The reaction between ethane & chlorine produces nine substitued products. briefly described the mechanism of free radical substitution that leads to the formation of (i) a monosubstitued chloroethane (ii) 2 disubstitued chloroethanes
Cl2 => 2Cl*
C2H6 + Cl* => C2H5* + HCl
C2H5* + Cl2 => C2H5Cl + Cl*
C2H5Cl + Cl* => C2H4Cl2
how do i explain?
!)n the presence of sunlight, Cl2 breaks down (homolytically) to form two chlorine radicals. This is the initiation step. I will use the * symbol to indicate a radical, in fact this should be expressed using a superscript dot. 2)The generated chlorine radical can then react with ethane to give an ethane radical (C2H5*) and HCl. 3)The ethane radical then reacts with another molecule of chlorine to generate the initial halogenated product and another chlorine radical which can react again. This is the propagation step. 4)The newly generated chlorine radical can now either react with another molecule of ethane (as above) or react with the halogenated product to give further halogenated products
thank you so much!
ur welcome
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