Question

A hydrocarbon, ‘X’(C₃H₄) decolourise bromine in carbon tetrachloride forming ‘Y’ . ‘X’ gave a red precipitate of a compound ‘Z’ with ammonical cuprous chloride and formed a carbonyl compound with dilute sulphuric acid in the presence of Hg²⁺ ions. X, Y, and Z are given by the set -

A. CH₂=C=CH₂,CH₂Br–CBr₂–CH₂Br,CH₃–C≡C–Cu
B. CH₃–C≡CH,CH₃–CBr₂–CHBr₂,CH₃–C≡C–Cu
C. CH₃–C≡CH,CH₃–CBr=CHBr,CH2Cu–C≡CH
D. CH₂=C=CH₂,CH₂Br–CH=CHBr,CH₃–C≡C–Cu

Answer 1

The hydrocarbon 'X' is likely an alkyne, satisfying the conditions to form a red precipitate with ammonical cuprous chloride and a carbonyl compound with Hg²⁺ ions. The correct set for X, Y, and Z is B: CH₃–C≡CH, CH₃–CBr₂–CHBr₂, and CH₃–C≡C–Cu respectively. Therefore, the correct option is B.

Step-by-step explanation:

The hydrocarbon 'X' given as C₃H₄, would have to exhibit behavior consistent with an alkyne to react with ammonical cuprous chloride to give a red precipitate, suggesting the hydrocarbon contains a terminal triple bond. Alkynes react with ammonical cuprous chloride to precipitate a copper acetylide, which is characteristically red. The carbonyl compound formed with dilute sulphuric acid in the presence of Hg²⁺ ions suggests an alkyne hydration reaction, leading to the formation of a ketone or aldehyde.

‘X’ decolorises bromine in carbon tetrachloride, forming compound ‘Y', which indicates the addition across the triple or double bond. Options A and D contain dienes which are unlikely to give a red precipitate with ammonical cuprous chloride. Option B and C contain an alkyne as ‘X’, which satisfies the red precipitate condition.

Therefore, the correct option is B: CH₃–C≡CH as 'X', CH₃–CBr₂–CHBr₂ as 'Y', and CH₃–C≡C–Cu as 'Z'.