ChemistryISC Class 12 · Haloalkanes and Haloarenes

SN1 and SN2 Mechanisms: ISC Class 12 Haloalkanes

Updated 15 October 2026 · Fact-checked

SN2 is a one-step substitution: the nucleophile attacks from the back as the halide leaves, rate = k[RX][Nu⁻], with inversion of configuration; it is fastest for methyl and primary halides. SN1 goes through a carbocation in a slow first step, rate = k[RX], gives racemisation, and is fastest for tertiary, allylic and benzylic halides.

Understand SN1 and SN2 Mechanisms

Why haloalkanes undergo substitution. The C–X bond is polar: carbon carries a partial positive charge and halogen a partial negative charge. A nucleophile (an electron-pair donor such as OH⁻, CN⁻, NH₃ or H₂O) attacks the electron-poor carbon, and the halide ion leaves. This is nucleophilic substitution. It happens by two mechanisms, named by Hughes and Ingold.

SN2 (substitution, nucleophilic, bimolecular). One step. The nucleophile approaches the carbon from the side opposite to the halogen (backside attack). In the transition state the carbon is partly bonded to both the incoming and the leaving group, the three other groups lie in one plane, and the carbon is momentarily five-coordinate. As the new bond forms, the C–X bond breaks and the three groups flip over like an umbrella in the wind. Example: $C H_{3} B r + O H^{-} \to C H_{3} O H + B r^{-}$ . Both the substrate and the nucleophile are in the rate-determining step, so rate = k[RX][OH⁻], second order. Because the nucleophile enters from the back, a chiral carbon undergoes inversion of configuration (Walden inversion). Bulky groups around the carbon block the backside approach (steric hindrance), so the SN2 order is CH₃X > 1° > 2° > 3°. Tertiary halides hardly react by SN2.

SN1 (substitution, nucleophilic, unimolecular). Two steps. Step 1 (slow, rate-determining): the C–X bond ionises to give a carbocation and a halide ion: $(C H_{3})_{3} C B r \to (C H_{3})_{3} C^{+} + B r^{-}$ . Step 2 (fast): the nucleophile attacks the planar carbocation: $(C H_{3})_{3} C^{+} + O H^{-} \to (C H_{3})_{3} C O H$ . Only the substrate is in the slow step, so rate = k[RX], first order, independent of [OH⁻]. The order of reactivity follows carbocation stability: 3° > 2° > 1° > CH₃X, because alkyl groups release electrons (+I effect and hyperconjugation) and spread the positive charge. Allylic and benzylic halides also react fast by SN1, even when primary, because their carbocations are stabilised by resonance. SN1 reactions are generally carried out in polar protic solvents (water, alcohols, acetic acid), which solvate and stabilise the ions formed.

Stereochemistry of SN1. The carbocation is planar ( $s p^{2}$ ), so the nucleophile can attack equally from either face. A single enantiomer of a chiral halide therefore gives a mixture of the two enantiomeric products: racemisation (in practice there is often some excess of the inverted product). A racemic mixture is optically inactive.

Effect of the halogen. For the same alkyl group, reactivity in both mechanisms is R–I > R–Br > R–Cl > R–F, because the C–X bond gets weaker as X gets larger, so iodide is the best leaving group.

Summary of differences.
SN2: one step; second order; transition state, no intermediate; inversion; favoured by 1°, strong nucleophiles, less hindered carbon.
SN1: two steps; first order; carbocation intermediate; racemisation; favoured by 3°, allylic, benzylic halides and polar protic solvents.

Classic identification. Of the C₄H₉Br isomers, 2-bromo-2-methylpropane (3°) hydrolyses by SN1 (rate depends only on [RBr]); 1-bromobutane and 1-bromo-2-methylpropane (1°) by SN2 (rate depends on [RBr] and [OH⁻]); 2-bromobutane (2°, chiral) can react either way.

SN1 and SN2 Mechanisms: formulas and key facts

SN2 rate law: $rate = k [R X] [N u^{-}]$; Second order; one step through a transition state.
SN1 rate law: $rate = k [R X]$; First order; slow ionisation to a carbocation.
SN2 reactivity order: $C H_{3} X > 1^{\circ} > 2^{\circ} > 3^{\circ}$; Controlled by steric hindrance to backside attack.
SN1 reactivity order: $3^{\circ} > 2^{\circ} > 1^{\circ} > C H_{3} X; allylic, benzylic fast$; Controlled by carbocation stability.
Leaving group order: $R - I > R - B r > R - C l > R - F$; Weaker C–X bond, better leaving group; same in both mechanisms.

How to solve SN1 and SN2 Mechanisms questions

ISC asks: identify the mechanism from rate data or structure, arrange halides in reactivity order, write the mechanism with arrows, and predict stereochemistry. Work through the substrate first.

1Classify the carbon bearing X: methyl, 1°, 2°, 3°, allylic or benzylic.
2If rate data are given: rate depending only on [RX] means SN1; on [RX] and [Nu⁻] means SN2.
3For SN2 order, look at crowding around the carbon (CH₃ > 1° > 2° > 3°); for SN1 order, look at carbocation stability (3° > 2° > 1°, with resonance for allylic and benzylic).
4When halogens differ, the iodide reacts fastest and the fluoride slowest.
5Draw SN2 as one step with a pentacoordinate transition state; draw SN1 as two steps with a planar carbocation.
6For a chiral carbon, SN2 gives the inverted product; SN1 gives (mostly) a racemic mixture.

SN1 and SN2 Mechanisms in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

SN1 hydrolysis ignores how much base is added

Suppose an effluent tank at a chemical plant contains a small amount of a tertiary alkyl bromide, which hydrolyses by the SN1 mechanism with a first order rate constant of $2.0 \times 1 0^{- 4}$ s⁻¹ at the tank temperature. An operator proposes doubling the NaOH added to destroy it twice as fast.

Calculate the half-life of the bromide and judge whether adding more NaOH helps.

Try it first, then show the solutionSolution

SN1: rate = k[RBr]; [OH⁻] does not appear in the rate law.
$t_{1/2} = \frac{0.693}{k} = \frac{0.693}{2.0 \times 1 0 ^{- 4}} = 3465$ s, about 58 minutes.
Doubling NaOH does not change k or the rate, because the slow step is ionisation of the C–Br bond.
To speed it up, raise the temperature (k increases), not the base concentration.

Answer: Half-life about 3465 s (58 min). More NaOH does not help; the rate depends only on [RBr] (and on temperature).

Memory hook: SN1 waits for the halide to leave on its own: the nucleophile is not in a hurry, so more of it changes nothing.

Tricks to remember SN1 and SN2 Mechanisms

Short memory aids. Say each one aloud once, then use it on a question the same day.

SN1: one molecule, one carbocation, one rate term
The '1' in SN1 means one species in the slow step, so one concentration in the rate law; the '2' in SN2 means two.
Use it when: Writing rate laws.
Tertiary takes its time alone
Tertiary halides are too crowded for backside attack, so they ionise by themselves (SN1). Primary halides are open at the back (SN2).
Use it when: Matching substrate to mechanism.
Umbrella in the wind
In SN2 the three groups flip over like an umbrella in a gust: inversion.
Use it when: Stereochemistry of SN2.

SN1 and SN2 Mechanisms: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Fastest SN1 substrate · 1 mark · MCQBoard level

Which of the following reacts fastest by the SN1 mechanism?
(a) CH₃Cl
(b) CH₃CH₂CH₂Cl
(c) (CH₃)₂CHCl
(d) C₆H₅CH₂Cl

Try it first, then show the solutionSolution

SN1 speed depends on carbocation stability.
The benzyl carbocation, C₆H₅CH₂⁺, is stabilised by resonance with the ring; it is more stable than the 2° and 1° ions offered.

Answer: (d) C₆H₅CH₂Cl.

Example 2: Mechanism of SN2 · 2 marks · Short answerBoard level

Write the SN2 mechanism for the reaction of bromomethane with hydroxide ion, showing the transition state, and give the rate law.

Try it first, then show the solutionSolution

$C H_{3} B r + O H^{-} \to C H_{3} O H + B r^{-}$ (½ mark).
OH⁻ attacks carbon from the side opposite Br. Transition state: HO···C···Br with the three H atoms in a plane perpendicular to the HO–C–Br axis, partial negative charge on O and Br (1 mark).
Br⁻ leaves and the H atoms flip (inversion). Rate = k[CH₃Br][OH⁻] (½ mark).

Answer: One-step backside attack via [HO···CH₃···Br]⁻; rate = k[CH₃Br][OH⁻].

Example 3: SN1 reactivity order · 3 marks · Short answerHarder board level

Arrange CH₃CH₂Br, (CH₃)₃CBr, CH₃Br and (CH₃)₂CHBr in increasing order of reactivity by the SN1 mechanism, and explain the order.

Try it first, then show the solutionSolution

CH₃Br < CH₃CH₂Br < (CH₃)₂CHBr < (CH₃)₃CBr (1 mark).
The slow step forms a carbocation; the faster it forms, the faster the reaction (1 mark).
Stability of the carbocations: CH₃⁺ < 1° < 2° < 3°, because each alkyl group releases electrons (+I effect and hyperconjugation) and disperses the positive charge (1 mark).

Answer: CH₃Br < CH₃CH₂Br < (CH₃)₂CHBr < (CH₃)₃CBr, following carbocation stability.

Example 4: Stereochemistry under two sets of conditions · 3 marks · Short answerHarder board level

(R)-2-Chlorobutane is converted to butan-2-ol (a) with concentrated NaOH in acetone, which favours SN2, and (b) by warming in water, which favours SN1. Predict the configuration and optical activity of the product in each case.

Try it first, then show the solutionSolution

(a) SN2: backside attack inverts the configuration. OH takes the priority position Cl had (OH > C₂H₅ > CH₃ > H), so the product is (S)-butan-2-ol, optically active (1½ marks).
(b) SN1: the planar sec-butyl carbocation is attacked from both faces, giving (R)- and (S)-butan-2-ol in nearly equal amounts (1 mark).
The product is (nearly) racemic and shows little or no optical rotation (½ mark).

Answer: (a) (S)-butan-2-ol, optically active (inversion). (b) Racemic butan-2-ol, optically inactive (racemisation).

Example 5: The four isomers of C₄H₉Br · 5 marks · Long answerHOTS / competency

Four isomeric bromides have the formula C₄H₉Br.
(a) Write the structures and IUPAC names of all four. [2]
(b) Which isomer is hydrolysed by aqueous KOH at a rate that does not depend on [OH⁻]? Write its rate law. [1]
(c) Which isomer is chiral? What happens to the optical activity of a pure enantiomer of it when it is hydrolysed by the SN1 route? [1]
(d) Which isomer reacts fastest by SN2, and why? [1]

Try it first, then show the solutionSolution

(a) CH₃CH₂CH₂CH₂Br, 1-bromobutane (1°); CH₃CH₂CHBrCH₃, 2-bromobutane (2°); (CH₃)₂CHCH₂Br, 1-bromo-2-methylpropane (1°); (CH₃)₃CBr, 2-bromo-2-methylpropane (3°) (2 marks).
(b) 2-Bromo-2-methylpropane reacts by SN1: rate = k[(CH₃)₃CBr] (1 mark).
(c) 2-Bromobutane (C-2 carries H, CH₃, C₂H₅ and Br). SN1 through a planar carbocation gives a (nearly) racemic alcohol, so optical activity is largely lost (1 mark).
(d) 1-Bromobutane: primary and unbranched, so least hindered for backside attack (1-bromo-2-methylpropane is primary but has a branch next to the reacting carbon) (1 mark).

Answer: (a) As listed. (b) 2-Bromo-2-methylpropane, rate = k[RBr]. (c) 2-Bromobutane; racemisation. (d) 1-Bromobutane, least steric hindrance.

Where students lose marks in SN1 and SN2 Mechanisms

The slips examiners see most often, and what to write instead.

Mistake: Mixing up which halides react by SN1 and SN2
ISC examiners found candidates unclear about the type of reaction given by tertiary and primary alkyl halides in a 2025 identification question.
Wrong working
Rate depends on [A] only, so A is 1-bromobutane.
Right working
Rate depends on [A] only: SN1, so A is tertiary, 2-bromo-2-methylpropane.
Do this: First-order rate means SN1 means tertiary; second-order rate means SN2 means primary.
Usually costs: 1 mark per compound.
Mistake: Drawing SN2 with a carbocation
Students write an intermediate for a one-step reaction.
Do this: SN2 has a transition state only (in square brackets with partial bonds), no intermediate.
Usually costs: 1 mark.
Mistake: Saying SN1 gives inversion
Students apply the SN2 result to every substitution.
Do this: SN1 goes through a planar carbocation and gives racemisation; only SN2 gives clean inversion.
Usually costs: 1 mark.
Mistake: Reversed reactivity orders
Students give 3° > 2° > 1° for SN2.
Do this: SN2: CH₃ > 1° > 2° > 3° (steric); SN1: 3° > 2° > 1° > CH₃ (carbocation stability).
Usually costs: 1 mark.

Exam tips

In mechanism answers, draw curved arrows from the nucleophile's lone pair to carbon and from the C–X bond to X.
Give the rate law with every mechanism; examiners often award a mark for it.
Use the words 'steric hindrance' for SN2 and 'stability of carbocation' for SN1 in reasoning answers.
Remember that allylic and benzylic halides are exceptions: primary yet fast by SN1.
For stereochemistry, state the product's configuration (R/S) only when you have checked priorities.

Practice questions on SN1 and SN2 Mechanisms

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
An SN2 reaction at a chiral carbon proceeds with:
(a) retention of configuration
(b) inversion of configuration
(c) racemisation
(d) formation of a carbocation
Hint
The nucleophile attacks from the back.
Answer
(b) inversion of configuration.
Q2 · 2 marks · Short answerBoard level
Write the two steps of the SN1 hydrolysis of 2-bromo-2-methylpropane and state which step is rate-determining.
Hint
Ionisation comes first.
Answer
Step 1 (slow): $(C H_{3})_{3} C B r \to (C H_{3})_{3} C^{+} + B r^{-}$ . Step 2 (fast): $(C H_{3})_{3} C^{+} + O H^{-} \to (C H_{3})_{3} C O H$ . Step 1 is rate-determining, so rate = k[(CH₃)₃CBr].
Q3 · 2 marks · Short answerHarder board level
Benzyl chloride, C₆H₅CH₂Cl, is a primary halide, yet it undergoes SN1 reactions readily. Explain.
Hint
Draw the carbocation.
Answer
Ionisation gives the benzyl carbocation, C₆H₅CH₂⁺, whose positive charge is delocalised into the benzene ring by resonance. This stabilised carbocation forms easily, so SN1 is fast.
Q4 · 3 marks · Short answerHarder board level
Explain how (a) the solvent and (b) the strength of the nucleophile influence whether a substitution follows SN1 or SN2.
Hint
Which mechanism forms ions in the slow step?
Answer
(a) Polar protic solvents (water, alcohols, acetic acid) solvate and stabilise the carbocation and halide ion formed in the slow step, favouring SN1; less polar solvents such as acetone do not, and favour SN2. (b) A strong nucleophile at high concentration speeds SN2, whose rate depends on [Nu⁻]; it has no effect on SN1, whose rate does not depend on the nucleophile.
Q5 · 1 mark · Assertion-reasonHarder board level
Assertion (A): The rate of hydrolysis of 2-bromo-2-methylpropane by aqueous NaOH does not change when [OH⁻] is doubled.
Reason (R): The rate-determining step is the ionisation of the C–Br bond to form a carbocation.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
Which species appear in the slow step?
Answer
(a).
Q6 · 3 marks · Short answerHarder board level
Arrange 1-chlorobutane, 1-iodobutane and 1-bromobutane in decreasing order of reactivity towards SN2 substitution by CN⁻, and explain using bond enthalpies.
Hint
Which C–X bond is weakest?
Answer
1-Iodobutane > 1-bromobutane > 1-chlorobutane. All are primary with the same steric hindrance, so the difference comes from the leaving group: the C–I bond is the longest and weakest (lowest bond enthalpy), so I⁻ leaves most easily; C–Cl is the strongest.
Q7 · 3 marks · Short answerHOTS / competency
(S)-3-Bromo-3-methylhexane is hydrolysed in aqueous ethanol. (a) By which mechanism? (b) Will the 3-methylhexan-3-ol formed be optically active? Explain.
Hint
C-3 is tertiary and carries four different groups.
Answer
(a) SN1, because the halide is tertiary and the solvent is polar protic. (b) The carbocation formed at C-3 is planar, so water attacks both faces about equally, giving both enantiomers of 3-methylhexan-3-ol: a (nearly) racemic, optically inactive product.
Q8 · 3 marks · Case-basedHOTS / competency
Read and answer.
Suppose a student studies two bromides, X (C₃H₇Br) and Y (C₄H₉Br), with KCN in aqueous ethanol. For X, doubling [CN⁻] doubles the initial rate, and the product is butanenitrile. For Y, doubling [CN⁻] has no effect on the rate.
(i) Identify X and its mechanism. [1]
(ii) Identify Y and write its rate law. [1]
(iii) Which of the two would react faster if the solvent were made more polar and protic? Explain. [1]
Hint
Butanenitrile is CH₃CH₂CH₂CN.
Answer
(i) X is 1-bromopropane, CH₃CH₂CH₂Br (primary), reacting by SN2: CN⁻ replaces Br to give CH₃CH₂CH₂CN. (ii) Y is 2-bromo-2-methylpropane, (CH₃)₃CBr; rate = k[(CH₃)₃CBr] (SN1). (iii) Y: a more polar protic solvent stabilises the carbocation and bromide ion formed in its slow step, speeding SN1.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Rate law for SN1?Show answer
rate = k[RX].
2.Order of SN2 reactivity?Show answer
CH₃X > 1° > 2° > 3°.
3.What is formed in the slow step of SN1?Show answer
A carbocation and a halide ion.
4.Which solvents favour SN1?Show answer
Polar protic solvents such as water, alcohols and acetic acid.
5.Stereochemical result of SN2?Show answer
Inversion of configuration.
6.Order of leaving ability of halides?Show answer
I⁻ > Br⁻ > Cl⁻ > F⁻.

SN1 and SN2 Mechanisms: frequently asked questions

What is the difference between SN1 and SN2?

SN2 is one step, second order, goes through a transition state and gives inversion; it suits primary halides. SN1 is two steps, first order, goes through a carbocation and gives racemisation; it suits tertiary, allylic and benzylic halides.

Why do tertiary halides not react by SN2?

Three bulky alkyl groups block the backside approach of the nucleophile (steric hindrance).

Why are tertiary carbocations the most stable?

Three alkyl groups release electrons by the +I effect and hyperconjugation, spreading the positive charge.

What is racemisation?

Conversion of an optically active compound into an equal mixture of enantiomers, which is optically inactive; it occurs in SN1 because the carbocation is planar.

Does the halogen affect the rate of substitution?

Yes. Iodides react fastest and fluorides slowest, because the C–I bond is the weakest.