What product could we expect when we let cyclohexene react with an equivalent of bromine in methanol at 0 °C? Is it just the trans-addition of bromine (rac.) or also the SN1-susbtitution on a carbon due to MeO? Maybe twice? See below:
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$\begingroup$ The first product (bromoether) is correct but the mechanism is wrong. Consequently, the second and third structures do not form. Have you tried googling "bromohydrin formation"? $\endgroup$– user55119Commented Sep 12, 2021 at 21:18
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$\begingroup$ Thank you. I just looked it up, so the bromoaddition gives us an intermediate where meoh can attack, but I wasn’t sure if it’ll really deprotonate $\endgroup$– user115162Commented Sep 12, 2021 at 23:49
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$\begingroup$ The oxygen of MeOH attacks the intermediate bromonium ion from the rear (SN2) then the proton is lost. With no strong base, no prior deprotonation of MeOH occurs. $\endgroup$– user55119Commented Sep 13, 2021 at 0:52
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$\begingroup$ Note that at 273K MeOH is not a strong enough nucleophile to attack a secondary alkyl bromide so the third product, the di-ether, will not be formed. $\endgroup$– WaylanderCommented Sep 13, 2021 at 8:18
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1 Answer
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It starts with the formation of brominium ion, as in direct bromination of alkenes. If you do not know the mechanism, you can refer to it here, at MasterOrganicChemistry.
But here, the $\ce{MeOH}$ can itself act as a potential nucleophile. It is similar to if $\ce{H2O}$ is used as a solvent - refer the formation of halohydrin here.
Since $\ce{MeOH}$ is available in larger concentrations, it more readily attacks the carbon. Markovnikov rule is followed for determining which carbon is attacked, but in this case, both are symmetrical.
The attack proceeds through $\mathrm{S_N2}$ mechanism, forming two enantiomers. After the $\ce{C-O}$ bond is formed, the $\ce{MeOH}$ loses $\ce{H+}$ to form ether and $\ce{HBr}$.
