Sometimes we're going to be asked to actually calculate the energy barrier in kilojoules per mole to rotating this molecule. Alright? And what we do is we are actually going to memorize 4 really common values. And by memorizing those 4 really common values of energy, we're going to be able to solve for what's called the barrier to rotation. Okay? So, what are these values that I'm talking about? Well, there's basically 4 really common interactions that you should commit to memory in terms of their energy level. Alright? And in terms of which one is most energetic and least energetic, you should already know that based on the fact that anti is going to be the least energetic and eclipsed is going to be the most energetic. Alright? But we actually need to know the values for this, so let's just talk about it. Alright? First of all, we have 2 eclipsed values. Notice that they say, well, actually, we have 3 eclipsed values, but we have, let's just start off with the highest one. The highest eclipsed value that you guys need to memorize would be 2 methyl groups perfectly overlapping in clips. Okay? If you have 2 methyl groups that are overlapping, that's 11 kilojoules per mole, which is a lot of energy to spend. Okay? 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? Eleven 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. Okay? So let's keep going down the list. So 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 less energetic because now we don't have 2 big groups. We just have one big group and an H. Okay? So there's obviously some torsional strain going on here, but it's not quite as bad as having 2 methyl groups. So then we've got H H. HH. And by the way, that one is 6. You just have to remember that one. 6. Then we've got HH. HH is the easiest one to happen for eclipse, but it still does cost some energy. So for an HH eclipse, 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 and 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 eclipsed interaction? And the reason is because the Gauche that we're talking about is actually methyl methyl Gauche. So this means that you have 2 large groups that are staggered, but they're so big that they're interacting in almost the same way as 2 Hs would if they were overlapping. Okay? These are your 4 values that you want to come into memory because guess what? For these types of questions, they're going to give you these values and you're going to have to solve for the unknown bond interaction by figuring out these values.
Barrier To Rotation - Online Tutor, Practice Problems & Exam Prep
Sometimes we’ll be asked to actually calculate the amount of energy a Newman Projection “spends” while rotating.
Important Barrier of Rotation Values
4 Values You Should Memorize
Video transcript
These are the values we’ll need so we can solve for the unknown interactions in these questions.
Determining energy cost of eclipsed interaction
Video transcript
Let's go ahead and look at an example. We're just going to do this as a worked example together. Here it goes. It says the barrier to rotation for the following molecule is 22 kilojoules per mole. Okay? So, that means that is the total amount of energy cost that it's going to take me to make these molecules or make these bonds go eclipse to each other. Okay? So what we want to do notice that right now it's staggered, 22 kilojoules per mole is what it's going to cost to make it go eclipse.
So what it's asking is to determine the energy cost associated with the eclipsing interaction between a bromine and a hydrogen atom. Okay? So, this is the way we think about it. First of all, did I make you memorize the value between H and bromine that are eclipsed? No. I could literally fill up a table of all the different atoms that could overlap, and I could make you memorize all those energies. But that's not going to be a really good use of your time. Okay? Especially because these questions will always give you some easy values that you can then back calculate the unknown value from. So all we do is we just look at each interaction, and we see how much each interaction is going to cost me. So overall, this rotation is going to cost me 22 kilojoules per mole. But we know that hey, if this H is going to overlap with this Br, that means that I'm also going to have an H and an H overlapping, and I'm also going to have a methyl and a methyl overlapping. So then I go ahead, and I get those formulas from my memory because I memorized them and I write them down. So H2 is 4 kilojoules per mole. Methyl-methyl is 11 kilojoules per mole. Okay?
Now we only have one more interaction that could be making my 22. The 22 is basically made out of 3 interactions summed together. Okay? So what that means is I basically have 4 + 11 + some unknown number x, and that's going to give me 22.
4 + 11 + x = 22
Does that make sense? So all I have to do is I just have to solve for x, and this is super easy. You can just use really easy math. So this is 15, so I'm going to subtract 15. So it's x equals 22 - fifteen. So that means that x equals 7.
x = 7
So that means that the energy cost for this interaction here is that HBr eclipsed is equal to 7 kilojoules per mole. Does that make sense? So all we did was we used the values that we're supposed to memorize, and we were able to figure out the unknown bond through that value. Cool. Right?
So now what I want you guys to do is do the next practice completely on your own. It's the same concept except it's a little bit harder because now you're going to have to draw out the compound yourself. So you're going to have to draw the Newman projection yourself and then you're going to have to work with it. You're going to have to play with it and figure out use the values that I told you to memorize to figure out the unknown bond. Alright? So go ahead and try to do
Now it’s time to put your knowledge to the test. Remember to draw the eclipsed version to know what the interactions are!
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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