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.

  • $\begingroup$ so steric hindrance is the reason. Then, what would you say about this dropbox.com/s/dukb3mewv2vy8g5/… $\endgroup$ May 9 '18 at 8:11
  • $\begingroup$ Goes by a different mechanism. The protonated ether in that case fragments to give tBu cation, this is very stable and drives the reaction $\endgroup$
    – Waylander
    May 9 '18 at 8:21
  • $\begingroup$ With excess HBr propan-1-ol will also react to form 1-bromopropane. $\endgroup$
    – user55119
    May 9 '18 at 15:20
  • $\begingroup$ In a real experiment yes, probably get measurable quantities of 2-bromopropane and propene too $\endgroup$
    – Waylander
    May 9 '18 at 15:26

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