The product of the following reaction is: CH_ | vedclass

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(A) The reaction shown is hydroboration-oxidation of $1-$pentene.$(i)$ The first step involves the addition of $BH_3$ across the double bond,which fol

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$1-$Pentanol

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(A) The reaction shown is hydroboration-oxidation of $1-$pentene.$(i)$ The first step involves the addition of $BH_3$ across the double bond,which follows anti-Markovnikov addition.(ii) The second step involves oxidation with $H_2O_2$ in the presence of $OH^-$,which replaces the boron atom with a hydroxyl group $(-OH)$ with retention of configuration.Therefore,$CH_3CH_2CH_2CH=CH_2$ reacts to form $CH_3CH_2CH_2CH_2CH_2OH$,which is $1-$pentanol.

The product of the following reaction is:
$CH_3CH_2CH_2CH=CH_2 \xrightarrow[(ii) H_2O_2, OH^-]{(i) BH_3/THF} \text{Product}$

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The product of the following reaction is: $CH_3CH_2CH_2CH=CH_2 \xrightarrow[(ii) H_2O_2, OH^-]{(i) BH_3/THF} \text{Product}$