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Question

(B) The reaction of $1$-methylcyclohex-$1$-ene with $Br_2$ in the presence of light $(h
u)$ is a free radical substitution reaction (allylic brominat

Vedclass · Accepted Answer

$1-$bromo$-1-$methylcyclohex$-2-$ene

Vedclass · Answer

(B) The reaction of $1$-methylcyclohex-$1$-ene with $Br_2$ in the presence of light $(h
u)$ is a free radical substitution reaction (allylic bromination).In this reaction,the bromine radical $(Br^\bullet)$ abstracts a hydrogen atom from the allylic position.The most stable free radical formed is the tertiary $(3^\circ)$ allylic radical at the $C-1$ position.This radical then reacts with $Br_2$ to form $1$-bromo-$1$-methylcyclohex-$2$-ene as the major product.

The major product of the following reaction is

Similar Questions

In the following reaction,$A$ and $B$ respectively are
$A \xrightarrow{HBr} C_2H_5Br \xrightarrow{B} A$

$A$ and $B$ respectively are:
$A \xrightarrow[(2) Zn-H_2O]{(1) O_3} \text{Ethane-}1,2\text{-dicarbaldehyde} + \text{Glyoxal}$
$B \xrightarrow[(2) Zn-H_2O]{(1) O_3} 5\text{-oxohexanal}$

The compound $B$ formed in the following sequence of reactions is:
$CH_3-CH_2-CH_2-OH$ $\xrightarrow{PCl_3} A$ $\xrightarrow{Alco. KOH} B$

Which of the following alkenes on oxidation by $KMnO_4$ in dil. $H_2SO_4$ forms adipic acid?

The stereochemistry of the product formed in the following reaction is:

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