question_answer

Two organic compounds X and Y on analysis gave the same percentage composition namely;\[C=(12/13)\times 100%\]and\[H=(1/13)\times 100%\]. However, compound X decolourises bromine water while compound Y does not. The two compounds X and Y may be respectively

A)  acetylene and ethylene

B)  acetylene and benzene

C)  ethylene and benzene

D)  toluene and benzene

E)  benzene and styrene

Correct Answer: B

Solution :

The percentage composition of compounds X and Y \[\begin{align}   & \,\,\,\,\,C~~~~~~~~~~~\,\,\,\,\,:~~~~~\,\,\,\,\,\,\,H \\  & \frac{12}{13}\times 100%\,\,\,\,\,\,:\,\,\,\,\,\,\,\,\,\,\,\frac{1}{13}\times 100% \\ \end{align}\] i.e., the ratio of masses of C and H in the organic compounds                 \[\begin{align}   & C~~~~~:~~~~~H \\  & \frac{12}{13}\,\,\,\,\,:\,\,\,\,\,\,\,\frac{1}{13} \\ \end{align}\] or \[12:1\] Thus, the empirical formula of the compounds X and V is CH and X decolorize bromine water while Y does not. Thus, X and Y must be acetylene\[(CH\equiv CH)\]and benzene\[({{C}_{6}}{{H}_{6}})\] respectively. \[\xrightarrow[{}]{B{{r}_{2}}\,water}\underset{1,\text{ }1,\text{ }2\text{ },2-tetrabromoethane}{\mathop{H-\underset{\begin{smallmatrix}  | \\  Br \end{smallmatrix}}{\overset{\begin{smallmatrix}  Br \\  | \end{smallmatrix}}{\mathop{C}}}\,-\underset{\begin{smallmatrix}  | \\  Br \end{smallmatrix}}{\overset{\begin{smallmatrix}  Br \\  | \end{smallmatrix}}{\mathop{C}}}\,-H}}\,\]