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During monochlorination of propane why is chloropropane the major product formed. As far as I know using free radical mechanism 2-chloropropane is the supposed to be formed as the secondary free radical is more stable than a primary free radical. But the calculations involving reactivity order of chlorination tell otherwise. Why is this so!?

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When a Cl· atom is knocking a propane molecule, it may touch it at the end of the molecule, where there are 6 H atoms. Or it may touch it by the middle, where there are are only 2 H atoms. The probability that the collision happens with the end carbon is at least 3 times higher than the other possibility. So the secondary fragment issues from this collision may be more stable than the primary fragment. It is much less frequent. Statistically speaking, more 1-choropropane than 2-chloropropane will be produced.

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The reason might be one of kinetic versus thermodynamic control. The actual halogenation step is stage 2 in this scheme: The Free-Radical Halogenation of Alkanes (reaction scheme from chem.libretexts.org)

The 2-radical might well be more stable, but that doesn't mean that it will react faster than the 1-radical. If the reaction rate constant for the 1-radical with chlorine is much faster then the 1-chloro- product will predominate.

More likely, though is that it is that the 1-radical is formed in preference over the 2-radical in step 1. The halogen can attack at one of three positions, so statistically speaking you'd expect more 1-radical over 2-radical intermediates. If this species then reacts immediately in step 2 without rearrangement then you'd get a predominance of the 1-product formed. (Thanks to Karl for pointing this out).

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    $\begingroup$ The stability of the radical is irrelevant in the mechanism you show. It only plays a role if there is an (intra or intermolecular) equillibrium between the two radicals (which doesn´t seem to be the case). And no, "sterical hindrance" is not given here. You have a methyl:methylene ratio of 6:2, thats what makes the difference! If anything, sterics make the recombination of two methylene radicals is less probable, which would increase the yield of sec. brompropane. $\endgroup$ – Karl Jan 2 at 10:44
  • $\begingroup$ Seems like you should write an answer to the question. $\endgroup$ – deadlyvices Jan 2 at 10:51
  • $\begingroup$ I wanted to leave that as an exercise for you. ;) Happy new year! $\endgroup$ – Karl Jan 2 at 11:02
  • $\begingroup$ And you! So it's a branching ratio issue? The methyl-methylene branching ratio ratio is 6:2 in step 1? $\endgroup$ – deadlyvices Jan 2 at 11:04
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    $\begingroup$ Thats the first and simplest explanation (see also the other answer below), and it fits the bill, so why look any further. All other possible effects, where the radical stability plays a role, make the sec. brompropane more likely, but that doesn´t seem to be the case. $\endgroup$ – Karl Jan 2 at 11:14

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