Calculating Energy of Newman Projection Rotation

Sometimes we’ll be asked to actually calculate the amount of energy a Newman Projection “spends” while rotating. 

Important Barrier of Rotation Values

Concept: 4 Values You Should Memorize

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Video Transcript

Sometimes we're going to be asked to actually calculate the energy barrier in kilojoules per mole to rotating this molecule. What we do is we are actually going to memorize four very common values. By memorizing those four really common values of energy, we're going to be able to solve for what's called the barrier to rotation.
What are these values that I'm talking about? Well, there's basically four really common interactions that you should commit to memory in terms of their energy level. And in terms of which one's most energetic and least energetic, you should already know that based on the fact that anti is going to be the least energetic and eclipse is going to be the most energetic. But we actually need to know the values for this, so let's just talk about it.
First of all, we have two eclipsed values. Notice that they say – well actually we have three eclipsed values, but let's just start off with the highest one. The highest eclipsed value that you guys need to memorize would be two methyl groups perfectly overlapping, eclipsed. If you have two methyl groups overlapping, that's 11 kilojoules per mole, which is a lot of energy to spend.
That energy that we're talking about is actually referred to as the energy cost. That basically means if I were to rotate these bonds eclipsed, how much energy would it cost me. 11 kilojoules per mole is kind of a lot, so you can imagine that that's not going to happen very often or sometimes it's going to want to prevent that as much as possible.
Let's keep going down the list. An interaction that is still not very favorable, but is a little bit easier to happen is eclipsed with a methyl and an H. This is going to be a little bit less energetic because now you don't have two big groups. You just have one big group and an H. There's obviously some torsional strain going on here, but it's not quite as bad as having two methyl groups.
Then we've got H/H. H/H – and by the way, that one is 6. You just have to remember that one's 6.
Then you've got H/H. H/H is the easiest one to happen for eclipsed, but it still does cost some energy. So for an H/H eclipsed that's going to be 4 kilojoules per mole. Again, just something that you should commit to memory.
Then finally we have gauche. We're not going to have to memorize all the different gauche combinations because that could take a long time. That could be very exhausting. But one that you should know is 3.8 kilojoules per mole.
Now notice that this number is actually very close to 4, so how can a gauche interaction be almost the same as an eclipse interaction. The reason is because the gauche that we're talking about is actually methyl/methyl gauche, so this means that you have a two large groups that are staggered but they're so big that they're interacting in almost the same way as two H's would if they were overlapping.
These are your four values that you want to commit to memory because – guess what? For these types of questions, they're going to give you these values and then you're going to have to solve for the unknown bond interaction by figuring out these values. 

These are the values we’ll need so we can solve for the unknown interactions in these questions. 

Example: The barrier to rotation for the following molecule is 22 kJ/mol . Determine the energy cost associated with the eclipsing interaction between a bromine and hydrogen atom.

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Now it’s time to put your knowledge to the test. Remember to draw the eclipsed version to know what the interactions are!

Problem: The barrier to rotation for 1,2 -dibromopropane along the C1—C2 bond is 28 kJ/mol. Determine the energy cost associated with the eclipsing dibromine interaction.

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