Which of the following intermediates is forme | vedclass

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

Question

(C) The addition of $HBr$ to propene in the presence of peroxide follows the anti-Markovnikov rule via a free radical mechanism.In the propagation ste

Vedclass · Accepted Answer

$CH_3-\dot{C}H-CH_2Br$

Vedclass · Answer

(C) The addition of $HBr$ to propene in the presence of peroxide follows the anti-Markovnikov rule via a free radical mechanism.In the propagation step,the bromine radical $(\dot{B}r)$ attacks the double bond of propene.There are two possibilities for the attack of the bromine radical:$1$. Attack at $C_1$ leads to the formation of a secondary free radical: $CH_3-\dot{C}H-CH_2Br$.$2$. Attack at $C_2$ leads to the formation of a primary free radical: $CH_3-CH(Br)-\dot{C}H_2$.Since a secondary free radical is more stable than a primary free radical,the intermediate $CH_3-\dot{C}H-CH_2Br$ is formed preferentially,leading to the major product $CH_3-CH_2-CH_2Br$.

Which of the following intermediates is formed to produce the major product in the addition of $HBr$ to propene in the presence of peroxide?

Similar Questions

$CH_2=CH-CH_2-CH_2-C(OH)(CH_2OH)-CH_2-CH_3$ $\xrightarrow[TsOH]{(CH_3)_2CO}$ $(A)$ $\xrightarrow[{(CH_3)_2S}]{O_3}$ $(B) + HCHO$. Product $(B)$ is

$CH_3-CH_2-CH_2-Br$ $\xrightarrow{\text{alk. KOH}} A$ $\xrightarrow[(ii) H_2O_2/OH^{-}]{(i) BD_3/THF} B$; $B$ is:

In the following reaction: $CH_3-C(CH_3)_2-CH=CH_2 \xrightarrow{H_2O/H^{+}} A + B$,the major product is:

The product $6-$oxoheptanal is formed by the reductive ozonolysis of:

Two alkenes,$X$ ($91\%$ yield) and $Y$ ($9\%$ yield),are formed when the following compound is heated:
The structures of $X$ and $Y$,respectively,are:

Vedclass Test Series

Exam Paper Generator

Online Exam Module