The order of reactivity of $C H_{3} C H O, C H_{3} C O C_{2} H_{5} a n d C H_{3} C O C H_{3}$ is:

C H_{3} C H O > C H_{3} C O C H_{3} > C H_{3} C O C_{2} H_{5}

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C_{2} H_{5} C O C H_{3} > C H_{3} C O C H_{3} > C H_{3} C H O

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C H_{3} C O C H_{3} > C H_{3} C H O > C_{2} H_{5} C O C H_{3}

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C H_{3} C O C H_{3} > C_{2} H_{5} C O C H_{3} > C H_{3} C H O

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Solution

The correct option is C $C H_{3} C H O > C H_{3} C O C H_{3} > C H_{3} C O C_{2} H_{5}$
Since order of reactivity depends upon stability of leaving group which is
$C H_{3} C O O^{-} > C H_{3} O^{-}$ and also on electrophilicity of carbonyl carbon. Since $C H_{3} C O C_{2} H_{5}$ has has less electrophilic due to more (+) inductive effect of alkyl group so it is less reactive.

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Similar questions

C H_{3} C H O

C H_{3} C O C_{2} H_{5}

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

C H_{3} C H_{2} C H O

C H_{3} C H O, C H_{3} C O C_{2} H_{5} and C H_{3} C O C H_{3}

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

K M n O_{4}

The compound $C H_{3} | C H_{3} - C = C H - C H_{3}$ on reaction with $N a I O_{4}$ in the presence of $K M n O_{4}$ gives

C H_{3} C H O

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

> C = O

C H_{3} C H O

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