I searched for this on Google but I only got research articles in the SERP. I know of the concept of hydrogen bonds, that they are a special class of van der Waals forces between highly electronegative atoms ($\ce{F/O/N/Cl})$ and the hydrogen atom. I am asking if deuterium also exhibits the same hydrogen bonding, and how does its "$\ce{D}$ bond" compare to the hydrogen's original "$\ce{H}$ bond"?
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asked Jan 13, 2018 at 9:43
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10$\begingroup$ Boiling point of D2O: 101.4°C. Does that answer your question? Also: en.wikipedia.org/wiki/Heavy_water $\endgroup$– KarlCommented Jan 13, 2018 at 10:02
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2$\begingroup$ Why does water have a Bp of 100°C? In comparision CH4: -161, NH3: -33, SH2: -60°C. $\endgroup$– KarlCommented Jan 13, 2018 at 10:05
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5$\begingroup$ Btw. hydrogen bonding is NOT a vdW interaction but an exeptionally strong form of polar interaction. And nitrogen and chlorine don't do hydrogen bonding, only fluorine and oxygen. Bp HF: +20°C, HCl: -85 $\endgroup$– KarlCommented Jan 13, 2018 at 10:14
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3$\begingroup$ What do you mean by "SERP"? $\endgroup$– paracetamolCommented Jan 13, 2018 at 10:27
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9$\begingroup$ To say hydrogen bonds only form with oxygen and fluorine is a gross oversimplification. Many common elements can be both hydrogen bond donors and acceptors: carbon, selenium, sulfur, chlorine, iodine, phosphorus, cobalt, iron, manganese, even hydrogen itself in dihydrogen bonds. Also a misconception: the hydrogen bonds with oxygen and fluorine must always be stronger than others which is not true: see, e.g., doi: 10.1039/C7CP05265K. $\endgroup$– Linear ChristmasCommented Jan 13, 2018 at 12:21
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3 Answers
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(Based on ron's answer here on the inductive effect accorded by deuterium - do give that a read, and consider upvoting it)
Since deuterium has a higher mass than protium, simple Bohr theory tells us that the deuterium 1s electron will have a smaller orbital radius than the 1s electron orbiting the protium nucleus.
The smaller orbital radius for the deuterium electron translates into a shorter (and stronger) $\ce{C-D}$ bond length.
See the bottom half of ron's answer for the proof
In your case, that would mean a shorter $\ce{N/O/F-D}$ bond.
A shorter bond has less volume over which to spread the electron density (of the 1 electron contributed by $\ce{H}$ or $\ce{D}$) resulting in a higher electron density throughout the bond, and, consequently, more electron density at the carbon end of the bond. Therefore, the shorter $\ce{C-D}$ bond will have more electron density around the carbon end of the bond, than the longer $\ce{C-H}$ bond.
The net effect is that the shorter bond with deuterium increases the electron density at carbon, e.g. deuterium is inductively more electron donating than protium towards carbon.
So we can expect the $\ce{N/O/F-D}$ bond to be smaller and more polar than the corresponding bond with protium.
Karl very astutely pointed out that the boiling point of heavy water is higher than that of normal water. In light of the inferences drawn earlier, I took the liberty of interpreting this fact as being (at least somewhat) indicative of deuterium permitting a stronger hydrogen bond than protium.
As Linear Christmas pointed out, the variations in mass across different isotopes is known to affect boiling point (as in the case of helium). This is likely to contribute to some degree, to the increase in boiling point in heavy water. However, I'm still of the opinion that the higher dipole moment of $\ce{D2O}$ over $\ce{H2O}$ is the main cause for the higher boiling point.
Finis
Hydrogen bonding with deutrium does occur and should be, in theory (drawn from ron's answer), stronger than that with protium. The fact that heavy water $\ce{D-O-D}$ is more polar (and hence has a higher boiling point) than regular water $\ce{H-O-H}$, appears to be a consequence of this.
answered Jan 13, 2018 at 10:19
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1$\begingroup$ I actually only wanted to point out that heavy water has practically the same high Bp. ;-) $\endgroup$– KarlCommented Jan 13, 2018 at 10:33
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3$\begingroup$ @GaurangTandon I used ron's answer to explain why the $\ce{N/O/F-D}$ bond is more polar than its protium counterpart. H-bonding is largely an electrostatic dipole-dipole interaction. So if the covalent bonds (that effect H-bonding) are more polar, then the H-bonding is stronger :-) $\endgroup$ Commented Jan 13, 2018 at 10:36
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1$\begingroup$ Do you know what's the reasoning for counting the much weaker N-H--N bond among the much stronger O-H--O or F-H--F hydrogen bonds? $\endgroup$– KarlCommented Jan 13, 2018 at 10:39
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1$\begingroup$ Not necessarily. There's no autoprotolysis in HCl, i think. Or? Does $\ce{H2Cl+}$ exist? $\endgroup$– KarlCommented Jan 13, 2018 at 11:26
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1$\begingroup$ What about masses ; ? I was given this question/exercise many years ago. I didn't solve it nor I did it ast a follow up. Qualitatively it should raise the boiling point even further (extent might be very small). $\endgroup$ Commented Jan 13, 2018 at 14:39
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In short:
Chemistry is about the last shell of electrons in the atoms. Nuclear differences are in the field of Physics. Deuterium is only a nuclear isotope of hydrogen, so no fundamental differences are to be expected in the chemical sphere.
Hydrogen bonding is a chemical property.
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$\begingroup$ Yes, that's also another perspective to look at it! $\endgroup$ Commented Jan 13, 2018 at 16:39
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$\begingroup$ Thank you. I do not want to discredit @paracetamol 's answer. Just to remember the fundamentals. Something my father, a Chemical Engineer, used to tell me. $\endgroup$ Commented Jan 13, 2018 at 16:48
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$\begingroup$ I think is not the best perspective tough. Not for fine details. Else we have to eliminate boiling point arguments as well and stop at rough level, such as D2O does react wit sodium and so on. (Of course it does). $\endgroup$ Commented Jan 13, 2018 at 16:54
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3$\begingroup$ Aha, this is something my teachers repeatedly sought to hammer into my brain :-) That nuclear differences don't affect the chemistry of substances is a general statement; it is usually (but not always) true. For instance, ron's answer (which is linked in my post), explains how hydrogen isotopes (which differ in the composition of their nucleus) produce different inductive effects, and the inductive effect is indeed a chemical phenomenon. O:) $\endgroup$ Commented Jan 13, 2018 at 17:31
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This is not at all an anwer, but the discussion above pointed to the 2011 IUPAC definition of hydrogen bonds, which I did not know about:
The hydrogen bond is an attractive interaction between a hydrogen atom from a molecule or a molecular fragment X–H in which X is more electronegative than H, and an atom or a group of atoms in the same or a different molecule, in which there is evidence of bond formation.
That definition paper from Pure and Applied Chemistry is accompanied by a technical report that gives history, examples, etc.
So, while hydrogen bonds are most well known for fluorine (FH-F), oxygen (OH-O) and nitrogen (NH-N) (and permutations), there is a huge zoo of other examples (e.g. reaction intermediates for proton transfers, etc) where this definition applies, and they include most of the periodic table except the electropositive metals.
(Any hydrogen bond of course also works with deuterium, just the stability and reaction dynamics are in most cases slightly different.)
