Question

Tetraethyl lead, $Pb{\left( {{C_2}{H_5}} \right)_4}$ initiates the chlorination of methane in the dark at $150^\circ C$ . Give reason.

Answer 1

Hint: Tetraethyl lead is a petro-fuel additive which was initially being mixed with gasoline as a patented octane rating booster allowing engine compression to be raised substantially. This led to the increased vehicle performance and fuel economy.

Complete answer:
When the tetraethyl lead $Pb{\left( {{C_2}{H_5}} \right)_4}$ initiates with chlorination of methane, here chlorination of methane is obtained when the mixture of chlorine and methane is exposed to flame and explodes producing carbon and hydrogen chloride. The chemical formula goes like,
$C{H_4} + C{l_2} \to C{H_3}Cl + HCl$
Above equation is initiated by the tetraethyl lead that is $Pb{\left( {{C_2}{H_5}} \right)_4}$ in the dark at $150^\circ C$ . Now the reason for this is tetraethyl lead $(Pb{\left( {{C_2}{H_5}} \right)_4})$ undergoes thermal homolysis of $C - Pb$bond at $150^\circ C$ . The free radical $C{H_3}C{H_2}^ * $ (Ethyl) then generates chlorine radical $C{l^ * }$ chlorination of methane in dark.
Homolytic cleavage produces free radical atoms with unpaired valence electrons. Basically, the dissociation of a molecule into two neutral fragments is called homolytic or homolysis. The thermally induced bond homolysis of peroxide is a common way of initiating radical reactions.
This is so far the explanation of the reason why tetraethyl lead $\left( {Pb{{\left( {{C_2}{H_5}} \right)}_4}} \right)$ initiates the chlorination of methane in the dark at $150^\circ C$ .
This is pretty much the information regarding this. All that we have to focus on is given step by step, after studying deep into this it comes easier to solve this faster in multi choice questions. It is all these basics that make up the complete equation. The main thing is to look properly during solving, hence it reduces the errors.

Note:
The chlorination of alkanes occurs by a series of homolytic bond cleavage and homogenic bond formation steps . when the mixture of chlorine and methane is exposed to ultraviolet rays then the substitution reaction occurs.