Select the reaction(s) that would result in t | vedclass

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

Question

(B) $(I)$ Reaction with $HBr$ in the presence of peroxide follows anti-Markovnikov addition to form $1-$bromopropane.$(II)$ Reaction with $HBr$ in $CC

Vedclass · Accepted Answer

$II$ and $III$

Vedclass · Answer

(B) $(I)$ Reaction with $HBr$ in the presence of peroxide follows anti-Markovnikov addition to form $1-$bromopropane.$(II)$ Reaction with $HBr$ in $CCl_4$ follows Markovnikov's addition to form $2-$bromopropane.$(III)$ Free radical bromination of propane $(CH_3-CH_2-CH_3)$ with $Br_2$ under $hv$ (light) preferentially forms $2-$bromopropane due to the higher stability of the secondary free radical intermediate.$(IV)$ Reaction with $Br_2$ in $CCl_4$ results in the formation of $1,2-$dibromopropane.Therefore,reactions $(II)$ and $(III)$ result in the formation of $2-$bromopropane.

Select the reaction$(s)$ that would result in the formation of $2-$bromopropane.
$(I) \ CH_3-CH=CH_2 + HBr \xrightarrow{\text{peroxide}}$
$(II) \ CH_3-CH=CH_2 + HBr \xrightarrow{CCl_4}$
$(III) \ CH_3-CH_2-CH_3 + Br_2 \xrightarrow{hv}$
$(IV) \ CH_3-CH=CH_2 + Br_2 \xrightarrow{CCl_4}$

Similar Questions

Which branched chain isomer of the hydrocarbon with molecular mass $72 \ u$ gives only one isomer of mono-substituted alkyl halide?

Which of the following compounds has the highest boiling point?

The melting points of alkanes with an odd number of carbon atoms are ............... compared to those with an even number of carbon atoms in the chain.

The molecular formula of an alkane having a weight of $11 \ g$ and a volume of $5.6 \ L$ at $STP$ is.......

In the reaction $CH_3-Br + 2Na + Br-CH_3 \to$,the product is called

Vedclass Test Series

Exam Paper Generator

Online Exam Module

Select the reaction$(s)$ that would result in the formation of $2-$bromopropane.$(I) \ CH_3-CH=CH_2 + HBr \xrightarrow{\text{peroxide}}$$(II) \ CH_3-CH=CH_2 + HBr \xrightarrow{CCl_4}$$(III) \ CH_3-CH_2-CH_3 + Br_2 \xrightarrow{hv}$$(IV) \ CH_3-CH=CH_2 + Br_2 \xrightarrow{CCl_4}$