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Product of reaction between cyclohexene and bromine in methanol at 273 K
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Product of reaction between cyclohexene and bromine in methanol at 273 K
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 $\ce{HBr}$.
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 $\ce{HBr}$.
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 $\ce{HBr}$.
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 $\ce{HBr}$.
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