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Is the +M or -I effect more dominant in when an alkyl (mono)halide is subject to free radical chlorination/bromination?

For example what will be the major product in case of free radical (mono)chlorination of 2-chloropropane? 2-chloropropane

Would it be different if we did bromination?

I have found a similar question: Stability of carbocations attached to halogen atoms which was marked as duplicate but the other question was only about fluorine. I want to know in case of chlorine.

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1 Answer 1

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TL;DR: Even though alpha radical of 2-chloropropane might be more stable, due to a higher number of hydrogens present at the beta position, 1,2-dichloropropane might be formed in majority.

Edit: Research indicates 2,2-dichloropropane is probably the major product of the reaction because the radical is stabilized by 6 hyperconjugative structures. The +R effect of $\ce{Cl}$ and +I effect of $\ce{Me}$ groups also makes the alpha hydrogen more labile. More details at the end of the answer :)


The details

Monochlorination of Propane

Firstly, (citing ChemLibreTexts) the monochlorination of propane yields 1-chloropropane in majority:

relative amounts of products in monochlorination of propane showing 1-chloropropane is obtained with higher yield compared to 2-chloropropane

Dichlorination of Propane

If we were to isolate 2-chloropropane and carry out the reaction further, there are two factors to consider:

  1. Number of hydrogens: only one hydrogen is present at the alpha position while 6 hydrogens are present at the beta position.
  2. Stability of Free Radical: we must consider the stabilities of radical formation at alpha and beta positions.

ChemLibreTexts avoids this problem by simply stating that four isomers are formed in the dichlorination of propane:

production of four isomers from dichlorination of propane

I searched a few textbooks, including 10th edition of Organic Chemistry by Solomons, T. W. G. and Fryhle, C.B. and 7th edition of Organic Chemistry by Morrison R. T., Boyd R. N. and Bhattacharjee S. K., where no discussion on this reaction is provided. Why hasn't this been discussed by anyone?

Nonetheless, we are expecting the formation two isomers, 1,2-dichloropropane and 2,2-dichloropropane.

My thoughts

Even though I expect the chlorine atom to stabilize the free radical at alpha position, the beta hydrogens outnumber this and, chlorine radical not being very selective, 1,2-dichloropropane might form in majority. Bromine radical, being more selective, might form 2,2-dibromopropane, or 2-bromo-2-chloro-propane upon bromination of 2-chloropropane.


A Research Study

Since OP was not satisfied with my (very informed) intuition, I thought to go out and find references for this. Now, this took some time, and it has not been explored by a lot of researchers, but I was able to find an excellent research article, namely "Cl Atom Initiated Photo-oxidation of Mono-chlorinated Propanes To Form Carbonyl Compounds: A Kinetic and Mechanistic Approach" by Kumar A. and Rajakumar B. of Department of Chemistry, Indian Institute of Technology Madras, Chennai, India, that, besides many other things, explores the lability of $\ce{C-H}$ bonds in 2-chloropropane. The following is from the cited study.

Figure 1: Naming of transition states for the abstraction of hydrogen by chlorine radical from 2-chloropropane

Naming of transition states for the abstraction of hydrogen by chlorine radical from 2-chloropropane

Figure 2: Transition states involved in the reaction of chlorine radical and 2-chloropropane

Transition states involved in the reaction of chlorine radical and 2-chloropropane

Figure 3: Products of the reaction of chlorine radical and 2-chloropropane

Products of the reaction of chlorine radical and 2-chloropropane

Figure 4: Relative energies of the transition states and products in the reaction of chlorine radical and 2-chloropropane

Relative energies of the transition states and products in the reaction of chlorine radical and 2-chloropropane

To quote the authors:

For the reaction of 2-chloropropane with Cl atoms, it is clear from the computational calculations that the abstraction of the H-atom will be either from −CH3 (TS1 or TS2 or TS3) or from −CHCl (TS4). The radical formed via TS1 is the primary radical, and it has one hyperconjugation structure. The radical formed via TS4 is a secondary radical, and it has six hyperconjugation structures. Thus, TS4 is more stabilized due to secondary radicals and has more hyperconjugation structures than TS1. TS4 is also the more kinetically favorable pathway than other pathways (TS1 or TS2 or TS3) because of the electron-donating resonance effect (+R) of the directly attached Cl atom along with the +I effect of the two methyl groups attached to the radical carbon, which increases the electron density and makes it more kinetically labile.

Thus, 2,2-dichloropropane is probably (because I wasn't able to find a study that answers the question exactly) the major product of the reaction because the radical is stabilized by 6 hyperconjugative structures. The +R effect of $\ce{Cl}$ and +I effect of $\ce{Me}$ groups also makes the alpha hydrogen more labile.

Hopefully, this goes to show that researching upon one's intuition is extremely important as not only it may contradict our intuition but also provide further insights into the problem.

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  • $\begingroup$ Why does Cl stabilise the radical formed if -I is more dominating? $\endgroup$
    – Valoruz
    May 7 at 7:59
  • $\begingroup$ @Valoruz because it has empty d-orbitals to accommodate the free-radical electron. -I effect acts on charged species, even if -I effect is acting, it will withdraw the free radical electron density towards the chlorine atom, stabilizing the system. From your question, you shouldn't know which effect is dominating, so don't make any assumptions. Nevertheless, since you have asked, I will look into the stability of substituted free radicals. $\endgroup$
    – ananta
    May 7 at 8:10
  • $\begingroup$ @Valoruz hope the latest edits answer your question. $\endgroup$
    – ananta
    May 7 at 13:37
  • $\begingroup$ Thank you for the detailed answer. But I just want to know one thing. Why doesn't the -I effect of Cl destabilise TS4? $\endgroup$
    – Valoruz
    May 8 at 16:23
  • $\begingroup$ @Valoruz perhaps this article would be useful. I am also reading it as I write this comment. $\endgroup$
    – ananta
    May 8 at 18:33

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