Answer
a)
$CH_{4} + Cl_{2} --h\nu --> CH_{3}Cl + HCl$
When methane reacts with $Cl_{2}$ in presence of sunlight, one H atom of $CH_{4}$ is substituted with one Cl atom of $Cl_{2}$ molecule and the end product will be methyl chloride ($CH_{3}Cl$) & HCl. This is an example of monosubstitution reaction where one atom is substituted by another atom.
b)
$CH_{4} + Cl_{2} --h\nu --> CH_{3}Cl + HCl$
$CH_{3}Cl + Cl_{2} --h\nu --> CH_{2}Cl_{2} + HCl$
$CH_{2}Cl_{2} + Cl_{2} --h\nu --> CHCl_{3} + HCl$
$CHCl_{3} + Cl_{2} --h\nu --> CCl_{4} + HCl$
c) Name of compounds :
$1. CH_{4} -> Methane$
$2. CH_{3}Cl -> methyl chloride$
$3. CH_{2}Cl_{2} -> Methylene chloride$
$4. CHCl_{3} -> Chloroform$
$5. CCl_{4} -> Carbon tetra chloride$
$6. HCl -> Hydrogen Chloride$
d) Halogenation reaction of the larger alkanes have limited value because for a larger alkane, there are so much possibilities of which 'H' atom will be substituted by the halogen atom. That's why there are lot of mixtures of haloalkanes produced by halogenation reaction of larger alkanes & yield of the desired product is very low.
Work Step by Step
Explanation of every question asked are given on the answer section above.
