Elimination reactions

β-Elimination (dehydrohalogenation). Label the halogen-bearing carbon the α-carbon and any carbon directly attached to it a β-carbon. When the haloalkane is heated with a concentrated, anhydrous (alcoholic) solution of a strong base — usually $KOH$ or $NaOH$ in ethanol — the base pulls off a hydrogen from a β-carbon while the halide ion leaves the α-carbon, and the two carbons close the gap with a new π bond:

where: $X$ is the halogen ($Cl$, $Br$ or $I$) on the α-carbon, the β-carbon is the neighbouring carbon that supplies the hydrogen removed by the base, "alc. KOH" is potassium hydroxide dissolved in ethanol rather than water, and $\Delta$ marks that heating drives the reaction. Losing $H$ and $X$ from adjacent carbons is why this is also called 1,2-elimination or dehydrohalogenation — literally, removing a hydrogen halide.

Why alcoholic, not aqueous. The solvent choice is not incidental. Water is polar enough to stabilise the ionic transition states of substitution ($S_N1/S_N2$), so aqueous $KOH$ mostly gives back the alcohol. Ethanol is less polar and lets the hydroxide ion act primarily as a base, snatching a β-hydrogen rather than attacking the α-carbon as a nucleophile. Anhydrous, alcoholic, heated conditions are therefore the standard recipe for pushing a haloalkane toward elimination rather than substitution.

Saytzeff's (Zaitsev's) rule. Many haloalkanes have β-hydrogens on more than one side, so more than one alkene is geometrically possible. Saytzeff's rule says the major product is the more substituted alkene — the one with more alkyl groups directly attached to the two double-bond carbons. More substituted alkenes are more stable, because alkyl groups donate electron density into the π system by hyperconjugation and induction, and under the thermal, base-driven conditions of dehydrohalogenation the reaction favours this more stable product.

The competition between substitution and elimination comes down to a simple tug-of-war for the same reagent's attention: does the hydroxide/alkoxide species attack the α-carbon (substitution) or pull off a β-hydrogen (elimination)? Three factors settle which pathway wins.

Substrate class controls steric access. A primary halide leaves its α-carbon wide open, so a base can approach and displace the halide directly — substitution ($S_N2$) usually dominates, though some elimination still happens. A tertiary halide crowds the α-carbon with three alkyl groups, blocking the backside attack an $S_N2$ pathway needs, so the base finds it far easier to reach around and pluck a β-hydrogen instead — elimination dominates.

Base strength and bulk matter. A small, strongly basic species reacts readily at carbon (substitution) unless something discourages it. A bulky base, such as potassium tert-butoxide, cannot easily squeeze in for backside attack no matter how open the α-carbon is, so it defaults to removing a β-hydrogen — favouring elimination even on primary halides.

Product stability sets the outcome among alkenes. Once elimination is happening, it can proceed in more than one direction if more than one β-carbon bears hydrogens. The transition state leading to the more substituted alkene is lower in energy, because the developing double bond is better stabilised by neighbouring alkyl groups (hyperconjugation and induction) — the same reasoning that makes tri-substituted alkenes more stable than di-substituted ones, and di- more stable than mono-substituted. Saytzeff's rule is this stability argument applied to predict the major product.

Problem: predict the major and minor alkene products when 2-bromobutane is heated with alcoholic KOH, then extend the reasoning to 2-bromo-2-methylbutane.

Step 1. Identify the α-carbon of 2-bromobutane, $CH_3-CHBr-CH_2-CH_3$: it is $C2$, bearing the bromine.

Step 2. Find every β-carbon with hydrogens available to eliminate. Here $C1$ (a $CH_3$ group) and $C3$ (a $CH_2$ group) both qualify, so elimination can proceed in two directions.

Step 3. Write both possible alkenes. Removing a β-hydrogen from $C1$ gives $CH_2{=}CH{-}CH_2{-}CH_3$ (but-1-ene, mono-substituted). Removing a β-hydrogen from $C3$ gives $CH_3{-}CH{=}CH{-}CH_3$ (but-2-ene, di-substituted).

Step 4. Apply Saytzeff's rule. The di-substituted but-2-ene is more stable than the mono-substituted but-1-ene, so but-2-ene is the major product (as a mixture of cis and trans isomers), with but-1-ene as the minor product.

Competitive stretch

Now take 2-bromo-2-methylbutane, $(CH_3)_2CBr-CH_2-CH_3$, a tertiary halide. Its α-carbon ($C2$) carries two methyl groups and one ethyl group, so β-hydrogens sit both on the methyls and on $C3$. Eliminating toward a methyl gives $CH_2{=}C(CH_3)-CH_2-CH_3$ (2-methylbut-1-ene, di-substituted); eliminating toward $C3$ gives $(CH_3)_2C{=}CH-CH_3$ (2-methylbut-2-ene, tri-substituted). Saytzeff's rule again picks the more substituted alkene: 2-methylbut-2-ene is the major product, and this example doubles as a reminder that a tertiary halide — sterically blocked from $S_N2$ substitution — reacts with alcoholic KOH mainly by elimination in the first place.

• Using aqueous KOH and still expecting elimination — aqueous conditions favour substitution to the alcohol; elimination needs the alcoholic (anhydrous) medium.
• Applying Saytzeff's rule by counting the alkene with more total carbons rather than counting substituents directly on the two double-bond carbons — chain length elsewhere in the molecule is irrelevant.
• Forgetting that a primary halide often has only one type of β-carbon, so there is nothing for Saytzeff's rule to choose between — it only applies when more than one elimination direction exists.
• Treating substitution and elimination as an either/or choice rather than competing pathways whose ratio shifts with substrate class, base strength/bulk, and solvent — both usually occur together to some extent.

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