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(C) The reaction is an ozonolysis of an alkene,$Me_2C=CHCH_3$.In the presence of $O_3$ followed by $H_2O$ (oxidative workup,as $Zn$ is absent),the dou

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$Me_2CO + CH_3COOH$

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(C) The reaction is an ozonolysis of an alkene,$Me_2C=CHCH_3$.In the presence of $O_3$ followed by $H_2O$ (oxidative workup,as $Zn$ is absent),the double bond is cleaved.The fragments formed are $Me_2C=O$ (acetone) and $CH_3CH=O$ (acetaldehyde).Since $Zn$ is absent,the aldehyde $(CH_3CHO)$ is further oxidized to the corresponding carboxylic acid,$CH_3COOH$.Therefore,the final products are $Me_2CO$ and $CH_3COOH$.

What are the products of the following reaction: $Me_2C=CHCH_3 \xrightarrow[(ii)\, H_2O]{(i)\, O_3}$?

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