Question
Why does tert-butyl bromide preferentially undergo SN1 rather than SN2 nucleophilic substitution? Explain based on mechanism and electronic/steric factors.
Solution — Step by Step
Tert-butyl bromide has the structure . The carbon bearing bromine (the electrophilic carbon, C-1) is a tertiary carbon — it is directly bonded to three other carbon groups (three methyl groups).
This tertiary nature is the key to everything that follows.
The SN2 mechanism requires the nucleophile to attack the back side of the C–Br bond simultaneously while the bond breaks (concerted, one-step). This requires physical access to the carbon from the back.
In tert-butyl bromide, the three methyl groups create severe steric crowding around the central carbon. Any nucleophile approaching from the back encounters intense van der Waals repulsion from the three bulky methyl groups. This steric shield makes the SN2 transition state extremely high in energy — effectively preventing the reaction.
Steric effect: Primary (least hindered) → Secondary → Tertiary (most hindered). SN2 rate order: methyl > primary > secondary; tertiary substrates essentially don’t undergo SN2.
The SN1 mechanism proceeds in two steps:
- Slow (rate-determining): Ionisation of C–Br bond → tertiary carbocation + Br⁻
- Fast: Nucleophile attacks the carbocation from either face
The tertiary carbocation is highly stabilised by three electron-donating methyl groups through hyperconjugation and inductive effect. The positive charge on carbon is partially dispersed over the three methyl groups, significantly lowering the energy of the intermediate.
Carbocation stability order: Tertiary > Secondary > Primary > Methyl
Since the rate-determining step produces the carbocation, a more stable carbocation means a lower activation energy — faster SN1 rate.
Unlike SN2 (which gives inversion of configuration — Walden inversion), SN1 via a planar carbocation intermediate leads to attack from both faces equally, giving a racemic mixture (approximately equal amounts of both enantiomers).
This loss of stereochemical information is a diagnostic test for SN1 — if a chiral substrate gives a racemic product, SN1 was the likely mechanism.
Why This Works
The fundamental competition between SN1 and SN2 comes down to two factors:
- Steric environment of the carbon bearing the leaving group (sterically hindered → SN2 blocked)
- Stability of the carbocation formed (stable → SN1 favoured)
For tertiary substrates, both factors point to SN1:
- SN2 is blocked by steric hindrance
- SN1 is favoured by stable tertiary carbocation
For primary substrates, both factors favour SN2: little steric hindrance, unstable primary carbocation.
Common Mistake
Students sometimes say “tert-butyl bromide can’t react at all” or “it only forms elimination products.” While E1 and E2 (elimination) do compete with SN1 in certain conditions (especially with bulky, strong bases), tert-butyl bromide CAN undergo SN1 with good nucleophiles (water, methanol) under appropriate conditions. The statement “tertiary substrates never undergo substitution” is incorrect — they undergo SN1, not SN2.
JEE question pattern: “Arrange in order of SN1/SN2 reactivity.” For SN1: more substituted → faster (better carbocation). For SN2: less substituted → faster (less steric hindrance). These are opposite trends — tertiary is fastest for SN1 and slowest for SN2.