Question

# Which is the wrong statement about oxymercuration-demercuration?(a)- In the first step, oxymercuration i.e., water and $Hg{{(OAc)}_{2}}$ add to the double bond.(b)- In the second step, demercuration i.e., $NaB{{H}_{4}}$ reduces ${{(-HgOAc)}_{2}}$ group to hydrogen(c)- The net reaction is the addition of water according to Markovnikov’s rule.(d)- Rearrangement takes place.

Hint: In all the steps there is a nucleophilic attack taking place. Oxymercuration-demercuration takes place in the presence of mercuric acetate. There is no formation of carbocation intermediates.

Oxymercuration-demercuration: Alkenes react with mercuric acetate, ${{(C{{H}_{3}}COO)}_{2}}Hg$ or $Hg{{(OAc)}_{2}}$ , to form oxymercuration products which upon reduction with $NaB{{H}_{4}}$ in basic medium gives alcohols. Thus,
Mechanism: Attack of $\pi -electrons$ of the double bond of the alkene on mercuric acetate gives an unsymmetrical $\pi -complex$ (I) with the expulsion of an acetate ion. Nucleophilic attack by ${{H}_{2}}O$ on the carbon atom of the $\pi -complex$ carrying the +ve charge gives II which subsequently loses a proton to give the oxymercuration product (III). Another nucleophilic attack on III by hydride ion $(:{{H}^{-}})$ from $NaB{{H}_{4}}$ on the carbon atom carrying $HgOCOC{{H}_{3}}$ group ultimately completes the addition with the expulsion of mercurous acetate.