A) \[C{{H}_{3}}C{{H}_{2}}CH(OH)C{{H}_{2}}C{{H}_{2}}C{{H}_{3}}\]
B) \[C{{H}_{3}}C{{H}_{2}}CD\left( OH \right)C{{H}_{2}}C{{H}_{2}}C{{H}_{3}}\]
C) \[C{{H}_{3}}C{{H}_{2}}CH\left( OD \right)C{{H}_{2}}C{{H}_{2}}C{{H}_{3}}\]
D) \[C{{H}_{3}}C{{H}_{2}}CD\left( OD \right)C{{H}_{2}}C{{H}_{2}}C{{H}_{3}}\]
Correct Answer: C
Solution :
[c] Remember that during reduction with\[NaB{{H}_{4}}\], a hydride ion is transferred by \[NaB{{H}_{4}}\] to carbonyl carbon and a proton (or deutron) is transferred from the solvent (in the second step) to carbonyl oxygen. \[\underset{3-Hexanone}{\mathop{{{C}_{2}}{{H}_{5}}-\overset{O}{\mathop{\overset{||}{\mathop{C}}\,}}\,-{{C}_{3}}{{H}_{7}}}}\,\xrightarrow{NaB{{H}_{4}}}{{C}_{2}}{{H}_{5}}-\overset{{{O}^{-}}}{\mathop{\overset{|}{\mathop{C}}\,}}\,H-{{C}_{3}}{{H}_{7}}\] \[\xrightarrow[(-O{{D}^{-}})]{{{D}_{2}}O}\underset{{}}{\mathop{{{C}_{2}}{{H}_{5}}-\overset{OD}{\mathop{\overset{|}{\mathop{C}}\,H}}\,-{{C}_{3}}{{H}_{7}}}}\,\]
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