Question

What is (are) true about free radical chlorination of an alkane?
(This question has multiple correct options.)
(A) Taking a large excess of alkane over chlorine minimizes the chances of dechlorination and polychlorination.
(B) Propagation is preferred over termination because the radicals are highly reactive.
(C) Major product is always formed at the carbon where the most stable free radical is formed in the intermediate step.
(D) Reaction is usually carried out in aqueous medium.

Answer 1

The Chlorination of Alkane proceeds through free radical Mechanism. The radical mechanism consists of 3 steps-Initiation, propagation, and termination. To start the reaction we need to provide an input of energy. Halogenation of an alkane requires the presence of light and heat for initiation.

Complete answer:
Let us consider an example of Chlorination of methane i.e. reaction with Chlorine the reaction occurs in the following sequence-
$C{H_4} + C{l_2} \to C{H_3}Cl \to C{H_2}C{l_2} \to CHC{l_3} \to CC{l_4}$
(Although we need Chloromethane as our major product, several other products like dichloromethane, trichloromethane and tetrachloromethane are also formed).
To answer the question let us study all the steps in a little detail-
(1)Initiation- This step of chlorination is the initiation that results in the formation of free radicals. The reaction requires energy and cannot occur in the absence of heat and light conditions. Thus we can say the reaction is energetically favourable in nature. (Free radicals are highly reactive species that cannot exist in isolation).
$Cl - Cl\xrightarrow[{Light}]{{U.V}}\mathop {Cl}\limits^ \bullet + \mathop {Cl}\limits^ \bullet$
(2)Propagation-The next step is propagation that occurs in 2 parts. In the first part chlorine radical reacts with Methane (or any alkane) to form Hydrochloric Acid and Methyl radical. This step is endothermic thus it is not energetically favourable.
$\mathop {Cl}\limits^ \bullet + C{H_4} \to HCl + \mathop {C{H_3}}\limits^ \bullet$
In the second part, chlorine atoms react with Methyl radical to form chloromethane and Chlorine radical again. This step is exothermic and energetically favourable.
$C{l_2} + \mathop {C{H_3}}\limits^ \bullet \to \mathop {Cl}\limits^ \bullet + C{H_3}Cl$
(3)Termination-This is the last step of the chlorination reaction. Here Chlorine radicals combine together, Methyl radicals combine to form Ethane, and Chlorine radical reacts with Methyl radical to form chloromethane.
$\mathop {Cl}\limits^ \bullet + \mathop {Cl}\limits^ \bullet \to Cl - Cl$
$\mathop {Cl}\limits^ \bullet + \mathop {C{H_3}}\limits^ \bullet \to C{H_3}Cl$
$\mathop {C{H_3}}\limits^ \bullet + \mathop {C{H_3}}\limits^ \bullet \to {C_2}{H_6}$
Now let us look at all the options given-
(A)The reaction just does not stop at the chloromethane product but rather forms di, tri, and tetra products. So to prevent this we use an excess of alkane. Also taking excess of Alkane allows the abstraction of hydrogen from alkane rather than alkyl halide preventing dechlorination. So the option is correct.
(B) This option is true since we all know how reactive free radicals are and so they react rather than recombining.
(C) This option is incorrect as we have already seen that the first propagation step is unfavourable and the second step that forms chlorine radical is favourable and uses the product obtained from the first step.
(D) This option is incorrect since the reaction occurs in the gaseous phase in presence of heat and light and does not require any aqueous medium.
So Options (A) and (B) are correct.

Note:
It is not only the chlorination that one can perform on Alkanes but we can perform Fluorination and Bromination as well. The halogenation reaction is not very selective. This reaction is useful to produce Chloroform, hexachlorobutadiene, etc. Alkanes are used since they are highly unreactive but Haloalkanes are highly reactive. Also, we use halogens and light as initiators.