Halogenation of alkanes is an example of

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

Question

(C) The halogenation of alkanes is a chain reaction that proceeds via a free-radical mechanism in the presence of sunlight $(h
u)$ or heat.The reacti

Vedclass · Accepted Answer

Free-radical substitution

Vedclass · Answer

(C) The halogenation of alkanes is a chain reaction that proceeds via a free-radical mechanism in the presence of sunlight $(h
u)$ or heat.The reaction involves three steps:$1$. Initiation: The halogen molecule $(X_2)$ undergoes homolytic fission to form halogen radicals $(X\cdot)$.$2$. Propagation: The halogen radical abstracts a hydrogen atom from the alkane $(R-H)$ to form an alkyl radical $(R\cdot)$ and $HX$. The alkyl radical then reacts with another halogen molecule to form the alkyl halide $(R-X)$ and regenerates a halogen radical.$3$. Termination: The reaction stops when two radicals combine to form a stable molecule.Since the reaction proceeds through the formation of free radicals,it is classified as a free-radical substitution reaction.The reaction sequence is as follows:$CH_4 + Cl_2 ightarrow CH_3Cl + HCl$ (chloromethane)$CH_3Cl + Cl_2 ightarrow CH_2Cl_2 + HCl$ (dichloromethane)$CH_2Cl_2 + Cl_2 ightarrow CHCl_3 + HCl$ (trichloromethane)$CHCl_3 + Cl_2 ightarrow CCl_4 + HCl$ (tetrachloromethane)

Similar Questions

What is the number of moles of $sp^2$ hybridized carbon atoms present in $n$ moles of isopentane?

The most strained cycloalkane is

$A$ sample of gasoline contains $81\%$ iso-octane and $19\%$ $n$-heptane. Its octane number will be

Consider the following reaction:
$H_3C-CH(D)-CH(CH_3)-CH_3 + Br^\bullet \to X + HBr$
Identify the structure of the major product $X$.

$Ph-CH_2-CH(D)-CH_3 \xrightarrow{Br_2/hv}$
Product of the above reaction will be:

Vedclass Test Series

Exam Paper Generator

Online Exam Module