In the following sequence of reactions CH_3CH | vedclass

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(C) The reaction sequence is as follows:$1$. $CH_3CH_2OH + P + I_2 ightarrow CH_3CH_2I$ ($A$ is ethyl iodide).$2$. $CH_3CH_2I + Mg \xrightarrow{	ex

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$n-$propyl alcohol

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(C) The reaction sequence is as follows:$1$. $CH_3CH_2OH + P + I_2 ightarrow CH_3CH_2I$ ($A$ is ethyl iodide).$2$. $CH_3CH_2I + Mg \xrightarrow{	ext{Ether}} CH_3CH_2MgI$ ($B$ is ethylmagnesium iodide).$3$. $CH_3CH_2MgI + HCHO ightarrow CH_3CH_2CH_2OMgI$ ($C$ is an intermediate alkoxide).$4$. $CH_3CH_2CH_2OMgI + H_2O ightarrow CH_3CH_2CH_2OH + Mg(OH)I$ ($D$ is $n-$propyl alcohol).Thus,the final product '$D$' is $n-$propyl alcohol.

In the following sequence of reactions $CH_3CH_2OH$ $\xrightarrow{P + I_2} A$ $\xrightarrow[Ether]{Mg} B$ $\xrightarrow{HCHO} C$ $\xrightarrow{H_2O} D,$ then compound '$D$' is

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