Two organic compounds A and B both containing only carbon and hydrogen, on quantitative analysis gave the same percentage composition by weight :
`C=(12//13)xx100% , H=(1//13)xx100%`
A decolourises bromine water but B does not. A and B respectively are

Options

A. `C_(2)H_(2) and C_(6)H_(6)`
B. `C_(6)H_(6) and C_(2)H_(2)`
C. `C_(2)H_(4) and C_(2)H_(6)`
D. `C_(2)H_(2) and C_(2)H_(6)`

Answer

A

Explanation

Determine the empirical formula of the two compounds
`C : H= ((12)/(13)xx100): ((1)/(13)xx100)=92.3 : 7.7`
`=(92.3)/(12): (7.7)/(1)=7.7 : 7.7 =1 : 1`
`therefore` The empirical formula of the two compounds is CH. Thus the molecular formula of A and B will be the simple multiple of CH, i.e. `C_(2)H_(2),C_(3)H_(3),C_(4)H_(4),C_(6)H_(6)`. etc.
Thus options (c ) and (d ) are discarded. Further since A decolourises bromine water and B does not, A and B should be `C_(2)H_(2)` (acetylene) and `C_(6)H_(6)` (benzene) respectively.