Why Do Radicals React Differently with Br?

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In organic chemistry, the stability of radicals and carbocations is crucial for predicting reaction products. Tertiary radicals are the most stable compared to primary and secondary radicals, which raises questions about reactivity. When considering reactions with bromine (Br), it is observed that Br tends to attach to the more stable tertiary radical rather than the less stable primary radical. This behavior can be explained by the fact that Br reacts preferentially with the most stable and abundant radical present in the reaction mixture. Additionally, secondary hydrogens are more reactive than primary hydrogens, which aligns with the observation that Br will often replace secondary hydrogens in reactions. Overall, the preference for Br to react with the most stable radical is due to the stability and abundance of that radical in the reaction environment, leading to the formation of major products that reflect these dynamics.
Nick tringali
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Alright so in organic chemistry we learn that for radicals and carbocations, the tertiary radical is the most stable (of primary and secondary). Okay, so why when we are predicting the major product of the reaction, let's take Br for example, the Br is going to attach to the tertiary radical. If tertiary is the "more stable" one, shouldn't Br seek out the less stable radical like a primary radical. fluorine is very unstable and fluorine will react with everything. A Nobel gas is very stable and will react with nothing. When looking at another slide on my powerpoint, it says that secondary hydrogens are way more reactive than primary hydrogens, this makes sense because when I am trying to find the major products of a reaction, I will place my Br where the secondary hydrogen was. The semantics of this makes no sense to me. Why is a Br going to attach on a "more stable" tertiary radical than a less stable primary radical. I hope I articulated this question well enough. Please try to explain this in layman's terms.
 
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In otherwords why is Br going to react with the most stable radical. But, Br is going to also take the place of the most reactive tertiary hydrogen. How does that make any sense.
 
Br will most often react with the carbocation/radical that is present in the largest amount - and the most stable one dominates the mixture.
 
What I know and please correct me: a macroscopic probe of raw sugar you can buy from the store can be modeled to be an almost perfect cube of a size of 0.7 up to 1 mm. Let's assume it was really pure, nothing else but a conglomerate of H12C22O11 molecules stacked one over another in layers with van de Waals (?) "forces" keeping them together in a macroscopic state at a temperature of let's say 20 degrees Celsius. Then I use 100 such tiny pieces to throw them in 20 deg water. I stir the...

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