Write a halogenation reaction of the following alkyne with Br₂ and name the product formed.

• A.
• B.
• C.
• D.

Answer: (C)

Verified step by step guidance

1. Identify the starting alkyne structure. In this case, it is a symmetrical internal alkyne with alkyl groups on both sides of the triple bond.
2. Recall that halogenation of an alkyne with Br\(\textsubscript{2}\) involves the addition of bromine atoms across the triple bond, converting it first into a dibromoalkene intermediate and then further to a tetrabromoalkane if excess Br\(\textsubscript{2}\) is present.
3. Write the first step of the reaction: the alkyne reacts with one equivalent of Br\(\textsubscript{2}\) to form a dibromoalkene. The triple bond is converted into a double bond, and two bromine atoms add across the former triple bond carbons.
4. Write the second step: if excess Br\(\textsubscript{2}\) is present, the dibromoalkene undergoes a second addition of Br\(\textsubscript{2}\), converting the double bond into a single bond and adding two more bromine atoms, resulting in a tetrabromoalkane.
5. Name the final product by identifying the longest carbon chain and numbering it to give the bromine substituents the lowest possible numbers. Use the prefix 'tetrabromo' to indicate four bromine atoms added, and name the compound as a dibromo-substituted alkane derived from the original alkyne.