c. How much more stable is the most stable staggered conformer th...

Question

c. How much more stable is the most stable staggered conformer than the most stable eclipsed conformer?

Accepted Answer

Hello, everyone. Let's do this problem. It says determine the energy difference in kilos per mole between the least stable staggered conformer and the least stable eclipse conformer for the compound given below 12 di bromo, Ethane. Note a Gauch conformer has two energy barriers. 22.6 kilojoules per mole and 39.7 kilojoules per mole. The anti conformer is 5.86 kilojoules per mole more stable than a Gauch conformer. So we are given a lot of information here. So let's break this down. First, we're going to draw our compound 12 Dibromm Ethane. Then we'll draw all of the eclipsed and staggered conformers for this compound. Then we can draw a potential energy diagram showing the energy of those different conformers. And then we'll use the information given to us in our potential energy diagram to determine the difference in energy that they are asking us for. All right. So let's get started first, let's draw 12 di bromo, Ethane. So Ethane means we have a two carbon chain and di bromo means we have two bromine atoms on carbons, one and two. So giving carbon one, a bromine carbon 2 a bromine and to satisfy the octets for each carbon, that means we need two hydrogen atoms on each carbon as well. So from left to right, we have C H two B R C H two B R. Now to draw our conformers, we're going to place an eyeball on the left side of this compound down carbon one. So down the carbon one to carbon two bond, and we're going to draw Newman projections and we'll show the staggered and eclipsed conformers by rotating that single carbon carbon bond. So let's start with the staggered conformers, which means that the groups on each carbon are not overlapping when we look down the carbon 1-2 bond in this way. So we can have three staggered conformers. So I'll draw three circles and we can draw the groups on carbon one at basically, if we think of it like a clock, the bromine can be at 12 o'clock, the hydrogen can be at four o'clock and the other hydrogen can be at seven o'clock, all right, or seven rather. And we'll draw the groups the same for all of these conformers on carbon one. And then we'll simply change their position on the back carbon. So the groups are not overlapping in a staggered conformer. So we're going to draw them in the windows. So we'll have one group at around 2:00, one group at around six o'clock and one group at around 10 o'clock So at two o'clock we'll put the bromine at them, six o'clock, we'll put the hydrogen 10 will 10 o'clock, we'll put the other hydrogen. So, in this staggered conformer, we'll call it conformer. A, the bromine or the largest groups are 60° apart. All right. So, now let's keep rotating that bond. And now the bromine atom will be at six o'clock, it'll be 180 degrees from the other bromine that dihedral angle. So for conformer B, that is 180 degrees. And our last conformer, our last staggered conformer conformer C, the bro will be at 10 o'clock and the hydrogens fill the other spaces. So that dihedral angle is 300° going clockwise, right? If we go counterclockwise, that's only 60°. But we count the dihedral angle going clockwise. All right. So those are our staggered conformers. Let's draw our eclipsed conformers. And in eclipse conformers, the groups are overlapping now. So before they were in windows and now they are going to overlap one another. So again, we can have three conformers. I'll draw the front carbon groups the same for each conformer Bromine at the top at 12 o'clock, hydrogen at four o'clock, hydrogen at seven o'clock ish eight o'clock. I guess it would be 8:00 if the other one is at 4:00. Now, let's draw the carbon two groups. So first, we can have the two largest groups right in front of each other So I'll draw the bromine also at 12 o'clock and then the hydrogens on carbon two directly behind the hydrogens on carbon one. And we'll call this conformer D and that dihedral angle between the two bromine atoms is going to be zero, right. They are in the same exact spot. Now, rotating the back carbon groups that bromine will be behind the 4:00 hydrogen on carbon one. And this is conformer E and that has a dihedral angle of 20 degrees. And lastly rotating one more time that bromine will now be behind that 8:00ish hydrogen on carbon one. So this is conformer F and that dihedral angle is 240°. All right. So now let's draw a potential energy diagram to reflect the energy of all of these conformers. So four hour diagram, we have the Y axis and the X axis, the y axis is labeled potential energy. So that is increasing as we go up the Y axis and on the X axis, we have the dihedral angle. So we start at zero and at 360 degrees and we go in increments of 60 degrees. So zero 60 degrees, 120°, 180°, 240° And 300°. So now we're going to plot the energy for the conformers that we just drew. So starting with a zero degree dihedral angle that will be conformer D right are eclipsed conformer. And do you think this will have really high energy or low energy, really high energy? Because those two large groups are directly overlapping each other. So that's actually going to be the highest energy conformer we have. And that'll be the same structure as 360 degrees. Right? Once we've rotated the bond completely, it will go back to conformer D. So we'll plot a point at the same point of the Y axis for zero degrees and 360 degrees. And first, we'll plot the points, we'll label this D, these two points D and then we'll connect all the points to show our curve. So at 60° our conformer is gauch, right? This is called a gauch conformer and it actually has the same energy as conformer C, right? Conformer A and C are both gauch and those will have lower energy than conformer D, right? Because those groups are, have a little bit of space between them. So we'll plot their points about a quarter of the way up the Y axis at 60° and 300°. And so that's conformer A at 60°, conformer c at 300°. Now, I'm actually going to jump to 180° because we plotted our highest point and conformer B, 180° is going to be our lowest energy. And that is the anti conformer. And that's because the bromine atoms are as far away from each other as they can get. Our two largest groups have a lot of space between each other. So that's going to be very stable and have just a little bit of energy or at least less than the other conformers. I'm going to move points A and C up a little bit. All right. So that leaves E and F which are eclipsed. So they'll be a little bit higher energy. But the bromine atoms are still further from each other than they are in conformer D, that's zero degrees. So they'll be higher energy but lower than conformer D higher than conformers A and C. So for 1:20 and 2:40, they'll be the same energy. So now that we've plotted all of our points, let's connect our curve D to A to E to B to F to C to D. OK. So I wanted to draw this curve so that we can use the information about the energy barriers and the energy amounts that they gave us. It's easier to me when it's represented by a graph and I can visualize the difference that they are looking for. So let's write the amounts that they gave us onto our graph. So it says the note says a Gauch conformer has two energy barriers, 22.6 kilojoules per mole and 39.7 kilojoules per mole. So let's look at our graph, our potential energy diagram. We have the gauch conformer at point A, I'm going to label that Gauch and C is our other gauch conformer and B is our anti conformer. So we'll label that as well. And if we notice to get to our gauch points, we need to overcome these big humps of energy, right? These big amounts of energy. So we'll draw a dotted line across the Y axis at point A, our gauch conformer, which is the same as point C, right. And we can draw a line between this dotted line in that maximum conformer point D and that maximum conformer point E. So these are the two energy barriers. So which amount goes for which one? Well, the hump between A and D is larger. So that will be the 0.7 kilo per mole barrier. And from point A to E that will be the 0.6 kilojoule per mulb barrier. And it also mentioned that the anti conformer is 5.86 kilojoules per mole more stable than a Gaus conformer. So we can continue this dotted line from A, we'll draw it all the way to C and we notice that below this line dips point B, right. So the line or the amount of energy between point B and this dotted line, the Gauss energy level is 5. kilojoules per mole. OK. So now getting to the question that the problem asks, determine the energy difference. So let's look at our graph, the energy difference between the least stable staggered conformer and the least stable eclipsed conformer. So which is the least stable staggered conformer? Let's actually look at our conformers that we drew. The least stable staggered conformer is going to be the gauch conformers and they are equal energy. So they are both the least stable. So we'll mark our graph A and C and it also asked about the least stable eclipsed conformer. So which eclipsed conformer is the least stable? That is the zero degree dihedral angle. So conformer D, so now that we've drawn this all on the diagram, we can see that the question is basically asking for the energy difference between D and A and they already gave that to us, right? It's 39.7 kilojoules per mole. So the energy difference in kilojoules per mole between the least stable staggered conformer and the least stable eclipse conformer is 39.7 kilojoules per mole, which makes C the correct answer for this problem. That's it for this one. I hope this video was helpful and thank you for watching.

c. How much more stable is the most stable staggered conformer than the most stable eclipsed conformer?

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