Q.When alkyl halides are treated with alcoholic KOH, the products are ________.

Alcoholic KOH with heat is the standard reagent for dehydrohalogenation (elimination) of a haloalkane to an alkene.

CH3-CH2-CH(Br)-CH3 is 2-bromobutane. Treating it with KOH dissolved in ethanol (alcoholic KOH), under heat, makes hydroxide act as a base rather than a nucleophile: it abstracts a beta-hydrogen and the C-Br bond breaks, eliminating HBr and forming the C=C double bond (E2 mechanism), giving the more substituted, Zaitsev alkene CH3-CH=CH-CH3 (2-butene).

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Alcoholic KOH favours elimination over substitution, so n-propyl bromide (CH3-CH2-CH2-Br) loses HBr to form propene (CH3-CH=CH2).

Whether a haloalkane undergoes substitution or elimination with KOH depends on the solvent. Aqueous KOH provides OH- ions, which are good nucleophiles and favour substitution (SN2), giving an alcohol. Alcoholic KOH, however, generates ethoxide ions (from the solvent) which are strong, bulky bases that preferentially abstract a beta-hydrogen atom from the substrate, in an E2 elimination, expelling the bromide ion and forming a carbon-carbon double bond.

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Alkyl halides heated with alcoholic KOH undergo elimination (dehydrohalogenation) to give alkenes (olefins), not alcohols.

When an alkyl halide is treated with KOH dissolved in alcohol (an alcoholic/non-aqueous medium), the hydroxide ion acts predominantly as a strong base rather than as a nucleophile. It abstracts a beta-hydrogen atom from the carbon adjacent to the carbon bearing the halogen, causing the elimination of H and X (HX) across the C-C bond. This forms a carbon-carbon double bond, producing an alkene …