The reaction sequence described involves the conversion of an alkyne to a ketone through hydroboration-oxidation followed by halogenation and subsequent elimination. Let's break it down step by step:

1. Hydroboration-Oxidation: The alkyne reacts with catecholborane. Hydroboration adds across the triple bond, followed by oxidation, which typically gives a ketone from an alkyne.
2. Bromination: The resultant ketone undergoes alpha-bromination using Br₂. This creates a bromoketone.
3. Elimination: Treatment with NaOMe favors the formation of an enone through elimination (E2 mechanism).

Applying these steps to the given structure:

The starting alkyne is a terminal alkyne, which through the described sequence results in the formation of a 2-enone structure.

Thus, the major product formed is the enone shown in the provided image. This is the result of the bromination followed by elimination yielding a conjugated enone.

Conclusion: The major product is the conjugated enone, which corresponds with the correct answer option provided.