applying Zaitsev's rule to the original cation gives a wrong answer that looks completely reasonable.
Where E1 actually shows up
E1 is rarely chosen on purpose with an alkyl halide. It matters most as the mechanism of acid-catalyzed dehydration, which turns an alcohol into an alkene by removing water.
Heat 2-methylbutan-2-ol with concentrated sulfuric acid. The acid protonates the OH, which turns a poor leaving group into water, the same trick SN1 uses to make an alcohol react. Water leaves in the slow step, giving the same tertiary cation that 2-bromo-2-methylbutane gave. From there nothing is new: a water molecule removes a β-hydrogen, and 2-methylbut-2-ene is again the major product.
The cation in a dehydration can rearrange, just as in the worked example above. A shift like that is the usual reason a dehydration gives an unexpected alkene as its major product.
Every step of a dehydration can also run backwards, just as ionization does in the common-ion effect of SN1. So the reaction is an equilibrium, not a one-way process.
Looking ahead (nothing on this page depends on it): adding a proton to an alkene gives back the cation, and water can then bond to it. That reverse reaction is the acid-catalyzed hydration of alkenes in Addition reactions.
The practical recipe pushes the equilibrium toward the alkene. The alkene is the lowest-boiling compound in the flask, so a hot reaction lets it distill out as it forms. Taking a product away pulls an equilibrium toward making more of it. This is Le Châtelier's principle, and it turns dehydration into a useful way to prepare alkenes. Alcohol reactions covers it in full.
The chapter's square, completed
| Substitution | Elimination | |
|---|---|---|
| Concerted | SN2 | E2 |
| Via cation | SN1 | E1 |
Read across a row, and the two mechanisms share a route, so they share conditions. SN2 and E2 both need a strong reagent, and neither rearranges. SN1 and E1 both need a weak reagent and a solvent that supports ionization, and both can rearrange. Read down a column, and the two mechanisms share an outcome. Reading the table this way is the quickest route into Substrate & solvent effects, where you choose between the four.
What carries forward
E1 is the mechanism of alcohol dehydration in Alcohols, Ethers & Related Chemistry. Its reverse, adding a proton to an alkene to give a carbocation, is the first step of Markovnikov addition in Alkenes & Alkynes. The same carbocation sits at the center of SN1, E1, alkene addition and alcohol dehydration, and seeing that turns four separate reactions into one idea.