A hydrocarbon, 'X' $(C_{3} H_{4})$ decolourises $B r_{2} / C C l_{4}$ solution forming 'Y'. 'X' gave a red precipitate of a compound 'Z' with ammoniacal cuprous chloride.
X, Y, and Z respectively are:

C H_{2} = C = C H_{2}, C H_{2} B r - C B r_{2} - C H_{2} B r, C H_{3} - C \equiv C^{-} C u^{+}

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C H_{3} - C \equiv C H, C H_{3} - C B r_{2} - C H B r_{2}, C H_{3} - C \equiv C^{-} C u^{+}

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C H_{3} - C \equiv C H, C H_{3} - C B r = C H B r, C u^{+} C H_{2}^{-} - C \equiv C H

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C H_{2} = C = C H_{2}, C H_{2} B r - C H = C H B r; C H_{3} - C \equiv C^{-} C u^{+}

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Solution

The correct option is B $C H_{3} - C \equiv C H, C H_{3} - C B r_{2} - C H B r_{2}, C H_{3} - C \equiv C^{-} C u^{+}$
The molecular formula of compound 'X' is $C_{3} H_{4}$ . It decolourises bromine solution and it gives red precipitate when reacts with ammoniacal cuprous chloride. This confirms the presence of terminal alkyne which has acidic proton.
The only possible structure with molecular formula $C_{3} H_{4}$ for terminal alkyne is $C H_{3} - C \equiv C H$ (X).

When bromine is treated with alkyne in $C C l_{4}$ , they gets added to alkyne to give 1, 1, 2, 2-tetrabromo propane (Y).

When 1-propyne (X) react with ammoniacal cuprous chloride it gives copper acetylide.

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