Organic Elimination Reactions

An elimination reaction removes two atoms or groups from adjacent carbon atoms of a molecule to form a double bond: haloalkanes lose HX when heated with a strong base, and alcohols lose water when heated with a strong acid. In brief: 2-bromobutane with ethanolic KOH gives mainly but-2-ene (the more substituted alkene, Zaitsev’s rule), and ethanol heated with concentrated H₂SO₄ at 170 °C gives ethene.

Elimination is the reverse of addition, and it is the main way to prepare alkenes. It competes with substitution whenever a haloalkane meets a nucleophile that is also a base, so it is important to know which conditions favor each. This article explains the two mechanisms (E1 and E2), the rule that predicts the product, the dehydration of alcohols and the competition with substitution, with a table and worked examples.

What are the types of elimination?

The most common is β-elimination: a hydrogen is removed from the carbon next to the carbon that carries the leaving group (the β-carbon), and the leaving group departs from the α-carbon, so a C=C bond forms:

• Dehydrohalogenation: R–CH₂–CH₂X + base → R–CH=CH₂ + HX.
• Dehydration of alcohols: R–CH₂–CH₂OH → R–CH=CH₂ + H₂O (acid catalysis).

What is the E2 mechanism?

E2 (elimination, bimolecular) is a one-step process. A strong base removes the β-hydrogen at the same time as the leaving group departs and the π bond forms. The rate depends on both the haloalkane and the base: rate = k[RX][base]. The best geometry has the H and the leaving group on opposite sides of the C–C bond in the same plane (anti-periplanar). E2 is favored by a strong base, such as ethoxide or hydroxide in ethanol, high concentration of base and heat.

What is the E1 mechanism?

E1 (elimination, unimolecular) shares the first step with S_N1: the leaving group departs to give a carbocation. In the second step, a base (often the solvent) removes a proton from the adjacent carbon. The rate depends only on the substrate: rate = k[RX]. E1 occurs with tertiary and secondary substrates in polar protic solvents, with a weak base and heat, and always competes with S_N1. The dehydration of alcohols with acid is an E1 reaction for secondary and tertiary alcohols.

Which alkene forms?

Zaitsev’s rule: when more than one alkene can form, the major product is usually the more highly substituted (more stable) alkene, that is, the one with more alkyl groups on the double bond carbons. For 2-bromobutane, loss of H from C3 gives but-2-ene (disubstituted, a mixture of trans and cis, trans is the major), and loss from C1 gives but-1-ene (monosubstituted). With ethoxide in ethanol, the mixture is about 80 % but-2-ene and 20 % but-1-ene.

Exceptions: with a bulky base such as potassium tert-butoxide, the less hindered hydrogen is removed and the less substituted alkene is the major product (Hofmann product). Elimination from quaternary ammonium hydroxides also follows the Hofmann rule.

How are alcohols dehydrated?

Heating an alcohol with concentrated sulfuric or phosphoric acid protonates the –OH group, so that it can leave as water. Tertiary alcohols dehydrate most easily, because they give the most stable carbocations.

At a lower temperature (about 140 °C), the primary alcohol with excess alcohol gives an ether by substitution, so temperature control is important in the preparation of ethene: 2C₂H₅OH → C₂H₅OC₂H₅ + H₂O at 140 °C, but C₂H₅OH → C₂H₄ + H₂O at 170 °C.

How does elimination compete with substitution?

A common example is the difference between warm aqueous KOH with bromoethane (substitution, giving ethanol) and hot ethanolic KOH (elimination, giving ethene). Because water solvates the hydroxide ion strongly, it acts as a nucleophile in water, whereas in ethanol it is a stronger base.

Worked examples

Example 1: predicting the major product

• 2-Bromo-2-methylbutane + ethoxide: the more substituted alkene, 2-methylbut-2-ene, is the major product (Zaitsev), with 2-methylbut-1-ene as a minor product.
• 2-Methylbutan-2-ol with acid (heat): the same alkenes, mainly 2-methylbut-2-ene.
• 2-Bromobutane + potassium tert-butoxide: mainly but-1-ene (Hofmann), since the bulky base removes the more accessible hydrogen.

Example 2: volume of gas from elimination

When 10.0 g of 1-bromopropane (M = 123.0 g/mol) is heated with excess ethanolic KOH, propene forms: C₃H₇Br + KOH → C₃H₆ + KBr + H₂O. n = 10.0 / 123.0 = 0.0813 mol. At 100 % yield, V = 0.0813 × 24.47 = 1.99 L at 25 °C and 1 atm. If only 60 % is obtained, 1.19 L.

Example 3: how many alkenes?

3-Bromo-3-methylpentane has β-hydrogens on the CH₂ groups of the two (equivalent) ethyl chains and on the methyl group, so two constitutional alkenes are possible: 3-methylpent-2-ene (trisubstituted, the major product, which exists as cis and trans isomers) and 2-ethylbut-1-ene (disubstituted, minor). Counting the cis and trans forms separately, there are three possible products.

Example 4: which alcohol dehydrates faster?

Compare butan-1-ol, butan-2-ol and 2-methylpropan-2-ol. The tertiary alcohol gives the most stable carbocation and dehydrates under the mildest conditions. The order of ease is tertiary > secondary > primary.

Who formulated the rules?

August Wilhelm von Hofmann studied the decomposition of quaternary ammonium hydroxides to alkenes in 1851. Alexander Zaitsev (Saytzeff), who had worked with Hermann Kolbe and later taught in Kazan, formulated in 1875 the rule on the direction of elimination that is named after him. The terms E1 and E2 came with the mechanistic studies of Hughes and Ingold in the 1930s, at the same time as S_N1 and S_N2.

Which mistakes are common?

• Mixing up Zaitsev and Markovnikov: Zaitsev concerns elimination (more substituted alkene), Markovnikov concerns addition.
• Ignoring the solvent: aqueous conditions favor substitution, alcoholic favor elimination.
• Forgetting a β-hydrogen: without one, no elimination is possible.
• Overlooking cis–trans isomers: but-2-ene exists as two stereoisomers.
• Using too low a temperature for dehydration: an ether may form instead.