A bromoalkane X reacts with magnesium in dry | vedclass

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Question

(C) The reaction sequence is as follows:$R-Br + Mg \xrightarrow{	ext{Dry ether}} R-MgBr (Y)$$R-MgBr + HCHO \xrightarrow{	ext{Hydrolysis}} R-CH_{2}OH

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$1-bromopropane$

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(C) The reaction sequence is as follows:$R-Br + Mg \xrightarrow{	ext{Dry ether}} R-MgBr (Y)$$R-MgBr + HCHO \xrightarrow{	ext{Hydrolysis}} R-CH_{2}OH$Given the product is $C_{4}H_{10}O$,which is a primary alcohol $(R-CH_{2}OH)$,we have $R+CH_{2} = C_{3}H_{7}$.Thus,$R$ is a propyl group $(C_{3}H_{7}-)$.Therefore,$X$ is $1-bromopropane$ $(CH_{3}CH_{2}CH_{2}Br)$:$CH_{3}CH_{2}CH_{2}Br + Mg \xrightarrow{	ext{Dry ether}} CH_{3}CH_{2}CH_{2}MgBr$$CH_{3}CH_{2}CH_{2}MgBr + HCHO ightarrow CH_{3}CH_{2}CH_{2}CH_{2}OMgBr$$CH_{3}CH_{2}CH_{2}CH_{2}OMgBr + H_{2}O ightarrow CH_{3}CH_{2}CH_{2}CH_{2}OH + Mg(OH)Br$

$A$ bromoalkane $X$ reacts with magnesium in dry ether to form compound $Y$. The reaction of $Y$ with methanal followed by hydrolysis yields an alcohol having molecular formula $C_{4}H_{10}O$. The compound $X$ is

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