The number of alkene(s) that can produce 2-butanol by the successive treatment of $\mathrm{B}_{2} \mathrm{H}_{6}$ in tetrahydrofuran solvent and alkaline $\mathrm{H}_{2} \mathrm{O}_{2}$ solution is $2 .$ The alkenes are $c i s$ -but -2 - ene and trans-but-2-ene. The hydration product is obtained in accordance with opposite Markownikoff's rule. In this reaction, addition takes place through the initial formation of $\pi$ -complex which changes into a cyclic four centre transition state with the addition of boron atom to the less hindered carbon atom.
$\mathrm{H}_{3} \mathrm{C}-\mathrm{HC}=\mathrm{CH}-\mathrm{CH}_{3} \frac{(\mathrm{i}) \mathrm{B}_{2} \mathrm{H}_{6}}{(\mathrm{ii}) \mathrm{H}_{2} \mathrm{O}_{2} \mathrm{OH}} \mathrm{CH}_{3}-\mathrm{CH}-\mathrm{CH}_{2} \mathrm{CH}_{3}$