Ethylene bromide (C2H4Br2) can undergo a reaction with zinc (Zn) to give an alkene.

The reaction between ethylene bromide and zinc is known as the dehalogenation reaction. In this reaction, the zinc metal acts as a reducing agent by donating electrons to the halogen atom (bromine) present in ethylene bromide. This results in the removal of the bromine atom from the molecule, leading to the formation of an alkene.

Here is the detailed explanation of the reaction:

1. Electrophilic addition of ethylene bromide: Ethylene bromide is an alkyl halide, and it can undergo electrophilic addition reactions. In the presence of a strong reducing agent like zinc, the bromine atom becomes the electrophile and attacks the double bond of ethylene. This leads to the formation of a cyclic intermediate with a positive charge on one of the carbon atoms.

2. Reduction by zinc: Zinc metal donates electrons to the cyclic intermediate, reducing the positive charge on the carbon atom. This results in the formation of a radical intermediate.

3. Elimination of bromine: The radical intermediate formed undergoes a rearrangement, leading to the elimination of the bromine atom. This results in the formation of an alkene. The double bond is formed between the two carbon atoms that were originally bonded to the bromine atom.

The reaction can be represented as follows:

C2H4Br2 + Zn → C2H4 + ZnBr2

In this reaction, ethylene bromide (C2H4Br2) reacts with zinc (Zn) to give ethylene (C2H4) and zinc bromide (ZnBr2). Therefore, the correct answer is option 'B' - Alkene.

It is important to note that the reaction conditions and the choice of reducing agent can influence the outcome of the reaction. In this specific case, the reaction between ethylene bromide and zinc results in the formation of an alkene. However, different alkyl halides and reducing agents can lead to the formation of other organic compounds such as alkanes or alkynes.