Haloalkanes and Haloarenes — Chemistry STD 12 Science — Question

Since D of D2O gets attached to the carbon atom to which MgBr is attached, C is

$\text{CH}_3\text{CHCH}_3\\\ \ \ \ \ \ \ \ \ |\\\ \ \ \ \ \ \ \text{MgBr}\\\text{Isopropylmagnesium bromide}$

Therefore, the compound R - Br is

$\text{CH}_3\text{CHCH}_3\\\ \ \ \ \ \ \ \ \ |\\\ \ \ \ \ \ \ \text{Br}\\2-\text{Bromopropane}$

$\text{CH}_3\text{CHCH}_3+\text{Mg}\xrightarrow[]{\text{dry ether}}\text{CH}_3\text{CHCH}_3\xrightarrow[]{\text{D}_2\text{O}}\text{CH}_3\text{CHCH}_3\\\ \ \ \ \ \ \ \ \ |\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |\\\ \ \ \ \ \ \ \ \text{Br}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{MgBr}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{D}\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (\text{C})$

When an alkyl halide is treated with Na in the presence of ether, a hydrocarbon containing double the number of carbon atoms as present in the original halide is obtained as product. This is known as Wurtz reaction. Therefore, the halide, R1-X, is

$\ \ \ \ \ \ \ \ \ \ \ \text{CH}_3\\\ \ \ \ \ \ \ \ \ \ \ \ \ |\\\text{CH}_3-\text{C}-\text{X}\\\ \ \ \ \ \ \ \ \ \ \ \ \ |\\\ \ \ \ \ \ \ \ \ \ \ \text{CH}_3\\\ \text{tert}-\text{Butylhalide}$

Therefore, compound D is

$\ \ \ \ \ \ \ \ \ \ \ \text{CH}_3\\\ \ \ \ \ \ \ \ \ \ \ \ \ |\\\text{CH}_3-\text{C}-\text{MgBr}\\\ \ \ \ \ \ \ \ \ \ \ \ \ |\\\ \ \ \ \ \ \ \ \ \ \ \text{CH}_3\\\ \text{tert}-\text{Butylmagnesiumbromide}$

And, compound E is

$\ \ \ \ \ \ \ \ \ \ \ \text{CH}_3\\\ \ \ \ \ \ \ \ \ \ \ \ \ |\\\text{CH}_3-\text{C}\\\ \ \ \ \ \ \ \ \ \ \ \ \ |\\\ \ \ \ \ \ \ \ \ \ \ \text{CH}_3\\2-\text{Methylpropane}$