Question

Indicate the expected structure of the organic product when ethylmagnesium bromide is treated with heavy water $\left(D_{2}O\right)$.

• A. $C_2{H}_5-C_2{H}_5$
• B. $C_2{H}_{5}OD$
• C. $C_2{H}_6$
• D. $C_2{H}_{5}D$

Answer 1

The correct option is D $C_2{H}_{5}D$

The expected structure of organic product when ethyl magnesium bromide is treated with heavy water is:

We know heavy water is $D_{2}O$ as heavy hydrogen is deuterium hence heavy water is $D_{2}O$.

Ethyl magnesium bromide is a Grignard reagent that is RMgX.

when the Grignard reagent is treated with $H_{2}O$ that is hydrolysis then it gives corresponding alkanes and RMg(OH)X where hydrogen of alkane and OH in RMg(OH)X comes from water that is $H_{2}O$.

Now when $H_{2}O$replaced by $D_{2}O$ then in alkane H is replaced by heavy hydrogen that is D and h in RMg(OH)X also replaced by D.

So reaction become as

$C_2{H}_{5}MgBr+D_{2}O\to C_2{H}_{5}D+Mg\left(OD\right)Br$

Structure of alkane is tetrahedral as both carbons are $s{p}^3$ hybridised.