The reaction of propylbenzene with bromine in | vedclass

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

Question

(B) The reaction of propylbenzene $(C_6H_5-CH_2-CH_2-CH_3)$ with bromine $(Br_2)$ in the presence of light $(h
u)$ and heat is a free radical substit

Vedclass · Accepted Answer

$C_6H_5-CH(Br)-CH_2-CH_3$

Vedclass · Answer

(B) The reaction of propylbenzene $(C_6H_5-CH_2-CH_2-CH_3)$ with bromine $(Br_2)$ in the presence of light $(h
u)$ and heat is a free radical substitution reaction.In alkylbenzenes,the benzylic hydrogen is the most reactive due to the stability of the resulting benzylic radical.The benzylic radical is stabilized by resonance with the benzene ring.Therefore,the bromine atom replaces the hydrogen atom at the benzylic position (the carbon atom directly attached to the benzene ring).Thus,the product formed is $1$-phenyl-$1$-bromopropane,which is $C_6H_5-CH(Br)-CH_2-CH_3$.

The reaction of propylbenzene with bromine in the presence of light and heat gives ...................

Similar Questions

The products $A$ and $B$ formed in the above reactions are

The most reactive molecules towards $SN^1$ reaction are:

The reactivity order of the following molecules towards $S_N1$ reaction is:
$(I)$ Allyl chloride
$(II)$ Chlorobenzene
$(III)$ Ethyl chloride

The major product of the reaction is:

$C_6H_6 + CH_3CH_2CH_2Cl + AlCl_3 \to X$ (Major product)
$X$ (Major product) $\xrightarrow{HNO_3 + H_2SO_4} Y$ (Major product) $\xrightarrow{Sn + HCl} Z$ $\xrightarrow{H_2SO_4 + SO_3} W$
$W$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module