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Nucleophilic Substitution: SN1 and SN2 Mechanisms

Nucleophilic substitution is a reaction in which a nucleophile replaces a leaving group such as halide ion from a haloalkane. SN1 occurs in two steps through a carbocation, while SN2 occurs in one concerted step.

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Student-friendly explanation

How to write this in exams

1
Start with the exact idea
Nucleophilic substitution is a reaction in which a nucleophile replaces a leaving group such as halide ion from a haloalkane. SN₁ occurs in two steps through a carbocation, while SN₂ occurs in one concerted step.
2
Then show how to use it
Mark the carbon bonded to halogen, classify it as methyl, 1°, 2°, or 3°, check nucleophile and solvent, choose SN₁ or SN₂, then state rate law and stereochemical result.
3
Add one concrete example
Tert-butyl bromide undergoes hydrolysis mainly by SN₁ because it forms a stable tertiary carbocation. Methyl bromide reacts with OH^- mainly by SN₂ because backside attack is easy and no stable carbocation is needed.
4
Avoid this incomplete answer
Saying tert-butyl bromide undergoes SN₂ with OH^- as the main path ignores steric hindrance at the tertiary carbon.

Definition

Example

Tert-butyl bromide undergoes hydrolysis mainly by SN₁ because it forms a stable tertiary carbocation. Methyl bromide reacts with OH^- mainly by SN₂ because backside attack is easy and no stable carbocation is needed.

Rule to remember

SN₁ rate law: rate = k[R-X]. SN₂ rate law: rate = k[R-X][Nu-]. Reactivity pattern for substrate: SN₁ generally 3° > 2° > 1°, while SN₂ generally methyl > 1° > 2° > 3°.

Memory hook

SN₁: one molecule in rate law, carbocation, racemisation. SN₂: two particles in rate law, backside attack, inversion.

Examples and method

Worked example

Predict the major mechanism for hydrolysis of (CH₃)₃CBr. The substrate is tertiary, so carbocation formation is feasible. Water is a weak nucleophile in a polar medium, so the reaction follows SN₁ and forms tert-butyl alcohol.

Method to apply

Mark the carbon bonded to halogen, classify it as methyl, 1°, 2°, or 3°, check nucleophile and solvent, choose SN₁ or SN₂, then state rate law and stereochemical result.

Diagram support

A comparison diagram is strongly useful: show two-step SN₁ with carbocation and one-step SN₂ backside attack with transition state.

How CBSE asks it

Questions ask for mechanism comparison, rate law, order of reactivity, stereochemical result, and reasons for racemisation or inversion.

Avoid common mistakes

Common confusion

Students often think stronger nucleophile always means SN₂. Substrate structure is equally important; a tertiary haloalkane is too crowded for normal SN₂ attack.

Common wrong answer

Saying tert-butyl bromide undergoes SN₂ with OH^- as the main path ignores steric hindrance at the tertiary carbon.

Exam tip

For mechanism questions, check substrate class first, then nucleophile strength, solvent, and stereochemical outcome.

Quick check

Why does 2-bromobutane give inversion in an SN₂ reaction?

In SN₂, the nucleophile attacks from the side opposite the leaving group, so the arrangement around the chiral carbon is inverted.

Answer writing and exam use

1-mark answer

2-mark answer

Nucleophilic substitution is a reaction in which a nucleophile replaces a leaving group such as halide ion from a haloalkane. SN₁ occurs in two steps through a carbocation, while SN₂ occurs in one concerted step. SN₁ rate law: rate = k[R-X]. SN₂ rate law: rate = k[R-X][Nu-]. Reactivity pattern for substrate: SN₁ generally 3° > 2° > 1°, while SN₂ generally methyl > 1° > 2° > 3°. Tert-butyl bromide undergoes hydrolysis mainly by SN₁ because it forms a stable tertiary carbocation. Methyl bromide reacts with OH^- mainly by SN₂ because backside attack is easy and no stable carbocation is needed.

3-mark answer

SN₁ reactions are favoured by tertiary substrates because the carbocation intermediate is more stable. They may lead to racemisation when the carbocation is planar and the nucleophile attacks from either side. SN₂ reactions are favoured by primary substrates because backside attack is less hindered; they usually give inversion of configuration. Leaving group ability and solvent also affect the rate. SN₁ rate law: rate = k[R-X]. SN₂ rate law: rate = k[R-X][Nu-]. Reactivity pattern for substrate: SN₁ generally 3° > 2° > 1°, while SN₂ generally methyl > 1° > 2° > 3°. Predict the major mechanism for hydrolysis of (CH₃)₃CBr. The substrate is tertiary, so carbocation formation is feasible. Water is a weak nucleophile in a polar medium, so the reaction follows SN₁ and forms tert-butyl alcohol. Questions ask for mechanism comparison, rate law, order of reactivity, stereochemical result, and reasons for racemisation or inversion. Saying tert-butyl bromide undergoes SN₂ with OH^- as the main path ignores steric hindrance at the tertiary carbon.

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