Just Started Methylene Blue

[Travis]

Correct, I should have said that but I meant it as all one step, where the radical absconds the hydrogen with an electron and replaces it with an O2. Thanks for the info, it is hard to find these basic measurements that should be available sometimes, isn't it?

I found this, which already has units of energy.

† ​

I suppose you can work backwards to see what pKa this gives, to see of the 9⋅kcal/mol simply represents the hydrogen abstraction. I would expect a value of between 15 and 25 since the pKa of 1,3‐cyclopentadiene is often heuristically stated as 'the most acidic diene' and has a value of ~18.⁽¹⁾ This radical is stabilized by resonance, making it more acid than most diene carbon acids. An article on olefins gives a range of 10–17,⁽²⁾ so the pKa should be something like that.. . .

[1] Cyclopentadiene:​

But it is only at 7.20? Better check his source [10].⁽³⁾

'Hydrogen abstraction from a bis-allylic methylene group by alkoxyl and alkylperoxyl radicals is favourable with Gibbs energies of -23 and -9 kcal/mol, respectively.' ―Koppenol⁽³⁾​

The other value yields a pKa of 18.40,* which is in the range of olefins⁽²⁾ and close to that of 1,3 cyclopentadiene.⁽¹⁾ The energy of the alkoxyl abstraction (Lipid₁–O· + H–Lipid₂) of −23⋅kcal/mol seems close to the spontaneous loss of the initial central hydrogen (pKa) of dienes in water.

'Thermodynamically, the hydrodioxyl radical should also be able to oxidize the singly allylic hydrogen in oleic acid, but no evidence for this reaction has been found [15].' ―Koppenol⁽³⁾​

So the energies involved in the chain reaction seem slightly different than the initial loss of the first hydrogen to water, unless this event happens to be characterized by abstraction with a rogue hydroxyl radical (H–O·). I would think this would be similar in energy to that of the alkoxy radical (Lipid₁–O·) abstraction of −23⋅kcal/mol since they are both single oxygen radicals. Of course, hydroxyl radicals can be formed through Fenton reactions and are thought to be common molecules within the cell (especially in those supplementing with iron, eating refined wheat flour, or low in antioxidants.)

'The data in Tables I and II indicate that an alkylperoxyl radical cannot abstract a non-allylic hydrogen from a hydrocarbon, ΔG = + 12 kcal.' ―Koppenol⁽³⁾​

The oxidation of saturated fatty acids is thermodynamically unfavorable, making coconuts a great choice.

[2] Hegedus, Louis S. "Palladium-assisted alkylation of olefins." Journal of the American Chemical Society (1980)
[3] Koppenol, W. H. "Oxyradical reactions: from bond‐dissociation energies to reduction potentials." FEBS letters (1990)
[†] Min, B. "Mechanism of lipid peroxidation in meat and meat products-A review." Food Science and Biotechnology (2005)
[*] First, to go from calories to joules:

(−9⋅kcal/mol)×(4.184⋅J/cal) = −37.656⋅kJ/mol​

Then the Gibbs Free Energy equation solving for Ka:

ΔG = −R⋅T⋅lnKa
(−ΔG/R⋅T) = lnKa
(−37.656·kJ/mol)/(8.3144598⋅J⋅mol⁻¹⋅K⁻¹)⋅(273.15) = lnKa​

The standard temperature in Kelvin is 273.15 degrees, and 'R' is Boltzmann's constant.

(−37.656·kJ/mol)/(8.3144598⋅J⋅mol⁻¹⋅K⁻¹)⋅(273.15°K) = lnKa
(−37.656·kJ)/(8.3144598⋅J)⋅(273.15) = lnKa
(−37.656·kJ)/(8.3144598⋅J)⋅(273.15) = lnKa
(16.581) = lnKa
ℯ¹⁶⋅⁵⁸¹ = Ka
15886813 = Ka
log(15886813) = pKa
7.20 = pKa = when starting with the value of −9⋅kcal/mol
18.40 = pKa = when starting with the value of −23⋅kcal/mol​

 

[Kyle M]

I found this, which already has units of energy.

View attachment 7806† ​

I suppose you can work backwards to see what pKa this gives, to see of the 9⋅kcal/mol simply represents the hydrogen abstraction. I would expect a value of between 15 and 25 since the pKa of 1,3‐cyclopentadiene is often heuristically stated as 'the most acidic diene' and has a value of ~18.⁽¹⁾ This radical is stabilized by resonance, making it more acid than most diene carbon acids. An article on olefins gives a range of 10–17,⁽²⁾ so the pKa should be something like that.. . .

View attachment 7807[1] Cyclopentadiene:​

But it is only at 7.20? Better check his source [10].⁽³⁾

'Hydrogen abstraction from a bis-allylic methylene group by alkoxyl and alkylperoxyl radicals is favourable with Gibbs energies of -23 and -9 kcal/mol, respectively.' ―Koppenol⁽³⁾​

The other value yields a pKa of 18.40,* which is in the range of olefins⁽²⁾ and close to that of 1,3 cyclopentadiene.⁽¹⁾ The lower pKa for the peroxyl abstraction (Lipid₁–O–O· + H–Lipid₂) is probably a result of the additional reactivity of the peroxyl group (probably why it's placed on eicosanoids, to destroy invaders such as bacteria). The energy of the alkoxyl abstraction (Lipid₁–O· + H–Lipid₂) of −23⋅kcal/mol seems close to the spontaneous loss of the initial central hydrogen (pKa) of dienes.

So the energies involved in the chain reaction seem a bit lower than the initial loss of the first hydrogen, unless this event happens to be characterized by an abstraction by a rogue hydroxyl radical (H–O·). I would think this would be similar in energy to that of the alkoxy radical (Lipid₁–O·) abstraction of −23⋅kcal/mol since they are both single oxygen radicals. Of course, hydroxyl radicals can be formed through Fenton reactions and are thought to be common molecules within the cell (especially in those supplementing with iron, eating refined wheat flour, or low in antioxidants.)

[2] Hegedus, Louis S. "Palladium-assisted alkylation of olefins." Journal of the American Chemical Society (1980)
[3] Koppenol, W. H. "Oxyradical reactions: from bond‐dissociation energies to reduction potentials." FEBS letters (1990)
[†] Min, B. "Mechanism of lipid peroxidation in meat and meat products-A review." Food Science and Biotechnology (2005)
[*] First, to go from calories to joules:

(−9⋅kcal/mol)×(4.184⋅J/cal) = −37.656⋅kJ/mol​

Then the Gibbs Free Energy equation solving for Ka:

ΔG = −R⋅T⋅lnKa
(−ΔG/R⋅T) = lnKa
(−37.656·kJ/mol)/(8.3144598⋅J⋅mol⁻¹⋅K⁻¹)⋅(273.15) = lnKa​

The standard temperature in Kelvin is 273.15 degrees, and 'R' is Boltzmann's constant.

(−37.656·kJ/mol)/(8.3144598⋅J⋅mol⁻¹⋅K⁻¹)⋅(273.15°K) = lnKa
(−37.656·kJ)/(8.3144598⋅J)⋅(273.15) = lnKa
(−37.656·kJ)/(8.3144598⋅J)⋅(273.15) = lnKa
(16.581) = lnKa
ℯ¹⁶⋅⁵⁸¹ = Ka
15886813 = Ka
log(15886813) = pKa
7.20 = pKa = when starting with the value of −9⋅kcal/mol
18.40 = pKa = when starting with the value of −23⋅kcal/mol​

My ultimate purpose was to show in raw numbers that the free energy available to the reaction in a mammalian body temperature is sufficient to initiate. I guess iron catalysis would lower the input more eh?

 

[Travis]

My ultimate purpose was to show in raw numbers that the free energy available to the reaction in a mammalian body temperature is sufficient to initiate. I guess iron catalysis would lower the input more eh?

Well, he's got the Gibbs Free Energy of the O₂ addition on the initial radical:

...and it's negative, so it would happen spontaneously. So to find the rate and energy of linoleic acid, water, and O₂ alone, you would of course need the pKa value of linoleic acid's dienic hydrogen in water. This should be ~20 but it's hard to find.

All of the numbers have already been calculated for the chain reaction itself, and also very likely for the hydroxyl‐initiated event—which of course depends mostly on iron. I think it would be cool to have both values because one could then compare the effects of iron against just the water + O₂ system, gauging the catalytic effect of iron.

The hydroxyl‐initiated event appears to be well characterized. Koppenol says this:

'The oxidizing species could be the hydroxyl radical [8] or a higher oxidation state of iron [9,10]. While the thermodynamic properties of hydrogen peroxide, superoxide and the hydroxyl radical are well documented [11].' ―Koppenol​

Well documented. So it appears just a matter of reading his citations and perhaps finding the second pKa of linoleic acid, but one might chance upon that while reading Koppenol's citations. It was written in 1990, so I think it would represent the state‐of‐the‐art in lipid peroxidation thermodynamics because this topic appears to have been figured‐out by that time. I'm sure that there's been a few more studies since then, but I would think that Koppenol's article—plus his citations (and the citations of his citations, and infinite regress)—would represent the majority of data on this topic.

If I were to do it I would start with Koppenol. From the Gibbs Free Energy you can get the pKa values—and from those you can calculate kinetic rates using various concentrations of lipid, oxygen, and iron.

 

[Kyle M]

So it appears just a matter of reading his citations and perhaps finding the second pKa of linoleic acid, but one might chance upon that while reading Koppenol's citations.

I'll dig there when next I could use the info.

 

[Mito]

I have not seen anything suggesting methylene blue is a methyl donor.

Peripheral Neuropathy, Rapidly Receding Gums + Bloods

 

[haidut]

Peripheral Neuropathy, Rapidly Receding Gums + Bloods

The studies do not provide much information on how many methyl groups MB donates. Even if it every molecule of it donates a methyl group, keep in mind that MB is taken in tiny doses (mg) compared to methyl donors like methionine (grams) daily. So, I doubt a few mg extra methyl groups daily would change much.

 

[Mito]

Even if it every molecule of it donates a methyl group, keep in mind that MB is taken in tiny doses (mg) compared to methyl donors like methionine (grams) daily. So, I doubt a few mg extra methyl groups daily would change much.

Similar to folate and B12?

 

[haidut]

Similar to folate and B12?

Yes, that's how I see it.