I have heard that the carbon–deuterium bond is stronger than the carbon–hydrogen bond. What are the possible reasons for it? Is this also the reason that C–H bonds participate more in hyperconjugation than C–D bonds? Please explain.

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    $\begingroup$ I sincerely recommend to look at the matter from a slightly different angle. C-D bonds are not stronger; for most purposes they are identical. Only if you measure the energies with really high precision, you might notice some very small, very insignificant difference, so small that most people don't even care which way is it. That's the difference you want to discuss. No, I don't think it has anything to do with hyperconjugation. My bet is on lower zero vibrational level for C-D because of higher equivalent mass. $\endgroup$ – Ivan Neretin Mar 9 '17 at 17:59
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    $\begingroup$ The lesser contribution of C-D to hyperconjugation is the cause of some secondary kinetic isotope effects; it is related to the very slightly different bond strengths. See for example p 12 of this MacMillan group meeting (The classic text, Anslyn/Dougherty, doesn't actually mention it. I haven't checked Lowry/Richardson yet.) Warning: organic chemists' explanations tend to be a bit handwavy. However, the KIE is indeed there for all to see, so it's not fake. $\endgroup$ – orthocresol Mar 9 '17 at 19:46
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    $\begingroup$ While it is true that the C-D strength will have only very minute effects on reactions, it is worth noting that many many systems if placed in D2O will switch any protium that leaves at a reasonable pace with a deuterium. And like the answer given below indicates, this is almost solely due to a lower zero point vibrational energy because electronically the systems are the same. $\endgroup$ – jheindel Mar 10 '17 at 1:11
  • $\begingroup$ @IvanNeretin No need to bet on it. That is exactly correct. :) You don't need to measure the energy with the precision you suggest those. The lower ZPE results in a measurable difference in the rate of bond cleavage. $\endgroup$ – Zhe Apr 5 '17 at 19:14
  • $\begingroup$ The question focuses on the difference in bond strength, and the answers explain this in terms of ZPE. But for proton transfer, quantum mechanical tunneling has to be considered. Especially at low temperatures, this makes it possible to have huge differences between reaction rates with H and with D -- many orders of magnitude, not just small effects. There is a nice open-acess review article on this sort of thing: Scheiner, "Calculation of isotope effects from first principles," doi.org/10.1016/S0005-2728(00)00058-X $\endgroup$ – Ben Crowell Oct 22 '19 at 23:25

Physics is better able to answer "how" questions than "why" questions, but here goes. The quantum mechanical description of the C–D system versus the C–H system gives the former a lower zero-point energy, which is the minimum energy the quantum system can attain. A good conceptual model of this is to consider the C–X system as two masses connected by a spring. When X = D, the system will vibrate more slowly than when X = H simply due to mass.

Since energy is proportional to the frequency of vibration, the energy of the C–D system is less (lower). This table lists the C–D bond dissociation energy as $\pu{341.4 kJ/mol}$ and C–H as $\pu{338 kJ/mol}$. Since this is the energy to break the bond, the C–D bond is stronger.

As far as your question about hyperconjugation; I'm surprised that C–H would have greater hyperconjugation than C–D, but it is not something that I know anything about. I guess I could speculate and wave my hands around and say that since the C–D bond is stronger, it keeps the electrons "closer" and so more confined to the sigma orbital, but that is just hot air.

Here's something which claims that the reason D does less hyperconjugation is because it vibrates less (see slide 15). So, it seems (if this is correct) that it's the smaller bond deformations that reduce the hyperconjugation for deuterated systems. Here is a pretty clear explanation of the kinetic isotope effect.

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    $\begingroup$ I thought I would try to do a quantitative check of this explanation. The first vibrational excitation of CH (methylidine) occurs at an inverse wavelength of 2840.2 cm^-1 ( webbook.nist.gov/cgi/inchi?ID=C3315375&Mask=1000 ). This means that for CH we have a phonon energy $E=\hbar\omega=0.35$ eV. The zero point energy will be almost entirely in the hydrogen or deuterium, not the carbon, due to the smaller mass, so we expect $E_H/E_D\approx\sqrt{2}$. Since the ZPE is half the phonon energy, the difference in bond energies should be $(1/2)(E_H-E_D)=(1/2)(1-1/\sqrt{2})E_H=0.05$ eV.[...] $\endgroup$ – Ben Crowell Oct 16 '19 at 20:11
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    $\begingroup$ [...] This seems to be in excellent agreement with the figure I've seen, which is that the C-D bond is stronger than the C-H bond by about 1.2 kcal/mol=.052 eV. $\endgroup$ – Ben Crowell Oct 16 '19 at 20:13

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