hv & UV

[Mstoothlady2012]

Can some1 explain the mechanism behind these two. I know what each one does, but I don't understand why they work differently even though they both form radical 😕

hv works with alkanes where it just replaces hydrogen with halogen - MARK

UV works with alkenes where it adds halogen ANTI-MARK. Also what would happen if Br2/UV is reacted with alkynes? would it stop after 1 addition or would it go all the way to alkane? This is basic stuff but confuses me alot.

They both form radical though right? then why this difference?

Also this might be stupid question but what is hv? light right?

Thanks!!🙂

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[DRHOYA]

Yeah, in alkanes, the Br will bond to the most stable. Ex: methylcyclohexane....the Br would add to the same carbon as the CH3. With alkynes, I think it depends on how much Br you have. It is just 1 rxn, then I'm pretty sure it would end with an alkene.

 

[PreDent2009]

The above post is correct about Br2/hv picking off the most stable hydrogen to form the most stable dot. Bromine favors 3 prime hydrogens on alkanes, or even better would go after the allyic site on alkenes or benzyllic site off an aryl group.

When reacting Br2/hv with an alkene, the most stable dot will be the allyic position or the carbon that is one away from the double bond. THis allows for resonance stabalization of the free radical. Now, depending on your reaction conditions you will either get the kinetic of thermodynamic product. With high temp, you will get the bromine on the position that puts the makes the most stable alkene system in the end. At low temps, the kinetic product will precede. You will get the dot that forms the fastest, which does not always give you the most stable alkene system.

When you are talking about the alkene and alkyne going to alkanes I think you are referring to Br2 forming a bridge and adding two bromines across the double or triple bond. In an alkene, you will get anti-addition of the bromines. In an alkyne you also get anti-addition, forming the trans alkene if you use 1 mole of Br2. If you use two moles of Br2 you will get neighboring geminal hydrogens so you will have 4 bromines attached to neighboring carbons as an end product. I do research for Organic and have tutored it for the past year and half and have never seen Br2/hv reacted with an alkyne. If you think about it you would not want add a free radical to a triple bond carbon or neighboring carbon. This would not give you resonance stabalization and with a terminal alkyne, plucking off that hydrogen to put a dot would most likely not happen. You never want a vicinal dot, or a dot that is located right on the carbon with triple or double bond.

Can some1 explain the mechanism behind these two. I know what each one does, but I don't understand why they work differently even though they both form radical 😕

hv works with alkanes where it just replaces hydrogen with halogen - MARK

UV works with alkenes where it adds halogen ANTI-MARK. Also what would happen if Br2/UV is reacted with alkynes? would it stop after 1 addition or would it go all the way to alkane? This is basic stuff but confuses me alot.

They both form radical though right? then why this difference?

Also this might be stupid question but what is hv? light right?

Thanks!!🙂

 

[DRHOYA]

The above post is correct about Br2/hv picking off the most stable hydrogen to form the most stable dot. Bromine favors 3 prime hydrogens on alkanes, or even better would go after the allyic site on alkenes or benzyllic site off an aryl group.

When reacting Br2/hv with an alkene, the most stable dot will be the allyic position or the carbon that is one away from the double bond. THis allows for resonance stabalization of the free radical. Now, depending on your reaction conditions you will either get the kinetic of thermodynamic product. With high temp, you will get the bromine on the position that puts the makes the most stable alkene system in the end. At low temps, the kinetic product will precede. You will get the dot that forms the fastest, which does not always give you the most stable alkene system.

When you are talking about the alkene and alkyne going to alkanes I think you are referring to Br2 forming a bridge and adding two bromines across the double or triple bond. In an alkene, you will get anti-addition of the bromines. In an alkyne you also get anti-addition, forming the trans alkene if you use 1 mole of Br2. If you use two moles of Br2 you will get neighboring geminal hydrogens so you will have 4 bromines attached to neighboring carbons as an end product. I do research for Organic and have tutored it for the past year and half and have never seen Br2/hv reacted with an alkyne. If you think about it you would not want add a free radical to a triple bond carbon or neighboring carbon. This would not give you resonance stabalization and with a terminal alkyne, plucking off that hydrogen to put a dot would most likely not happen. You never want a vicinal dot, or a dot that is located right on the carbon with triple or double bond.

Awesome explanation!

One other question. Say, you have a diene hydrocarbon. At low temps, it is under kinetic control, and at high temps under thermodynamic control.......correct? Also, at low temps, what does it mean to do the 1,2 addition, and at high temps 1,4 addition?? Thanks man.

 

[PreDent2009]

When you read in textbooks about the 1,2 or 1,4 addition it is refering to the kinetic of thermodynamic product. Lets say you have 2,4 hexadiene reacting with Br2/hv. The kinetic product will be the 3-bromo-1,4-hexadiene. This gives the most stable dot but not the most stable alkene system since you lost conjugation. The thermo product will be 1-bromo-2,4-hexadiene. If you draw it out you will see that since you have higher temp you are able to better stabalize the 1 prime allyic dot, get the bromine on the primary carbon, while still being able to maintain a conjugated system. Hope the explanation helps.

Awesome explanation!

One other question. Say, you have a diene hydrocarbon. At low temps, it is under kinetic control, and at high temps under thermodynamic control.......correct? Also, at low temps, what does it mean to do the 1,2 addition, and at high temps 1,4 addition?? Thanks man.