Question

${\left({CH}_3\right)}_{3}CMgBr$ on reaction with $D_{2}O$ produces:

• A. ${\left({CH}_3\right)}_{3}CD$
• B. ${\left({CH}_3\right)}_{3}COD$
• C. ${\left({CD}_3\right)}_{3}CD$
• D. ${\left({CD}_3\right)}_{3}OD$

Answer 1

The correct option is A ${\left({CH}_3\right)}_{3}CD$

${\left({CH}_3\right)}_{3}CMgBr+D_{2}O⟶{\left({CH}_3\right)}_{3}C-D+MgBrOH$

The $C-Mg$ Bond donates its electron pair to a devterium atom, forming ${\left({CH}_3\right)}_{3}CD+MgBrOH$.

This occurs because the $C$ that is bonded to $Mg$ is strongly nucleophilic and can attack a region of positive charge density such as devterium atom in a water molecule.