Question

Indicate the expected structure of the organic product when ethyl magnesium bromide is
treated with heavy water (\[{D_2}O\])
\[\begin{array}{*{20}{l}}
{A:{C_2}{H_5} - {C_2}{H_5}} \\
{B:{C_2}{H_5}OD} \\
{C:{C_2}{H_6}} \\
{D:{C_2}{H_5}D}
\end{array}\]

Answer 1

Hint:\[{D_2}O\] i.e. deuterium oxide, is also known as heavy water. It is a compound made up of deuterium and oxygen.Deuterium is a heavy isotope of hydrogen, denoted by either \[^2H\]or \[D\].
On the other hand, ethyl magnesium bromide is a Grignard reagent.Grignard reagent is basically a chemical compound having the generic formula of \[R - Mg - X\], while, \[X\] refers to a halogen and R refers to an organic group, primarily an alkyl or aryl group.

Complete answer:When Grignard reagent (\[R - Mg - X\]) is treated with water, it leads to a hydrolysis reaction yielding corresponding alkanes as well as \[RMg\left( {OH} \right)X\]. In this reaction, hydrogen of alkane along with hydroxyl group i.e. \[OH\] in \[RMg\left( {OH} \right)X\] actually comes from water i.e. \[{H_2}O\].
Now in the present question, water (i.e. \[{H_2}O\]) is being replaced with heavy water (i.e. \[{D_2}O\]). Therefore correspondingly in alkane also, H will be replaced with heavy hydrogen that is \[D\] and \[OH\] in \[RMg\left( {OH} \right)X\] will also get replaced with \[OD\]. Thus, the final reaction of ethyl magnesium bromide with heavy water can be written as follows:

\[{C_2}{H_5}MgBr + {D_2}O \to {C_2}{H_5}D + Mg\left( {OD} \right)Br\]

The structure of the compounds in reaction are also depicted below:

Hence, the expected structure of the organic product i.e. alkane is tetrahedral (since both carbon are sp3 hybridized) when ethyl magnesium bromide is treated with heavy water (\[{D_2}O\]).

Therefore, the correct answer is Option D.

Note: Since deuterium’s atomic mass is greater as compared to that of protium (hydrogen), molar mass of \[{D_2}O\] is more than that of \[{H_2}O\]. Thus, \[{D_2}O\] possess slightly different physical and chemical properties in comparison to \[{H_2}O\]. \[{D_2}O\] is non-radioactive owing to the deuterium being a stable isotope. \[{D_2}O\] can also be produced via hydrogen sulfide-water chemical exchange method, electrolysis and water distillation.