A species such as OH⁻ or CH₃CH₂O⁻ can behave in two different ways when it meets an alkyl halide. It can use its lone pair to attack the carbon bearing the leaving group, producing an SN2 substitution reaction. Or it can behave as a base and remove a hydrogen from an adjacent carbon, producing an E2 elimination reaction.
The important question is therefore not simply whether the reagent is a nucleophile or a base. Many reagents are both. Instead, we have to ask which pathway is kinetically easier for the particular combination of substrate, attacking species and reaction conditions.
Try these five questions before reading the chapter. Focus on the structure of the substrate and on what the attacking species is trying to reach.
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Why E2 and SN2 can compete
Both E2 and SN2 are concerted, bimolecular reactions. Their rate depends on collisions involving both the organic substrate and the attacking species.
If the same species can behave as both Nu⁻ and B⁻, the competition is therefore determined largely by the relative magnitudes of kSN2 and kE2. Those rate constants depend on molecular structure, steric hindrance, basicity, nucleophilicity, solvent and temperature.
The reaction does not “choose” substitution or elimination according to a memorised label. The faster pathway is the one with the more favourable activation barrier under those particular conditions.
Two different targets
SN2: the nucleophile must approach the α-carbon from the side opposite the leaving group. This close backside approach is extremely sensitive to steric hindrance.
E2: the base removes a β-hydrogen from a carbon adjacent to the α-carbon while the leaving group departs and the C=C bond forms.
This difference in target explains almost all of the substrate effect. A crowded α-carbon is disastrous for SN2, but the β-hydrogen needed for E2 may still remain relatively exposed.
Substrate effects: how crowded is the α-carbon?
The α-carbon is the carbon directly bonded to the leaving group. As more carbon groups are attached to this carbon, backside approach becomes increasingly difficult.
Methyl substrates: the extreme SN2 case
A methyl halide, CH₃X, contains no alkyl groups surrounding the reaction centre. Backside attack is therefore exceptionally accessible, making SN2 very favourable.
There is another important point: a methyl halide has no β-carbon and therefore no β-hydrogen. E2 is impossible.
CH₃X cannot undergo E2 because there is no β-carbon from which a β-hydrogen can be removed.
Primary substrates: SN2 usually has the structural advantage
In a normal primary alkyl halide, only one carbon group surrounds the α-carbon. A small nucleophile can therefore approach from the backside relatively easily.
For this reason, an unhindered primary substrate reacting with a good, relatively small nucleophile commonly undergoes SN2.
For example, 1-bromopropane reacting with CN⁻ is strongly suited to SN2 substitution. The carbon bearing Br is primary and readily accessible to nucleophilic attack.
However, “primary means SN2” is not an absolute rule. A strong bulky base can make E2 much more competitive even for a primary substrate.
Secondary substrates: the true competition zone
Secondary substrates sit between the two extremes. The α-carbon is hindered enough that SN2 has become noticeably more difficult, but it is not completely blocked.
At the same time, β-hydrogens are normally available for elimination. Consequently, relatively small changes in the attacking species or conditions can shift the major pathway.
Good nucleophile but relatively weak base: SN2 becomes more favourable.
Strong base, especially with heat: E2 becomes more favourable.
Strong bulky base: E2 is usually strongly favoured.
Tertiary substrates: SN2 is effectively blocked
At a tertiary α-carbon, three carbon groups surround the reaction centre. A nucleophile cannot approach closely enough along the required backside trajectory for a normal SN2 displacement.
A strong base can nevertheless abstract an accessible β-hydrogen. If such a hydrogen is available, E2 is therefore strongly favoured over SN2.
For a 3° alkyl halide, rule out SN2 first. Then ask whether the reagent and structure allow E2.
This statement concerns the E2-versus-SN2 competition. Under different conditions, particularly with weak nucleophiles in ionising solvents, SN1 and E1 pathways may also become relevant.
A useful primary-substrate exception: β-branching
The label “primary” describes only the α-carbon. A primary substrate can still be sterically hindered by branching on the neighbouring β-carbon. The classic extreme is a neopentyl halide.
The carbon bonded to Br is primary, but the adjacent carbon is extremely bulky. Backside approach is therefore much slower than the simple classification “1°” might suggest.
Base effects: strength and size
Once the substrate has been classified, the next question is whether the attacking species is better suited to attacking carbon or removing a proton.
Nucleophilicity and basicity are related, but they are not identical. Nucleophilicity describes the kinetic tendency to attack an electrophilic atom. Basicity describes the thermodynamic tendency to accept H⁺.
Good nucleophile, relatively weak base
Species such as I⁻ and CN⁻ can be effective nucleophiles while being much less strongly basic than alkoxide ions. When the substrate is sufficiently unhindered, this combination favours substitution rather than elimination.
I⁻: good nucleophile, weak base → commonly favours SN2.
CN⁻: strong nucleophile but considerably less basic than OH⁻ or RO⁻ → commonly favours SN2 with accessible substrates.
Small strong bases and nucleophiles
OH⁻, CH₃O⁻ and CH₃CH₂O⁻ are both strong bases and good nucleophiles. Consequently, the substrate becomes particularly important.
| Substrate | With a small strong base / nucleophile | Main reason |
|---|---|---|
| Methyl | SN2 | No β-carbon exists, so E2 is impossible. |
| Primary | Often mainly SN2 | Backside attack is relatively unhindered. |
| Secondary | SN2 / E2 competition | Backside attack is hindered and β-H abstraction is accessible. |
| Tertiary | E2 if a β-H is available | SN2 backside attack is effectively blocked. |
Bulky strong bases
A bulky base behaves differently because its own substituents create steric hindrance. Potassium tert-butoxide is the standard example.
The tert-butoxide ion is strongly basic, but its bulky methyl groups make it comparatively poor at squeezing into the crowded region around an electrophilic carbon.
An exposed β-hydrogen is often easier to reach. As a result, replacing a small nucleophile/base with a bulky strong base can shift the reaction strongly towards E2.
A bulky base does not make E2 chemically inevitable. It changes the relative rates by making close nucleophilic attack at carbon less favourable.
The essential E2 requirement: a β-hydrogen
Before predicting E2, always inspect the substrate itself. The leaving group is attached to the α-carbon. A carbon immediately adjacent to the α-carbon is a β-carbon.
During E2, the base removes a hydrogen from a β-carbon. The electrons from the C–H bond form the new π bond between Cα and Cβ while the leaving group departs.
No suitable β-hydrogen = no E2, regardless of how strong or bulky the base is.
Why anti-periplanar geometry matters
E2 is not merely a matter of having any β-hydrogen somewhere in the molecule. The C–H bond being broken and the C–X bond being broken must be aligned appropriately.
The most favourable arrangement is anti-periplanar: the β-C–H bond and the Cα–X bond lie in the same plane but point in opposite directions. This alignment allows efficient overlap as the new π bond develops.
Flexible acyclic molecules can often rotate into the required conformation. In rigid cyclic systems, however, this geometrical requirement can determine whether elimination is possible at all.
Cyclohexane: the trans-diaxial consequence
In a cyclohexane system, the anti-periplanar requirement normally means the leaving group and the β-hydrogen must both occupy axial positions and point in opposite directions. This is commonly described as a trans-diaxial arrangement.
Thus, simply spotting a β-hydrogen in a skeletal formula is not always sufficient for a rigorous E2 analysis. Its three-dimensional orientation may matter.
A reliable decision process for E2 vs SN2
Instead of memorising a large collection of disconnected rules, work through the competition in the same order every time.
Step 1: classify the α-carbon
Decide whether the carbon bearing the leaving group is methyl, 1°, 2° or 3°. This tells you how difficult backside SN2 attack will be.
Step 2: check whether E2 is structurally possible
Look at the β-carbons. Is there at least one β-hydrogen? If not, E2 can be ruled out immediately.
Step 3: classify the attacking species
Decide whether it is mainly:
- a good nucleophile but relatively weak base;
- a small strong base and good nucleophile; or
- a strong bulky base.
Step 4: combine substrate and reagent effects
| Situation | Likely outcome | Reason |
|---|---|---|
| Methyl substrate | SN2 | E2 is impossible because there is no β-carbon. |
| 1° + good small nucleophile | Mainly SN2 | α-carbon is accessible. |
| 1° + bulky strong base | E2 becomes strongly competitive and may dominate | Bulky base is poor at close carbon attack. |
| 2° + good weakly basic nucleophile | SN2 favoured | Reduced tendency to abstract β-H. |
| 2° + small strong base | E2 / SN2 competition | Both carbon attack and β-H abstraction are viable. |
| 2° + bulky strong base | Mainly E2 | Steric bulk strongly suppresses substitution. |
| 3° + strong base + β-H | E2 | SN2 is sterically blocked. |
Other factors that can tip a close competition
Substrate structure and reagent identity normally provide the strongest first prediction. When the competition is close, however, solvent and temperature can shift the balance.
Temperature
Higher temperatures commonly make elimination more competitive relative to substitution. This is why heating a secondary substrate with a strong base often increases the proportion of alkene formed.
Temperature should nevertheless be treated as a secondary factor rather than a rule that overrides molecular structure. Heating does not make E2 possible when no β-hydrogen exists.
Polar aprotic solvents
Polar aprotic solvents do not strongly hydrogen-bond to anionic nucleophiles. Anions can therefore remain relatively reactive towards electrophilic carbon, which often favours SN2 when the substrate is accessible.
Polar protic solvents
Polar protic solvents can strongly solvate anions through hydrogen bonding, reducing their nucleophilicity. This can suppress SN2 relative to what might otherwise be expected.
Treat solvent and temperature as tiebreakers. Substrate steric hindrance, base strength and base size usually provide the first and most important prediction.
Base size can also affect which alkene forms
Base size influences more than the competition between substitution and elimination. Once E2 is occurring, different β-hydrogens may lead to different alkene products.
A relatively small base can often abstract the hydrogen that leads to the more substituted alkene. This frequently gives the Zaitsev product.
A bulky base may instead remove the most accessible β-hydrogen, even when that produces the less substituted alkene. Increased formation of this product is commonly described as Hofmann selectivity.
Smaller strong base: more substituted alkene is often favoured when steric access is not a major problem.
Bulky strong base: abstraction of the less hindered β-hydrogen may become faster, increasing formation of the less substituted alkene.
E2 vs SN2 summary
| Factor | Favours SN2 | Favours E2 |
|---|---|---|
| α-carbon | Methyl or unhindered 1° | Increasing substitution, especially 3° |
| Attacking species | Good nucleophile with lower basicity | Strong base |
| Size of attacking species | Small | Bulky |
| β-hydrogen | Not required | Essential |
| Temperature | Lower temperature can help substitution in close cases | Higher temperature often increases elimination |
| Polar aprotic solvent | Often enhances anionic nucleophilicity | Not normally the main E2-driving factor |
Common Mistakes
| Common mistake | Why it is wrong | Better approach |
|---|---|---|
| Assuming every strong nucleophile must give SN2. | Many strong nucleophiles are also strong bases. With a hindered substrate they may favour E2 instead. | Consider substrate substitution and base strength together. |
| Allowing SN2 at a tertiary carbon. | The required backside trajectory is sterically blocked. | Rule out normal SN2 immediately for a 3° substrate. |
| Predicting E2 without checking for a β-hydrogen. | The π bond cannot form unless a β-C–H bond supplies the electron pair. | Identify α- and β-carbons before choosing the mechanism. |
| Saying a bulky base “always” gives E2. | Steric bulk favours E2, but structural requirements still apply. | Say that a bulky strong base strongly favours E2 when elimination is structurally possible. |
| Assuming every primary substrate reacts rapidly by SN2. | Severe β-branching can hinder backside attack even though the α-carbon itself is primary. | Inspect the local three-dimensional environment, not just the 1° label. |
| Ignoring anti-periplanar geometry. | E2 requires suitable orbital alignment as the π bond forms. | Check the orientation of β-C–H and C–X bonds, especially in cyclic systems. |
What you should be able to do
Classify the substrate
Identify methyl, primary, secondary and tertiary alkyl halides from a skeletal structure.
Identify the competing targets
Explain why SN2 attacks the α-carbon whereas E2 removes a β-hydrogen.
Judge the attacking species
Distinguish a good weakly basic nucleophile, a small strong base and a bulky strong base.
Check whether E2 is possible
Identify suitable β-hydrogens and explain the anti-periplanar requirement.
Predict E2 versus SN2
Combine substrate steric hindrance with nucleophile/base properties to predict the dominant pathway.
Handle exceptions
Recognise methyl substrates, β-branched primary substrates and cases where no β-hydrogen is available.
These ten questions test whether you can combine substrate structure, nucleophilicity, basicity, steric hindrance and β-hydrogen requirements.
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Frequently asked questions
Questions students often ask about E2 vs SN2
Once the basic substrate rules are clear, most remaining questions come down to why a reagent behaves as a nucleophile in one reaction but as a base in another.
Can OH⁻ act as both a nucleophile and a base?
Yes. OH⁻ can attack an electrophilic carbon as a nucleophile or remove a β-hydrogen as a base. With an unhindered primary substrate, SN2 is often favoured. With a secondary substrate, substitution and elimination compete. With a tertiary substrate, normal SN2 is blocked and a strong-base E2 pathway becomes much more important.
Why can a tertiary substrate undergo E2 but not SN2?
SN2 requires the nucleophile to approach very close to the α-carbon from the backside. Three carbon groups surrounding a tertiary centre prevent this approach. E2 instead requires the base to reach a β-hydrogen, which may remain accessible even though the α-carbon itself is crowded.
Does a primary alkyl halide always undergo SN2?
No. Primary substrates generally favour SN2 because their α-carbons are accessible, but a strong bulky base can make E2 much more competitive. Severe branching close to the α-carbon can also slow SN2 considerably.
Does tert-butoxide always give E2?
No. tert-Butoxide strongly favours E2 because it is both strongly basic and bulky, but the substrate must still contain a suitable β-hydrogen. If no β-hydrogen exists, E2 is impossible regardless of base strength.
Why is a secondary substrate the hardest case to predict?
A secondary α-carbon is hindered enough to slow SN2 but not hindered enough to eliminate it completely. At the same time, β-hydrogens are usually available for E2. Reagent basicity, nucleophilicity, steric size, solvent and temperature can therefore all influence the result.
Does a stronger base always mean faster E2?
Strong basicity generally increases the ability to remove a β-hydrogen, so it favours E2. However, reaction rate still depends on substrate structure, leaving-group ability, molecular geometry and reaction conditions. In particular, no base can carry out E2 if an appropriate β-hydrogen is absent.
Why does E2 need an anti-periplanar hydrogen?
The C–H bond being broken and the C–leaving-group bond being broken must have suitable orbital alignment while the new π bond forms. The anti-periplanar arrangement provides especially effective overlap and is therefore the preferred geometry for E2.
Is “primary = SN2, tertiary = E2” a good rule?
It is a useful starting pattern but not a complete rule. Primary substrates can undergo E2 with sufficiently strong bulky bases, and tertiary substrates undergo E2 only when an appropriate β-hydrogen and sufficiently basic reagent are present. Mechanism prediction works best when substrate and reagent are analysed together.
The easiest way to understand E2–SN2 competition is to stop thinking of a reagent as permanently labelled “nucleophile” or “base”. Instead, look at the two possible targets. If the α-carbon is easy to reach, substitution becomes attractive. If carbon attack is crowded but a β-hydrogen is accessible to a strong base, elimination becomes attractive. Once you combine substrate structure with base strength and size, most E2-versus-SN2 predictions become systematic rather than memorised.