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In the previous lesson, you learned that halogenoalkanes undergo nucleophilic substitution. Now we examine the two different mechanisms by which this can occur: SN2 and SN1. The distinction between these pathways is one of the most important concepts in A-Level organic chemistry.
SN2 stands for Substitution, Nucleophilic, 2nd order (bimolecular). It is a one-step mechanism:
The nucleophile (e.g., OH⁻) attacks the δ+ carbon from the opposite side to the halogen leaving group. This is called backside attack.
In the mechanism diagram:
Both bond-making (C-OH) and bond-breaking (C-X) happen at the same time in a single concerted step. There is no intermediate — only a transition state in which the carbon is partially bonded to both the incoming nucleophile and the departing halogen.
Curly arrow detail: The curly arrow from OH⁻ starts from a lone pair on the oxygen atom and points to the δ+ carbon. This represents the lone pair being donated to form the new C-O bond. The second curly arrow starts from the C-Br bond and points to the Br atom. This represents the bonding electrons transferring entirely to bromine as it leaves as Br⁻. Both arrows are drawn simultaneously to emphasise the concerted nature of the step.
In a primary halogenoalkane, the carbon bearing the halogen has only one alkyl group attached. This means there is relatively little steric hindrance (physical blocking) around the δ+ carbon. The nucleophile can approach easily for backside attack.
In a tertiary halogenoalkane, three bulky alkyl groups surround the carbon, physically blocking the nucleophile from approaching. SN2 is therefore extremely slow or impossible for tertiary substrates.
SN1 stands for Substitution, Nucleophilic, 1st order (unimolecular). It is a two-step mechanism:
Step 1: Formation of the carbocation (slow, rate-determining step)
The C-X bond breaks heterolytically — both bonding electrons go to the halogen, which departs as X⁻. The carbon is left with a positive charge (carbocation).
In the mechanism diagram: a curly arrow from the C-X bond to the X atom, showing the bond breaking. Only one curly arrow is needed for this step.
This is the slow step because energy is required to break the C-X bond without any assistance from the nucleophile.
Step 2: Nucleophilic attack (fast)
The nucleophile (e.g., OH⁻) rapidly attacks the planar carbocation. A curly arrow is drawn from the lone pair on OH⁻ to the C⁺.
Because the carbocation is planar (sp² hybridised with an empty p orbital), the nucleophile can attack from either side with equal probability.
The key to SN1 is the formation of a stable carbocation. Tertiary carbocations are the most stable due to the inductive effect of three electron-donating alkyl groups. Primary carbocations are very unstable and rarely form. Therefore:
| Feature | SN2 | SN1 |
|---|---|---|
| Number of steps | 1 (concerted) | 2 (via carbocation) |
| Intermediate | None (transition state only) | Carbocation |
| Rate equation | Rate = k[RX][Nu⁻] | Rate = k[RX] |
| Stereochemistry | Inversion of configuration | Racemisation |
| Favoured by | Primary halogenoalkanes | Tertiary halogenoalkanes |
| Effect of nucleophile | Strong nucleophile speeds reaction | Nucleophile strength does not affect rate |
| Steric effects | Hindered by bulky groups | Not affected (nucleophile not in rate-determining step) |
Secondary halogenoalkanes can undergo either SN1 or SN2, depending on conditions:
The distinction between SN1 and SN2 can be determined experimentally by studying reaction kinetics. This is a key topic because it demonstrates how experimental evidence supports mechanistic proposals.
To distinguish SN1 from SN2, a chemist carries out a series of experiments varying the concentration of one reactant at a time while keeping the other constant, then measures the initial rate of reaction.
Experiment 1: Double [halogenoalkane], keep [nucleophile] constant → rate doubles. Experiment 2: Keep [halogenoalkane] constant, double [nucleophile] → rate doubles.
Conclusion: Rate depends on both concentrations. Rate = k[RX][Nu⁻]. This is second-order kinetics, confirming SN2. Both the halogenoalkane and the nucleophile are involved in the single rate-determining step.
Experiment 1: Double [halogenoalkane], keep [nucleophile] constant → rate doubles. Experiment 2: Keep [halogenoalkane] constant, double [nucleophile] → rate unchanged.
Conclusion: Rate depends only on [halogenoalkane]. Rate = k[RX]. This is first-order kinetics, confirming SN1. Only the halogenoalkane is involved in the rate-determining step (the slow ionisation step). The nucleophile participates only in the fast second step, so its concentration does not affect the overall rate.
| Experiment | [RX] / mol dm⁻³ | [OH⁻] / mol dm⁻³ | Rate / mol dm⁻³ s⁻¹ |
|---|---|---|---|
| 1 | 0.10 | 0.10 | 2.0 × 10⁻⁴ |
| 2 | 0.20 | 0.10 | 4.0 × 10⁻⁴ |
| 3 | 0.10 | 0.20 | 4.0 × 10⁻⁴ |
Analysis: From experiments 1→2, doubling [RX] doubles the rate. From experiments 1→3, doubling [OH⁻] doubles the rate. Therefore Rate = k[RX][OH⁻]. This is SN2.
If experiment 3 had given rate = 2.0 × 10⁻⁴ (unchanged), that would indicate SN1.
Beyond kinetics, the stereochemical outcome provides further evidence:
SN2 — Inversion: If you start with a single enantiomer of a chiral halogenoalkane, SN2 gives pure inversion. The product is the opposite enantiomer. This is because backside attack flips the arrangement of groups around the carbon, like an umbrella inverting in the wind.
SN1 — Racemisation: If you start with a single enantiomer, SN1 gives a racemic mixture (50:50 mix of both enantiomers). This is because the planar carbocation intermediate has no chirality — the nucleophile attacks equally from both sides.
If you observe a product that is optically inactive (racemic) from an optically active starting material, this supports SN1. If you observe complete inversion of optical rotation, this supports SN2.
The type of halogenoalkane is the primary factor determining the mechanism. Primary substrates undergo SN2 because the carbon is accessible and primary carbocations are too unstable for SN1. Tertiary substrates undergo SN1 because the carbon is too sterically hindered for SN2 but forms a stable tertiary carbocation. Understanding these mechanisms allows you to predict products, rates, and stereochemical outcomes.
Edexcel 9CH0 specification Topic 7 — Organic Chemistry I, sub-topic 7.1 covers the two mechanisms of nucleophilic substitution: SN1 (unimolecular, two-step, via a carbocation intermediate) and SN2 (bimolecular, one-step concerted, via a five-coordinate transition state). The specification expects candidates to (a) draw both mechanisms with curly arrows, (b) explain why primary substrates prefer SN2 (steric access, lack of cation stability) and tertiary substrates prefer SN1 (cation stability, steric crowding for SN2), (c) predict stereochemical outcomes (Walden inversion for SN2, racemisation for SN1), and (d) link the mechanism to kinetic order (rate = k[RX][Nu] for SN2 vs rate = k[RX] for SN1) — refer to the official specification document for exact wording. Sub-topic 7.1 is examined directly on Paper 2 (mechanism with curly arrows, kinetic order) and Paper 3 (synoptic with chirality, drug stereochemistry, Walden inversion). The data booklet does not provide rate constants; candidates work from mechanism principles.
Question (8 marks):
(a) Draw the full SN2 mechanism for the reaction of OH⁻ with bromomethane (CH3−Br). Include the transition state. (4)
(b) Draw the full SN1 mechanism for the reaction of OH⁻ with 2-bromo-2-methylpropane ((CH3)3C−Br). Identify the intermediate. (4)
Solution with mark scheme:
(a) SN2 mechanism for OH⁻ + CH3Br:
M1 — curly arrow from OH⁻ lone pair to C of CH3Br.
M1 — curly arrow from C-Br σ bond to Br (showing simultaneous bond breaking).
M1 — transition state drawn with HO and Br on opposite faces of the C, partial bonds (δ− on both); the three H atoms in the plane of C (sp2-like geometry at TS).
A1 — product: methanol (CH3OH) + Br⁻; configuration inverts (Walden inversion), but for CH3Br itself there is no chirality so no stereochemical consequence is observed.
(b) SN1 mechanism for OH⁻ + (CH3)3CBr:
M1 — Step 1: curly arrow from C-Br σ bond to Br; gives tertiary carbocation (CH3)3C+ + Br⁻.
M1 — Tertiary carbocation drawn explicitly with + on the central C.
M1 — Step 2: curly arrow from OH⁻ lone pair to C+; product is 2-methylpropan-2-ol.
A1 — racemisation: if the C+ were chiral (it is not in this case, by symmetry), the planar cation would be attacked from either face, giving a racemic mixture of products. For (CH3)3CBr, the substrate has no chirality, so this point is illustrative.
Total: 8 marks (M6 A2).
Question (6 marks): Two reactions are studied:
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