Halohydrin formation from propene with Br2/H2O yields which product?

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Multiple Choice

Halohydrin formation from propene with Br2/H2O yields which product?

Halohydrin formation with Br2/H2O runs through a bromonium ion intermediate, and water opens the ring at the more substituted carbon. This places the hydroxyl on the more substituted carbon and the bromine on the less substituted carbon, with anti addition. For propene, the double bond is between a terminal CH2 (less substituted) and the secondary carbon (more substituted). Water attacks the secondary carbon, giving OH there, while bromine ends up on the terminal carbon. The product is CH2Br-CH(OH)-CH3, i.e., 1-bromo-2-propanol.

Halohydrin formation with Br2/H2O runs through a bromonium ion intermediate, and water opens the ring at the more substituted carbon. This places the hydroxyl on the more substituted carbon and the bromine on the less substituted carbon, with anti addition.

For propene, the double bond is between a terminal CH2 (less substituted) and the secondary carbon (more substituted). Water attacks the secondary carbon, giving OH there, while bromine ends up on the terminal carbon. The product is CH2Br-CH(OH)-CH3, i.e., 1-bromo-2-propanol.

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