Question

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

Answer 1

Answer: (B)

The percentage composition of compounds X and Y
$\begin{array}{rl}& C:H\\ & \frac{12}{13}\times 100percent:\frac{1}{13}\end{array}$ $\times$ $100percent$ i.e.,
the ratio of masses of C and H in the organic compounds
$\begin{array}{rl}& C:H\\ & \frac{12}{13}:\frac{1}{13}\end{array}$ 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.
$\stackrel{B{r}_{2}water}{\to }\underset{1,1,2,2-tetrabromoethane}{H-\underset{\begin{array}{c}|\\ Br\end{array}}{\stackrel{\begin{array}{c}Br\\ |\end{array}}{C}}-\underset{\begin{array}{c}|\\ Br\end{array}}{\stackrel{\begin{array}{c}Br\\ |\end{array}}{C}}-H}$