question_answer

This conversion can be achieved by

A) \[M{{e}_{2}}CO/{{H}^{\oplus }},\,{{H}_{3}}{{O}^{\oplus }},KMn{{O}_{4}}/HO\Theta \]

B) \[M{{e}_{2}}CO/{{H}^{\oplus }},\,KMn{{O}_{4}}/{{H}_{3}}{{O}^{\oplus }}\]

C) \[KMn{{O}_{4}}\,NaI{{O}_{4}},\,{{H}_{3}}{{O}^{\oplus }},M{{e}_{2}}CO/{{H}^{\oplus }}\]

D) \[KMn{{O}_{4}}/NaI{{O}_{4}},\,M{{e}_{2}}CO/{{H}^{\oplus }},{{H}_{3}}{{O}^{\oplus }}\]

Correct Answer: B

Solution :

[b]