Identify Z in the series: CH_2 = CH_2 xrighta | vedclass

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(C) Step $1$: $CH_2 = CH_2$ reacts with $HBr$ to form ethyl bromide $(X = CH_3CH_2Br)$.Step $2$: Hydrolysis of $CH_3CH_2Br$ yields ethanol $(CH_3CH_2O

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$CHI_3$

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(C) Step $1$: $CH_2 = CH_2$ reacts with $HBr$ to form ethyl bromide $(X = CH_3CH_2Br)$.Step $2$: Hydrolysis of $CH_3CH_2Br$ yields ethanol $(CH_3CH_2OH)$.Step $3$: Ethanol reacts with $I_2$ in the presence of $Na_2CO_3$ (Iodoform test conditions) to produce iodoform $(CHI_3)$,which is a yellow precipitate.Therefore,$Z$ is $CHI_3$.

Identify $Z$ in the series:
$CH_2 = CH_2$ $\xrightarrow{HBr} X$ $\xrightarrow[Hydrolysis, Na_2CO_3]{I_2 (excess)} Z$

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Identify $Z$ in the series: $CH_2 = CH_2$ $\xrightarrow{HBr} X$ $\xrightarrow[Hydrolysis, Na_2CO_3]{I_2 (excess)} Z$