The alkene R-CH=CH2 reacts readily with B2 H6 and formed the product B which on oxidation with alkaline H2 O2 produces

Question

The alkene R-CH=CH2 reacts readily with B2 H6 and formed the product B which on oxidation with alkaline H2 O2 produces (A)  R-CH2- CHO (B) R-CH2-CH2- OH (C)  (D)

Tardigrade Admin · Accepted Answer

Here, (1/2) mole of B 2 H 6 react with alkene by Markownikoff's addition and form trialkyl Borone called Hydroboration, H 2 O 2 / OH -gives oxidation. So, trialky borone oxidise in alcohols and reaction is also called Hydroboration-oxidation. 3 R - CH = CH 2 xrightarrow B 2 H 6 underset'B' ( R - CH 2- CH 2)3- B xrightarrow H 2 O 2 / H +3 R - CH 2- CH 2- OH

Q. The alkene $R - C H = C H_{2}$ reacts readily with $B_{2} H_{6}$ and formed the product $B$ which on oxidation with alkaline $H_{2} O_{2}$ produces

$R - C H_{2} - C H O$

$R - C H_{2} - C H_{2} - O H$

Q. The alkene R−CH=CH2​ reacts readily with B2​H6​ and formed the product B which on oxidation with alkaline H2​O2​ produces

R−CH2​−CHO

R−CH2​−CH2​−OH

Q. The alkene $R - C H = C H_{2}$ reacts readily with $B_{2} H_{6}$ and formed the product $B$ which on oxidation with alkaline $H_{2} O_{2}$ produces

$R - C H_{2} - C H O$

$R - C H_{2} - C H_{2} - O H$