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Chemistry · When a base takes a hydrogen

Elimination, and choosing the pathway

A nucleophile can attack carbon, or a base can take a hydrogen. Elimination is the second path, and exams love asking which one wins.

JEE Main: high priority · JEE Advanced: high priority · about 20 minutes

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β-Elimination

CH3CH2Br+KOHethanol→CH2=CH2+KBr+H2O

The base removes a hydrogen from the β-carbon, next to the one holding X. The two names below are not in NCERT, but JEE questions use them.

E2 of 2-bromobutane

HCH3 HBrCH3 HEtO− 

Three pairs move at the same time: base to H, C-H to the new π bond, and C-Br to bromide. H and Br must lie anti-periplanar, as drawn.

CH3 CHCHCH3 EtOH+Br− 

But-2-ene forms, as the Zaitsev product. There is no intermediate and no rearrangement.

E1 of tert-butyl bromide in hot ethanol

(CH3 )3 CBrslow(CH3 )3 C+ +Br− 

The same first step as SN1, so the same rate law: rate =k[RX].

H3 CC+ CH2 HCH3 EtOH(CH3 )2 CCH2 +EtOH2 + 

A weak base is enough here, because the cation is so keen to lose a proton. Heat favours this over adding ethanol.

E2E1
StepsOneTwo, through a carbocation
Ratek[RX][Base]k[RX]
BaseStrongWeak or none
GeometryH and X anti-periplanarNo special need
RearrangementNoPossible
Substrate order3∘>2∘>1∘3∘>2∘>1∘
BrcdHabH and Br at 180°Base takes HBr leaves together
E2 needs H and the leaving group anti-periplanar.

Watch the animation: E2 needs the anti arrangement

Which alkene?

Zaitsev's rule: the base prefers the β-hydrogen that gives the alkene with more alkyl groups on the double bond, because that alkene is more stable. 2-Bromobutane gives mainly but-2-ene. A very bulky base such as (CH3)3CO− cannot reach the crowded hydrogen, so the less substituted alkene increases. This is called the Hofmann product.

Watch the animation: Which hydrogen goes? Zaitsev or Hofmann

Geometry in rings

In a cyclohexane chair, H and X are anti-periplanar only when both are axial. A chloride held equatorial by a large group cannot react by E2 until the ring flips.

Explore in 3D: Axial chloride in a chair

JEE Advanced: E2 is stereospecific

Because H and X must be anti, one diastereomer gives one alkene geometry. Build the molecule in the anti arrangement, then remove H and X. The groups that end up on the same side stay on the same side of the double bond. So meso-1,2-dibromo-1,2-diphenylethane with base gives the (E)-bromoalkene, in which the two phenyl groups are cis.

Aqueous and alcoholic KOH are not the same reagent. Aqueous base tends to substitute. Alcoholic base tends to eliminate.

Choosing the pathway

SubstrateGood nucleophile, weak base (I−, CN−, RS−)Strong base (OH−, RO−)Weak nucleophile (water, alcohol)
Methyl, primarySN2SN2, or E2 with a bulky baseVery slow
SecondarySN2E2 mostlySN1 with E1
TertiarySN1 with E1E2SN1 with E1

This table is a working summary based on Clayden, Organic Chemistry. Real reactions give mixtures, so read it as the main path.

Try the simulator: Predict the pathway

Higher temperature favours elimination, because it makes more molecules. An ethanol solvent is less polar than water, which also favours elimination.

Worked example: one halide, four conditions

What is the main organic product from 2-bromo-2-methylbutane with (a) ethanol alone, cool, (b) sodium ethoxide, (c) potassium tert-butoxide, (d) ethanol alone, hot?

Show the solution
  • (a) Tertiary, weak nucleophile: SN1 gives 2-ethoxy-2-methylbutane, with some alkene.
  • (b) Strong small base: E2 gives 2-methylbut-2-ene, the Zaitsev alkene.
  • (c) Bulky base: E2 gives 2-methylbut-1-ene, the Hofmann-type alkene.
  • (d) Heat moves it towards elimination: more E1 alkene, mainly 2-methylbut-2-ene.