The shape of the lesson
By the end of the lesson, learners can
- Two mechanisms: Step through both mechanisms in the simulator, first with bromoethane and then with 2-bromo-2-methylpropane.
- Why the structure decides: Compare the rates of the three classes in the hydrolysis simulator.
- SN2: one step, transition state, rate = k[RX][Nu].
- SN1: two steps, carbocation, rate = k[RX].
- Tertiary: steric hindrance plus stable carbocation.
- Arrows from lone pairs and bonds.
One picture every idea returns to
SN2 (substitution, nucleophilic, bimolecular) happens in one step: the nucleophile attacks the δ+ carbon from the side opposite the halogen while the C–X bond breaks, through a transition state with partial bonds. It is typical of primary halogenoalkanes, and its rate depends on both reactants. SN1 happens in two steps: the C–X bond breaks heterolytically (slowly) to give a carbocation, which the nucleophile then attacks (quickly). It is typical of tertiary halogenoalkanes, because the tertiary carbocation is stabilised by three alkyl groups and the crowded carbon blocks a backside attack. Its rate depends only on the halogenoalkane.
Forty-five minutes, phase by phase
| Time | Phase | What happens in the room | Grouping |
|---|---|---|---|
| 0 to 5 min | Hook: predict | Open with the question in the title, One step or two?. Hydroxide swaps for bromine in every halogenoalkane, but not always the same way. Put the figure on the board: it is the one picture every idea in the lesson returns to. | Whole class, then pairs |
| 5 to 17 min | Teach and check: Two mechanisms | Step through both mechanisms in the simulator, first with bromoethane and then with 2-bromo-2-methylpropane. Project tab 2 of the lab (Halogen compounds lab, SN2 and SN1) and let a student drive it while the class predicts. | Whole class, then pairs |
| 17 to 30 min | Teach and check: Why the structure decides | Compare the rates of the three classes in the hydrolysis simulator. Project tab 3 of the lab (Halogen compounds lab, rate of hydrolysis) and let a student drive it while the class predicts. | Frayer Model |
| 30 to 36 min | Check | The four Stage 1 questions from the student lesson, answered on mini whiteboards or by show of fingers. Re-teach the one most get wrong. | Individual, then whole class |
| 36 to 43 min | Plenary | The exam-style questions from Stage 2, one at a time: thirty seconds alone, then a random call. The feedback lines in the key give the examiner's wording. | Individual, random call |
| 43 to 45 min | Exit and homework | Set the worksheet below (6 questions, 16 marks). Learners self-assess against the objectives before they leave. | Individual |
Protect the parts that carry the learning
The sequence carries about fifty minutes of material, so the lab is driven by one student while the class predicts rather than explored freely. Times are a guide.
Protected: the check at 30 minutes. It is the only individual read on whether the key idea landed, and it is not cut.
Cut first if the build overruns: the last check in Why the structure decides, never the check or the plenary.
In a 60 minute block: run worksheet questions 1 to 3 in class as a Rally Coach set and model both the hardest worked example and its check live.
Frayer Model
Pairs fill a four-box grid for the key term: definition, characteristics, examples and non-examples. The non-examples box is where the misconceptions surface, so give it the most time.
The full structure, with its variations and the other twelve, is on the cooperative-learning page.
What to head off, and how
| Trap learners fall into | Teaching move that pre-empts it |
|---|---|
| Curly arrows start from the lone pair on OH −, not from the minus sign, and the arrow from the C–X bond must point to the halogen. | In S N 2 draw the transition state in square brackets with partial bonds and an overall negative charge; S N 2 has no carbocation. |
SN1 and SN2 mechanisms: worksheet
6 questions, 16 marks. Written in the style of the papers; no past paper question is reproduced.
Describe the SN2 mechanism for the reaction of bromoethane with hydroxide ions.
Describe the SN1 mechanism for the hydrolysis of 2-bromo-2-methylpropane.
Explain why 2-bromo-2-methylpropane reacts by SN1 but 1-bromobutane reacts by SN2.
Write the rate equations for the SN1 and SN2 hydrolysis of a halogenoalkane RBr.
Explain why SN1 hydrolysis of an optically active halogenoalkane gives a racemic product.
Predict the order of rate of hydrolysis of 1-bromobutane, 2-bromobutane and 2-bromo-2-methylpropane with water, and give a reason.
Answer key and marking notes
- C–Br is polar: Cδ+, Brδ−.
- Curly arrow from a lone pair on OH− to the δ+ carbon, approaching opposite Br.
- Curly arrow from the C–Br bond to Br; transition state [HO···CH2(CH3)···Br]−.
- Products: CH3CH2OH and Br−.
- Step 1 (slow): C–Br breaks heterolytically; curly arrow from the bond to Br, giving (CH3)3C+ and Br−.
- Step 2 (fast): curly arrow from a lone pair on OH− to C+.
- Product: (CH3)3COH.
- The tertiary carbocation from 2-bromo-2-methylpropane is stabilised by three electron-releasing alkyl groups.
- Its crowded carbon also blocks backside attack (steric hindrance).
- 1-bromobutane would give an unstable primary carbocation, and its carbon is unhindered, so the one-step SN2 route is preferred.
- SN1: rate = k[RBr].
- SN2: rate = k[RBr][OH−].
- The carbocation intermediate is planar (trigonal).
- OH− can attack equally from either face, giving equal amounts of both enantiomers.
- 2-bromo-2-methylpropane > 2-bromobutane > 1-bromobutane.
- Tertiary forms the most stable carbocation, so the SN1 pathway is fastest.
8 multiple-choice questions on this lesson
From the topic question bank. Assign them in the app for a marked score in your gradebook, or run the topic as a class game.
The diagram shows the SN2 reaction between hydroxide ions and bromomethane. Which statement about the species in square brackets is correct?
- Ait is a transition state in which the C-O bond is partly formed and the C-Br bond is partly broken, with an overall charge of 1−
- Bit is a carbocation intermediate that can be isolated before OH− attacks
- Cit is a neutral molecule, because the negative charge has been lost from the oxygen
- Dit is the product, methanol, with bromine still attached as a spectator
In the mechanism of the reaction between OH− and a halogenoalkane, where should the first curly arrow start and finish?
- Afrom a lone pair on the oxygen atom of OH− to the δ+ carbon atom bonded to the halogen
- Bfrom the negative sign on OH− to the carbon atom bonded to the halogen
- Cfrom the hydrogen atom of OH− to the halogen atom
- Dfrom the carbon atom to the oxygen atom of OH−
Bromoethane, a primary halogenoalkane, is hydrolysed by aqueous OH−. What is the rate equation?
- Arate = k[RX][OH−], second order overall, because both species are in the single step
- Brate = k[RX], first order, because primary halogenoalkanes react by SN1
- Crate = k[OH−], first order, because the nucleophile controls the rate
- Drate = k[RX]2, second order, because two molecules of RX collide
The diagram shows an SN1 mechanism. Why does 2-bromo-2-methylpropane react this way rather than by SN2?
- Athe tertiary carbocation is stabilised by the electron-donating inductive effect of three alkyl groups, and those groups block attack from the back
- Bthe tertiary carbocation is the least stable carbocation, so it forms and reacts quickly
- Cthe C-Br bond in a tertiary halogenoalkane is non-polar, so OH− cannot attack it
- Dtertiary halogenoalkanes have no δ+ carbon, so the halogen must leave on its own
Which statement correctly compares the SN1 and SN2 mechanisms?
- ASN2 is one step through a transition state; SN1 is two steps through a carbocation intermediate
- BSN2 is one step through a carbocation; SN1 is two steps through a transition state only
- Cboth mechanisms go through a carbocation, but SN1 is faster
- DSN1 is used by primary halogenoalkanes and SN2 by tertiary ones
A single enantiomer of 3-bromo-3-methylhexane, CH3CH2CBr(CH3)CH2CH2CH3, is hydrolysed by aqueous OH−. The product alcohol is optically inactive. Why?
- Athe reaction is SN1; the planar carbocation is attacked equally from both faces, giving a racemic mixture
- Bthe reaction is SN2; inversion of configuration always gives an optically inactive product
- Cthe product has no chiral centre, because OH replaces the methyl group
- Dthe reaction is SN1, and the carbocation keeps its original shape so the configuration is retained
1-bromobutane reacts with OH− by SN2. In a second experiment the concentration of 1-bromobutane is doubled and the concentration of OH− is tripled. By what factor does the initial rate change?
- A×6
- B×2
- C×5
- D×3
A sample of (R)-2-bromobutane is warmed with a high concentration of aqueous OH−. The butan-2-ol formed is optically active and has the opposite configuration. What does this show about the mechanism?
- Athe secondary halogenoalkane reacted by SN2: OH− attacked from the side opposite Br in one step, inverting the configuration
- Bit reacted by SN1, because a carbocation always gives a single inverted product
- Cit reacted by SN2 through a planar carbocation, which OH− attacked from the back
- Dno substitution took place; the bromine and the OH group simply exchanged positions on the same side
Quiz key
In SN2 the new bond forms as the old bond breaks, in a single step. At the top of the energy barrier the carbon is partly bonded to both O and Br (dashed partial bonds) and has five groups around it; the 1− charge of OH− is spread over the O and Br, so the whole species carries 1−. It lasts only an instant and cannot be isolated. A carbocation forms only in SN1; describing SN2 as one step via a carbocation is a classic error. The charge is not lost: it moves from O towards Br and leaves as Br−. It is not the product: the C-Br bond has not yet fully broken.
A curly arrow always starts at a pair of electrons (a lone pair or a bond) and points to where that pair goes. The oxygen of OH− has lone pairs, and one of them forms the new bond to the electron-deficient δ+ carbon. The minus sign is a charge label, not an electron pair, so an arrow starting from it scores no credit; this is a classic error. The H of OH− has no lone pair and does not attack anything. An arrow from carbon to oxygen points the wrong way: the nucleophile donates the electrons, the carbon accepts them.
A primary halogenoalkane cannot form a stable carbocation (a primary carbocation has only one alkyl group to stabilise it), so it reacts by SN2. The one and only step involves both RX and OH−, so the rate is first order in each and second order overall: rate = k[RX][OH−]. rate = k[RX] describes SN1, which tertiary halogenoalkanes follow; assigning SN1 to a primary halogenoalkane is the classic error. The rate cannot depend on OH− alone, because RX is in the step too. Two RX molecules do not react with each other; the collision is between RX and OH−.
Each alkyl group pushes electron density towards the positive carbon (positive inductive effect), so a carbocation with three alkyl groups is the most stable type, and the slow heterolysis of C-Br to (CH3)3C+ is feasible. At the same time the three bulky CH3 groups crowd the carbon (steric hindrance), so OH− cannot approach from the side opposite Br, which SN2 needs. Saying the tertiary carbocation is the least stable reverses the order (tertiary > secondary > primary); an unstable ion would not form. The C-Br bond is polar in every halogenoalkane, and the carbon is δ+ in all of them, so those two statements are false.
In SN2 bond making and bond breaking happen together, so there is one step and one transition state, with two species (RX and the nucleophile) in it. In SN1 the C-X bond first breaks heterolytically (slow, one species) to give a carbocation intermediate, which is then attacked by the nucleophile (fast). Putting a carbocation into SN2 is a classic error: no intermediate forms in a one-step reaction. Only SN1 has a carbocation. The assignment is the other way round: primary halogenoalkanes use SN2 and tertiary ones use SN1, while secondary can go either way.
The bromine is on a tertiary carbon, so the reaction is SN1. Loss of Br− gives a trigonal planar carbocation (bond angle 120°), so OH− is equally likely to attack from above or below the plane. Equal amounts of the two enantiomers form: a racemic mixture, which does not rotate plane-polarised light. SN2 gives inversion, but a single inverted enantiomer is still optically active, and a tertiary centre is too hindered for SN2 anyway. OH replaces Br, not CH3, so the carbon in the product still carries four different groups and is chiral. A planar carbocation cannot remember the original configuration, so retention is wrong.
For SN2, rate = k[RX][OH−], first order in each reactant. Doubling [RX] doubles the rate and tripling [OH−] triples it, so the new rate is 2 × 3 = 6 times the original. ×2 is the SN1 answer, using rate = k[RX]; a primary halogenoalkane does not react by SN1. ×5 adds the factors instead of multiplying them. ×3 uses only the hydroxide term, as if the rate were independent of [RX].
In SN2 the nucleophile must approach from the side directly opposite the leaving group, and as the C-Br bond breaks the other three groups flip over, like an umbrella turning inside out. A single, inverted, optically active enantiomer is therefore the signature of SN2; a high [OH−] favours this route for a secondary halogenoalkane. SN1 goes through a planar carbocation and would give a racemic (inactive) mixture, not a single inverted product. SN2 does not involve a carbocation at all: describing it as passing through one is a classic error. Attack on the same side as Br would give retention, not inversion, and the OH has clearly replaced the Br, so substitution did occur.
What to project
The student lesson embeds these tabs. Open them full screen on the projector; every slider has a drag handle on the stage.
Slides for this lesson
A projectable deck for every pack is on the way. Until then, present the student lesson full screen: its sections, figure and lab tabs follow the same order as the plan above.