$ (CH_3 )_3 CMgCl on reaction with D_2O $ produces
The reaction of $ (CH_3 )_3CMgCl $ with $ D_2O $ results in the replacement of the MgCl group with a deuterium atom (D). Therefore, the product will be $ (CH_3)_3CD $. This is because the Grignard reagent $ (CH_3 )_3CMgCl $ will react with $ D_2O $ to form $ (CH_3)_3CD $ and Mg(OD)Cl.