The major product of the following reaction is





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Hint: Alkynes can be easily prepared by the process of dehydrohalogenation of alkenes and alkanes. It undergoes elimination reactions for that purpose. Dehydrohalogenation refers to the removal of hydrogen halides present in the compounds.

Complete step by step solution:
-The given compound is a dibromide and can undergo elimination reaction of two types. One is dehalogenation and the other is dehydrohalogenation.
-Alkynes can easily be prepared from dichlorides by the process of hydrohalogenation. It is a process in which a hydrogen atom is removed from 1 carbon atom and a halide atom is removed from the other carbon atom present at the anti position to form a pi-bond between the two carbon atoms.
-Both the given reagents in the figure shown are dehydrohalogenating agents and so the reaction that occurs here in both the stages will be dehydrohalogenation elimination and no other reaction.
-Here we have 2 bromine atoms. So the process can occur in 2 steps. First 1 hydrogen and 1 bromine will get removed from opposite directions to form an alkene and then the process will be repeated so as to form alkyne. This way we can form alkyne from the given dichloride.
-The chemical equation can be shown as follows-

Therefore the correct option is A.

Note: Dehalogenation is different from dehydrohalogenation. In the former, the halogen escapes as a gas and in the form of its molecule with the formula ${{X}_{2}}$ where X represents the halogens. In the latter, HX gas is liberated after the completion of the reaction where X is the single atom of halogen.