Halohydrin Practice Problems
Halohydrins are formed by the reaction of alkenes with bromine in water. A different product is produced in a similar reaction when ethanol is used instead of water. Determine the mechanism for the formation of this product.
Identify the product(s) formed when the given compound undergoes a reaction with Cl2 and H2O.
Illustrate the product(s) formed when the given compound undergoes a reaction with Br2 and EtOH.
Determine the product(s) of the reaction below, noting the relative stereochemical outcome. Draw both enantiomers to indicate any racemic mixtures.
Provide an arrow-pushing mechanism for the given reaction that explains how the halohydrin is formed in a regioselective and stereospecific manner.
What are the products formed when the given alkene compound reacts with Br2 and EtOH.
Determine the reagent(s) needed for 1-ethyl-4,5-dimethylcyclohex-1-ene to form a racemic mixture of 2-chloro-1-ethyl-4,5-dimethylcyclohexan-1-ol.
Explain why, in the second step of halohydrin formation, water attacks the bromonium ion's carbon as opposed to the ion itself.
Determine the product of the reaction of the following compound with bromine if the formula of the product is C8H15BrO. Propose a mechanism for the formation of the product.
Show how to accomplish the following synthetic transformation.
2-methylbut-2-ene → 3-chloro-2-methylbutan-2-ol
Hint: The electrophilic nature of a halonium ion drives its opening. Carbon bearing more positive charge will be attacked by the weak nucleophile.
Suggest a mechanism for the addition of bromine water to cyclohexene. Your proposed mechanism should show why the -Br and -OH substituents have anti relationship in the addition product and how each enantiomer is formed in the reaction.
For each of the following reactions, determine the major product(s). Include stereochemistry in your answer.
(i) 1-methylcyclopentene + Cl2/H2O
(ii) 2-methylpent-2-ene + Br2/H2O