Question

What happens when
(i) ethyl magnesium bromide is treated with heavy water ? (ii) methyl magnesium chloride reacts with ethanoic acid ?

Answer 1

${D_2}O$ 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. Carbonation of Grignard reagent is the reaction of Grignard reagent with $C{O_2}$ . So, when $C{O_2}$ reacts with organometallic reagents (Grignard reagents), we are familiar with the attack of $R$ - anion on carbonyl groups, i.e., an acid salt is formed. This acid salt is further hydrolyzed in the presence of acid to form a carboxylic acid. These acids can then be used for various purposes.

Complete answer:
Grignard reagents react with carbon dioxide in two stages, in the first stage, we will get an addition of the Grignard reagent to the carbon dioxide. Dry carbon dioxide is bubbled through a solution of the Grignard reagent in ethoxyethane. So, when Methyl magnesium bromide is treated with carbon dioxide, we know that $C{O_2}$ has two $C = O$ bonds. The anion attacks the carbon and forms an acid salt. Carbon dioxide gets added to the Methyl magnesium bromide in the reaction shown below $C{H_3}MgBr + C{O_2} \to C{H_3}COOMgBr$ The product is then hydrolysed or it is reacted with water in the presence of a dilute acid. Typically, we add dilute sulphuric acid or dilute hydrochloric acid to the solution formed by the reaction with the $C{O_2}$ . A carboxylic acid is produced with one more carbon than the original Grignard reagent. In this case, acetic acid is formed as the product $C{H_3}COOMgBr \to C{H_3}COOH + MgBrOH$ So, when Methyl magnesium bromide is treated with carbon dioxide and hydrolysed, the product is Acetic acid. Now in the present question, water (i.e. ${H_{2O}}$ ) 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 $RMg(OH)X$ will also get replaced with $OD$ . Thus, the final reaction of ethyl magnesium bromide with heavy water can be written as follows:
${C_2}{H_5}MgBr + {D_2}O \to {C_2}{H_5}D + Mg(OD)Br$
Alkane is tetrahedral (since both carbon are sp3 hybridized) when ethyl magnesium bromide is treated with heavy water .

Note:
A Grignard reagent of Grignard compound is a chemical compound with the generic formula $R - Mg - X$ , where $\;X$ is a halogen and $R$ is an organic group, it can be an alkyl or aryl normally. Since deuterium’s atomic mass is greater as compared to that of protium (hydrogen), the 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.