Styrene on reaction with reagent X gave Y, wh | vedclass

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(C) Styrene $(C_6H_5CH=CH_2)$ reacts with $HBr$ in the presence of peroxide $((C_6H_5CO)_2O_2)$ via anti-Markovnikov addition to give $Y$ $(C_6H_5CH_2

Vedclass · Accepted Answer

$HBr / (C_6H_5CO)_2O_2 ; C_6H_5CH_2CHO$

Vedclass · Answer

(C) Styrene $(C_6H_5CH=CH_2)$ reacts with $HBr$ in the presence of peroxide $((C_6H_5CO)_2O_2)$ via anti-Markovnikov addition to give $Y$ $(C_6H_5CH_2CH_2Br)$.Hydrolysis of $Y$ gives $C_6H_5CH_2CH_2OH$, and subsequent oxidation gives $Z$ ($C_6H_5CH_2CHO$, phenylacetaldehyde).$Z$ $(C_6H_5CH_2CHO)$ is an aldehyde, so it gives a positive $2,4-DNP$ test.However, it does not contain the $CH_3CO-$ group, so it does not give the iodoform test.Therefore, $X$ is $HBr / (C_6H_5CO)_2O_2$ and $Z$ is $C_6H_5CH_2CHO$.

Styrene on reaction with reagent $X$ gave $Y$, which on hydrolysis followed by oxidation gave $Z$. $Z$ gives a positive $2,4-DNP$ test but does not give the iodoform test. What are $X$ and $Z$ respectively?

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