$\text{Zn}$ dust mediates reductive elimination of vicinal dihalides to form alkenes. $\text{HBr}$ addition to a simple alkene yields the alkyl halide via carbocation formation (Markovnikov rule).
To solve this problem, let's analyze the chemical reaction step-by-step.
The given compound is a dibromo compound, specifically a vicinal dibromide. This means that two bromine atoms are attached to adjacent carbon atoms in the cyclohexane ring.
The first step of the reaction involves the use of zinc (Zn) with heat (Δ). Zinc is commonly used for dehalogenation reactions, where it helps to remove the bromine atoms by forming zinc bromide (ZnBr2).
During this step, a dehalogenation takes place, resulting in the formation of a double bond between the two carbon atoms that initially held the bromine atoms, leading to the generation of a cyclohexene derivative.
The second step involves the use of hydrobromic acid (HBr). This step is typically intended for the Markovnikov addition of HBr across a double bond in an alkene, which is not applicable here as the dehalogenated product is not further reacted with HBr beyond the reaction conditions given. The main significance here is confirmation of (A) as a stable product.
Thus, the major product (A) of this reaction would be cyclohexene, where the two bromine atoms have been removed, and a double bond is formed.
