Carbon-Deuterium bonds

[kkentm]

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Why are deuterium-carbon bonds stronger than hydrogen-carbon bonds?

 

[MD Odyssey]

Why are deuterium-carbon bonds stronger than hydrogen-carbon bonds?

Fair warning - I wanted to explain this without having to invoke quantum mechanics, but I ultimately couldn't. I'll post an additional response afterwards using an analogy which I have that might work better.

The amount of energy required to break a chemical bond is a function of the zero-point energy of the system, in this case, a C-H or C-D bond. In their ground state, any quantum mechanical system retains some finite amount of energy - its translational kinetic energy may be zero, but it will still have a certain amount of vibrational kinetic energy. This is true, even at absolute zero.

In quantum mechanical systems, energy is quantized and one can use the Schrodinger equation to calculate what those are for various potentials. In this case, if you model the C-X bond as a quantum oscillator, you can calculate the allowable energy levels and you get that

This means, that there are n energy levels and they all come in half integer multiples of Planck's constant and the frequency. Some more math leads you to this relationship for the frequency

where k is the spring constant (we modeled this as a simple harmonic oscillator, remember) and μ is the reduced mass of the system. I calculated the reduced mass of both systems and for C-H it was 0.923 u and for C-D it was 1.71 u. So, you can see from this that, an increase in the reduced mass of the oscillator leads to a decrease in the frequency of oscillation. And, since the energy level of the oscillator is proportional to the frequency, we can see that ground state energy level for C-D is lower than it is for C-H.

Since the ground state of the oscillator is lower, more energy has to be put in to separate it, which results in a larger bond energy for C-D than for C-H.
​

 

[BilalL]

Interesting explanation MDOdyssey

 

[kkentm]

Fair warning - I wanted to explain this without having to invoke quantum mechanics, but I ultimately couldn't. I'll post an additional response afterwards using an analogy which I have that might work better.

The amount of energy required to break a chemical bond is a function of the zero-point energy of the system, in this case, a C-H or C-D bond. In their ground state, any quantum mechanical system retains some finite amount of energy - its translational kinetic energy may be zero, but it will still have a certain amount of vibrational kinetic energy. This is true, even at absolute zero.

In quantum mechanical systems, energy is quantized and one can use the Schrodinger equation to calculate what those are for various potentials. In this case, if you model the C-X bond as a quantum oscillator, you can calculate the allowable energy levels and you get that

This means, that there are n energy levels and they all come in half integer multiples of Planck's constant and the frequency. Some more math leads you to this relationship for the frequency

where k is the spring constant (we modeled this as a simple harmonic oscillator, remember) and μ is the reduced mass of the system. I calculated the reduced mass of both systems and for C-H it was 0.923 u and for C-D it was 1.71 u. So, you can see from this that, an increase in the reduced mass of the oscillator leads to a decrease in the frequency of oscillation. And, since the energy level of the oscillator is proportional to the frequency, we can see that ground state energy level for C-D is lower than it is for C-H.

Since the ground state of the oscillator is lower, more energy has to be put in to separate it, which results in a larger bond energy for C-D than for C-H.
​

whoa, its almost as if i shouldnt have asked lol. thanks for the very in depth explanation though. so basically the higher the mass, the lower the frequency of oscillation and thus the lower the energy state?

 

[MD Odyssey]

whoa, its almost as if i shouldnt have asked lol. thanks for the very in depth explanation though. so basically the higher the mass, the lower the frequency of oscillation and thus the lower the energy state?

Precisely. If you look at a block oscillating on a spring, you get the same sort of behavior, with the frequency depending upon the mass of the block in the same way as the quantum oscillator depends upon the reduced mass of the system.

The big difference is that a classical simple harmonic oscillator can have a zero frequency and thus no energy at all. But, for quantum systems, we get the surprising result that zero energy isn't possible - even at absolute zero - therefore, the oscillator has zero-point energy.

You're to be commended for asking an insightful question - we're so used to thinking of simple models for covalent bonds, but at the end of the day, anything other than the simplest systems, like molecular hydrogen, gets more complicated.

Even the simplest systems, like the hydrogen atom, are just approximations - a proper quantum mechanical description of an electron in a central Coulomb potential isn't sufficient. If you want to get really serious, you have to include relativistic effects. But that's not precisely accurate either - since the particles involved have spin, one has to also account for spin-orbit coupling as well. The simplest quantum mechanical systems are exceedingly complicated and there doesn't seem to be an end to the onion either.