Knowing that a halogenoalkane can undergo elimination is only the beginning. At a deeper level, we want to know how the C–H and C–X bonds are broken, whether an intermediate exists, and why one alkene may be produced more readily than another.
This is where kinetics becomes especially useful. A proposed mechanism is not merely a sequence of curly arrows: it must also be consistent with the experimentally observed rate law. When kinetic evidence is combined with an energy profile, substrate structure, stereochemistry and product distribution, we can build a much stronger picture of what is occurring at the molecular level.
Try these five questions before reading the chapter. Focus on the relationship between the rate law, the reaction pathway and the structure of the products formed.
- 5 diagnostic questions
- Energy profiles and structures where useful
- Instant explanation for every answer option
Using kinetics to investigate an elimination mechanism
A mechanism describes a sequence of elementary molecular events. Because each elementary step involves particular reacting species, the proposed mechanism should be consistent with the observed rate law.
This makes kinetics one of the most useful experimental tools for distinguishing E1 from E2. Both mechanisms can transform a halogenoalkane into an alkene, but they predict different concentration dependences.
A rate law is experimental evidence for a mechanism. It should agree with the species involved in the rate-determining kinetic event, but a rate law by itself does not prove every detail of a proposed mechanism.
Reaction order and molecularity are not the same idea
Reaction order comes from the experimentally determined rate law. Molecularity, on the other hand, describes how many reacting particles participate in a particular elementary step.
For example, the slow ionisation step of an E1 reaction contains only one substrate molecule. This elementary event is therefore unimolecular, and the experimentally observed reaction is normally first-order in the substrate.
In E2, the base and substrate participate together in the concerted elimination step. The reaction is therefore bimolecular at the elementary level and normally first-order in each reactant.
| Feature | E1 | E2 |
|---|---|---|
| Number of elementary steps | Two | One |
| Intermediate | Carbocation | None |
| Species in rate-determining kinetic event | Substrate | Substrate + base |
| Typical rate law | rate = k[RX] | rate = k[RX][B⁻] |
| Overall order | First-order | Second-order |
The E1 mechanism: elimination through a carbocation
E1 stands for elimination unimolecular. Its defining feature is that the leaving group departs before the β-hydrogen is removed. The reaction therefore passes through a carbocation intermediate.
Step 1: ionisation of the substrate
The C–X bond undergoes heterolytic fission. The leaving group takes the bonding electron pair and a carbocation is formed.
Breaking the C–X bond and creating separated ions is energetically demanding. In the usual E1 mechanism this ionisation is the slow, rate-determining step.
Step 2: removal of a β-hydrogen
A base removes a proton from a carbon adjacent to the positively charged carbon. The electrons from the β-C–H bond then form the π bond of the alkene.
Because the carbocation already exists before deprotonation occurs, the base is not involved in the slow ionisation step.
E1 rate law: rate = k[RX]
The rate is normally first-order in the halogenoalkane and independent of base concentration within the E1 kinetic regime.
It is important not to overinterpret this statement. The base still participates chemically in the reaction. It simply does not appear in the rate law because it acts after the rate-determining step. Changing the base or reaction medium may still alter competing reactions or the overall product distribution.
The E1 energy profile
Because E1 contains two elementary steps, its energy profile contains two transition states separated by a local energy minimum corresponding to the carbocation intermediate.
The first maximum is TS1‡, where C–X bond breaking is well advanced. After crossing this barrier, the system reaches the carbocation intermediate.
The second maximum is TS2‡, associated with removal of the β-hydrogen and formation of the C=C bond. In the usual E1 case, the first activation barrier is larger because carbocation formation is the rate-determining step.
Carbocation stability matters in E1
E1 is favoured when the substrate can form a sufficiently stable carbocation. Tertiary carbocations are generally more stable than secondary carbocations because alkyl groups stabilise positive charge through inductive effects and hyperconjugation.
This is why tertiary halogenoalkanes are much more likely than ordinary primary halogenoalkanes to undergo an E1 pathway.
E1 can involve carbocation rearrangement
Once a carbocation has formed, a hydride shift or alkyl shift may sometimes generate a more stable carbocation before elimination occurs. The final alkene may therefore have a carbon skeleton that appears rearranged relative to the original substrate.
A rearranged product is particularly useful mechanistic evidence because a conventional E2 reaction has no free carbocation intermediate available to undergo such a rearrangement.
The E2 mechanism: a concerted elimination
E2 stands for elimination bimolecular. Unlike E1, there is no carbocation intermediate. Bond breaking and bond making occur together in one concerted elementary step.
As the base removes a β-hydrogen, the electrons from the C–H bond begin to form the new C=C π bond while the C–X bond simultaneously breaks and the leaving group departs.
E2 does not mean that every bond changes to exactly the same extent at one instant. It means there is only one elementary step and one transition state, with no discrete intermediate between reactant and product.
The E2 rate law
Both the base and substrate participate in the transition state of the elementary E2 step. Consequently, a simple E2 reaction normally follows:
The reaction is first-order in RX and first-order in B⁻, giving an overall reaction order of two.
Therefore, if [RX] is doubled while [B⁻] remains constant, the rate doubles. If both concentrations are doubled, the rate increases by a factor of four.
The E2 energy profile
A concerted mechanism contains only one elementary step, so the energy profile has one maximum corresponding to a single transition state.
There is no valley between reactants and products because no intermediate exists. The top of the barrier represents the E2 transition state, in which the C–H bond and C–X bond are being broken while the C=C π bond is forming.
Why E2 requires the correct geometry
E2 is not controlled only by which β-hydrogen is easiest for the base to reach. The reacting bonds must also have the correct spatial relationship.
The most favourable arrangement places the β-C–H bond and the C–X bond anti-periplanar: the two bonds lie approximately in the same plane but point in opposite directions, corresponding to a dihedral angle close to 180°.
Anti-periplanar requirement: efficient E2 elimination requires favourable alignment between the breaking C–H bond, the developing p orbitals and the departing C–X bond.
Why anti-periplanar is favourable
As the C–H σ bond breaks, its electron pair must develop into the new π bond. At the same time, electron density associated with the C–X bond is displaced onto the leaving group.
The anti-periplanar arrangement gives the required orbital alignment while also reducing eclipsing interactions compared with the corresponding syn-periplanar arrangement.
E2 can therefore be stereospecific
Because only particular conformations have the required geometry, different stereoisomers of the same substrate can produce different alkene stereoisomers.
This is an important distinction from simply saying that elimination “removes any β-hydrogen”. A β-hydrogen may exist but still be poorly aligned for E2 reaction in the relevant conformation.
Cyclohexanes: the trans-diaxial requirement
In a cyclohexane chair, the anti-periplanar requirement is normally satisfied when the leaving group and β-hydrogen are both axial and point in opposite directions.
If the leaving group is equatorial, a ring flip may first be required to place it axial. The availability of an axial β-hydrogen can therefore determine whether a particular elimination pathway is possible.
Regioselectivity: which β-hydrogen is removed?
A halogenoalkane may contain β-hydrogens on more than one carbon. Removing hydrogen from different β-carbons places the new C=C bond in different positions, producing constitutional isomers.
The reaction is therefore regioselective if one possible alkene is formed preferentially over another.
Zaitsev’s rule
With many ordinary substrates and relatively unhindered bases, the more substituted alkene is the major elimination product. This is known as the Zaitsev or Saytzeff product.
More substituted alkenes are generally lower in energy because alkyl substitution stabilises the π system through effects including hyperconjugation.
Zaitsev tendency: when other factors are comparable, elimination commonly favours formation of the more substituted alkene.
For 2-bromobutane, abstraction of hydrogen from the appropriate β-carbon can produce but-2-ene rather than but-1-ene. But-2-ene is the more substituted alkene and is normally favoured under typical Zaitsev conditions.
Do not turn Zaitsev’s rule into an absolute rule
Product stability is important, but elimination selectivity is determined by the activation barriers of competing pathways. Steric effects, stereochemical accessibility, leaving-group behaviour and the structure of the base can all alter which β-hydrogen is removed most rapidly.
It is therefore better to regard Zaitsev's rule as a strong general trend rather than an instruction that overrides every mechanistic factor.
The Hofmann product
The Hofmann product is the less substituted alkene. It can become favoured when steric effects make the hydrogen leading to the Zaitsev alkene relatively difficult for the base to abstract.
A classic example involves a large, sterically hindered base such as tert-butoxide, (CH₃)₃CO⁻. Although it is strongly basic, its size can make approach to a crowded β-hydrogen less favourable.
The base may instead abstract a more exposed β-hydrogen, even though this leads to the less substituted alkene.
Zaitsev product: more substituted alkene; commonly favoured with relatively small, unhindered bases.
Hofmann product: less substituted alkene; its proportion is often increased by bulky bases or by other structural factors that favour removal of the less hindered β-hydrogen.
A useful example with a bulky base
This does not mean a bulky base always gives 100% Hofmann product. It means steric effects can change the relative activation energies of the competing elimination pathways sufficiently for the less substituted product to become more important or even predominant.
Regioselectivity in E1 is influenced by the carbocation
E1 regioselectivity must be considered after the carbocation has formed. Removal of different β-hydrogens can again produce different alkenes, and formation of a more substituted alkene is frequently favoured.
However, E1 has an extra complication that E2 does not: the carbocation itself may rearrange before elimination.
In E1, always ask whether a carbocation rearrangement is possible before deciding which alkene products can form.
A hydride shift or alkyl shift may create a more stable carbocation at a different position. Subsequent β-deprotonation can then produce an alkene that could not have arisen from simple elimination at the original carbocation centre.
E2 versus SN2: why the same reagent can give different reactions
A negatively charged reagent may act as both a nucleophile and a base. Consequently, substitution and elimination frequently compete.
For a nucleophile to undergo SN2, it must attack the carbon bearing the leaving group from the backside. For E2, the base instead approaches a β-hydrogen.
Substrate structure, reagent structure, solvent and temperature therefore determine which pathway is faster.
Effect of substrate structure
| Substrate | SN2 tendency | E2 tendency with a strong base |
|---|---|---|
| Methyl | Very favourable | Impossible because there is no β-carbon or β-hydrogen |
| Primary | Usually favourable | Possible, especially with a strong bulky base |
| Secondary | Possible | Possible; competition is often important |
| Tertiary | Strongly hindered and effectively unavailable | Often very favourable with a strong base |
Do not confuse steric hindrance in SN2 with steric hindrance in E2. A tertiary substrate is too crowded for normal SN2 backside attack but can undergo E2 very readily because the base attacks a β-hydrogen rather than the tertiary carbon itself.
Effect of the base or nucleophile
A strong, sterically hindered base is often particularly effective at promoting E2 because proton abstraction is favourable while nucleophilic attack at carbon is sterically difficult.
A good nucleophile that is relatively weakly basic is more likely to favour substitution, especially with an unhindered primary substrate.
Bulky strong base: tends to increase E2 relative to SN2 and may also increase Hofmann regioselectivity.
Good nucleophile with weaker basicity: tends to increase substitution relative to elimination when the substrate permits SN2 attack.
Effect of temperature
Increasing temperature often increases the proportion of elimination products relative to substitution products. Elimination commonly has a more favourable entropy contribution because the products possess greater dispersal of matter and energy.
As temperature increases, the −TΔS contribution becomes increasingly important. This is one reason heating commonly favours elimination.
However, “heat means elimination” is not an absolute rule. The relative activation barriers of the competing pathways, together with substrate structure, base strength, nucleophilicity and solvent, remain important.
Putting E1 and E2 side by side
| Feature | E1 | E2 |
|---|---|---|
| Meaning | Elimination unimolecular | Elimination bimolecular |
| Mechanistic steps | Two | One |
| Carbocation intermediate | Yes | No |
| Typical rate law | rate = k[RX] | rate = k[RX][B⁻] |
| Energy profile | Two maxima and one intermediate minimum | One maximum and no intermediate minimum |
| Strong base required? | No | Usually favoured by a strong base |
| Carbocation rearrangement | Possible | Not part of the E2 mechanism |
| Anti-periplanar requirement | Not the defining requirement | Strong stereoelectronic requirement |
| Substrate commonly favoured | Substrates capable of forming stable carbocations | Secondary or tertiary substrates with strong base; primary substrates can react with sufficiently strong or bulky bases |
How to identify an elimination mechanism from experimental evidence
An examination question may not simply tell you whether a reaction is E1 or E2. Instead, you may be given kinetic data, product structures or stereochemical evidence and asked to infer the mechanism.
Evidence for E1
- The rate is proportional to [RX] but not to base concentration.
- The energy profile contains two transition states and one intermediate.
- The substrate can form a relatively stable carbocation.
- Rearranged products may be observed.
- SN1 products may compete because both pathways can arise from the same carbocation.
Evidence for E2
- The rate depends on both [RX] and [base].
- The energy profile contains one transition state and no intermediate.
- A strong base promotes reaction.
- No carbocation rearrangement occurs.
- Product stereochemistry reflects an anti-periplanar requirement.
Suppose doubling [RX] doubles the rate, while independently doubling [B⁻] also doubles the rate. The experimental rate law is therefore rate = k[RX][B⁻]. This strongly supports a kinetic step containing both species and is consistent with E2 rather than E1.
Why one piece of evidence is rarely enough
Chemists prefer a mechanism that explains several observations simultaneously. A rate law may indicate how many reactants participate in the kinetically important event, while stereochemistry may reveal the geometry of the transition state and rearrangement products may reveal the existence of a carbocation.
The strongest mechanistic argument is therefore a consistent picture in which kinetics, energetics and molecular structure all support the same pathway.
Common Mistakes
| Common mistake | Why it is wrong | Better approach |
|---|---|---|
| Assuming a tertiary halogenoalkane cannot undergo E2 because it is sterically hindered | Steric hindrance prevents SN2 backside attack at the α-carbon, but E2 involves abstraction of a β-hydrogen. | Remember that tertiary substrates often undergo E2 very readily in the presence of a strong base. |
| Saying the base is not involved in an E1 reaction | The base does participate in deprotonation; it simply does not participate in the rate-determining ionisation step. | Say that base concentration does not normally appear in the E1 rate law. |
| Assuming Zaitsev's rule always determines the major product | Bulky bases, stereochemical constraints and substrate structure can alter regioselectivity. | Use Zaitsev's rule as a trend, then check steric and stereoelectronic factors. |
| Drawing an intermediate at the maximum of an energy profile | An energy maximum corresponds to a transition state, not an intermediate. | Place intermediates at local minima between transition states. |
| Drawing two peaks for E2 | E2 is concerted and contains only one elementary step. | Draw one TS‡ maximum and no intermediate. |
| Ignoring the anti-periplanar requirement in E2 | A β-hydrogen may exist but may not have the geometry required for efficient elimination. | Check the relative orientation of the β-C–H and C–X bonds before predicting the product. |
What you should be able to do
Distinguish E1 from E2 kinetically
Use experimentally determined rate laws to identify whether the base participates in the kinetically important step.
Interpret energy profiles
Recognise transition states, intermediates and the number of elementary steps from an energy profile.
Explain E1
Describe carbocation formation, β-deprotonation and the possibility of carbocation rearrangement.
Explain E2
Describe concerted C–H bond breaking, C=C formation and C–X bond breaking through a single transition state.
Apply the anti-periplanar requirement
Use molecular geometry to determine whether a particular β-hydrogen can participate effectively in E2.
Predict regioselectivity
Compare Zaitsev and Hofmann products and explain the effect of base steric bulk.
Analyse E2 versus SN2 competition
Consider substrate substitution, base strength, nucleophilicity, steric hindrance and temperature.
Use multiple forms of mechanistic evidence
Combine kinetics, product structure, stereochemistry and energy profiles rather than relying on a single observation.
These ten questions test whether you can connect kinetics, energy profiles, stereochemistry, regioselectivity and reaction competition to a coherent mechanism.
- 10 mastery questions
- Structures and energy profiles where useful
- Instant explanation for every answer option
Frequently asked questions
Questions students often ask about E1 and E2 mechanisms
Once the two mechanisms are placed side by side, most remaining difficulties come from deciding how kinetics, structure and stereochemistry fit together.
Why does the base not appear in the E1 rate law?
The base participates after the slow ionisation step. The rate-determining step involves only the halogenoalkane breaking its C–X bond to form a carbocation, so the usual rate law is rate = k[RX]. This does not mean that the base is chemically unnecessary; it removes the β-proton in the subsequent step.
Why does E2 give only one peak on an energy profile?
E2 is concerted. β-C–H bond breaking, C=C bond formation and C–X bond breaking occur within the same elementary step. There is therefore one transition state and no intermediate, producing a single maximum on the energy profile.
Does E2 always give the Zaitsev product?
No. The more substituted Zaitsev alkene is commonly favoured with small bases, but bulky bases can preferentially remove a more accessible β-hydrogen and increase formation of the less substituted Hofmann alkene. Anti-periplanar geometry can also determine which β-hydrogen is actually able to react.
Why can a tertiary halogenoalkane undergo E2 but not SN2?
SN2 requires a nucleophile to approach the carbon bearing the leaving group from the backside, which is severely hindered in a tertiary substrate. In E2, however, the base attacks a β-hydrogen instead. Tertiary halogenoalkanes can therefore undergo E2 very readily in the presence of a strong base.
Can the rate law alone prove that a reaction follows E2?
A rate law such as rate = k[RX][B⁻] provides strong support for a pathway in which RX and the base participate in the kinetically important event, and it is consistent with E2. However, a complete mechanistic argument is stronger when kinetic evidence is supported by stereochemistry, product distribution, isotope effects or other structural evidence.
E1 and E2 become much easier to distinguish once you stop viewing them as two reaction names and instead connect their molecular events to observable evidence. E1 leaves the signature of a carbocation intermediate and first-order kinetics, whereas E2 leaves the signature of a concerted, bimolecular and stereoelectronically controlled transition state. From there, Zaitsev versus Hofmann selectivity and E2 versus SN2 competition become consequences of the same underlying mechanistic picture.