Imagine you have a molecule that is "unstable" in a specific way — it carries a halogen atom (like Cl, Br, I) on one carbon and a hydrogen atom on the neighbouring carbon. If you treat it with a strong base, the base can pull off that hydrogen, and simultaneously the halogen leaves as a negative ion. The two carbons that lost these atoms now form a double bond between them. That's the core idea: dehydrohalogenation is the elimination of H and X (halogen) from adjacent carbons, producing an alkene.
The name itself tells you what happens: dehydro (removal of hydrogen) + halogenation (removal of halogen). So you are literally removing a hydrogen halide (HX) from the molecule.
The Precise Reaction
A haloalkane (alkyl halide) is treated with alcoholic KOH (potassium hydroxide dissolved in ethanol). The KOH acts as a strong base. The reaction follows this general pattern:
R−CH2−CH2−XKOHalcoholicR−CH=CH2+KX+H2O
For example, bromoethane gives ethene:
CH3−CH2−BrKOHalcoholicCH2=CH2+KBr+H2O
Watch out
Aqueous KOH (KOH in water) does not cause elimination — it gives substitution (an alcohol). The alcoholic medium is essential because it keeps the base strong enough to pull off the hydrogen, and it does not favour the competing substitution reaction.
Why Alcoholic KOH and Not Aqueous?
In water, the hydroxide ion (OH−) is heavily solvated — surrounded by water molecules — which reduces its basic strength. In ethanol, the solvation is weaker, so OH− is a much stronger base. A strong base is needed to abstract the β-hydrogen (the hydrogen on the carbon next to the one bearing the halogen). The reaction proceeds via a one-step concerted mechanism (E2) where the base pulls the H, the halogen leaves, and the double bond forms — all at once.
General dehydrohalogenation (beta-elimination): a base removes the hydrogen on the beta-carbon while the halogen X leaves from the alpha-carbon, forming a C=C double bond
Saytzeff's Rule — Which Alkene Forms?
When the haloalkane has more than one possible β-hydrogen (i.e., the carbon next to the halogen is attached to two different sets of hydrogens), more than one alkene can form. Saytzeff's rule tells you which one is the major product:
In dehydrohalogenation, the alkene with the more substituted double bond (the one with more alkyl groups attached to the double-bonded carbons) is the major product.
Why? More substituted alkenes are more stable (hyperconjugation and inductive effects). The reaction favours the pathway that leads to the more stable alkene.
NCERT's own example uses 2-bromopentane, and the preference is just as clear there:
Saytzeff's rule illustrated with 2-bromopentane: alcoholic KOH gives pent-2-ene (81 percent, the more substituted major product) and pent-1-ene (19 percent, minor)
Example: 2-bromobutane has two possible β-hydrogens: …
Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.
KEAM 2026Set pha-2026-0418F4 marksMCQ
Q.2-Methylpropene is obtained when sodium methoxide reacts with
(A) n-butyl bromide
(B) n-propyl bromide
(C) sec-butyl bromide
(D) isopropyl bromide
(E) tert-butyl bromide
›Reveal solutionSolution
Strong base CH3ONa on tert-butyl bromide favours E2 elimination, giving 2-methylpropene (CH3)2C=CH2.
Sodium methoxide is a strong, bulky base, so with a tertiary halide it promotes elimination over substitution. tert-Butyl bromide (CH3)3C−Br loses HBr to form (CH3)2C=CH2, which is 2-methylpropene (isobutylene). Primary/secondary halides would give d …
Q.The correct order of the rate of β-elimination reaction among the alkyl halides is
(A) Secondary > Tertiary > Primary
(B) Tertiary > Primary > Secondary
(C) Tertiary > Secondary > Primary
(D) Primary > Tertiary > Secondary
(E) Primary > Secondary > Tertiary
›Reveal solutionSolution
Ease of dehydrohalogenation increases with the number of alkyl groups on the halogen-bearing carbon: tertiary > secondary > primary.
In β-elimination (dehydrohalogenation) of alkyl halides, the reactivity increases with substitution at the α-carbon. Tertiary halides eliminate fastest because they form the most stable (most substituted, Saytzeff) alkene and, via E1, the most s …
