The number of alkene (s) which can produce 2-butanol by the successive treatment of (i) B2H6 in tetrahydrofuran solvent and (ii) alkaline H2O2 solution is

Question

The number of alkene (s) which can produce 2-butanol by the successive treatment of (i) B2H6 in tetrahydrofuran solvent and (ii) alkaline H2O2 solution is (A) 1 (B) 2 (C) 3 (D) 4

Tardigrade Admin · Accepted Answer

The number of alkene(s) that can produce 2 -butanol by the successive treatment of

B_{2} H_{6}

in tetrahydrofuran solvent and alkaline

H_{2} O_{2}

solution is

2.

The alkenes are cis-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

π

-complex which changes into a cyclic four centre transition state with the addition of boron atom to the less hindered carbon atom.

Q. The number of alkene (s) which can produce 2-butanol by the successive treatment of (i) B2​H6​ in tetrahydrofuran solvent and (ii) alkaline H2​O2​ solution is

1

2

3

4

Q. 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