SolutionsElectrophilic Addition of Halogens

1. Intramolecular bromine and double elimination

Predict the result of the following reactions:

An unsaturated alcohol with Br₂ gives C₉H₁₇BrO; 1,2-dimethylcyclohexene with 1) Br₂ and 2) t-BuOK in t-BuOH gives C₈H₁₂

The OH group opens the bromonium ion intramolecularly to form a tetrahydropyran bearing a CH₂Br; the dibromide of 1,2-dimethylcyclohexene eliminates two HBr to give 1,2-dimethylidenecyclohexane

In the first case the nucleophile that opens the bromonium ion is the molecule's own OH, which attacks the more substituted carbon and forms a six-membered ring. In the second, the bulky base removes the hydrogens of the methyl groups (Hofmann-type elimination).

2. Cycloalkenes and the meso form

Explain why the reaction of cycloalkenes such as cyclobutene, cyclopentene or cyclohexene with Br₂ never gives a meso form, but a racemic mixture.

Addition of bromine to cyclohexene: the bromonium ion forms on either face and bromide opens it from the opposite side; the two enantiomers (S,S) and (R,R) of trans-1,2-dibromocyclohexane are obtained, which do not interconvert with the cis isomer

Bromine addition takes place with anti stereochemistry, giving the racemic mixture of the trans isomer. The meso form is the cis isomer, which cannot be obtained by an anti addition.

3. Halohydrins

Give the product of the reaction of propene with Br₂ in water and explain its regiochemistry.

1-Bromopropan-2-ol. Water, present in large excess, opens the bromonium ion at the secondary carbon, which best supports the partial positive charge of the transition state.

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