Propagation steps of fluoridation vs Iodination

[510586]

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Can someone explain the reasoning behind why F2 has a very high exothermicity in the propagating steps while I2 has a very endothermic first propagating step (thus reacting slowest)?

 

[Typical Average Student]

I2 is more likely to form a termination step with itself before it can even propagate another compound. Thus, you will not see I2 form a radical, it's an exception to know.

 

[510586]

I2 is more likely to form a termination step with itself before it can even propagate another compound. Thus, you will not see I2 form a radical, it's an exception to know.

Chlorine and Br act the same though I think. Fluorine appears to be the exception right? Also, how come fluorine doesn't want to form a termination step with itself?

 

[Typical Average Student]

Chlorine and Br act the same though I think. Fluorine appears to be the exception right? Also, how come fluorine doesn't want to form a termination step with itself?

Br2 and Cl2 are ideal for radical formation, you will see them the most. F2 forms a transient bond so it'll release its attached atom nicely. I'm not too sure about the deep detail about why it happens, maybe someone else can fill in.

 

[510586]

Do you think it would be rude to tag orgoman ?

 

[cdp1210]

It has to do with the different in bond dissociation energies between each and their respective reactivities for removing a hydrogen atom from the alkane to form the radical. I don't think the specifics matter. What is important to understand is:
1. F>Cl>Br>I in propagation reactivitiy
2. F and Cl propagation is exothermic therefore they can form a radical on primary, secondary, or tertiary
3. Br and Cl propagation is endothermic therefore they can only form radical on tertiary (Iodine is so unreactive and endothermic it won't even form a radical)

You will only really see Br and Cl used to make radical though because fluorine is too violent a reaction to be useful (and Iodine unreactive)

 

[510586]

It has to do with the different in bond dissociation energies between each and their respective reactivities for removing a hydrogen atom from the alkane to form the radical. I don't think the specifics matter. What is important to understand is:
1. F>Cl>Br>I in propagation reactivitiy
2. F and Cl propagation is exothermic therefore they can form a radical on primary, secondary, or tertiary
3. Br and Cl propagation is endothermic therefore they can only form radical on tertiary (Iodine is so unreactive and endothermic it won't even form a radical)

You will only really see Br and Cl used to make radical though because fluorine is too violent a reaction to be useful (and Iodine unreactive)

Awesome thanks

 

[DAT Destroyer]

Can someone explain the reasoning behind why F2 has a very high exothermicity in the propagating steps while I2 has a very endothermic first propagating step (thus reacting slowest)?

510586


The important concept here is to understand that Fluorine is the MOST reactive !!!! Up until very recently, most organic chemists, myself included avoided Fluorine Chemistry because reactions are EXPLOSIVE !!!!! If you looked at a textbook, you would note the highly exothermic propagation steps !!!!!

This is why F2 is so dangerous. Recently electrophilic Fluorine was able to be incorporated as F+ into a double bond system safely with reagents such a Selectfluor. Many cancer drugs could possibly now be built and studied with Fluorine.

For the DAT exam.....make sure you can recognize an initiating, propagating, and terminating step. Also make sure you are clear on radical reaction reactivity F2>>Cl2>Br2>>>>>I2. Overall Iodine is super slow.....reaction thermodynamics are too endothermic.....F2....too exothermic.

Also understand that Br2 although not as reactive as Cl2 is HIGHLY selective for tertiary, allylic, and benzylic positions !!!!

Hope this helps.....always a pleasure to answer questions.

Dr. Jim Romano

 

[Incis0r]

Can someone explain the reasoning behind why F2 has a very high exothermicity in the propagating steps while I2 has a very endothermic first propagating step (thus reacting slowest)?

There is also an explanation that cites the Hammond Postulate. Since Br propagation is endothermic, the transition state resembles the products much more than the reactants (i.e. it has more radical character); this allows the radical character to be more stabilized on the tertiary carbon.

 

[510586]

510586


The important concept here is to understand that Fluorine is the MOST reactive !!!! Up until very recently, most organic chemists, myself included avoided Fluorine Chemistry because reactions are EXPLOSIVE !!!!! If you looked at a textbook, you would note the highly exothermic propagation steps !!!!!

This is why F2 is so dangerous. Recently electrophilic Fluorine was able to be incorporated as F+ into a double bond system safely with reagents such a Selectfluor. Many cancer drugs could possibly now be built and studied with Fluorine.

For the DAT exam.....make sure you can recognize an initiating, propagating, and terminating step. Also make sure you are clear on radical reaction reactivity F2>>Cl2>Br2>>>>>I2. Overall Iodine is super slow.....reaction thermodynamics are too endothermic.....F2....too exothermic.

Also understand that Br2 although not as reactive as Cl2 is HIGHLY selective for tertiary, allylic, and benzylic positions !!!!

Hope this helps.....always a pleasure to answer questions.

Dr. Jim Romano

Thank you! It does help

 

[510586]

It has to do with the different in bond dissociation energies between each and their respective reactivities for removing a hydrogen atom from the alkane to form the radical. I don't think the specifics matter. What is important to understand is:
1. F>Cl>Br>I in propagation reactivitiy
2. F and Cl propagation is exothermic therefore they can form a radical on primary, secondary, or tertiary
3. Br and Cl propagation is endothermic therefore they can only form radical on tertiary (Iodine is so unreactive and endothermic it won't even form a radical)

You will only really see Br and Cl used to make radical though because fluorine is too violent a reaction to be useful (and Iodine unreactive)

Sorry to bump this old post, but it wouldn't be called f2 if we talk about propagation right? Because it is only in f2 step during initiation?

 

[cdp1210]

As in fluorine gas? Yeah. I was just referring to the elements involved in the reaction. They would all be in their standard state so F2, Cl2, Br2, I2.

 

[510586]

As in fluorine gas? Yeah. I was just referring to the elements involved in the reaction. They would all be in their standard state so F2, Cl2, Br2, I2.

No I meant isn't propagation more like F radical and something becoming another radical pair? F2 is only used in initiation right? Or would fluorine gas F2 also propagate, and if so, how?