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 $\ce{HCl}$?

  • $\begingroup$ Can you propose mechanisms for your reactions? $\endgroup$
    – ssavec
    Nov 5, 2013 at 7:01

2 Answers 2

  • Assuming i) a primary alcohol $\ce{R-OH}$ to be converted into the corresponding primary alkyl halogenide $\ce{R-X}$ ii) by $\ce{HX}$ acid only, I would expect a decrease of reactivity to be observed in this line: $\ce{HI} > \ce{HBr} > \ce{HCl} > \ce{HF}$, with $\ce{HF}$ of lowest reactivity. This is based on the decrease of the nucleophilicity of the anions of these acids.
  • Of course, halogenations with these four acids are reactions of equilibria, and these may be driven forward by increasing the concentration of the reactants, for example. And I do recall two-step protocols where the alcohol is activated by sulfuric acid before adding hydrobromic acid. But in both variants, rather $\ce{H3O+}$ then a neutral $\ce{OH}$ would leave the molecule. (However secondary and tertiary alcohols that are of higher reactivity than primary ones under these acidic conditions, are more prone to isomerization, too. Simultaneously, as hinted by ssavec, instead of a mechanism of type $\mathrm{S_{N}1}$, the of $\mathrm{S_{N}2}$ may become more favorable.)
  • For the preparation of alkyl chlorides, the addition of (dry) $\ce{ZnCl2}$ in presence of $\ce{HCl}$ is sometimes used. More easy to handle are routes relying on $\ce{PCl3}$ or $\ce{SOCl2}$, for example.

$\ce{HBr}$ is a stronger acid than $\ce{HCl}$ (pKa -9 vs. -7), meaning that $\ce{Br-}$ feels more comfortable in dissociated form than $\ce{Cl-}$. In other words, $\ce{Br-}$ is a better leaving group. I think this has a direct relationship to reactivity via $S_N1$.

  • $\begingroup$ One reason for using HBr/H2SO4 is to absorb the water produced in the reaction with primary alcohol [Price, C. C., et al. J. Org. Chem, 1946, 11, 281]. HBr is often generated in situ from NaBr/H2SO4 . Gaseous HBr is employed without H2SO4 [Organic Syntheses, Coll. Vol. 2, p.246 (1943); Vol. 15, p.24 (1935); DOI:10.15227/orgsyn.015.0024]. Sulfuric acid is a weaker acid (pKa -3/2) than HBr (pKa -9). $\endgroup$
    – user55119
    Nov 25, 2017 at 22:45

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