4. Alkanes and Cycloalkanes
Barrier To Rotation
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
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Sometimes we're gonna be asked to actually calculate the energy barrier in kilograms per mole to rotating this molecule. Alright? And what we do is we're gonna actually gonna memorize four really common values and 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. Okay, so 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, alright? And in terms of which one's most energetic and least energetic. You should already know that based on the fact that an anti is gonna be the least energetic and Eclipse is going to be the most energetic. All right, But we actually need to know the values for this. So let's just talk about it, All right? First of all, we have to eclipse values notice that they say, Well, actually have three eclipse values, but we have Let's just start off with the highest one. The highest eclipse value that you guys need to memorize would be to methyl groups. Perfectly overlapping eclipse. Okay, if you have two methyl groups that are overlapping. That's 11 Kill a Jules 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 costs. That basically means if I were to rotate these bonds eclipse, how much energy would it cost me? 11 kg. Promote is kind of a lot. So you can imagine that that's not gonna happen very often. Or sometimes it's gonna want to prevent that as much as possible. Okay, so let's keep going down the list. So in interaction that it's still not very favorable, but is a little bit easier toe happen is eclipsed with a metal. And in H okay, this is gonna be a little bit more a little bit less energetic, because now we don't have two big groups. We just have one big group and in h. Okay, so there's obviously some torch it'll strain going on here, but it's not quite as bad as you know, as having two methyl groups. Okay, so then we've got HHS, And by the way, that one is six. You just have to remember that 16 Then you got HHS the easiest one toe happen for Eclipse, but it still does cost some energy. So for an hh eclipse, that's gonna be four. Kill a jewel for more again. Just something that you should commit to memory. Then finally we have Gosh, okay, we're not gonna have to memorize all the different gosh combinations because that could take a long time. That could be very exhausting, but one that you should know is 3.8 kg per mole. Now, notice that this number is actually very close to four. So how can a gosh interaction be almost the same as in eclipse interaction? And the reason is because the gosh that we're talking about is actually metal metal. Gosh, so this means that you have to large groups that are staggered, but they're so big that they're interacting, and almost the same way as to h is worth would if they were overlapping. Okay, these air, your four values that you want to come into memory because guess what? For these types of questions. They're gonna give you these values, and then you're gonna 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.
Determining energy cost of eclipsed interaction
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So let's go ahead and look at an example, all right? And we're just gonna do this as a worked example together. Okay, here it goes. It says the barrier to rotation for the following molecule is 22 killer jewels promote. Okay, so that means that that is the total amount of energy cost that it's gonna take me to make these molecules or make these bonds going Clips to each other. Okay, so what we wanna do notice that right now it's staggered 22 kg from was what it's gonna cost to make it go eclipse. So what it's asking is determined the energy costs associated with the eclipsing interaction between a bro mean and a hydrogen atom. Okay, so this is the way we think about about it. First of all, did I make you memorize the value between H and bro? Mean better eclipse? No. Okay. I could literally I could I could fill up a table off all the different atoms that could overlap, and I could make you memorize all those energies, but that's not gonna 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 we see how much is each interaction gonna cost me? So overall, this rotation is gonna cost me 22 kg per mole. But we know that Hey, if this h is gonna overlap with this br, that means that I'm also gonna have to have an agent and h overlapping. And I'm also gonna have to have a metal in a method overlapping. So then I go ahead and get those formulas from my memory because I memorized them and I write them down. So hh is 4 kg for more metal. Metal is 11 kg from all. Okay, now we only have one mawr interaction that could be making my 20 to the 22 is basically made out of three interactions. Some together. Okay, So what that means is, I basically have four plus 11 plus some unknown number X, and that's going to give me 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 could just use a really easy math. So this is 15. So I'm going to subtract 15. So it's X equals 20 to minus 15. So that means the X equal seven. Okay, So that means that the energy costs for this interaction here is that h br eclipsed is equal to seven. Kill it, Jules, for more. Does that make sense? So all we did was he used the values that were supposed to memorize and were able to figure that 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 gonna have to draw out the compound yourself. So you have to draw the new and projection yourself, and then you're gonna have to work with it. You're going to play with it and figure out use the values on that. I told you to memorize, to figure out the unknown bond. Alright, so go ahead, try to do it
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.
Additional resources for Barrier To Rotation
PRACTICE PROBLEMS AND ACTIVITIES (1)