Question

draw the structure of all products of the mechanism below.

Explanation:

Step1: Identify reaction type

This is an elimination reaction where the two bromine atoms and adjacent hydrogens are removed.

Step2: Determine the double - bond formation

The carbon - carbon triple bond will become a double - bond as the bromine atoms leave with adjacent hydrogens.

Step3: Draw the product structure

The product will have a carbon - carbon double - bond and the bromine atoms are eliminated as $HBr$. The structure of the product is $CH_3CH = CHCH_3$ (assuming the reaction conditions lead to the formation of a stable alkene product).

Answer:

$CH_3CH = CHCH_3$