CH_3-CH=CH-CH_3 xrightarrow[R_2O_2, Delta]{HB | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The reaction proceeds via a free radical mechanism.In the first step,the bromine radical $(Br^{\bullet})$ attacks the $sp^2$-hybridized carbon ato

Vedclass · Accepted Answer

Racemic

Vedclass · Answer

(A) The reaction proceeds via a free radical mechanism.In the first step,the bromine radical $(Br^{\bullet})$ attacks the $sp^2$-hybridized carbon atom of the symmetric alkene $CH_3-CH=CH-CH_3$.This attack results in the formation of a planar radical intermediate,$CH_3-\dot{C}H-CH(Br)-CH_3$.In the subsequent step,the hydrogen atom from $HBr$ can attack either face of the planar radical with equal probability.This leads to the formation of both enantiomers of $2-bromobutane$ in equal amounts,resulting in a racemic mixture.

$CH_3-CH=CH-CH_3 \xrightarrow[R_2O_2, \Delta]{HBr} \text{ (Anti-Markownikoff's addition)}$
Comment on the optical activity of the products.

Similar Questions

The percentage composition of an organic compound $A$ is: carbon = $85.71 \%$ and hydrogen = $14.29 \%$. Its vapour density is $14$. Consider the following reaction sequence:
$A$ $\xrightarrow{Cl_2/H_2O} B$ $\xrightarrow[(ii) H_3O^+]{(i) KCN/EtOH} C$
Identify $C$.

Propane is obtained from propene by which method?

What is the product formed when glycerol reacts with excess $HI$?

Which of the following compounds has the minimum heat of hydrogenation?

Identify the final product $Y$ in the following reaction sequence:
$CH_2=CH_2$ $\xrightarrow{O_2/Ag} X$ $\xrightarrow{473 \ K} Y$

Vedclass Test Series

Exam Paper Generator

Online Exam Module

$CH_3-CH=CH-CH_3 \xrightarrow[R_2O_2, \Delta]{HBr} \text{ (Anti-Markownikoff's addition)}$Comment on the optical activity of the products.