Question

Two isomers 'A' and 'B' with molecular formula C₄H₈ give different products on oxidation with KMnO₄ in acidic medium. Isomer 'A' on reaction with KMnO₄/H⁺ results in the effervescence of gas and gives ketone. The compound 'A' is:

Options

• But-1-ene

• cis-But-2-ene

• trans-But-2-ene

• 2-methyl propene

Answer 1

2-methyl propene

Explanation:

Two isomers with molecular formula C₄H₈ are:

$$\begin{array}{c}\mathrm{H}{\phantom{A}}_3\mathrm{C}\phantom{........................................}\\ \mathrm{\setminus }\phantom{....................................}\\ \mathrm{C}=\mathrm{CH}{\phantom{A}}_2\mathrm{and}\mathrm{CH}{\phantom{A}}_3-\mathrm{CH}{\phantom{A}}_2-\mathrm{CH}=\mathrm{CH}{\phantom{A}}_2\\ /\phantom{....................................}\\ \mathrm{H}{\phantom{A}}_3\mathrm{C}\phantom{.........................................}\end{array}$$

(i) $$\begin{array}{c}\phantom{}\mathrm{H}{\phantom{A}}_3\mathrm{C}\phantom{..............}\mathrm{H}{\phantom{A}}_3\mathrm{C}\phantom{....................}\\ \phantom{}\mathrm{\setminus }\phantom{.................}\mathrm{\setminus }\phantom{...............}\\ \mathrm{C}=\mathrm{CH}{\phantom{A}}_2\underset{\mathrm{H}{\phantom{A}}^{+}}{\overset{\mathrm{KMnO}{\phantom{A}}_4}{\to }}\mathrm{C}=\mathrm{O}+\mathrm{CO}{\phantom{A}}_2\\ \phantom{}/\phantom{................}/\phantom{...............}\\ \phantom{.}\mathrm{H}{\phantom{A}}_3\mathrm{C}\phantom{.............}\mathrm{H}{\phantom{A}}_3\mathrm{C}\phantom{.....................}\end{array}$$

(ii) $$\mathrm{CH}{\phantom{A}}_3-\mathrm{CH}{\phantom{A}}_2-\mathrm{CH}=\mathrm{CH}{\phantom{A}}_2\underset{\mathrm{H}{\phantom{A}}^{+}}{\overset{\mathrm{KMnO}{\phantom{A}}_4}{\to }}\mathrm{CH}{\phantom{A}}_3-\mathrm{CH}{\phantom{A}}_2\mathrm{COOH}+\mathrm{CO}{\phantom{A}}_2$$

So, Isomer (A) is (H₃C)₂ C = CH₂ as it gives ketone and CO₂.