R'-Cl xrightarrow{Na/text{Ether}} 2,3-text{di | vedclass

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(D) The given reaction is a Wurtz reaction,where two molecules of an alkyl halide $(R'-Cl)$ react with sodium in the presence of dry ether to form a s

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$isopropyl$

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(D) The given reaction is a Wurtz reaction,where two molecules of an alkyl halide $(R'-Cl)$ react with sodium in the presence of dry ether to form a symmetric alkane.The reaction is: $2R'-Cl + 2Na \xrightarrow{	ext{Dry ether}} R'-R' + 2NaCl$.The product formed is $2,3-	ext{dimethylbutane}$,which has the structure $CH_3-CH(CH_3)-CH(CH_3)-CH_3$.By splitting this symmetric alkane at the center,we get two identical fragments of $CH_3-CH(CH_3)-$. Thus,the alkyl group $R'$ is the isopropyl group,$CH_3-CH(CH_3)-$.Therefore,$R'-Cl$ is isopropyl chloride.

$R'-Cl \xrightarrow{Na/\text{Ether}} 2,3-\text{dimethylbutane}$. What is $R'$ in the above reaction?

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