JEE Main · 2007 · 29 Apr 2007 · Offline · Hydrocarbons

To reach $2,2$-dibromopropane, $\mathrm{C}{\mathrm{H}}_3\mathrm{CB}{\mathrm{r}}_2\mathrm{C}{\mathrm{H}}_3$, both bromine atoms must settle on the central carbon of a three-carbon chain. Adding two moles of hydrogen bromide to propyne, $\mathrm{C}{\mathrm{H}}_3\mathrm{C}\equiv \mathrm{CH}$, does exactly that: by Markovnikov's rule each $\mathrm{HBr}$ places its bromine on the more substituted middle carbon, first giving $\mathrm{C}{\mathrm{H}}_3\mathrm{CBr}=\mathrm{C}{\mathrm{H}}_2$ and then the geminal dibromide. The addition of two equivalents of hydrogen bromide to propyne is thus the route that yields the compound.