The number of alkene(s) which can produce 2-b | vedclass

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(B) The reaction described is hydroboration-oxidation, which follows anti-Markovnikov addition of water across the double bond.To obtain $2-butanol$ $

Vedclass · Accepted Answer

$2$

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(B) The reaction described is hydroboration-oxidation, which follows anti-Markovnikov addition of water across the double bond.To obtain $2-butanol$ $(CH_{3}CH(OH)CH_{2}CH_{3})$, the starting alkene must be $1-butene$ $(CH_{3}CH_{2}CH=CH_{2})$ or $2-butene$ $(CH_{3}CH=CHCH_{3})$.For $1-butene$: The anti-Markovnikov addition of $H_{2}O$ yields $1-butanol$.For $2-butene$ ($cis$ or $trans$): The anti-Markovnikov addition of $H_{2}O$ yields $2-butanol$ because both carbons of the double bond are equivalent.Thus, only $2-butene$ (both $cis$ and $trans$ isomers) produces $2-butanol$ as the product.Therefore, the number of such alkenes is $2$.

The number of alkene$(s)$ which can produce $2-butanol$ by the successive treatment of $(i)$ $B_{2}H_{6}$ in tetrahydrofuran solvent and $(ii)$ alkaline $H_{2}O_{2}$ solution is:

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