Question
Explain the SN1 mechanism for the hydrolysis of tert-butyl bromide with aqueous NaOH. Why does the reaction proceed through a carbocation intermediate? What is the stereochemical outcome?
(JEE Main 2023 pattern; NEET asks mechanism identification)
Solution — Step by Step
The C–Br bond breaks heterolytically without any help from the nucleophile. The leaving group (Br⁻) departs on its own, forming a carbocation.
This is the slow step — it determines the overall rate. Since only the substrate is involved, the rate depends only on the substrate concentration: .
The nucleophile (OH⁻) attacks the planar carbocation from either side — top or bottom — because the carbocation is sp² hybridised with a vacant p-orbital perpendicular to the plane.
Because the nucleophile can attack from both faces of the flat carbocation, we get a mixture of both configurations — if the substrate were chiral, we would get racemisation (roughly 50:50 mix of R and S products).
In practice, slight excess inversion is sometimes observed because the leaving group partially shields one face as it departs.
flowchart TD
A["(CH₃)₃C-Br<br/>Substrate"] -->|"Slow: C-Br bond breaks"| B["(CH₃)₃C⁺ + Br⁻<br/>Carbocation formed"]
B -->|"Fast: OH⁻ attacks from top"| C["Product with retention"]
B -->|"Fast: OH⁻ attacks from bottom"| D["Product with inversion"]
C --> E["Racemic mixture<br/>(~50:50 R and S)"]
D --> E
style B fill:#ffeb3b,stroke:#333Why This Works
The reason tert-butyl bromide prefers SN1 over SN2 comes down to carbocation stability. A tertiary carbocation is stabilised by hyperconjugation from nine C–H sigma bonds and by the +I effect of three methyl groups.
The more stable the carbocation, the lower the activation energy for Step 1, and the faster the SN1 pathway. Primary substrates cannot form stable carbocations, so they go SN2 instead.
The first-order kinetics () arise because the nucleophile is not involved until after the rate-determining step. You could double the concentration of OH⁻ and the rate would not change — that is the hallmark of SN1.
Alternative Method
For MCQs, use this shortcut to identify SN1: 3° substrate + weak nucleophile + polar protic solvent = SN1. If any of these three conditions are met strongly, SN1 is favoured. You do not need to draw the full mechanism — just check these three factors.
Common Mistake
Students write the SN1 mechanism as a single concerted step — showing the nucleophile attacking while the leaving group departs. That is SN2, not SN1. The defining feature of SN1 is the two-step process: first the leaving group leaves (forming a carbocation), THEN the nucleophile attacks. If your mechanism shows both happening simultaneously, you have drawn SN2 by mistake.