Hydrocarbons - Result Question 36
####38. A hydrocarbon ’ $A$ ’ on chlorination gives ’ $B$ ’ which on heating with alcoholic potassium hydroxide changes into another hydrocarbon ’ $C$ ‘. The latter decolourises Baeyer’s reagent and on ozonolysis forms formaldehyde only. ’ $A$ ’ is
(a) Ethane
(b) Butane
(c) Methane
(d) Ethene
[1998]
Show Answer
Solution:
- (a) Since hydrocarbon $C$ gives only $CH_2 O$, on ozonolysis, $C$ should be $CH_2=CH_2$, Hence going backward, A should be ethane. Thus, the reactions are
(A)
$ \begin{equation*} \underset{\text{ (C) }}{CH_2=CH_2} \xrightarrow[\Delta]{O_3 / H_2 O} \underset{(D)}{HCHO} \tag{B} \end{equation*} $
