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Idle wonderings

  1. Mar 19, 2005 #1
    Is the chemistry of [tex]^2H[/tex] very different from that of [tex]^1H[/tex]? The nucleus is heavier, but electronically its the same so it should behave the same chemically... maybe the heavier nucleus of deuterium is less mobile and thus less acidic in C-D bonds? In particular I'm wondering about hydrogen bonding and whether the strength is significantly different with deuterium. Thanks!
  2. jcsd
  3. Mar 19, 2005 #2


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    I found this with a google search

    http://www.absoluteastronomy.com/encyclopedia/d/de/deuterium.htm [Broken]

    As for different acidity, I would say it's probably the same, at least for organics. Organic acidity is based almost entirely on how stable the resulting anion would be if hydrogen leaves. If hydrogen leaves methane, it creates an incredibly strong nucleophile and Lewis base, so that obviously won't happen. If hydrogen leaves an ammonium group, it creates a stable amine, so that is still just as likely to happen.
    I don't know much about inorganic acids so I can't really say anything about those.
    Last edited by a moderator: May 1, 2017
  4. Mar 20, 2005 #3


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    Strictly speaking, the proton form of an organic acid will be slightly more acidic than the corresponding deuterium species because the bond you are breaking is a little stronger. You can take advantage of this in NMR experiments if you have an alcohol in your molecule. If you add some D2O to the sample, the proton on the alcohol will exchange with the D2O and the NMR signal from the alcohol proton will disappear.

    You can also take advantage of the different bond energies to study reaction mechanisms. If you suspect that the rate limiting step in the mechanism involves cleaving a particular C-H (or other R-H) bond, you can synthesize the molecule where the H in question is replaced with D. If the C-H bond is indeed cleaved in the rate limiting step then the D version of the substrate molecule will react slower than the H version. The theoretical maximum difference in rate is a factor of 7 (that is, the H form reacts 7 times faster than the D form), but usually the observed number is more in the range of 3-4. This is called a deuterium isotope effect.

    I'm not familiar with any particular experiments that deal with hydrogen bonding, but I would suspec that hydrogen bonding would be slightly worse when D is concerned because it involves breaking (or at least weakening) a stronger covalent bond. However, I'm not sure how measurable this effect would be.
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