# 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 ?

## 2 Answers

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 $$\ce{HBr}$$.

• You haven't answered the main question in the body. Try to add that too. – Zenix Jan 6 at 11:22

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!

• @andselisk please tell him how to edit answers in future – Zenix Jan 6 at 15:21