Propan-1-ol can be prepared from propene by | vedclass

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Question

(C) The conversion of propene $(CH_3-CH=CH_2)$ to propan$-1-$ol $(CH_3-CH_2-CH_2-OH)$ requires an anti-Markovnikov addition of water across the double

Vedclass · Accepted Answer

$B_2H_6$ followed by $H_2O_2/OH^\Theta$

Vedclass · Answer

(C) The conversion of propene $(CH_3-CH=CH_2)$ to propan$-1-$ol $(CH_3-CH_2-CH_2-OH)$ requires an anti-Markovnikov addition of water across the double bond.This is achieved via the Hydroboration-Oxidation $(HBO)$ reaction.In this process,propene reacts with diborane $(B_2H_6)$ to form an alkylborane intermediate,which is then oxidized by hydrogen peroxide $(H_2O_2)$ in the presence of an aqueous base $(OH^\Theta)$ to yield the primary alcohol.Therefore,the correct reagent sequence is $B_2H_6$ followed by $H_2O_2/OH^\Theta$.

Propan$-1-$ol can be prepared from propene by

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