Can we use (KBr + H2SO4) instead of (NaBr +H2SO4) in reaction with alcohol to generate alkyl bromide?

I have read somewhere that $\ce{H2SO4}$ is used to generate $\ce{HBr},$ and at the other place it is written that it is there to protonate the hydroxyl group. Which fact should I believe ?




Yes, we can use $\ce{KBr + H2SO4}$ instead of $\ce{NaBr + H2SO4}$ because they both give $\ce{HBr}$ which is used in the preparation of alkyl bromide.
The basic reaction is as follows:
$$\ce{ROH + HBr -> RBr + H2O}$$
The above-mentioned compounds $\ce{NaBr}$ and $\ce{KBr}$ are used only for the preparation of $\ce{HBr}$.


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H2SO4 is used to protonate the OH group and form OH2+ group which is a good leaving group.


OH2+ leaves the substrate and forms a carbocation.


Then (Br-) anion attacks on the carbocation so formed.


HBr NaBr KBr all liberate (Br-) anion in the solvent.


So all will give the same product.


H2SO4 has the task to generate H+ that protonates the alcohol.


Your book might have tried to explain it like H2SO4 first forms HBr. Ultimately HBr will dissosiate to give H+ which does the protonation.


Rest over to you. You may believe both or none!


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