Free Radical Reactions

This video provides a basic introduction into free radical reactions. It explains the reactivit-selectivity principle between Chlorine and Bromine and it explains how to calculate the relative percent yields of each monochlorinated product. It discusses heterolytic and homolytic cleavage as it relates to the formation of radicals as well as how to identify initiation, propagation and termination steps in a radical reaction.
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Access The Full 1 Hour 32 Minute Video: bit.ly/3BeeOB3

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can you do a video on how to set-up calculation for the % of given monobromoalkane product formed

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why did you not use the other Chlorine radical instead of the Chlorine molecule for the second propagation step?

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I love you, thank you so much for this!!!!
Btw, in 10:33 isn't it more correct to say we get Hydrogen Chloride out of this reaction rather than Hydrochloric Acid, considering all of these components that are reacting are in gas form?

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At 15:00 did he mean to say Br was more reactive than Cl?

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no he's right; Chlorine is more reactive than bromine bcuz in group 7 reactivity decreases as you go down the group

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28:05

[@codycasanova2579]

Cl and Br both overall are exothermic, but their first steps are both endothermic (though Br is much more endothermic), so this is not a great graph to use. F on the other hand is super exothermic all the way through so it would be better to use this for the rxn proceeding vs energy graph. So, although Cl is less endothermic, it will still more so resemble the product but still be less selective.

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What about the ratios of Flourine and Iodine???
Please 🙏 reply

[@sociallyawkward3398]

fluorine reacts readily and hence is least selective in nature,so it just attaches to the carbon free radical with the least steric hindrance,as in if i have propane and i make it undergo fluorination,the major product would be 1-fluoropropane rather than 2-fluoropropane
in case of Iodine,it's very selective in nature and therefore bonds with the carbon free radical which is tertiary in nature(it's more stable due to hyperconjugation),so if i make propane undergo iodination,it's major product would be 2-iodopropane

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fluorination in itself is highly exothermic(literally forms bomb),so you wouldnt come across it in most reactions

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