Answer
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Hint: - In this reaction we always get a mixture of products whatever the reaction proportions of methane and chlorine we use. The whole process is simply governed by chance.
Complete answer:
When methane reacts with chlorine a substitution reaction occurs and the organic product is chloromethane.
$C{H_4}\; + \;C{l_2} \to C{H_3}Cl\; + \;HCl$
But the reaction doesn't stop there, and further reactions go on where all the hydrogens in the methane get replaced by chlorine atoms. That means that we can get chloromethane, dichloromethane, trichloromethane or tetrachloromethane.
$C{H_4}\; + \;C{l_2} \to C{H_3}Cl\; + \;HCl$
$C{H_3}Cl + \;C{l_2} \to C{H_2}C{l_2}\; + \;HCl$
$C{H_2}C{l_2}\; + \;C{l_2} \to CHC{l_3}\; + \;HCl$
$CHC{l_3}\; + \;C{l_2} \to CCl\; + \;HCl$
It is not simple to control the product which we get by various proportions of methane and chlorine we use. If we use enough chlorine, we will eventually get $CC{l_4}$, but other proportions will always lead to a mixture of products.
The mechanisms behind this process is formation of multiple substitution products like dichloromethane, trichloromethane and tetrachloromethane can be explained in just the same sort of way as the formation of the original chloromethane. We just have to look at the collisions as the reaction progresses.
Having produced some chloromethane in no way can be prevented from being hit by chlorine radicals, and similarly for dichloromethane and trichloromethane.
Note: - We must not skip these equations. We must look carefully at each one so that we understand what is happening during the reaction, and can relate it to what has gone before. Each and every step is very important to understand the full reaction.
Complete answer:
When methane reacts with chlorine a substitution reaction occurs and the organic product is chloromethane.
$C{H_4}\; + \;C{l_2} \to C{H_3}Cl\; + \;HCl$
But the reaction doesn't stop there, and further reactions go on where all the hydrogens in the methane get replaced by chlorine atoms. That means that we can get chloromethane, dichloromethane, trichloromethane or tetrachloromethane.
$C{H_4}\; + \;C{l_2} \to C{H_3}Cl\; + \;HCl$
$C{H_3}Cl + \;C{l_2} \to C{H_2}C{l_2}\; + \;HCl$
$C{H_2}C{l_2}\; + \;C{l_2} \to CHC{l_3}\; + \;HCl$
$CHC{l_3}\; + \;C{l_2} \to CCl\; + \;HCl$
It is not simple to control the product which we get by various proportions of methane and chlorine we use. If we use enough chlorine, we will eventually get $CC{l_4}$, but other proportions will always lead to a mixture of products.
The mechanisms behind this process is formation of multiple substitution products like dichloromethane, trichloromethane and tetrachloromethane can be explained in just the same sort of way as the formation of the original chloromethane. We just have to look at the collisions as the reaction progresses.
Having produced some chloromethane in no way can be prevented from being hit by chlorine radicals, and similarly for dichloromethane and trichloromethane.
Note: - We must not skip these equations. We must look carefully at each one so that we understand what is happening during the reaction, and can relate it to what has gone before. Each and every step is very important to understand the full reaction.
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