[nanoboy]

Hi there,

I gathered that radicals can be formed by hydrogen abstraction on a hydrocarbon molecule.  If radicals were to be formed from alcohol (by UV irradiation for instance), what would the effect of the chain length of the alcohol be on the yield of radicals formation?

Would the chain length also have an effect on the radical propagation?

My wild guess: 
(1) Since longer chain alcohol (e.g. methanol compared to ethanol) contains more hydrogen atom, the probability of hydrogen abstraction taking place would be higher?

Thanks very much in advance.


P.S.  I've just gained some insights about radicals stability from this thread:

http://www.chemicalforums.com/index.php?topic=27447.0

[spirochete]

Could you clarify your question?  Is there a specific problem you're trying to solve that relates to it?  If I'm understanding you correctly you're saying that a radical would be formed simply by irradiating an alcohol.  Typically you need some kind of initiator molecule containing bonds with especially low dissociation energies such as peroxides or halogen molecules.  Regular hydrocarbons or alcohols won't form radicals by themselves, even if you heat them up or shine UV light on them.

I gathered that radicals can be formed by hydrogen abstraction on a hydrocarbon molecule.  If radicals were to be formed from alcohol (by UV irradiation for instance), what would the effect of the chain length of the alcohol be on the yield of radicals formation?

Would the chain length also have an effect on the radical propagation?

My wild guess: 
(1) Since longer chain alcohol (e.g. methanol compared to ethanol) contains more hydrogen atom, the probability of hydrogen abstraction taking place would be higher?

The first thing that comes to mind is that you may be more likely to have several different products, for example during free radical halogenation.  Several locations for hydrogen abstraction means several possible products.  It is possible to select strongly for the product formed from the most stable radical using bromine (re: hammond postulate), but you will still get some unwanted side products.  And I personally don't know any way to get a major product which comes from a less stable radical intermediate.

I'm not sure what you mean by "the yield of radical formation."  Radicals are unstable intermediates that are almost never isolated.  Perhaps you mean would the reaction rate be faster?  Possibly yes, but the increased rate of reaction might be due to formation of unwanted products. 

Does that at least start to answer your questions?

[nanoboy]

Thanks for the answer, I've gained some better understandings of radicals from it.  Now I'll clarify better the system I'm dealing with:

Alcohol has long been known to possess a mild reducing power.  Some literatures out there have demonstrated making metal nanoparticles colloid in an alcohol matrix containing dissolved metal salts. 

Some other studies add certain polymers which contain hydroxyl end groups together into the alcohol + metal salts matrix.  The hydroxyl group has been accounted for the increased reducing power.  Heating the system can accelerate the reduction rate as well.

From my experience, by UV irradiation, the metal reduction also accelerates.  There can be a possibility that the reducing power of hydroxyl was enhanced by UV light, or some other reactions happened in the matrix that abstracts hydrogen from the polymer chain (which served as an intermediate), resulting in some radical products, which in turn reduced the dissolved metal ions?

One example of such polymers I used is poly(ethylene glycol) with a methoxy end group (-CH2CH2-OH) termination. 

So coming back to my first question here, would the chain length of the solvent (ethanol, methanol...) have an effect on the rate of radicals formation (which results in faster metal ion reduction).  I suppose the chain length of the polymer (the PEG) or the methoxy end group may well be more important to the chain length of the alcohol solvent.

Thanks again.

[spirochete]

Your question is admittedly more advanced than I'd thought, but I'm glad I could help with the basics.  Hopefully somebody else can give a full answer.