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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.

Answer Verified Verified
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.

Complete answer:
Oxymercuration-demercuration is a reaction which is used for the synthesis of alcohols.
This reaction takes place according to Markovnikov’s rule which states that reagents would add to the unsymmetrical alkenes in such a way that the negative part of the adding molecule goes to the carbon atom of the double bond which has lesser number of hydrogen atoms.
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,

This two-step procedure is called oxymercuration-demercuration or oxymercuration-reduction and gives alcohol corresponding to Markovnikov’s addition of water to alkenes without any rearrangement since carbocation is not the intermediates.
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.

Hence, option (d) is incorrect.

Note: Hydration of alkene occurs through a carbocation intermediate. But hydroboration- oxidation and oxymercuration-reduction do not involve carbocation intermediates and hence always give unexpected or unarranged alcohols.
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