Q.Suggest two different methods, with equations, to prepare propene, $\text{CH}_3\text{CH}{=}\text{CH}_2$, in the laboratory.

Method 1 -- dehydrohalogenation of an alkyl halide. Heating 1-bromopropane with potassium hydroxide dissolved in ethanol (alcoholic KOH favours elimination over the substitution that aqueous KOH would favour) eliminates $\text{HBr}$ across the C1--C2 bond: $\text{CH}_3\text{CH}_2\text{CH}_2\text{Br} \xrightarrow{\text{alc. KOH},\Delta} \text{CH}_3\text{CH=CH}_2 + \text{HBr}$. Method 2 -- acid-catalysed dehydration of an alcohol. Heating propan-1-ol with concentrated sulphuric acid removes a molecule of water across the same two carbons: $\text{CH}_3\text{CH}_2\text{CH}_2\text{OH} \xrightarrow{\text{conc. H}_2\text{SO}_4,\Delta} \text{CH}_3\text{CH=CH}_2 + \text{H}_2\text{O}$ (alumina at $\approx 350\,°\text{C}$ works equally well as the dehydrating agent). Both methods remove two atoms/groups from adjacent carbons to generate the double bond; since propan-1-ol/1-bromopropane have only one possible direction of elimination (the leaving group is on a t …

Recall the two standard elimination routes to an alkene (from a halide, from an alcohol), identify which atoms are removed from which adjacent carbons in each, and check that the leaving grou …

Using AQUEOUS KOH instead of alcoholic KOH for the halide route (aqueous conditions favour substitution to the alcohol instead of elimination to the alkene); forgetting the concentrated-acid/high-temperature condition needed fo …