$CH_3-C(CH_3)=CH-CH_3 \xrightarrow{Hg(OAc)_2/EtOH} (A) (90\%)$. The product $(A)$ of the above reaction is:

  • A
    $CH_3-C(CH_3)(HgOAc)-CH(OEt)-CH_3$
  • B
    $CH_3-C(CH_3)(OEt)-CH(HgOAc)-CH_3$
  • C
    $CH_3-C(CH_3)(OH)-CH(HgOAc)-CH_3$
  • D
    $CH_3-C(CH_3)(HgOAc)-CH(OH)-CH_3$

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