JEE Main · 2021 · 16 Mar 2021 · Shift 2 · Hydrocarbons

The double bond survives the reaction and chlorine has entered both carbons next to it, so this is substitution at the allylic $s{p}^3$ positions and not addition across the $\pi$ bond. Substitution of that kind proceeds through chlorine atoms, which are generated by homolysis of $\mathrm{C}{\mathrm{l}}_2$ under ultraviolet light; the allylic radical left behind is resonance stabilised, which is why those positions are attacked in preference to the others. Anhydrous aluminium chloride in the dark would instead drive the ionic addition of chlorine across the double bond, and hydrogen chloride would add to give a monochloride with the double bond gone. The reagent and condition are chlorine under ultraviolet light.