ChemistryCBSE Class 12 · Haloalkanes and Haloarenes

SN1 and SN2 Mechanisms of Haloalkanes

Updated 15 October 2026 · Fact-checked

Haloalkanes undergo nucleophilic substitution by two mechanisms. SN2 is a one-step, second-order reaction (rate $= k[RX][Nu^-]$) with backside attack and inversion; it is fastest for methyl and primary halides. SN1 is a two-step, first-order reaction (rate $= k[RX]$) through a carbocation; it is fastest for tertiary, allylic and benzylic halides.

Understand SN1 and SN2 Mechanisms

In a haloalkane the C–X bond is polar: carbon carries a partial positive charge because halogens are more electronegative. A nucleophile (an electron-rich species such as $O H^{-}$ , $C N^{-}$ , $N H_{3}$ ) attacks this carbon and the halide leaves as $X^{-}$ : $N u^{-} + R - X \to R - N u + X^{-}$ . This is nucleophilic substitution. With different nucleophiles it gives alcohols ( $O H^{-}$ ), ethers ( $R^{'} O^{-}$ ), nitriles ( $C N^{-}$ ), amines ( $N H_{3}$ ), esters ( $R^{'} C O O^{-}$ ) and hydrocarbons (hydride from $L i A l H_{4}$ ).

SN2 (substitution, nucleophilic, bimolecular). The rate depends on both reactants: rate $= k [R X] [N u^{-}]$ , so it is second order. Hughes and Ingold studied $C H_{3} C l + O H^{-} \to C H_{3} O H + C l^{-}$ . The nucleophile approaches the carbon from the side opposite the halogen. As the new C–O bond forms, the C–Cl bond breaks, all in one step. In the transition state the carbon is bonded partly to both OH and Cl, and the three other groups lie flat in a plane. When Cl leaves, the three groups flip over like an umbrella in the wind, so the configuration is inverted. Because the nucleophile must reach the back of the carbon, bulky groups block it. Reactivity for SN2: $C H_{3} X > 1° > 2° > 3°$ . Tertiary halides hardly react by SN2.

SN1 (substitution, nucleophilic, unimolecular). It is usually carried out in polar protic solvents such as water, alcohol or ethanoic acid. NCERT's example is 2-bromo-2-methylpropane with hydroxide: $(C H_{3})_{3} C B r + O H^{-} \to (C H_{3})_{3} C O H + B r^{-}$ . The rate depends only on the haloalkane: rate $= k [R X]$ , first order. The reason is a two-step path. Step 1 (slow): the C–Br bond ionises to give a carbocation and $B r^{-}$ ; the solvent helps by solvating both ions. Step 2 (fast): the nucleophile attacks the flat carbocation. Since the slow step involves only RX, $[O H^{-}]$ does not appear in the rate law. The more stable the carbocation, the faster the reaction. Carbocation stability is $3° > 2° > 1° > C H_{3}^{+}$ (alkyl groups release electrons by the inductive effect and hyperconjugation), so the SN1 order is $3° > 2° > 1° > C H_{3} X$ , the opposite of SN2.

Allylic and benzylic halides react fast by SN1 because their carbocations are stabilised by resonance: the positive charge spreads over the double bond or the benzene ring. They also react well by SN2 when primary.

Effect of the halogen. For the same alkyl group, the C–X bond becomes weaker down the group, and the halide becomes a better leaving group, so reactivity is $R - I > R - B r > R - C l > R - F$ in both mechanisms.

Ambident nucleophiles. Cyanide and nitrite can attack through two different atoms. KCN is ionic, so the free $C N^{-}$ attacks mainly through carbon (the C–C bond is stronger than C–N), giving an alkyl cyanide (nitrile). AgCN is mainly covalent, so only the nitrogen lone pair is free, giving an isocyanide. Similarly KNO₂ gives an alkyl nitrite ( $R - O - N = O$ ) and AgNO₂ gives a nitroalkane ( $R - N O_{2}$ ).

Secondary halides can follow either path; the solvent and the nucleophile decide. A strong nucleophile at high concentration in an aprotic solvent favours SN2; a weak nucleophile in a polar protic solvent favours SN1.

SN1 and SN2 Mechanisms: formulas and key facts

SN2 rate law and example: $rate = k [C H_{3} C l] [O H^{-}]; C H_{3} C l + O H^{-} \to C H_{3} O H + C l^{-}$; One step, transition state, inversion of configuration.
SN1 rate law and example: $rate = k [(C H_{3})_{3} C B r]; (C H_{3})_{3} C B r + O H^{-} \to (C H_{3})_{3} C O H + B r^{-}$; Slow ionisation to a carbocation, then fast attack.
Reactivity orders: $S N 2 : C H_{3} X > 1° > 2° > 3°; S N 1 : 3° > 2° > 1° > C H_{3} X$; Allylic and benzylic halides: high SN1 reactivity (resonance).
Effect of halogen: $R - I > R - B r > R - C l > R - F$; Weaker C–X bond, better leaving group.
Ambident nucleophiles: $R - X K C N R - C N; R - X A g C N R - N C; K N O_{2} R - O N O; A g N O_{2} R - N O_{2}$; Ionic salts attack through the stronger-bond atom; silver salts through the free lone pair.

How to solve SN1 and SN2 Mechanisms questions

Board questions on this topic are reasoning and ordering questions, plus mechanism write-ups for 3 marks. Use this checklist.

1Classify the carbon bearing the halogen: methyl, 1°, 2°, 3°, allylic or benzylic.
2Decide the likely mechanism: methyl/1° → SN2; 3°, allylic, benzylic → SN1; 2° depends on conditions.
3For SN2 order, rank by steric hindrance around the carbon (fewer and smaller groups react faster; β-branching also slows SN2).
4For SN1 order, rank by carbocation stability (3° > 2° > 1°; resonance-stabilised cations are very stable).
5If the alkyl group is the same, rank by the halogen: I > Br > Cl > F.
6For mechanism questions, write the rate law, show the transition state (SN2) or the carbocation (SN1), and state the stereochemical result.
7For ambident nucleophiles, check whether the reagent is ionic (K⁺ salt) or covalent (Ag salt) before writing the product.

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.

Tertiary halides hydrolyse fast by SN1

Suppose a student keeps two reagent bottles, 1-bromobutane and 2-bromo-2-methylpropane, in a humid lab. After some weeks, moist litmus held over the opened 2-bromo-2-methylpropane bottle turns red, while the 1-bromobutane bottle shows almost no change.

Explain why the tertiary halide reacts with water vapour so much faster. Write the equation for the reaction and state the mechanism.

Try it first, then show the solutionSolution

Water is a weak nucleophile, so substitution must proceed through a carbocation.
2-Bromo-2-methylpropane ionises to the tertiary carbocation $(C H_{3})_{3} C^{+}$ , which is stabilised by three electron-releasing methyl groups (inductive effect and hyperconjugation).
The carbocation is captured by water: $(C H_{3})_{3} C B r + H_{2} O \to (C H_{3})_{3} C O H + H B r$ . The HBr makes the vapour acidic.
1-Bromobutane would give an unstable primary carbocation, so SN1 is very slow, and water is too weak a nucleophile for fast SN2.

Answer: The tertiary halide reacts by SN1 through a stable tertiary carbocation: $(C H_{3})_{3} C B r + H_{2} O \to (C H_{3})_{3} C O H + H B r$ ; the primary halide cannot.

Memory hook: Tertiary halides fall apart on their own (SN1); primary halides wait to be pushed (SN2).

Tricks to remember SN1 and SN2 Mechanisms

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

1 = one molecule in the slow step
SN1: one species (RX) in the rate law, two steps. SN2: two species in the rate law, one step. The numbers swap.
Use it when: Matching rate law and number of steps.
Tertiary Tears apart, Primary gets Pushed
3° halides ionise (SN1); 1° halides are pushed out from the back (SN2).
Use it when: Choosing the mechanism.
K is Kind to carbon, Ag gives Nitrogen
KCN → R–CN (via C); AgCN → R–NC (via N). Same for KNO₂ (R–ONO) and AgNO₂ (R–NO₂).
Use it when: Ambident nucleophile products.

SN1 and SN2 Mechanisms: solved examples

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

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

Which of the following reacts fastest by the SN1 mechanism?
(a) $C H_{3} B r$ (b) $C H_{3} C H_{2} B r$ (c) $(C H_{3})_{2} C H B r$ (d) $(C H_{3})_{3} C B r$

Try it first, then show the solutionSolution

SN1 rate depends on carbocation stability; the tertiary carbocation $(C H_{3})_{3} C^{+}$ is the most stable.

Answer: (d) $(C H_{3})_{3} C B r$ .

Example 2: SN1 versus SN2 · 2 marks · Very short answerBoard level

Give two differences between SN1 and SN2 reactions.

Try it first, then show the solutionSolution

Kinetics: SN1 is first order (rate $= k [R X]$ ); SN2 is second order (rate $= k [R X] [N u^{-}]$ ) (1 mark).
Path: SN1 occurs in two steps through a carbocation and gives racemisation; SN2 occurs in one step through a transition state with inversion of configuration. (Also: SN1 order 3° > 2° > 1°; SN2 order 1° > 2° > 3°.) (1 mark)

Answer: Order (first vs second) and mechanism (two steps via carbocation with racemisation vs one step via transition state with inversion).

Example 3: Ordering by mechanism · 3 marks · Short answerHarder board level

Arrange in order of increasing reactivity: (a) towards SN2: $(C H_{3})_{3} C B r$ , $C H_{3} B r$ , $(C H_{3})_{2} C H B r$ , $C H_{3} C H_{2} B r$ ; (b) towards SN1: $C H_{3} C H_{2} C H_{2} B r$ , $(C H_{3})_{3} C B r$ , $C H_{3} B r$ , $(C H_{3})_{2} C H B r$ . Justify each order in one line.

Try it first, then show the solutionSolution

(a) $(C H_{3})_{3} C B r < (C H_{3})_{2} C H B r < C H_{3} C H_{2} B r < C H_{3} B r$ (1 mark).
Reason: SN2 needs backside attack; more alkyl groups around the carbon mean more steric hindrance (½ mark).
(b) $C H_{3} B r < C H_{3} C H_{2} C H_{2} B r < (C H_{3})_{2} C H B r < (C H_{3})_{3} C B r$ (1 mark).
Reason: SN1 rate follows carbocation stability, $3° > 2° > 1° > C H_{3}^{+}$ (½ mark).

Answer: (a) 3° < 2° < 1° < methyl. (b) methyl < 1° < 2° < 3°.

Example 4: Mechanism and kinetics · 3 marks · Short answerHarder board level

(a) Describe the mechanism of the reaction of chloromethane with hydroxide ion. (b) How would the rate change if $[O H^{-}]$ were doubled? (c) How would the rate of hydrolysis of 2-bromo-2-methylpropane by aqueous $O H^{-}$ change if $[O H^{-}]$ were doubled? Explain.

Try it first, then show the solutionSolution

(a) SN2: $O H^{-}$ attacks C from the side opposite Cl; in the transition state $H O \dots C \dots C l$ the C–O bond is forming while C–Cl breaks, and the three H atoms lie in a plane; Cl⁻ leaves and the configuration inverts: $C H_{3} C l + O H^{-} \to C H_{3} O H + C l^{-}$ (1½ marks).
(b) Rate $= k [C H_{3} C l] [O H^{-}]$ , so doubling $[O H^{-}]$ doubles the rate (½ mark).
(c) This tertiary halide reacts by SN1: rate $= k [(C H_{3})_{3} C B r]$ . The slow step is ionisation, which does not involve $O H^{-}$ , so the rate is unchanged (1 mark).

Answer: (a) One-step SN2 with backside attack and inversion. (b) Doubles. (c) No change (SN1, $O H^{-}$ not in rate law).

Example 5: Ambident nucleophiles and allylic reactivity · 5 marks · Long answerHOTS / competency

(a) Bromoethane gives ethanenitrile with KCN but ethyl isocyanide with AgCN. Explain. (b) Write the products of bromoethane with KNO₂ and with AgNO₂. (c) Allyl chloride ( $C H_{2} = C H C H_{2} C l$ ) is hydrolysed much faster than 1-chloropropane by the SN1 route. Explain with the structures involved.

Try it first, then show the solutionSolution

(a) $C N^{-}$ is ambident (can attack through C or N). KCN is ionic and gives free $C N^{-}$ ; attack through carbon forms the stronger C–C bond, giving $C H_{3} C H_{2} C N$ (1 mark).
AgCN is mainly covalent (Ag–C bonded), so carbon's electrons are tied up; the nitrogen lone pair attacks, giving $C H_{3} C H_{2} N C$ (1 mark).
(b) KNO₂ gives ethyl nitrite, $C H_{3} C H_{2} - O - N = O$ ; AgNO₂ gives nitroethane, $C H_{3} C H_{2} N O_{2}$ (1 mark).
(c) Allyl chloride ionises to the allyl carbocation $C H_{2} = C H - C H_{2}^{+}$ , which is stabilised by resonance with $^{+} C H_{2} - C H = C H_{2}$ : the positive charge is spread over two carbons (1 mark).
1-Chloropropane would give the primary carbocation $C H_{3} C H_{2} C H_{2}^{+}$ with no resonance, which is much less stable, so its SN1 rate is very low (1 mark).

Answer: (a) Ionic KCN attacks via C; covalent AgCN via N. (b) Ethyl nitrite and nitroethane. (c) The allyl carbocation is resonance-stabilised; the propyl cation is not.

Where students lose marks in SN1 and SN2 Mechanisms

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

Mistake: Writing the SN1 rate law with the nucleophile
Wrong working
SN1: rate $= k [R X] [O H^{-}]$
Right working
SN1: rate $= k [R X]$
Do this: Only species in the slow step appear; for SN1 that is RX alone.
Usually costs: 1 mark.
Mistake: Reversing the reactivity orders
Do this: SN2 follows steric hindrance (methyl fastest); SN1 follows carbocation stability (tertiary fastest).
Usually costs: 1-2 marks in ordering questions.
Mistake: Drawing front-side attack in SN2
Do this: Show the nucleophile approaching at 180° to the leaving group, with the three groups flattening in the transition state.
Usually costs: 1 mark in mechanism answers.
Mistake: Giving isocyanide with KCN
Do this: KCN (ionic) → cyanide; AgCN (covalent) → isocyanide.
Usually costs: 1 mark.

Exam tips

In mechanism questions, always write the rate law, the intermediate or transition state, and the stereochemical outcome.
Use the terms 'steric hindrance', 'carbocation stability', 'inductive effect', 'hyperconjugation' and 'resonance' exactly.
For ordering questions, write the order first and then a one-line reason.
Quote NCERT's examples: $C H_{3} C l + O H^{-}$ for SN2 and $(C H_{3})_{3} C B r + O H^{-}$ for SN1.
When both SN1 and SN2 are possible (2° halides), mention the effect of solvent and nucleophile strength.

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
The rate of an SN2 reaction depends on:
(a) the concentration of the haloalkane only
(b) the concentration of the nucleophile only
(c) the concentrations of both the haloalkane and the nucleophile
(d) neither
Hint
Bimolecular.
Answer
(c).
Q2 · 2 marks · Very short answerBoard level
Why do tertiary haloalkanes react by the SN1 mechanism rather than SN2?
Hint
Two reasons: one for SN1, one against SN2.
Answer
They ionise to a stable tertiary carbocation (stabilised by three alkyl groups through the inductive effect and hyperconjugation), favouring SN1. The three bulky groups block backside attack, so SN2 is very slow.
Q3 · 3 marks · Short answerHarder board level
Write the main organic product of: (a) $C H_{3} C H_{2} B r$ + KCN (ethanolic) (b) $C H_{3} C H_{2} B r$ + AgCN (c) $C H_{3} C H_{2} C H_{2} C l$ + NaI (dry acetone)
Hint
(c) is the Finkelstein reaction.
Answer
(a) $C H_{3} C H_{2} C N$ (propanenitrile). (b) $C H_{3} C H_{2} N C$ (ethyl isocyanide). (c) $C H_{3} C H_{2} C H_{2} I$ (1-iodopropane), NaCl precipitates.
Q4 · 1 mark · Assertion-reasonHarder board level
Assertion (A): Benzyl chloride is highly reactive towards the SN1 reaction.
Reason (R): The benzylic carbocation is stabilised by resonance with the benzene ring.
(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
The charge spreads into the ring.
Answer
(a).
Q5 · 2 marks · Very short answerHarder board level
In each pair, which compound undergoes SN2 faster? Give a reason. (a) 1-bromobutane or 1-bromo-2-methylpropane (b) 1-chlorobutane or 1-iodobutane
Hint
Steric hindrance; leaving group.
Answer
(a) 1-Bromobutane: the methyl branch on the next carbon in 1-bromo-2-methylpropane hinders backside attack. (b) 1-Iodobutane: the C–I bond is weaker and iodide is a better leaving group.
Q6 · 3 marks · Short answerHarder board level
Why are SN1 reactions usually carried out in polar protic solvents such as water or ethanol? Draw the energy profile of an SN1 reaction in words.
Hint
Solvation of ions in the slow step.
Answer
The slow step breaks C–X to form a carbocation and $X^{-}$ . Polar protic solvents stabilise both ions by solvation (hydrogen bonding to $X^{-}$ , dipole attraction to $R^{+}$ ), lowering the energy of the first step. Energy profile: a high first hump (ionisation) leading to a dip (carbocation intermediate), then a lower second hump (attack of nucleophile) down to products.
Q7 · 3 marks · Short answerHOTS / competency
For the reaction of a haloalkane RBr with $O H^{-}$ , doubling [RBr] doubles the rate, while doubling $[O H^{-}]$ has no effect. (a) Write the rate law and identify the mechanism. (b) Is RBr more likely to be bromomethane or 2-bromo-2-methylpropane? (c) What would you expect if the reaction were carried out with an optically active RBr in which the C–Br carbon is the chiral centre?
Hint
First order in RBr only.
Answer
(a) Rate $= k [R B r]$ ; SN1. (b) 2-Bromo-2-methylpropane (tertiary, forms a stable carbocation). (c) The planar carbocation can be attacked from either face, so a racemic (or largely racemised) product forms.
Q8 · 4 marks · Case-basedHOTS / competency
Read and answer.
Suppose a student reacts three bromides, $C H_{3} B r$ , $(C H_{3})_{2} C H B r$ and $(C H_{3})_{3} C B r$ , separately (i) with aqueous ethanol alone (a weak nucleophile, polar protic) and (ii) with sodium iodide in dry acetone (a good nucleophile, polar aprotic). She times how fast each reaction occurs.
(a) Which bromide reacts fastest in (i)? Name the mechanism. (1 mark)
(b) Which bromide reacts fastest in (ii)? Name the mechanism. (1 mark)
(c) Why is 2-bromopropane slow in both? (1 mark)
(d) Write the equation for the fastest reaction in (ii). (1 mark)
Hint
Solvent and nucleophile choose the path.
Answer
(a) $(C H_{3})_{3} C B r$ , SN1 (stable 3° carbocation; weak nucleophile, protic solvent). (b) $C H_{3} B r$ , SN2 (least hindered; strong nucleophile, aprotic solvent). (c) Its 2° carbocation is less stable than 3° (slow SN1) and it is more hindered than methyl (slow SN2). (d) $C H_{3} B r + N a I \to C H_{3} I + N a B r$ (NaBr precipitates in acetone).

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.Order of SN2 reactivity of alkyl halides?Show answer
$C H_{3} X > 1° > 2° > 3°$ .
2.Intermediate in SN1?Show answer
A carbocation.
3.Stereochemical result of SN2?Show answer
Inversion of configuration.
4.Product of R–X with AgNO₂?Show answer
Nitroalkane, $R - N O_{2}$ .
5.Order of reactivity of halides for the same alkyl group?Show answer
$R - I > R - B r > R - C l > R - F$ .
6.Why does allyl chloride react fast by SN1?Show answer
Its carbocation is resonance-stabilised.

SN1 and SN2 Mechanisms: frequently asked questions

What is the difference between SN1 and SN2 in Class 12?

SN1 is first order and occurs in two steps through a carbocation; SN2 is second order and occurs in one step through a transition state with inversion. SN1 suits tertiary halides; SN2 suits methyl and primary halides.

Why is SN2 fastest for methyl halides?

The nucleophile must attack from the back of the carbon; methyl carbon has only small hydrogen atoms around it, so there is least steric hindrance.

Why does tertiary butyl bromide follow SN1?

It forms a stable tertiary carbocation, and its bulky methyl groups block the backside attack needed for SN2.

Which solvents favour SN1?

Polar protic solvents such as water, alcohols and ethanoic acid, which stabilise the carbocation and the leaving halide ion by solvation.

Why does KCN give alkyl cyanides but AgCN gives isocyanides?

KCN is ionic, so the free cyanide attacks through carbon; AgCN is covalent, so only the nitrogen lone pair is available to attack.