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From what I know, the energy in ATP comes from the phosphate chain, which can be cleaved for energy using water(more specifically the energy comes the from the formation of new molecules upon cleaving the chain, as the cleavage itself takes energy.), and then the cleaved phosphates can be added back into the chain to release water and take energy. I have been spending a little bit trying to find a replacement for the phosphates, but I come up empty, So is there some replacement to the phosphate chain (that ideally won’t react with F2 gas)?

Some Clarity to the question:

I am looking for some molecule that can be put into a chain and will be cleaved by the addition of a simple molecule[that’s not so reactive in fluorine environment it’ll explode] to release energy [the energy will probably come from then formation of new bonds after cleaving the chain] and un-cleaved to release the same simple molecule and take energy

Edit: The cells’ solvent is liquid hydrogen fluoride(HF) at about -30 to -50 Celsius.

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    $\begingroup$ I'll put my thinking cap on. In the meantime, the google search phrase you want is 'reversible fluorination'. I can tell you now, there is bound to be some metal ions that bind poorly with fluorine and can be displaced by oxygen, at least in enzymes if not in bulk. $\endgroup$ Jul 4, 2022 at 4:20
  • $\begingroup$ Possibly acetates? See this: link.springer.com/article/10.1007/s00253-021-11608-0 $\endgroup$ Jul 4, 2022 at 4:24
  • $\begingroup$ I am not sure I understand your request: the title seems to be asking for fluorine based alternatives, while the body for non fluorine based ones? $\endgroup$
    – L.Dutch
    Jul 4, 2022 at 7:38
  • $\begingroup$ Well, I meant an alternative that won’t like, explode or react quickly in a fluorine environment. $\endgroup$
    – KaffeeByte
    Jul 4, 2022 at 13:55
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    $\begingroup$ What's your solvent? A big selling point of ATP is that it's essentially a fuel that burns in water (rather than needing O2 to react with). Your chemical needs to be able to take up or release a molecule of the solvent in a reaction with a large free energy difference but which does not occur rapidly without a catalyst. $\endgroup$ Jul 4, 2022 at 20:23

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Nobody knows, but that won't keep me from making a wild guess!

The obvious first thing to try is short fluorophosphine chains, like $P_3F_5$.You could turn that into a $-P_3F_4$ group ($R-PF-PF-PF_2$) attached to an appropriate carrier molecule. That's stable enough to not immediately dissociate like polyfluoronitrogens, and maybe you could get energy out of it by reacting with $HF$? To produce $R-PF-PFH + PF_3$, or $R-PF-PF_2 + F_2HP$. You'd get better solubility and more consistent chemical behavior by allowing a hydrogen substitution, so you end up with $R-PF-PF-PFH + HF \Leftrightarrow R-PF-PFH + F_2HP$.

I am disinclined to bother trying to figure out the energetics of such a reaction, though, because it has some significant problems; phosphate is just a generally useful ion to have around. It's charged, and participates in a bunch of catalytic reactions and it can form multiple bonds to build bridges between other molecular groups--like nucleic acid backbones. Phosphines are polar but uncharged, and don't have the same kind of functional diversity. It all comes down to fluorine just not being a good substitute for oxygen. However, if there is oxygen around (which there should be, because it's more abundant than fluorine and fluorine will displace it from water and silicates), then fluorophosphines will react with oxygen to form difluorphosphates ($F_2PO_2^-$), which is hydrolytically unstable in water (it wants to form regular phosphates and more hydrofluoric acid), but it is stable in hydrofluoric acid. These could probably be chained to produce something much more similar to terrestrial phosphate chains. Rather than

$R-O-PO(OH)-O-PO(OH)_2 + H_2PO_3 \Leftrightarrow R-O-PO(OH)-O-PO(OH)-O-PO(OH)_2 + H_2O$

you'd get

$R-O-POF-O-POF_2 + F_2PO(OH) \Leftrightarrow R-O-POF-O-POF-O-POF_2 + HF$

In which we replace, not the oxygens, but specifically the hydroxyl groups with fluorines. Synthesis occurs by removing a fluorine from the chain and a hydrogen from the fluorphosphoric acid, using them to form new HF, and forming a new ester bond between the oxygen from which the hydrogen was removed and the phosphorus from which the fluorine was removed. Hydrolysis is the reverse.

I have no idea how to look up the actual energetics of such a molecule--I would be surprised if anyone has ever actually synthesized and studied it--but on general principle it should be slightly more rigid and have comparable if not slightly higher bond energy than Earthling phosphate chains.

And just to throw it out there, since I'm looking at it now, maybe you could do something useful with the nitrogen difluoride radical ('cause unlike polyfluorophosphines, polyfluoronitrogens like to dissociate).

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  • $\begingroup$ I am certain that the reaction would be more energetic than regular ATP, which is perfect in this case as most if not all the bonds that fluorine makes are stronger than their oxygen counterparts, so the extra energy is good in breaking these strong bonds. When you say that "difluorophosphates are stable in hydrofluoric acid" did you mean the liquid state of HF, or the solution of HF in water? because water burns in the atmosphere of this fluorine planet. $\endgroup$
    – KaffeeByte
    Jul 10, 2022 at 19:03
  • $\begingroup$ @KaffeeByte Should be stable in both situations, precisely because water burns in fluorine--HF won't displace oxygen from difluorophosphates, because if it did, it would produce water, which wants to react with fluorophosphines to make more HF! $\endgroup$ Jul 11, 2022 at 1:55
  • $\begingroup$ This is probably a stupid question and I am probably overlooking something obvious, but does the oxygen radical actually participate in the reaction(bond with something), or is it there just to make the reaction happen, and not so that it bonds with anything? $\endgroup$
    – KaffeeByte
    Jul 11, 2022 at 14:13
  • $\begingroup$ @KaffeeByte Yes, it bonds with phosphorus. $\endgroup$ Jul 11, 2022 at 19:23
  • $\begingroup$ I'm sorry, I am not that good at chemistry, but the thing I am confused about is that on the left side of the reaction, (R−O−POF−O−POF2+HF2PO2) there are 6 oxygens, but on the right side of the reaction, (R−O−POF−O−POF−O−POF2+HF+O) there are 7 oxygens, why is that? Where is the missing oxygen on the left side of the reaction? $\endgroup$
    – KaffeeByte
    Jul 11, 2022 at 23:29
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Nitrogen trifluoride.

nf3

https://en.wikipedia.org/wiki/Nitrogen_trifluoride

It proved to be far less reactive than the other nitrogen trihalides nitrogen trichloride, nitrogen tribromide and nitrogen triiodide, all of which are explosive. Alone among the nitrogen trihalides it has a negative enthalpy of formation.

Enthalpy of formation is -31kJ/moles. The formation (starting with ammonia) gives off a little energy when fluorinated. Wikipedia says "It is a rare example of a binary fluoride that can be prepared directly from the elements only at very uncommon conditions, such as an electric discharge.". That is good because it means under ordinary conditons NH3 is safe from getting fluorinated. I propose that in your biology, an electric discharge or catalysts which can produce something similar are used to produce your NF3 from NH3 and harvest energy, then regenerate the NH3.

Reaally your energy storage molecule is not the NF3 but good old ammonia. A mellow way to get a little and give a little would be for one of the fluorines to reversibly trade places with a hydrogen. Ammonia has 3 hydrogens, NF3 has 3 fluorides and so you can have intermediates mirroring ATP ADP and AMP. Replacing one hydrogen with one fluoride should give you about 10 kJ/moles. Thus ammonia would be the most energetic molecule (probably ammonium fluoride in an HF solution) and you would replace H with F as you need energy. Instead of making sugar, your autotrophs make ammonia.

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  • $\begingroup$ But NH3 reacts with HF to make (NH4)F I think, then (NH4)F reacts with HF again to make (NH4)(H2F), how could this ammonia be transported to other parts of the cell? $\endgroup$
    – KaffeeByte
    Jan 24 at 15:50
  • $\begingroup$ @KaffeeByte NH4 F is ammonium fluoride - an ionic solution. Not sure about anything invoking H2F because I think H2F is not real. For your NH2F you will be altering covalent bonds - peeling an H off the ammonia and replacing it with one or more F. $\endgroup$
    – Willk
    Jan 24 at 18:31

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