Alkyl halides are best prepared from alcohols ($\ce{R-OH}$, where $\ce{R}$ is alkyl group), which are easily accessible. The hydroxyl group of an alcohol is replaced by halogen on reaction with concentrated halogen acid ($\ce{HX}$, where $\ce{X}$ is halogen), to get alkyl halides.
Usually we don't need boiling the reactants. But, for the preparation of alkyl bromide requires constant boiling of alcohol with $\ce{HBr}$ ($48\%$). Assume that alcohol is tertiary, so that, we can avoid using catalyst.
What I thought was, alcohol can develop inductive effect, and thus, can get detached with $\ce{OH}$, so that, the bromine from $\ce{HBr}$ can add to $\ce{R}$ group to form alkyl bromide. So, I thought there is no requirement of boiling of reactants, as like we do for preparation of alkyl chloride. Then, why do we need boiling in case of $\ce{HBr}$ and not in case of $\ce{HCl}$?