The reaction of one mol of bromine with ethyne yields:

B r C H_{2} - C H_{2} B r

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B r C H = C H B r

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B r_{2} C H - C H B r_{2}

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C H_{3} - C H_{2} B r

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Solution

The correct option is B $B r C H = C H B r$
Halogens, especially chlorine and bromine on reaction with alkynes readily produces tetra-halogen derivatives of alkane. The reaction is carried out in inert solvent like carbon tetrachloride.

Due to the presence of two n-bonds, each molecule of the alkyne can react with two molecules of the halogen.
One mole of Bromine with ethyne gives 1,2-dibromoethene.
$H C \equiv C H + B r_{2} C C l_{4} - -- \to H (B r) C = C (B r) H$

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