Nucleophilic Substitution: SN1 and SN2 Mechanisms

Nucleophilic Substitution — The Central Reaction

The polar C-X bond means haloalkanes are attacked by nucleophiles (electron-rich species: OH-, CN-, NH$_3$, OR-, etc.). The nucleophile replaces the halide (a good leaving group). This is nucleophilic substitution, and it happens by two distinct mechanisms: SN2 and SN1.

$$\text{Nu}^- + \text{R-X} \rightarrow \text{R-Nu} + \text{X}^-$$

Which mechanism operates depends mainly on the structure of the alkyl halide (1°/2°/3°), the nucleophile, and the solvent. Understanding the difference is the most important skill in this chapter.

The SN2 Mechanism (Bimolecular)

SN2 = Substitution Nucleophilic Bimolecular. It is a single-step reaction in which the nucleophile attacks the carbon from the side opposite to the leaving group (backside attack) as the C-X bond breaks.

Key features:

• One step, through a transition state in which both the nucleophile and the leaving group are partially bonded to carbon.
• Second-order kinetics: rate = k[R-X][Nu-] — both the halide and the nucleophile appear in the rate law (hence "bimolecular").
• Inversion of configuration at the carbon (like an umbrella turning inside out) — the Walden inversion.
• Reactivity order: CH$_3$-X > 1° > 2° > 3°. Steric hindrance decides this: bulky groups around the carbon block the backside attack, so methyl and primary halides react fastest and tertiary slowest.

Worked logic: in CH$_3$Br, the small methyl group leaves the backside open, so SN2 is fast. In (CH$_3$)$_3$CBr, the three methyls block the approach, so SN2 is essentially impossible — it goes by SN1 instead.

The SN1 Mechanism (Unimolecular)

SN1 = Substitution Nucleophilic Unimolecular. It is a two-step reaction that goes through a carbocation intermediate.

• Step 1 (slow, rate-determining): the C-X bond breaks heterolytically to give a carbocation and a halide ion.
• Step 2 (fast): the nucleophile attacks the planar carbocation.

Key features:

• First-order kinetics: rate = k[R-X] — only the halide appears in the rate law (the slow step does not involve the nucleophile), hence "unimolecular."
• Racemisation: the planar carbocation can be attacked from either face, giving a racemic mixture (a mix of both enantiomers).
• Reactivity order: 3° > 2° > 1° > CH$_3$-X — governed by carbocation stability (3° carbocations are the most stable, so they form fastest).

Allylic and benzylic halides react very fast by SN1 because their carbocations are resonance-stabilised.

Key Point — the great contrast:

• SN2: one step, rate ∝ [RX][Nu], inversion, 1° fastest (steric).
• SN1: two steps (carbocation), rate ∝ [RX], racemisation, 3° fastest (carbocation stability).

[JEE Tip] To predict the mechanism: primary halide + strong nucleophile → SN2; tertiary halide (or allylic/benzylic) + polar protic solvent → SN1. Secondary halides can go by either, depending on conditions.

Solved Examples

Example 1: Predict the mechanism

By which mechanism does CH$_3$CH$_2$Br react with OH-, and why?

Solution: SN2. CH$_3$CH$_2$Br is a primary halide with little steric hindrance, so the nucleophile can attack the backside in a single step. (Primary halides favour SN2.)

Example 2: Tertiary halide mechanism

By which mechanism does (CH$_3$)$_3$CBr undergo hydrolysis, and why?

Solution: SN1. The tertiary carbon is too hindered for backside attack, but it forms a stable 3° carbocation, so it reacts via the two-step SN1 mechanism.

Example 3: SN2 reactivity order

Arrange CH$_3$Br, CH$_3$CH$_2$Br, (CH$_3$)$_2$CHBr, (CH$_3$)$_3$CBr in decreasing order of SN2 reactivity.

Solution: Steric hindrance increases from methyl to tertiary, so SN2 rate decreases: CH$_3$Br > CH$_3$CH$_2$Br > (CH$_3$)$_2$CHBr > (CH$_3$)$_3$CBr.

Example 4: SN1 reactivity order

Arrange the same four bromides in decreasing order of SN1 reactivity.

Solution: Carbocation stability increases from methyl to tertiary, so SN1 rate increases: (CH$_3$)$_3$CBr > (CH$_3$)$_2$CHBr > CH$_3$CH$_2$Br > CH$_3$Br.

Example 5: Kinetics distinguishes the mechanisms

A substitution reaction is found to be first order (rate = k[R-X]). Which mechanism is operating?

Solution: SN1. First-order kinetics (rate depending only on the halide) indicates the slow step is the unimolecular formation of a carbocation.

Example 6: Why allylic halides are reactive in SN1

Why does an allylic halide (e.g. CH$_2$=CH-CH$_2$-Cl) react rapidly by SN1?

Solution: Its carbocation is resonance-stabilised (the positive charge is delocalised over the allylic system), so it forms easily, accelerating the rate-determining step of SN1.

Example 7: Effect of nucleophile concentration

In an SN2 reaction, what happens to the rate if the nucleophile concentration is doubled?

Solution: The rate doubles. SN2 is second order: rate = k[R-X][Nu-], so it is first order in the nucleophile.

Example 8: Effect of nucleophile in SN1

In an SN1 reaction, what happens to the rate if the nucleophile concentration is doubled?

Solution: The rate is unchanged. SN1 is first order: rate = k[R-X] — the nucleophile is not in the rate-determining step.

Example 9: Faster hydrolysis

Which is hydrolysed faster by aqueous KOH (SN1): C$_6$H$_5$CH$_2$Cl (benzyl chloride) or CH$_3$CH$_2$CH$_2$Cl (n-propyl chloride)?

Solution: Benzyl chloride — it forms a resonance-stabilised benzylic carbocation, so it undergoes SN1 much faster than the primary n-propyl chloride.

Example 10: Solvent effect

Why do polar protic solvents (like water) favour SN1?

Solution: Polar protic solvents stabilise the carbocation and the leaving halide ion through solvation, lowering the energy of the rate-determining ionisation step — thereby favouring the SN1 pathway.