Question

Two organic compounds $\underset{–}{R}$ and $\underset{–}{S}$, both containing only $C$ and $H$ yield on analysis, the same percentage composition by weight.
$C=\left(12/13\right)\times 100$% and $H=\left(1/13\right)\times 100$%.
$\underset{–}{R}$ decolourises $B{r}_2$ - water but $\underset{–}{S}$ does not.

• A. $C_2{H}_2,C_6{H}_6$
• B. $C_6{H}_6,C_2{H}_2$
• C. $C_2{H}_4,C_2{H}_6$
• D. $C_2{H}_2,C_3{H}_8$

Answer 1

Answer: (A)

$C_2{H}_2,C_6{H}_6$

According to question the compound have equal % composition of mole. So, component carbon$\left(C\right)$ and the component $H$ have same in number.

$\to C_n{H}_n$ type of compound

Now,

According to bromine water reaction see in fig.

But as this is question of physical chemistry, we know the $C_n{H}_n$ type of molecules should be obtained but in options only $\left(A\right)$ option has only $C_n{H}_n$ like atoms.

So $\left(A\right)$ is correct answer.