$[C H_{3}]_{2} C = C H C O C H_{3}$ can be oxidized to $[C H_{3}]_{2} C = C H C O O H$ by?

K M n O_{4}

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NaOI

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Chromic acid

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Cu at

300^{0}

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Solution

The correct option is B $K M n O_{4}$
$(C H_{3})_{2} - C = C H - O | | C - C H_{3} K M n O_{4} - --------- \to (o x i d i s e n g a g e n t) (C H_{3})_{2} C H = C H - O | | C - O H$

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Similar questions

(C H_{3})_{2} C = C H C O C H_{3}

(C H_{3})_{2} = C H C O O H

The oxidation numbers of C - 1 and C - 2 in propene $C H_{3}^{2} C H =^{1} C H_{2}$ are respectively

K M n O_{4} / O H^{-}

C H_{3} - C H | - C H_{3} C O O H

C H_{3} - O | | C - C H_{2} C H_{2} C H_{3}

C H_{3} - C H_{3} | C | C H_{3} - C H_{2} B r C_{2} H_{5} O H \to

C H_{3} - C \equiv C - C H_{3}

C H_{3} - C O - C O - C H_{3}

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