CH_3-C(CH_3)=CH-CH_3 xrightarrow{Hg(OAc)_2/Et | vedclass

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Question

(B) The reaction is an alkoxymercuration reaction. First,a cyclic mercurinium ion intermediate is formed across the double bond. The nucleophile $(EtO

Vedclass · Accepted Answer

$CH_3-C(CH_3)(OEt)-CH(HgOAc)-CH_3$

Vedclass · Answer

(B) The reaction is an alkoxymercuration reaction. First,a cyclic mercurinium ion intermediate is formed across the double bond. The nucleophile $(EtOH)$ attacks the more substituted carbon atom (Markovnikov's rule) because it can better stabilize the partial positive charge in the transition state. Therefore,the $-OEt$ group attaches to the tertiary carbon,and the $-HgOAc$ group attaches to the secondary carbon. The final product is $CH_3-C(CH_3)(OEt)-CH(HgOAc)-CH_3$.

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

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