I am thinking that maybe only hydrogen shifts are possible and methyl shifts are not allowed in such situation? Is that the reasoning? but then again in the second example the bromide could shift to a secondary position after doing a hydrogen shift.

two problems with an alcohol reacting with HBr

  • $\begingroup$ Consider the preferred pathway for a 2º alcohol vs a 1º alcohol. $\endgroup$
    – ringo
    May 6, 2016 at 8:36
  • $\begingroup$ Predicting the pathway for these Wagner-Meerwein rearrangements is rather hard, isn't it? Usually you get a lot of side products. $\endgroup$
    – Karl
    May 6, 2016 at 11:08

1 Answer 1


You need to consider the possible and plausible reaction mechanisms for both substrates. They are both nucleophilic substitutions in first approximation, so we need to decide whether $\mathrm{S_N1}$ or $\mathrm{S_N2}$ mechanisms are more plausible.

Primary electrophilic carbons are very likely to undergo $\mathrm{S_N2}$ reactions while $\mathrm{S_N1}$ is strongly disfavoured since primary carbocations are less stable than secondary or tertiary ones.

SN1-mechanism on a primary carbon

Secondary electrophilic carbons can react according to both mechanisms, technically. However, both mechanisms are not ideal: the steric strain is rather large so $\mathrm{S_N2}$ reactions are slower and secondary carbocations are still rather unstable compared to tertiary ones, so $\mathrm{S_N1}$ reactions are slower.

In this case, we have the possibility to stabilise the less stable second intermediate in the scheme shown below by a Wagner-Meerwein rearrangement. We can let both hydrogen or methyl migrate, but the hydrogen migration gives us the most stable carbocation (a tertiary one) so it will prevail. Multiple other migration paths will likely lead to minor products.

SN1-mechanism on a secondary carbon

The primary carbon will not, as stated above, form a primary carbocation since it is too unstable. Hence only the bimolecular substitution will occur. If no primary carbocation is generated, no Wagner-Meerwein rearrangement can occur. And the mechanism of an $\mathrm{S_N2}$ reaction does not allow bromide’s (or anything else’s) migration.

The secondary carbon could undergo bimolecular substitution, too. The reaction rate will probably be slower than that of fragmentation and rearrangement, though, since the latter are unimolecular processes. And other shifts (e.g. methyl followed by hydride) are possible but again slower since you need to shift twice. All in all, a higher proportion of side-products may be expected.

In no case will the bromide shift after substitution; any shift would need to occur before.

  • $\begingroup$ You should perhaps make it clear that there can be no rearrangement during SN2. It is only implied in your answer. I believe this is what the OP was stuck on. $\endgroup$
    – Karl
    May 6, 2016 at 14:40
  • 1
    $\begingroup$ @Karl Like this? $\endgroup$
    – Jan
    May 6, 2016 at 18:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.