Q. 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
Solution:
The percentage composition of compounds X and Y
$ \begin{align} & C~~~~~:~~~~~H \\ & \frac{12}{13} \times 100 \,percent \,\,\,\,\,:\,\,\,\,\,\,\,\frac{1}{13} \\ \end{align} $ $\times$ $100 \,percent $ 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 $ 100 $
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}}\, $
