# Why do radical inhibitors inhibit a reaction by creating a "more stable" radical, but for monohalogenation, we WANT to form the most stable radical?

According to my notes, radical inhibitors tend to turn a more reactive radical into a less reactive radical by making it more stable. For instance, the hydroquinone reacts with a radical species, forming a more resonance stabilized radical, which is less reactive.

However, when deciding to abstract a proton to create the major mono-halogenated product, we must choose a proton that has will create the most stable radical. In another words, the weakest C-H bond is broken first to form the most stable radical, which reacts to form the major mono-halogenated product. To me, this implies that greater radical stability increases rate of the monohalogenation reaction. And this contradicts the idea of the "less reactive, more stable radical" in radical inhibitor mechanism.

What am I missing? Any help will be appreciated.

• Just take the more stable radical formed in a reaction as an inhibitor for the formation of the other possible ones and the conflict should disappear. Also don't take this type of assertions as full 0/1. Reactions can give messy/mixed products mixtures. Anyway, you don't need a "stable" radical, just one that is more stable of the others. It can be rather unstable & reactive, but it gives the majority product. Nov 15, 2021 at 8:34

Radical inhibitors like dihydroquinone ($$\ce{H_2Q}$$) are quenchers which inhibit the radical reaction by quenching an active radical ($$\ce{R^.}$$) to $$\ce{R-H}$$ and itself gets converted into a resonance stabilized non-reactive monohydoquinone ($$\ce{HQ^.}$$) radical species. Once the $$\ce{R^.}$$ gets converted to $$\ce{R-H}$$ the reaction stops as the formed radical, $$\ce{HQ^.}$$ is not reactive enough to participate in the reaction further.
But in the case of mono-halogenated product formation, we are making a reactive intermediate species, $$\ce{R^{.}}$$ in the first step which will react further with the halogen ($$\ce{Cl_2}$$) in the second step to give the mono-chloro derivative ($$\ce{R-Cl}$$). The first step, i.e., $$\ce{R^.}$$ formation is the slowest and rate determining step, therefore activation energy for the radical intermediate formation gets lowered when more stable radical is formed (See: Hammond's postulate). Hence the rate of mono-halogenated product formation will increase when more stable radical is formed in the first step of the reaction which will give rise to major product if other possibility of hydrogen abstraction exists.
• Hello, please have a look at FAQ: How can I format math/chemistry expressions on Chemistry Stack Exchange?; in particular, you should use $\ce{H2O}$ $\ce{H2O}$, not just H_2O $H_2O$, which is incorrect as chemical element symbols are not supposed to be italicised. Nov 15, 2021 at 9:36