Mrthfx

When we say a conformer or compound is of "higher energy" than another, are we quantifying energy in terms of the bond strength? Or are we going off of how much energy it takes to break the bond? I don't think we can just assign energies to each compound... so what is this based off of?

One might answer that its based off of stability. Well, what is "Stability?" Is it how rigid/strong the bonds are? Is it how reactive the compound is?

Often times i find myself reading explanations with circular language on this topic. A higher energy conformer means lower stability or A stable compound is one that has low energy. I'm finding these definitions to be crazy ambiguous, and don't thoroughly define what these are.

If someone can clearly explain this in terms that a first year undergraduate student can understand, you have earned my gratitude.

As a chemist, I would agree that textbooks are not clear about the term stability and energy. It is not your fault and undergraduate organic chemistry books make the situation worse. I cannot recall the text, it was an old book, however it clearly said that stability of a compound does not mean anything. We should always ask, stability with respect to what? Are we talking about thermodynamic stability or kinetic stability? For example H2+O2 mixture is not stable. This statement is meaningless because we are not saying stability with respect to what? This mixture should form water right away but people actually waited for 30 years and did not find much water in a mixture of H2 and O2. Kinetically this mixture is not stable and it will react violently once you provide the right activation energy. When you are reading about stability, ask this question yourself.

In the same vein, the term energy should also be clarified by authors. As a crude analogy is that you can think of energy as the potential energy of the whole molecule when the authors talk about conformers. Have a overall picture of the molecule. Imagine two springs, A) one which is less stretched and the other one (B) is more stretched. You can say spring B has more (potential) energy than (A). An eclipsed conformer has higher potential energy (overall, no need to think locally) than a staggered one.

The answer is complicated because there are different kinds of stability and when we talk about energy, we also can mean different kinds of energy.

When we speak of energy, we generally refer to a free energy. Your normal laboratory setup is an open-air, constant-pressure system, so we use Gibbs free energy. If you were working with bomb calorimeters, you'd want to use Helmholtz free energy for constant-volume processes.

Note importantly, that energies are relative. "What is the energy of this compound?" is a meaningless question. And when we speak of "high energy," we mean high relative to some reference point. That's precisely why most examinations of energy in chemistry focus on a change, most commonly $Delta G$, the change in Gibbs free energy. Gibbs free energy is a tool for measuring changes in internal energy (most commonly bond-breaking and bond-making) while compensating for the entropic costs of those changes and pressure-volume work done by the system.

At standard state, the Gibbs free energy change for a reaction or transformation can describe the relative preference of the "reactants" versus the "products" via an equilibrium constant $K$.

$$Delta G^{circ} = -RT ln K$$

This is a way to measure one kind of stability, thermodynamic stability.

Considering $ce{A <=> B}$, $ce{A}$ is thermodynamically stable relative to $ce{B}$ if the free energy change of this process is positive (i.e., $Delta G > 0$), and vice versa.

Unfortunately, thermodynamics only tells half the story for a reaction. There's also the kinetic piece, which determines how fast a reaction proceeds, that is, the rate of the reaction. Rates are determined by the relative energy of a fleeting transition state, captured by $Delta G^{ddagger}$. For a reaction that is highly spontaneous, if the reactant is kinetically stable (that is $Delta G^{ddagger}$ has a large value), then reaction will still proceed very slowly. This is kinetic stability.

Finding minimums: a general view

When minimizing energy you are searching (using numerical methods) for the minimum of Energy). You are at the minimum when the derivative(Jacobian) is equal to zero. However, there are a couple unstable places that also have a derivative of zero, namely saddle points and maximums (top of a circle). We are after the minimum (bottom of a circle).

Think of a pendulum. It has two spots where the derivative is zero... when it is all the way up, and all the way down, but only when it is in the all the way down position is it stable. When the Hessian (second derivative) is positive definite, you are at the "bottom of the pendulum swing" and at a minimum. Pendulum up the derivative is zero, but this is a maximum.

There is also the problem that often in energy landscapes it looks like a rollercoaster...

We need to find not just the a minimum that has a derivative of zero,but the lowest of all minimums present. This is a real tough problem since numerical solvers stop once the derivative is zero. They don't know there are other minimums. There are methods to try to get around this, but it is a very active area of research and a huge prize awaits whoever can guarantee a global minimum is found.

A thought on energy

What is this energy we are minimizing? We are minimizing the energy compared to a reference point. Everything is always with respect to a reference point.

I am not a chemist, and I don't know much Quantum mechanics but I will try to give a chemistry answer...This is my impression of what is done in the scenario you are asking about. I perform minimizations on Free Energy, but that is related but not quite what you are after. In Quantum mechanics the reference point is when all nuclei and valence electrons are infinitely apart. Another common reference point in chemistry, physics and engineering is the ideal gas, which is when all molecules are infinitely apart from each other but atoms (and electrons) are still bonded to each other.

Given a reference point which we define to have "zero" energy, for each conformer we would calculate the energy to form all of the covalent bonds in the molecule, which is to say, calculate the energy to pull the valence electrons and nuclei in from infinitely far apart and put them in the exact geometry you want. Do this for many geometries. The one that took the lowest energy to form will be the stablest of the ones you tried (There may be others you did not try though! you may be in the wrong loop of the rollercoaster still). It takes different amounts of work to put two nuclei different distances apart etc. In practice, QM programs do all of this for us, and they are getting quite fast but it can still take a very long time. It is a hard thing to calculate.