Comment on Can I lick it?

• nickwitha_k@lemmy.sdf.org ⁨2⁩ ⁨weeks⁩ ago

  It would be more likely a secondary or tertiary effect. That is, H• radicals ripped away from their parent molecules would leave •OH, •R, and •RNH radicals. These are unstable and highly reactive, “desiring” to have that stable electron configuration. Likely, this will result in elections being shifted to bring in more stable species, like OH-. Overall, we’re looking at effectively a deprotonation of the saliva, with extra intermediary steps to stabilize the radicals.

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  • threelonmusketeers@sh.itjust.works ⁨2⁩ ⁨weeks⁩ ago

    Interesting. Given that H• is a neutral species, what would cause the preference for the creation of stable negative species (freeing up H+) over the creation of stable positive species (freeing up OH-)?

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    • nickwitha_k@lemmy.sdf.org ⁨2⁩ ⁨weeks⁩ ago

      Neutral as far as pH is concerned, yes. However, radicals tend to be very reactive due to their valance not being full. I am a bit rusty, TBH, as I’m about a decade and a half out of uni but, the best way to predict the products of the reaction is to look at the high-level of the equation:

      H• (excess) + H••OH + H••R + H••N-R -> H2(g) + •OH + •R + •N-R

      All of the products of the initial reaction here are radicals except for the H2 molecules. They all are going to further react to form more stable species with full valances, with possible exception being the molecular hydrogen. Because the elemental hydrogen is introduced as a radical rather than protons (H+ ions) in the solution, the final products are likely to be more negatively charged, neutral, and/or have some interesting hydrogen additions, especially in the hydrocarbons and amino acids.

      For example, there could be reactions like: R• + •OH + •N-R -> R-OH + HO-N-R

      Overall, however, the amount of free hydrogen/protons is likely to be reduced as they are effectively removed from solution as hydrogen gas.

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