Elimination: E2 and E1, and the competition with substitution

A nucleophile that attacks carbon in an SN2 reaction is also, almost always, a base. Instead of attacking the electrophilic carbon, it can remove a proton from the carbon next door. When it does, the leaving group departs, a π bond forms between the two carbons, and the product is an alkene. This is β-elimination, the reverse of the additions in lessons 11 and 12, and it is the main way of making alkenes in the laboratory. Like substitution, elimination has a concerted bimolecular pathway, E2, and a stepwise unimolecular one, E1. Substitution and elimination compete in nearly every reaction of a haloalkane with a nucleophile, and predicting which wins is one of the central practical skills of this course.

The vocabulary of elimination

The carbon bearing the leaving group is the α carbon, and its neighbours are β carbons. In a β-elimination the base removes a hydrogen from a β carbon (a β-hydrogen) and the leaving group departs from the α carbon, giving a C=C between them. Overall, H and X are lost from adjacent carbons; when X is a halogen the reaction is a dehydrohalogenation, and when an alcohol loses water it is a dehydration. If a substrate has β-hydrogens on more than one carbon, more than one alkene can form, and the question of which is favoured, regioselectivity, arises.

The E2 mechanism

The reaction of 2-bromopropane with sodium ethoxide in ethanol gives propene with a rate proportional to the concentrations of both the substrate and the base, rate = k[RBr][EtO⁻]. The mechanism is E2: elimination, bimolecular. It is one concerted step in which three things happen at once: the base removes the β-hydrogen, the C–H bonding pair moves in to form the π bond, and the leaving group departs with its bonding pair. Three curved arrows describe it.

Evidence for the concerted mechanism includes the kinetic isotope effect. Replacing the β-hydrogens with deuterium slows an E2 reaction by a factor of about 3 to 7, because the C–D bond, with its lower zero-point energy, is harder to break, and the effect appears only if the C–H bond is broken in the rate-determining step. The E2 reaction is also stereospecific, which is the most useful clue to its geometry.

Anti-periplanar geometry

For the C–H σ bond to become the π bond smoothly as the C–LG bond breaks, the orbitals must line up. The best arrangement is anti-periplanar: the H and the leaving group lie in the same plane, on opposite sides of the C–C bond, at a dihedral angle of 180°, as in the anti conformation of butane. In this geometry the filled σ C–H orbital overlaps directly with the back lobe of the empty σ* C–LG orbital. A syn-periplanar arrangement, at 0°, is possible but much less favourable, and other angles do not work.

The anti-periplanar requirement has visible consequences. When the β carbon is a stereocentre, the geometry of the alkene follows from the configuration of the substrate: the two diastereomers of 2-bromo-3-phenylbutane give different stereoisomers of 2-phenylbut-2-ene, each giving only one. In cyclohexanes the effect is dramatic. A ring hydrogen and a leaving group on adjacent carbons are anti-periplanar only when both are axial, trans-diaxial. If the leaving group is equatorial, no adjacent C–H is anti to it, and E2 cannot occur until the ring flips to put the leaving group axial. Menthyl chloride, whose bulky groups lock the chloride equatorial in the more stable chair, eliminates about 200 times more slowly than its diastereomer neomenthyl chloride, whose chloride is axial in the preferred chair, and the two give different alkenes.

Regioselectivity: Zaitsev and Hofmann

When a substrate can eliminate toward two different β carbons, the more substituted, more stable alkene usually predominates. This is Zaitsev's rule, from the Russian chemist Alexander Zaitsev (1875). 2-Bromobutane with sodium ethoxide gives about 80 per hundred but-2-ene (mostly trans) and 20 per hundred but-1-ene. Alkene stability increases with the number of alkyl groups on the double bond, as heats of hydrogenation show (lesson 11), and the E2 transition state, which has partial double-bond character, shares some of that stability.

A bulky base reverses the preference. Potassium tert-butoxide finds it hard to reach a hydrogen on an internal, more hindered carbon and takes one from the less hindered terminal CH₃ instead. The less substituted alkene, called the Hofmann product after August Wilhelm von Hofmann's eliminations of quaternary ammonium salts in the 1850s, then predominates: 2-bromo-2-methylbutane with tert-butoxide gives mainly 2-methylbut-1-ene, while with ethoxide it gives mainly 2-methylbut-2-ene. Poor leaving groups such as fluoride and trimethylamine also favour the Hofmann product. With choice of base, the chemist can choose the alkene.

Among Zaitsev products, the E (trans) isomer is usually favoured over Z, because the alkyl groups in the transition state leading to the cis alkene crowd each other. The reaction is therefore also stereoselective.

The E1 mechanism

Tertiary haloalkanes in polar protic solvents with weak bases, the conditions of SN1, also give alkenes. The mechanism is E1: elimination, unimolecular. It shares its slow first step with SN1, ionization to a carbocation, and the rate is first order, k[RX]. In the fast second step, a weak base, usually the solvent, removes a proton from a carbon next to the carbocation. Because the carbocation is planar and the proton is removed afterwards, E1 has no anti-periplanar requirement and is not stereospecific. It follows Zaitsev's rule, giving the more stable alkene, and because it passes through a carbocation it can be accompanied by rearrangements. E1 and SN1 always occur together; heating favours E1, because elimination produces two particles from one and has a more positive entropy change, and higher temperature magnifies the TΔS term.

Acid-catalysed dehydration of alcohols is the most important E1 reaction. Heating a secondary or tertiary alcohol with concentrated sulfuric or phosphoric acid protonates the OH, water leaves to give a carbocation, and loss of a proton gives the alkene, which is often distilled out as it forms to drive the equilibrium. Tertiary alcohols dehydrate at lower temperatures than secondary ones, and primary alcohols need harsh conditions and proceed by an E2-like path.

Other eliminations

Two eliminations in a row turn a dihalide into an alkyne. A vicinal dibromide, made by adding bromine to an alkene (lesson 11), loses two molecules of HBr when heated with excess sodium amide, first giving a vinyl bromide and then the triple bond. The sequence alkene → dibromide → alkyne is a standard way to convert one functional group into the other, and lesson 12 uses alkynes extensively. The second elimination needs the very strong base because the C–H and C–Br bonds of a vinyl bromide are both harder to break.

A third mechanism, E1cB (elimination, unimolecular, conjugate base), operates when the β-hydrogen is unusually acidic and the leaving group poor. The base first removes the proton to form a carbanion, often an enolate stabilized by a neighbouring carbonyl group, and the leaving group departs in a second step. It is the reverse order of events from E1. E1cB eliminations complete the aldol condensation in CHEM-263 and appear in several steps of metabolism.

Living cells perform eliminations constantly, usually dehydrations catalysed by enzymes that hold the substrate in the right geometry and supply acidic and basic groups at exactly the right places. In glycolysis, the enzyme enolase removes water from 2-phosphoglycerate to give phosphoenolpyruvate, and in the citric acid cycle fumarase interconverts fumarate and malate by adding and removing water across a C=C. The fumarase reaction is strictly stereospecific, an anti addition and elimination, just like an E2 in the laboratory.

On the largest scale of all, alkenes are made industrially not by these laboratory eliminations but by steam cracking, in which ethane, propane or naphtha is heated briefly to about 850 °C with steam. Radical reactions break C–C and C–H bonds and eliminate hydrogen, giving ethylene, propylene and butadiene. The process is an elimination in the broad sense, driven by the large entropy gain of making several small molecules from one, which becomes dominant at high temperature, exactly the effect that favours E1 over SN1 in a heated flask.

Substitution versus elimination

The decisive factors are the structure of the substrate and the basicity and size of the nucleophile, with temperature as a tie-breaker. A practical scheme:

1. Methyl substrates cannot eliminate (no β carbon): SN2 only.

2. Primary substrates give SN2 with good nucleophiles that are weak or moderate bases (I⁻, Br⁻, CN⁻, N₃⁻, RS⁻, RCOO⁻) and even with strong unhindered bases such as hydroxide and methoxide. Only a strong, bulky base such as tert-butoxide gives mainly E2.

3. Secondary substrates give SN2 with good nucleophiles that are weak bases, especially in polar aprotic solvents, but mainly E2 with strong bases such as hydroxide and alkoxides. With weak nucleophiles in protic solvents they react slowly by SN1 and E1.

4. Tertiary substrates give E2 with any strong base, and SN1 with E1 with weak bases and nucleophiles in protic solvents; heating favours E1.

5. Higher temperature favours elimination over substitution in all cases.

import math
RT = 2.479            # kJ/mol at 25 C

def ratio_from_ddG(ddG):
    return math.exp(ddG / RT)

def share(major_over_minor):
    return 100 * major_over_minor / (1 + major_over_minor)

# Zaitsev : Hofmann ratios as barrier differences
for label, pct in [('2-bromobutane, ethoxide', 80), ('2-bromo-2-methylbutane, tert-butoxide (Hofmann)', 72)]:
    r = pct / (100 - pct)
    print(f'{label:46s} {pct}:{100 - pct}  ->  ddG++ = {RT * math.log(r):.1f} kJ/mol')

def predict(substrate, base_strength, bulky, solvent, hot=False):
    """A rule-of-thumb predictor of the main pathway."""
    if substrate == 'methyl':
        return 'SN2'
    if substrate == 'primary':
        return 'E2' if (base_strength == 'strong' and bulky) else 'SN2'
    if substrate == 'secondary':
        if base_strength == 'strong':
            return 'E2'
        return 'SN2' if solvent == 'aprotic' else ('E1' if hot else 'SN1 (+E1)')
    if substrate == 'tertiary':
        if base_strength == 'strong':
            return 'E2'
        return 'E1' if hot else 'SN1 (+E1)'

cases = [('1-bromopropane + CN-, DMSO', 'primary', 'weak', False, 'aprotic'),
         ('2-bromopropane + EtO-, EtOH', 'secondary', 'strong', False, 'protic'),
         ('2-bromopropane + N3-, DMF', 'secondary', 'weak', False, 'aprotic'),
         ('tert-butyl bromide, EtOH, 25 C', 'tertiary', 'weak', False, 'protic'),
         ('tert-butyl bromide, EtOH, 80 C', 'tertiary', 'weak', False, 'protic', True),
         ('isobutyl bromide + tBuO-', 'primary', 'strong', True, 'protic')]
for c in cases:
    print(f'{c[0]:32s} -> {predict(*c[1:])}')

2-bromobutane, ethoxide                        80:20  ->  ddG++ = 3.4 kJ/mol
2-bromo-2-methylbutane, tert-butoxide (Hofmann) 72:28  ->  ddG++ = 2.3 kJ/mol
1-bromopropane + CN-, DMSO       -> SN2
2-bromopropane + EtO-, EtOH      -> E2
2-bromopropane + N3-, DMF        -> SN2
tert-butyl bromide, EtOH, 25 C   -> SN1 (+E1)
tert-butyl bromide, EtOH, 80 C   -> E1
isobutyl bromide + tBuO-         -> E2

Rules of thumb like these are a starting point, not a substitute for thinking about each case. Real reactions often give mixtures, and the proportions depend on temperature, solvent and concentration. But the logic, crowding at carbon, basicity and size of the reagent, stability of possible carbocations, and the entropy gained in elimination, will predict the main product correctly in the great majority of cases.