Imagine that you are given only the overall reaction RX + Nu⁻ → RNu + X⁻. The equation tells you which bonds disappear and which bonds form, but it says nothing about what happens between reactants and products. Does Nu⁻ attack while X⁻ is leaving, or does X⁻ leave first to generate a carbocation?
This is where physical organic chemistry becomes powerful. Instead of trying to watch individual molecules react, we use measurable evidence — especially the rate law, the shape of the energy profile and the stereochemistry of the products — to reconstruct the molecular events occurring during the reaction.
Try these five questions before reading the chapter. Focus on what the experimental evidence tells you about the molecular mechanism rather than simply memorising the names SN1 and SN2.
- 5 diagnostic questions
- Kinetic and energy-profile reasoning
- Instant explanation for every answer option
1. What kinetics tells us about a reaction mechanism
A reaction mechanism is a sequence of elementary steps. The overall chemical equation is obtained by adding those steps together, but the overall equation does not normally reveal the rate law. The rate law must be determined experimentally and then interpreted in terms of a proposed mechanism.
For nucleophilic substitution, we can write the overall change generally as:
There are two fundamentally different ways in which this overall change can occur. The nucleophile and leaving group can participate simultaneously in one elementary event, or the C–X bond can break first to form an intermediate that is attacked afterwards.
The rate equation is experimental evidence about the species involved in the rate-controlling part of the mechanism.
Reaction order is read from the rate equation
If an experimentally determined rate law is Rate = k[A]m[B]n, then the reaction is order m with respect to A, order n with respect to B, and order m + n overall.
| Experimental change | Observed rate change | Deduction |
|---|---|---|
| [A] doubled | Rate unchanged | Zero order in A |
| [A] doubled | Rate doubles | First order in A |
| [A] doubled | Rate quadruples | Second order in A |
The rate constant also contains useful information
The rate constant, k, is the proportionality constant connecting the reaction rate to the concentration terms in the rate law. At a particular temperature, a larger k corresponds to a faster reaction when the relevant concentrations are the same.
The units of k depend on the overall order of the reaction.
| Overall order | Example rate law | Units of k |
|---|---|---|
| First order | Rate = k[RX] | s⁻¹ |
| Second order | Rate = k[RX][Nu⁻] | dm³ mol⁻¹ s⁻¹ |
2. SN2 kinetics: one concerted molecular event
The notation SN2 means substitution, nucleophilic, bimolecular. The word “bimolecular” refers to the fact that two reacting species — the substrate and the nucleophile — are involved in the same elementary substitution step.
There is no carbocation intermediate. As the nucleophile begins to form a bond to carbon, the carbon–leaving-group bond is breaking at the same time.
The SN2 rate law
The reaction is first order with respect to RX and first order with respect to Nu⁻, giving an overall order of two. Doubling either concentration while keeping the other constant doubles the rate.
If [RX] is doubled and [Nu⁻] is tripled, an ideal SN2 rate becomes 2 × 3 = 6 times larger, provided temperature and other relevant conditions remain unchanged.
Why the rate law makes molecular sense
For an SN2 event to occur, a nucleophile and a substrate molecule must come together in a suitable orientation. Increasing the concentration of either reactant increases the frequency with which the required reacting pair can form.
One SN2 transition state, no intermediate, and a second-order rate law all describe the same concerted molecular picture.
3. SN1 kinetics: ionisation before nucleophilic attack
The notation SN1 means substitution, nucleophilic, unimolecular. In the simple SN1 mechanism, the leaving group departs first and a carbocation is formed. The nucleophile attacks only afterwards.
The first step requires heterolytic cleavage of the carbon–leaving-group bond. When this ionisation is the rate-determining step, only the concentration of the substrate appears in the simple rate law.
The SN1 rate law
The reaction is first order overall. Increasing [RX] increases the rate, but changing nucleophile concentration does not alter the rate predicted by this simple rate equation.
SN2: the nucleophile participates in the substitution transition state, so [Nu⁻] appears in the rate equation.
SN1: the rate-determining ionisation occurs before nucleophilic attack, so the nucleophile does not appear in the simple rate equation.
Do not confuse molecularity with overall reaction order
For these simple mechanisms, the names happen to agree with the commonly observed overall orders: SN1 commonly gives a first-order rate equation and SN2 a second-order rate equation.
However, the “1” and “2” refer to the molecularity of the kinetically important substitution process rather than being obtained simply by looking at the overall chemical equation.
4. Energy profiles reveal transition states and intermediates
A reaction-energy profile plots potential energy against reaction progress. A maximum corresponds to a transition state, while a local minimum between two maxima corresponds to an intermediate.
The SN2 energy profile
SN2 is concerted, so there is one elementary step and therefore one transition state. The nucleophile–carbon bond is partly formed while the carbon–leaving-group bond is partly broken.
The following diagram shows a representative exothermic SN2 profile. The precise relative energies of reactants and products depend on the particular reaction, but the mechanistically important feature is the single energy maximum.
One elementary step → one transition state → one energy maximum.
The SN1 energy profile
In the simple SN1 mechanism there are two elementary substitution steps. The first transition state leads to the carbocation intermediate; a second transition state is crossed when the nucleophile reacts with that intermediate.
The representative profile below is also drawn overall exothermic. Notice that the first activation barrier is considerably larger than the second, consistent with slow ionisation followed by relatively rapid nucleophilic attack.
Two elementary steps → two transition states → two energy maxima, with the carbocation intermediate in the minimum between them.
Compare activation barriers, not simply peak heights
For a multistep reaction, the activation energy for each elementary step is measured from the energy of the species immediately preceding that transition state.
For the first SN1 step, the relevant barrier is from the reactants to TS1‡. For the second step, the relevant barrier is from the carbocation intermediate to TS2‡.
Do not compare only the absolute vertical positions of TS1‡ and TS2‡. Compare each transition state with the species immediately before it.
5. The Hammond postulate connects transition states to stable species
A transition state cannot normally be isolated, yet we often need to reason about its structure. The Hammond postulate provides a link between the transition state and the stable species close to it in energy.
The structure of a transition state resembles the stable species — reactant, intermediate or product — that is closest to it in energy.
Exothermic elementary step: the transition state is relatively early and tends to be more reactant-like.
Endothermic elementary step: the transition state is relatively late and tends to be more product-like.
Applying Hammond to SN1 ionisation
The first SN1 step forms a relatively high-energy carbocation and is normally endothermic. Its transition state therefore has substantial carbocation-like character: the C–LG bond is significantly weakened and positive charge is developing at carbon.
Anything that stabilises the carbocation therefore tends also to stabilise this carbocation-like transition state. The activation barrier for ionisation falls, so the SN1 reaction becomes faster.
Carbocation stability does more than describe the intermediate after it has formed. Through Hammond reasoning, it also helps explain why the activation barrier for forming that carbocation changes.
6. The steady-state approximation in SN1
A carbocation intermediate is formed in one elementary step and consumed in a later one. Because the carbocation is highly reactive, its concentration may remain very small throughout most of the reaction.
After a short initial period, the rate at which the carbocation is formed can become approximately equal to the rate at which it is consumed. Consequently, its concentration changes only slowly with time.
Steady-state approximation: for a reactive intermediate such as R⁺, its rate of formation is taken to be approximately equal to its rate of consumption, so its concentration remains small and approximately constant during much of the reaction.
The important H3 idea is conceptual. You should be able to explain why a reactive intermediate does not accumulate to a large concentration and why its concentration can be treated as approximately constant.
Steady state does not mean equilibrium
At equilibrium, forward and reverse processes occur at equal rates. In a steady-state treatment, an intermediate can be continuously formed and consumed while the overall reaction continues in the forward direction.
Steady state: the concentration of an intermediate is approximately constant because its rates of formation and consumption are approximately equal.
Equilibrium: opposing reactions occur at equal rates and there is no net macroscopic change in composition.
Where the first-order SN1 rate law comes from
In the simple SN1 mechanism, unimolecular ionisation of RX is the slow step. Its frequency depends on the concentration of RX:
The steady-state approximation helps us describe the low concentration of the reactive carbocation intermediate. It should not be memorised as though it were itself the reason for the simple first-order rate law.
7. Stereochemistry provides another fingerprint of mechanism
Rate measurements tell us which concentrations affect reaction speed. Stereochemistry gives another type of evidence: it reveals the three-dimensional geometry involved as bonds form and break.
SN2: backside attack and inversion
In an SN2 reaction, the nucleophile approaches the reacting carbon from the side opposite the leaving group. This is called backside attack.
As the nucleophile forms its bond and the leaving group departs, the three other groups attached to the reacting carbon move through to the opposite arrangement. The tetrahedral centre therefore undergoes inversion of configuration, often called Walden inversion.
SN2 is stereospecific: backside attack produces inversion at the reacting tetrahedral carbon.
Inversion does not automatically mean R becomes S
“Inversion” describes the geometrical change at the reacting carbon. The labels R and S are assigned separately using Cahn–Ingold–Prelog priorities.
Replacing the leaving group can change those priorities. Therefore, an inverted product does not necessarily follow the simple written rule “R becomes S” or “S becomes R”. If an R/S descriptor is required, assign the product priorities again.
SN1: attack on an approximately planar carbocation
In a simple carbocation, the positively charged carbon is approximately trigonal planar. Once the leaving group has departed, the nucleophile can in principle approach from either face of this planar centre.
If attack from the two faces gives enantiomeric products, attack from both sides produces a tendency towards racemisation.
Why SN1 often gives incomplete racemisation
The simplest model assumes a completely free carbocation, but the leaving group does not necessarily move far away immediately after C–LG bond cleavage.
The positively charged carbon and leaving-group ion can remain close together for a short time as an ion pair. The leaving-group ion can partially block the face from which it departed, making attack from the opposite face more likely.
SN1 tends towards racemisation, but ion-pair interactions can result in more inversion product than retention product.
Therefore, saying “SN1 always gives an exact 50:50 racemic mixture” is too absolute.
8. Substituent effects push SN1 and SN2 in opposite directions
Changing the groups attached to the reacting carbon affects both electronic stability and steric accessibility. These effects help explain the familiar differences between methyl, primary, secondary and tertiary substrates.
Methyl is a separate category from primary. In CH₃Br, the carbon bonded to Br is attached to zero other carbon atoms. In a primary haloalkane such as CH₃CH₂Br, the carbon bonded to Br is attached to one other carbon atom.
SN1: carbocation stability is important
The first SN1 step develops positive charge at the reacting carbon. Alkyl groups help stabilise this electron-deficient centre through electron donation and hyperconjugative effects.
The methyl carbocation, CH₃⁺, is especially unstable because there are no alkyl groups attached to the positively charged carbon. A primary carbocation has one alkyl group, a secondary carbocation has two, and a tertiary carbocation has three.
The same broad trend is therefore expected in SN1 reactivity for comparable simple alkyl substrates.
More stable carbocation → lower barrier to ionisation → faster SN1 reaction.
SN2: steric hindrance is important
SN2 requires the nucleophile to approach the reacting carbon from behind the C–LG bond. Alkyl groups surrounding this carbon obstruct that approach.
Bromomethane is particularly favourable for SN2 because the carbon bearing Br is surrounded only by hydrogen atoms. There are no alkyl groups around that carbon to hinder the approaching nucleophile.
A primary substrate contains one alkyl group next to the reacting carbon and is therefore slightly more hindered than a methyl substrate. Secondary and tertiary substrates become progressively more crowded.
A tertiary substrate is therefore extremely unfavourable for ordinary SN2 substitution at the tertiary carbon.
Branching close to the reacting carbon also matters
Primary or secondary classification is useful, but it does not describe the entire three-dimensional environment. Branching close to the reacting carbon can make backside approach more difficult even when the carbon bearing the leaving group is formally primary.
Resonance-stabilised carbocations are important exceptions
The simple tertiary > secondary > primary trend applies to ordinary alkyl carbocations. If positive charge can be delocalised by resonance, such as in benzylic or allylic systems, carbocation formation can be much more favourable than the simple degree of substitution alone suggests.
Do not rank an unfamiliar SN1 substrate using only the words primary, secondary and tertiary. Check for resonance stabilisation too.
9. Nucleophile and leaving-group effects
Substrate structure is only part of the story. Nucleophile properties and leaving-group ability can also change the relative rates of substitution.
Nucleophile strength strongly affects SN2
The nucleophile is directly involved in the SN2 transition state. A more reactive nucleophile can form the new bond to carbon more readily.
Its concentration also appears directly in the rate law:
Increasing [Nu⁻] therefore increases the rate of the SN2 pathway.
Nucleophile concentration is absent from the simple SN1 rate law
In SN1, nucleophilic attack occurs after the slow ionisation step. Changing nucleophile concentration therefore does not alter the rate of that preceding unimolecular step.
A better leaving group makes C–LG bond cleavage easier
In both mechanisms, the carbon–leaving-group bond must ultimately break. A good leaving group is one that can accommodate the electron pair after departure relatively well.
For comparable alkyl halides, a commonly useful trend is:
Iodoalkanes therefore generally undergo nucleophilic substitution more readily than corresponding bromoalkanes or chloroalkanes under otherwise comparable conditions.
10. Competition between SN1 and SN2
A substrate does not carry an intrinsic label saying that it must react by SN1 or SN2. The observed mechanism reflects competition between alternative pathways and their activation barriers under the reaction conditions.
| Factor | Favours SN1 | Favours SN2 |
|---|---|---|
| Substrate | Ability to form a stable carbocation; tertiary and suitable resonance-stabilised substrates are favourable | Low steric hindrance; methyl and primary substrates are especially favourable |
| Nucleophile | Does not appear in the simple rate-determining ionisation step | A stronger and/or more concentrated nucleophile increases the bimolecular pathway |
| Leaving group | Better leaving group facilitates ionisation | Better leaving group also facilitates bond breaking in the concerted transition state |
| Stereochemistry | Racemisation may occur, often with some excess inversion because of ion-pair effects | Backside attack gives inversion |
| Kinetic evidence | Rate = k[RX] | Rate = k[RX][Nu⁻] |
Methyl substrates
Methyl halides such as CH₃Br are a special case. The reacting carbon is attached to no other carbon atoms, so steric hindrance is minimal. This makes methyl substrates particularly favourable for SN2.
In contrast, an SN1 pathway would require formation of CH₃⁺, an exceptionally unstable carbocation. SN1 is therefore strongly unfavourable for ordinary methyl substrates.
Primary substrates
A normal primary carbocation is very unstable, so direct SN1 ionisation is usually unfavourable. At the same time, the reacting carbon is relatively accessible to backside attack. Primary substrates therefore tend to favour SN2.
Tertiary substrates
A tertiary substrate represents the opposite situation. The corresponding carbocation is much more stable than a primary carbocation, while the reaction centre is highly crowded. SN1 is therefore favoured over SN2 in this comparison.
Secondary substrates require genuine comparison
Secondary substrates lie between the two extremes. Carbocation formation is more feasible than for primary substrates, yet backside attack is still possible. The nucleophile and leaving group can therefore have an important influence on which mechanism contributes more strongly.
For a secondary substrate, do not guess from the word “secondary”. Compare the competing kinetic pathways.
Use several pieces of evidence together
A reaction is first order in RX, independent of nucleophile concentration, and gives substantial racemisation at a stereogenic reacting carbon. These observations consistently support an SN1 mechanism.
A reaction is first order in RX and first order in Nu⁻ and produces stereospecific inversion at the reacting carbon. These observations consistently support an SN2 mechanism.
11. SN1 and SN2 at a glance
| Feature | SN1 | SN2 |
|---|---|---|
| Main substitution steps | Two | One |
| Intermediate | Carbocation | None |
| Transition states | Two | One |
| Typical rate law | Rate = k[RX] | Rate = k[RX][Nu⁻] |
| Overall order | First | Second |
| Typical units of k | s⁻¹ | dm³ mol⁻¹ s⁻¹ |
| Effect of [Nu⁻] on simple rate law | No effect | Rate increases with [Nu⁻] |
| Stereochemistry | Racemisation, possibly with excess inversion because of ion-pair interactions | Backside attack and inversion |
| Substrate preference | Substrates able to form relatively stable carbocations | Methyl and other sterically unhindered substrates |
| Key physical idea | Carbocation stability | Steric accessibility |
Common Mistakes
| Common mistake | Why it is wrong | Better approach |
|---|---|---|
| Treating CH₃Br as a primary haloalkane | The carbon bonded to Br in CH₃Br is attached to zero other carbon atoms. A primary carbon must be attached to one other carbon atom. | Treat methyl as a separate category from primary: methyl > primary > secondary ≫ tertiary for typical SN2 reactivity. |
| Saying SN1 always gives exactly 50:50 products | Ion-pair interactions can partly shield one face of the carbocation. | Say that a planar carbocation tends towards racemisation, but incomplete racemisation may occur. |
| Saying SN2 inversion automatically changes R into S | R/S descriptors depend on the CIP priority order in the product. | Recognise geometric inversion first, then assign the product R/S configuration separately if required. |
| Saying the nucleophile is unimportant in SN1 | The nucleophile is essential for product formation; it simply reacts after the rate-determining ionisation step. | Connect the rate equation directly to the slow elementary step. |
| Treating steady state as equilibrium | A steady-state intermediate can be continually formed and consumed while the overall reaction proceeds. | State that its concentration is approximately constant because its rates of formation and consumption are similar. |
| Saying SN2 reactivity is simply 1° > 2° > 3° | Methyl substrates are even less hindered than primary substrates. | Use methyl > primary > secondary ≫ tertiary as the useful basic steric trend. |
| Ranking SN1 substrates only by primary, secondary and tertiary | Resonance stabilisation can significantly stabilise benzylic or allylic carbocations. | Check both alkyl substitution and possible resonance stabilisation. |
| Looking only at the absolute height of peaks in a multistep profile | Each activation barrier begins from the species immediately before that transition state. | Compare each local minimum with the following transition-state maximum. |
What you should be able to do
Interpret rate laws
Deduce individual and overall reaction orders and predict how changing concentration affects rate.
Compare SN1 and SN2 kinetics
Explain why SN1 commonly follows Rate = k[RX] whereas SN2 follows Rate = k[RX][Nu⁻].
Read energy profiles
Identify transition states, intermediates and activation barriers from one-step and multistep profiles.
Apply the Hammond postulate
Relate the structure of a transition state to the stable species closest to it in energy.
Explain steady state
Explain why a reactive carbocation intermediate may remain at a low, approximately constant concentration.
Predict stereochemistry
Explain inversion in SN2 and racemisation with possible excess inversion in SN1.
Apply substituent effects
Distinguish methyl, primary, secondary and tertiary substrates and relate their structures to carbocation stability and steric hindrance.
Judge SN1/SN2 competition
Combine substrate structure, nucleophile behaviour, leaving-group ability, kinetics and stereochemical evidence.
These ten questions combine experimental rate data, energy profiles, stereochemistry and structural reasoning. The aim is to identify a mechanism from evidence rather than from a single memorised rule.
- 10 mastery questions
- Kinetic, energetic and stereochemical evidence
- Instant explanation for every answer option
| Experiment | [RX] / mol dm⁻³ | [Nu⁻] / mol dm⁻³ | Relative rate |
|---|---|---|---|
| 1 | 0.10 | 0.10 | 1 |
| 2 | 0.20 | 0.10 | 2 |
| 3 | 0.10 | 0.20 | 2 |
Frequently asked questions
Questions students often ask about SN1 and SN2 kinetics
Most questions in this topic become easier once the rate law, energy profile and stereochemistry are treated as different pieces of evidence describing the same molecular mechanism.
Is CH₃Br a primary bromoalkane?
No. CH₃Br is bromomethane and belongs to the separate methyl category. The carbon bonded to Br is attached to zero other carbon atoms. In a primary haloalkane such as CH₃CH₂Br, the carbon bonded to Br is attached to one other carbon atom.
Why is SN2 second order if it occurs in only one step?
“One step” and “first order” are different ideas. The single SN2 step involves both RX and Nu⁻, giving Rate = k[RX][Nu⁻]. The reaction is therefore second order overall.
Why is the nucleophile absent from the simple SN1 rate equation?
The slow step occurs before nucleophilic attack. RX first ionises to generate a carbocation, so increasing nucleophile concentration does not speed up that preceding unimolecular ionisation step.
Does SN1 always produce an exact 50:50 racemic mixture?
No. A completely free planar carbocation gives the idealised tendency towards equal attack from both faces. In reality, the leaving-group ion can remain nearby and partly shield one face, often giving some excess inversion.
Does inversion in SN2 always mean R becomes S?
No. Inversion refers to the geometry at the reacting carbon. R and S depend on the Cahn–Ingold–Prelog priority order in the product, and that priority order can change when the leaving group is replaced.
What exactly is the intermediate on an SN1 energy profile?
It is the carbocation formed after heterolytic cleavage of the carbon–leaving-group bond. On the energy profile it appears as the local minimum between TS1‡ and TS2‡.
Why is a transition state a peak but an intermediate a valley?
A transition state is an energy maximum along an elementary reaction coordinate. An intermediate lies at a local minimum between two activation barriers and has a finite lifetime before undergoing the next elementary step.
How does the Hammond postulate explain faster SN1 reactions for more stable carbocations?
Carbocation formation is normally endothermic, so the corresponding transition state is relatively carbocation-like. Anything that stabilises the carbocation therefore tends also to stabilise that transition state, lowering the activation barrier.
Is the steady-state approximation the same as equilibrium?
No. In steady state, an intermediate is formed and consumed at approximately equal rates, so its concentration remains roughly constant. The overall reaction can still continue towards products.
Why do tertiary substrates favour SN1 but hinder SN2?
Tertiary substitution stabilises the carbocation required for SN1, but the three alkyl groups also crowd the reacting carbon and obstruct the backside approach required for SN2.
Why are secondary substrates harder to classify?
A secondary centre is moderately hindered but still accessible to SN2 attack, while a secondary carbocation is more stable than a primary one but less stable than a tertiary one. Neither pathway automatically dominates, so other factors matter more.
What is the fastest way to distinguish SN1 from SN2 in an examination question?
Start with the experimental rate law. Rate = k[RX] supports the simple SN1 picture, whereas Rate = k[RX][Nu⁻] supports SN2. Then use substrate structure, energy profile and stereochemistry to confirm the conclusion.
The most useful way to learn SN1 and SN2 is not as two disconnected lists of facts. The rate law tells you which species participate in the kinetically important event, the energy profile shows how many elementary steps and intermediates are involved, the Hammond postulate explains how structural changes affect activation barriers, and stereochemistry reveals the geometry of reaction. Once these pieces are connected, SN1/SN2 competition becomes a problem that can be reasoned through rather than a collection of rules to memorise.