The product of the following reaction is: CH_ | vedclass

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(A) The given reaction is Oxymercuration-Demercuration $(OMDM)$. $1$. This reaction involves the Markovnikov addition of water ($H$ and $OH$) across t

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$CH_3-C(CH_3)_2-CH(OH)-CH_3$

Vedclass · Answer

(A) The given reaction is Oxymercuration-Demercuration $(OMDM)$. $1$. This reaction involves the Markovnikov addition of water ($H$ and $OH$) across the double bond. $2$. Crucially,no carbocation rearrangement occurs in this process because a bridged mercurinium ion intermediate is formed. $3$. The $-OH$ group adds to the more substituted carbon of the double bond,and the $-H$ atom adds to the less substituted carbon. $4$. Therefore,$3,3$-dimethyl-$1$-butene $(CH_3-C(CH_3)_2-CH=CH_2)$ reacts to form $3,3$-dimethyl-$2$-butanol $(CH_3-C(CH_3)_2-CH(OH)-CH_3)$.

The product of the following reaction is: $CH_3-C(CH_3)_2-CH=CH_2 \xrightarrow[(ii) NaBH_4]{(i) Hg(OAc)_2, H_2O}$

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