Q.• How are alkenes prepared from alkyl halides ?
• Which is stronger base from the following ?
i. aq. KOH ii. alc. KOH

When an alkyl halide bearing at least one beta-hydrogen is boiled with alcoholic KOH, it eliminates a beta-hydrogen and the alpha-halogen together to form an alkene -- a dehydrohalogenation (beta-elimination, or 1,2-elimination) reaction, C(beta)H-C(alpha)X + base -> C=C + base-H+ + X-. (The carbon bearing the halogen is the alpha-carbon; any carbon attached to it is a beta-carbon, and hydrogens on that carbon are beta-hydrogens.) When two or more non-equivalent beta-hydrogens are available, a mixture of alkene products forms, and Saytzeff's empirical rule predicts which one predominates: the major product is the alkene with the greater number of alkyl groups on its doubly-bonded carbons, because more highly substituted alkenes are inherently more stable (R2C=CR2 > R2C=CHR > R2C=CH2). For example 2-bromobutane with alcoholic KOH gives mainly but-2-ene (more substituted) over but-1-ene (less substituted). Elimination always competes with substitution when an a …