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Question
Identify A, B, C, D, E, R and R1 in the following:
Solution
Since D of D2O gets attached to the carbon atom to which MgBr is attached, C is
\[\begin{array}{cc}
\ce{CH3CHCH3}\\
|\phantom{..}\\
\phantom{...}\ce{\underset{Isopropyl Magnesium bromide}{MgBr}}\
\end{array}\]
Therefore, the compound R−Br is
\[\begin{array}{cc}
\ce{CH3CHCH3}\\
|\phantom{..}\\
\ce{\underset{2-Bromopropane}{Br}}\
\end{array}\]
\[\begin{array}{cc}
\ce{CH3CHCH3 + Mg ->[dry ether] CH3CHCH3 ->[D2O] CH3CHCH3}\\
|\phantom{..........................}|\phantom{.................}|\phantom{..}\\
\ce{Br}\phantom{.......................}\ce{\underset{(C)}{MgBr}\phantom{.............}\ce{D}}\phantom{..}\\
\end{array}\]
When an alkyl halide is treated with Na in the presence of ether, a hydrocarbon containing double the number of carbon atoms present in the original halide is obtained as a product. This is known as Wurtz reaction. Therefore, the halide, R1−X, is
\[\begin{array}{cc}
\phantom{.....}\ce{CH3}\\
\phantom{...}|\\
\ce{CH3 - C - X}\\
\phantom{...}|\\
\phantom{......}\ce{\underset{tert-Butylhalide}{CH3}}\
\end{array}\]
Therefore, compound D is
\[\begin{array}{cc}
\phantom{.}\ce{CH3}\\
|\phantom{..}\\
\ce{CH3 - C - MgBr}\\
|\phantom{..}\\
\phantom{.}\ce{\underset{tert-Butylmagnesiumbromide}{CH3}}\\
\end{array}\]
And, compound E is
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