What are the products of the reaction of methoxypropane with HBr? The products of the reaction of methoxypropane ($\ce{CH3CH2CH2OCH3}$) with $\ce{HBr}$ are?


In my opinion, the products should be: bromopropane ($\ce{CH3CH2CH2Br}$) + methanol ($\ce{CH3OH}$). The first step is the cleavage of the ether. This can happen via the formation of a propyl carbonation or a methyl carbocation. Since the propyl carbocation is more stable, bromine will react with the same to form bromopropane (the other product thus formed is methanol)

But the products given in the solution were propanol ($\ce{CH3CH2CH2OH}$) + bromomethane ($\ce{CH3Br}$). So, what is wrong in my reasoning? And what is the correct reasoning?



Source: NCERT (India) 12th
The first step is protonation of the oxygen of the ether. This cationic species is then subject to nucleophilic attack by Br-. There are two potential sites of attack i.e. the carbons bonded to the protonated oxygen. Bromine is quite a bulky nucleophile, and of the two sites, the methyl group carbon is less hindered so this is attacked in preference, leading to bromomethane and propan-1-ol as your products.