Organic Reaction Mechanisms: Step-by-Step Worked Examples (2)

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Tags Organic Mechanisms

Question

Write the mechanism for the SN1 reaction of (CH3)3C-Br(\text{CH}_3)_3\text{C-Br} with water. Explain why SN1 is favoured here.

Solution — Step by Step

(CH3)3C-Br(\text{CH}_3)_3\text{C-Br} is tert-butyl bromide — a tertiary alkyl halide. Three methyl groups attached to the carbon bearing Br.

The C-Br bond breaks heterolytically. Both bonding electrons leave with bromine, giving a tertiary carbocation:

(CH3)3C-Br→slow(CH3)3C++Br−(\text{CH}_3)_3\text{C-Br} \xrightarrow{\text{slow}} (\text{CH}_3)_3\text{C}^+ + \text{Br}^-

This is the rate-determining step. Rate =k[(CH3)3CBr]= k[(\text{CH}_3)_3\text{CBr}] — first order in substrate, zero order in nucleophile.

Water attacks the carbocation:

(CH3)3C++H2O→fast(CH3)3C-OH2+(\text{CH}_3)_3\text{C}^+ + \text{H}_2\text{O} \xrightarrow{\text{fast}} (\text{CH}_3)_3\text{C-OH}_2^+

A second water molecule removes a proton:

(CH3)3C-OH2++H2O→(CH3)3C-OH+H3O+(\text{CH}_3)_3\text{C-OH}_2^+ + \text{H}_2\text{O} \to (\text{CH}_3)_3\text{C-OH} + \text{H}_3\text{O}^+

Final product: tert-butyl alcohol.

Final answer: SN1 mechanism via tertiary carbocation; product is tert-butyl alcohol.

Why This Works

SN1 is favoured when:

  1. Substrate forms a stable carbocation: tertiary > secondary > primary. Hyperconjugation and inductive donation from the three methyl groups stabilise the t-butyl cation.
  2. Polar protic solvent: water solvates and stabilises the ions.
  3. Weak nucleophile: water is a poor nucleophile, so SN2 (which needs a strong nucleophile attacking simultaneously) doesn’t compete.

For primary substrates, SN2 wins because the carbocation would be unstable. For methyl halides, only SN2 is possible.

Alternative Method

Energy diagram approach: the rate-determining step has a high transition state corresponding to ionisation. The activation energy is offset by the stability of the resulting carbocation. Tertiary cations have the lowest activation energy of any alkyl carbocation.

Common Mistake

Students draw the mechanism with water attacking simultaneously with Br leaving — that’s SN2, not SN1. In SN1, the carbocation forms first (slow step), then water attacks (fast step). The two steps are separate.

Also: students forget step 3 (deprotonation). The protonated alcohol (CH3)3C-OH2+(\text{CH}_3)_3\text{C-OH}_2^+ is an intermediate, not the product. You must remove the proton to get the neutral alcohol.

JEE/NEET love asking the rate law: SN1 is first order, SN2 is second order. If the question gives you rate =k[substrate]= k[\text{substrate}] only, it’s SN1. If rate =k[substrate][Nu]= k[\text{substrate}][\text{Nu}], it’s SN2.

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