1.In the following reaction:XH3C CH3\underset{\text{H}_3\text{C}\;\;\;\;\text{CH}_3}{\text{X}}H3CCH3X →1. NaBH4 2. D3O+\xrightarrow{1.\;\text{NaBH}_4\;\; 2.\;\text{D}_3\text{O}^+}1.NaBH42.D3O+X and Y are:a.X = OD\text{OD}OD, Y = HOD\text{HOD}HODb.X = DOH\text{DOH}DOH, Y = HOD\text{HOD}HODc.X = DOD\text{DOD}DOD, Y = same as Xd.X = HOH\text{HOH}HOH, Y = same as XLogin to continueOnly logged in users canattempt or see the solution.
