# What is the difference between "bromine" and "bromine water" in their reaction with an alkene?

In question 16 of the A-Level Chemistry 2013 exam, there is this reaction scheme:

$$\ce{CH3CH=CH2 ->[?] CH3-CHBr-CH2Br}$$

The question asks for both the name of the reagent and the product. I know that the product is 1,2-dibromopropane, but am confused about the reagent. I was told that the correct answer was "bromine", whereas "bromine water" was a wrong answer. Why is this so?

From ChemGuide:

Using bromine water as a test for alkenes

If you shake an alkene with bromine water (or bubble a gaseous alkene through bromine water), the solution becomes colourless. Alkenes decolourise bromine water.

The chemistry of the test

This is complicated by the fact that the major product isn't 1,2-dibromoethane. The water also gets involved in the reaction, and most of the product is 2-bromoethanol.

In essence, the electrophilic pi bond in the presence of bromine and water attacks diatomic bromine as usual, forming the usual bromonium ion intermediate.

After this step, however, we have two competing nucleophiles which can open the ring: bromide ion and water. The water is likely present in great excess, making it the more likely attacker (statistically).

Of course, the desired reaction below to form a dibromide also occurs. But it is in competition with other reactions such as the one listed above.

So usually electrophilic halogenation of alkenes takes place in an inert medium such as methylene chloride or carbon tetrachloride. These aren't inclined to act as nucleophiles, so they don't participate in ring opening, leaving the $$\ce{Br-}$$ ion to do its thing unhindered.