In the following reaction C is,CH_3-CH=CH_2 + | vedclass

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(D) Step $1$: The reaction of propene with $HCl$ in the presence of organic peroxide follows the anti-Markovnikov addition rule. $CH_3-CH=CH_2 + HCl \

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Ethane

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(D) Step $1$: The reaction of propene with $HCl$ in the presence of organic peroxide follows the anti-Markovnikov addition rule. $CH_3-CH=CH_2 + HCl \xrightarrow{	ext{Peroxide}} CH_3-CH_2-CH_2Cl$ ($A$ is $1$-chloropropane).Step $2$: $1$-chloropropane reacts with moist $Ag_2O$ (which acts as $AgOH$) to undergo nucleophilic substitution. $CH_3-CH_2-CH_2Cl + AgOH ightarrow CH_3-CH_2-CH_2OH + AgCl$ ($B$ is $n$-propanol).Step $3$: $n$-propanol $(CH_3-CH_2-CH_2OH)$ reacts with $EtMgBr$ (Ethylmagnesium bromide). Since $EtMgBr$ is a strong base,it reacts with the acidic hydrogen of the alcohol to form an alkane. $CH_3-CH_2-CH_2OH + CH_3CH_2MgBr ightarrow CH_3CH_3 + CH_3CH_2CH_2OMgBr$. The product $C$ is ethane $(CH_3CH_3)$.

In the following reaction $C$ is,$CH_3-CH=CH_2 + HCl$ $\xrightarrow[\text{Peroxide}]{\text{Organic}} A$ $\xrightarrow[\text{Moist}]{\text{Ag}_2\text{O}} B$ $\xrightarrow{\text{EtMgBr}} C$

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