Which of these compounds represents the major monochlorination isomer formed in the following reaction?

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According to selectivity theory, the answer should be (c). However, the given answer is (b).


c is the least likely.

Alkane reactivity order is tertiary > secondary > primary > methyl reference here

I would be interested to see an explanation of why b predominates over a

  • $\begingroup$ Sure. We know that halogenation of alkanes involves free-radical intermediates. Now, free-radicals, though they are neutral, require one more electron to complete their octet. So, they get stabilized by electron releasing effects, hyperconjugation (or no bond resonance) being one of them. More the number of alpha hydrogens, more the number of C-H sigma bonds that can get delocalized and stabilize the electron hungry free-radical. Here, (b) has seven alpha hydrogens, whereas (a) has six. So the intermediate for product (b) will be more stable (though the difference won't be very big I think). $\endgroup$ – Anurag B. Jan 28 at 13:14
  • $\begingroup$ Also, there are two 2-degree hydrogens at position (c) ... wouldn't that make it more reactive than one 3-degree hydrogen at position (a), (b) or (d) because chlorine is not very selective? drive.google.com/file/d/1ZuSYdmoCEaKCiL0bxTl1PqwbbwSQEzCj/… (my notes - page 4) $\endgroup$ – Anurag B. Jan 28 at 13:22
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    $\begingroup$ @AnuragB. Still more selective then that. All this is an exercise in futility though as probably any of these products could be major depending on how exactly reaction was run - chemistry isn't some rules of thumb put together as textbooks might suggest. $\endgroup$ – Mithoron Jan 28 at 17:49
  • $\begingroup$ @Mithoron "Still more selective than* that." - Source? $\endgroup$ – Anurag B. Jan 28 at 18:24
  • $\begingroup$ @AnuragB. chemistry.stackexchange.com/questions/26949/… $\endgroup$ – Mithoron Jan 29 at 1:33

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