CH_3-CH=CH_2 xrightarrow[{(low , conc.)}]{Br_ | vedclass

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

Question

(B) The reaction of propene with $Br_2$ at low concentration in the presence of light $(h
u)$ or high temperature follows a free radical substitution

Vedclass · Accepted Answer

$CH_2=CH-CH_2-Br$

Vedclass · Answer

(B) The reaction of propene with $Br_2$ at low concentration in the presence of light $(h
u)$ or high temperature follows a free radical substitution mechanism at the allylic position.This is known as allylic bromination.The hydrogen atom attached to the allylic carbon ($CH_3$ group in propene) is replaced by a bromine atom.The product $(A)$ is allyl bromide: $CH_2=CH-CH_2-Br$.

$CH_3-CH=CH_2 \xrightarrow[{(low \, conc.)}]{Br_2/h\nu} (A)$; Product $(A)$ of the reaction is

Similar Questions

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

But$-2-$ene on reaction with alkaline $KMnO_4$ at elevated temperature followed by acidification will give:

The number of grams of bromine that will completely react with $5.0 \ g$ of pent$-1-$ene is $...... \times 10^{-2} \ g$. (Atomic mass of $Br = 80 \ g/mol$) [Nearest Integer]

When ethylene bromide is treated with $Zn$,we get

Predict the major alkene product formed upon heating the following ester:

Vedclass Test Series

Exam Paper Generator

Online Exam Module