Skip to main content

Dehydrohalogenation of t-butyl bromide; E1 vs E2

Short answer

Tertiary butyl bromide undergoes E1 via slow carbocation formation then fast base removal, yielding 2-methylpropene; Rate = k[RX]. Unlike E1, E2 is a concerted one-step mechanism with rate = k[RX][base], requiring strong base and anti-coplanar H and X.

Explain the reaction mechanism for the dehydrohalogenation of tertiary butyl bromide. Differentiate between E¹ and E² mechanism.

Dehydrohalogenation of tert-butyl bromide

tert-Butyl bromide, (CH₃)₃C–Br, is a tertiary halide. When heated with alcoholic KOH (or in a polar protic solvent such as ethanol), it undergoes E1 elimination through a stable 3° carbocation to give 2-methylpropene.

Mechanism: E1 (two steps).

Step 1 (slow): ionisation to carbocation
 (CH₃)₃C—Br  ---slow--->  (CH₃)₃C⁺ + Br⁻

Step 2 (fast): base removes a β-hydrogen
 OH⁻ + H—CH₂—C⁺(CH₃)₂ ---fast---> H₂O + CH₂=C(CH₃)₂

Overall: (CH₃)₃C–Br + KOH(alc) → (CH₃)₂C=CH₂ + KBr + H₂O (2-methylpropene)

  • Rate = k[RX]: first order (unimolecular), as only the halide takes part in the slow step.
  • The carbocation is an intermediate, so rearrangement is possible, and E1 competes with SN1.

Difference between E1 and E2

BasisE1E2
StepsTwo steps, carbocation intermediateOne step, concerted
Rate lawRate = k[RX], first orderRate = k[RX][Base], second order
MolecularityUnimolecularBimolecular
Substrate3° > 2° (stable carbocation)3° > 2° > 1°
BaseWeak base is enoughStrong, concentrated base needed
SolventPolar proticLess polar / alcohol
RearrangementPossibleNot possible
StereochemistryNo special geometryAnti-periplanar H and X
Example(CH₃)₃CBr in ethanolCH₃CH₂Br + alc. KOH

✓ Answer checked by Claude Opus 5.5 (high effort)

All Engineering Chemistry chapter 10 questions →