1-Propanol can be prepared from propene by | vedclass

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Question

(C) The preparation of $1-$propanol from propene is achieved via hydroboration-oxidation.Reaction: $CH_3-CH=CH_2$ $\xrightarrow{B_2H_6, THF} (CH_3-CH_

Vedclass · Accepted Answer

$B_2H_6 - THF$ and $H_2O_2/OH^{-}$

Vedclass · Answer

(C) The preparation of $1-$propanol from propene is achieved via hydroboration-oxidation.Reaction: $CH_3-CH=CH_2$ $\xrightarrow{B_2H_6, THF} (CH_3-CH_2-CH_2)_3B$ $\xrightarrow{H_2O_2, OH^{-}} 3 CH_3-CH_2-CH_2OH$.Option $A$ $(H_2O/H^{\oplus})$ follows Markovnikov's rule,yielding propan$-2-$ol.Option $B$ ($Hg(OAc)_2/H_2O$ and $NaBH_4$) is oxymercuration-demercuration,which also follows Markovnikov's rule,yielding propan$-2-$ol.Option $C$ ($B_2H_6 - THF$ and $H_2O_2/OH^{-}$) is hydroboration-oxidation,which follows anti-Markovnikov's rule,yielding $1-$propanol.Therefore,option $C$ is correct.

$1-$Propanol can be prepared from propene by

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