Why are free radicals are so reactive? They can break almost any bond, including $\ce{C-H}$ bonds, which are fairly stable. Don’t they have an activation energy or something? Do they not also experience steric effects when colliding with reactants ?
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$\begingroup$ They do experience steric effects. $\endgroup$– JanDec 6, 2016 at 23:46
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$\begingroup$ Related: chemistry.stackexchange.com/questions/8586/… $\endgroup$– Klaus-Dieter WarzechaDec 7, 2016 at 5:53
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Radicals are so reactive because they require so much energy to form. When we speak about radical reactivity, "more reactive" generally means a more exothermic hydrogen atom abstraction step. The more exothermic this step, the more the transition state resembles the reactants (Hammond's postulate), with a correspondingly lower character of the carbon-centered radical product. This makes the reaction less sensitive to the stability of the carbon-centered radical.
This being said, some radicals do have an activation energy, and some don't. The $E_a$ is essentially zero for all hydrogen atom abstractions by fluorine atom, and the free energy barrier arises solely from the pre-exponential factor of the Arrhenius equation ($k=Ae^{-E_{a}/(RT)}$), which is near 13 for the halogenations listed below.
The activation energies for abstraction by chlorine atoms are also exceedingly small, but there is a slight preference for more highly substituted $\ce{C-H}$ bonds. Lastly, the activation energies for abstraction by bromine atoms are substantial, clearly showing a preference for abstraction of 3°, 2°, and 1° $\ce{C-H}$ bonds in that order. It is because of these differences in activation energies that we observe the varying regioselectivities of the different halogenation reactions$^{[1]}$.
$[1]$ Anslyn, E. V.; Dougherty, D. A. Modern Physical Organic Chemistry; University Science: Sausalito, CA, 2006.