In the reaction sequence Me_3CCH=CH_2 xrighta | vedclass

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(D) The reaction $Me_3CCH=CH_2 \xrightarrow{Q} Me_3CCH_2CH_2OH$ represents an anti-Markovnikov addition of water, which is achieved by hydroboration-o

Vedclass · Accepted Answer

$B_2H_6, H_2O_2 / \overline{O}H ; Hg(OAc)_2, NaBH_4 / \overline{O}H$

Vedclass · Answer

(D) The reaction $Me_3CCH=CH_2 \xrightarrow{Q} Me_3CCH_2CH_2OH$ represents an anti-Markovnikov addition of water, which is achieved by hydroboration-oxidation using $B_2H_6, H_2O_2 / \overline{O}H$.The reaction $Me_3CCH=CH_2 \xrightarrow{R} Me_3C-CH(OH)-CH_3$ represents a Markovnikov addition of water without rearrangement, which is achieved by oxymercuration-demercuration using $Hg(OAc)_2, NaBH_4 / \overline{O}H$.Therefore, $Q$ is $B_2H_6, H_2O_2 / \overline{O}H$ and $R$ is $Hg(OAc)_2, NaBH_4 / \overline{O}H$.

In the reaction sequence $Me_3CCH=CH_2 \xrightarrow{Q} Me_3CCH_2CH_2OH$ and $Me_3CCH=CH_2 \xrightarrow{R} Me_3C-CH(OH)-CH_3$, $Q$ and $R$ are respectively:

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