SATHEE: Hydrocarbons - Result Question 36

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

(d) Ethene

[1998]

Show Answer

Solution:

(a) Since hydrocarbon $C$ gives only $C H_{2} O$ , on ozonolysis, $C$ should be $C H_{2} = C H_{2}$ , Hence going backward, A should be ethane. Thus, the reactions are

(A)

$\begin{matrix} (B) & \underset{(C)}{C H_{2} = C H_{2}} \to_{Δ}^{O_{3} / H_{2} O} \underset{(D)}{H C H O} \end{matrix}$

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Hydrocarbons - Result Question 36

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