Two organic compounds $R - -$ and $S - -$ , both containing only $C$ and $H$ yield on analysis, the same percentage composition by weight.
$C = (12 / 13) \times 100$ % and $H = (1 / 13) \times 100$ %.
$R - -$ decolourises $B r_{2}$ - water but $S - -$ does not.

C_{2} H_{2}, C_{6} H_{6}

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C_{6} H_{6}, C_{2} H_{2}

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C_{2} H_{4}, C_{2} H_{6}

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C_{2} H_{2}, C_{3} H_{8}

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Solution

The correct option is C $C_{2} H_{2}, C_{6} H_{6}$
According to question the compound have equal % composition of mole. So, component carbon $(C)$ 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 $(A)$ option has only $C_{n} H_{n}$ like atoms.
So $(A)$ is correct answer.

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C_{2} H_{4}

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C_{6} H_{6}

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List - A	List - B
(Molecule)	(Pure and hybrid orbitals involved)
A) $C_{2} H_{6}$	1) 12, 18
B) $C_{2} H_{4}$	2) 6, 4
C) $C_{2} H_{2}$	3) 6, 6
D) $C_{6} H_{6}$	4) 6, 8

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I \to C_{2} H_{2},

I I \to C_{2} H_{4},

I I I \to C_{2} H_{6}

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{CH}_{4}

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