Question

The characteristic chemical reaction of Alkanes is-
A.Substitution reaction
B.Addition reaction
C.Subtraction reaction
D.Displacement reaction

Answer 1

Alkanes are hydrocarbons containing carbon-carbon single bonds. They are saturated hydrocarbon. They are almost non-polar of the covalent nature of ${\text{C - C}}$and ${\text{C - H}}$ bonds. Also, there is very little electro-negativity difference between carbon and hydrogen bonds so the molecules are held only by the weak Vander Waals force.

Complete step-by-step answer: As we already know that alkanes are saturated compounds and ${\text{C - C}}$and ${\text{C - H}}$bonds are nonpolar so they are very less reactive. Hence they do not react with acids, bases, oxidizing, and reducing agents. But they do undergo replacement or substitution reactions.
Substitution reaction is a reaction in which an atom or a group of atoms in a molecule is replaced by some other atom or group of atoms. Alkanes undergo substitution reaction in which one or more H atoms are replaced with the different atoms. The characteristic reaction of alkanes is Free radical substitution reaction is generally a chain reaction completed in the following three steps-
Chain initiation -The attacking reactant splits into two identical parts each with one unpaired electron called free radical.
For example- In halogenation, a Chlorine molecule splits into two chlorine atoms or free radicals by absorbing the heat energy. This step starts the reaction so it is called the Chain initiation step.
$Cl - Cl\xrightarrow[{523 - 773K}]{{h\nu }}\mathop {Cl}\limits^ \bullet + \mathop {Cl}\limits^ \bullet$
Chain Propagation - The free radical then attacks the main reactant and produces another free radical. The new free radicals react with free radicals to form the product and a new free radical. So one free radical is consumed and the other is produced. This process is repeated.
In halogenation, chlorine free radical attacks methane and forms HCl and produces methyl free radical.
$C{H_4} + \mathop {Cl}\limits^ \bullet \to \mathop C\limits^ \bullet {H_3} + HCl$
This methyl free radical further reacts with the Chlorine molecule to form methyl chloride and produces free radical chlorine.
$\mathop C\limits^ \bullet H + Cl - Cl \to C{H_3}Cl + \mathop {Cl}\limits^ \bullet$
Chain Termination - The reaction stops when any two free radicals form such reactions which remove the reactive species (free radicals) which are needed for propagation.
In halogenation, the chlorine free radical can also abstract H atoms from methyl chloride which will produce a substituted methyl free radical.
$C{H_3}Cl + \mathop {Cl}\limits^ \bullet \to \mathop C\limits^ \bullet {H_2}Cl + HCl$
The new free radical will react with chlorine molecules to form methylene chloride.
$\mathop C\limits^ \bullet {H_2}Cl + Cl - Cl \to C{H_2}C{l_2} + \mathop {Cl}\limits^ \bullet$
This stops the reaction.

The correct answer is option A.

Note: Alkanes can also undergo oxidation or reduction reactions. It does not undergo an elimination reaction because it is a neutral compound and it cannot release hydride easily. And for the elimination reaction, there must be a leaving group present in the compound which is a weak base.