Final answer:
Propene reacts with bromine at low temperatures to form 1,2-dibromopropane, which is a dibrominated alkane with the same carbon skeleton as the original alkene. This reaction is a halogenation process, and the disappearance of the brownish red color of bromine is a test for alkenes.
Step-by-step explanation:
When propene reacts with bromine at low temperatures, a halogenation reaction occurs which results in the addition of bromine across the double bond of propene. This leads to the formation of a dibrominated alkane with the same carbon skeleton as the alkene. As alkenes readily undergo halogenation, the bromine, which is originally brownish red in solution, will lose its color, indicating that the reaction has taken place. This can be used as a qualitative test for the presence of alkenes.
Furthermore, the reaction with bromine at low temperatures would likely proceed without requiring high temperatures or ultraviolet light that are necessary for the radical substitution reactions with alkanes.
The expected product from this reaction is 1,2-dibromopropane, as each carbon atom from the double bond in propene binds to a bromine atom. Since the reaction involves an alkene and a halogen, it is an example of an electrophilic addition reaction, which is common for alkenes.