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

Question

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

Accepted Answer

Hello everyone. Let's do this problem. It says determine the energy difference in kilojoules per mole between the most stable eclipse conformer and the most stable staggered conformer for 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 the Gauss conformer. So this is a lot to take in. Let's break it down. First, we can draw the structure of 12 di bromo, Ethane. Then we'll use that to draw our conformers or our Newman projections, the three staggered conformers and three eclipsed conformers. Then in order to show the energy difference, I like to draw the potential energy diagram and plot all the points of our conformers. And then we can apply the question and see from our potential energy diagram, the energy difference that they are wanting from us. OK. So let's start by drawing 12 Di Bromo, Ethane. So Ethane tells us we have two carbons and di BROMO tells us we have two bromine atoms on carbons, one and two. So to satisfy the Octa of these carbons, we need two hydrogens on each carbon as well. So reading from left to right, we have C H two B R and C H two B R. So that is the structure of Di Bromo Ethane. Now, we want to draw our conformers, which we draw as Newman projections. So we're going to place an eyeball down the carbon 1 to 2 bond. So on the left side, facing the right hand side towards carbon one and then carbon two behind carbon one. And we're going to start by drawing our staggered conformers, which is when the groups on each carbon are not overlapping one another. So there is some space between the groups on carbon one and carbon two. And we can have three conformers for staggered. So I'll draw three circles and we'll draw carbon one the same for all three structures. So we'll draw the bromine at 12 o'clock, we'll draw the two hydrogens at, let's say four o'clock and eight o'clock repeating this for the other two staggered conformers. Now, we'll just change the groups on carbon number two. So for conformer A, we'll start at 60 degrees, which is the dihedral angle between the two largest groups. And in this case, that'll be bromine. So the bromine on carbon two will be at about two o'clock, the hydrogen for carbon two will be at six o'clock and the hydrogen for carbon two will be at the second hydrogen will be at around 10 o'clock. All right. So now rotating that back carbon will have the same angles, but the groups will be in different positions. So now the hydrogens will be at the top and that bromine will be at six o'clock. So this is conformer B and that dihedral angle is now degrees and continuing to rotate that bromine clockwise. For our last staggered conformer, we'll have the bromine at 10 o'clock in the hydrogens at two and six o'clock. So that is conformer C and the dihedral angle will be 300° if we're going clockwise, right. So those are the staggered conformers. Let's draw the eclipsed conformers, which is when the groups are overlapping one another. So again, we can draw three conformers and we'll draw the carbon one groups the same as we did for the staggered conformers, bro at 12 o'clock, hydrogens at four and eight o'clock drawing that three times. So for our first eclipsed conformer, now, instead of having the groups on carbon two in the windows between the groups on carbon one, they're going to be right in front of one another. So that bromine atom on carbon two will be right behind the bromine atom on carbon one and same for the hydrogens. So the dihedral angle for this conformer, which we can call conformer D will be zero degrees. All right, rotating that bromine clockwise, putting it behind the hydrogen at 4:00, this gives us a dihedral angle of 120° for conformer E and lastly rotating one more time to have that bromine behind the hydrogen at eight o'clock that gives us conformer F which has a dihedral angle of 240°. So now let's use a potential energy diagram to look at the energy of all of these conformers. And generally, we know that staggered conformers are going to be lower energy because there are more, there is more space between the groups on carbons one and two. Whereas in the eclipse conformers, those groups are kind of right on top of each other. So they're going to be higher in energy and less stable. So let's draw our Y axis and our X axis, the y axis is labeled potential energy and that is increasing as we go up the y axis, our X axis is labeled for the dihedral angles and we'll start at 0° and go to 360° in increments of 60. So we have zero degrees, 60°, 120° 180° halfway across 240° 300 degrees And 360°. All right. So let's plot our points based on their energy. So starting at zero degrees, that is an eclipse structure, right? Our brom means are directly overlapping each other. So that's going to be actually the highest energy conformer we have, right, because there is basically no space between our largest groups. So we'll plot d very high energy at zero degrees. So, since we're starting with our maximum, which will also be the same at 360°, right, that'll be the same exact structure. Once we just rotated it all the way around, let's go to our lowest energy. Where will our minimum be? That'll be when the structure is anti. I'm going to write that label anti for conformer B. When bromine, the bromine groups are 180 degrees apart. So that'll have low energy At 180°. And that is conformer B. So now let's go to 60°, which is our staggered conformer. And that is actually called a Gauch conformer. And it is the same for conformer C. That is also Gauss where the Bromine atom, the Bromine atoms are 60° apart. And those are going to be lower energy, but higher than 180, lower than D or lower than any eclipsed conformer. So we can draw 60° and 300° about a quarter of the way up our, our Y axis. So higher than conformer A and C, all right. And lastly, conformers E N F are eclipsed. So those will be high energy once again, but lower energy than D because the Bromine atoms have a little bit more space and those will also be equal to one another. So we'll draw equal high energy points for ENF at 1:20 and 2 40. OK. Now let's connect all of our points. It's going to be an oscillating wave D A E B F CD. Oops. All right. So let's go to, let's actually. So let's label our potential energy diagram with the information about energy that was given to us in the problem. So the problem mentioned that a gauch conformer has two energy barriers. Where are our gauch conformers? On the potential energy diagram, we have points A and see and I'll label the anti one as well. Point B. OK. So two energy barriers for our Gauch conformers, 22.6 kilojoules per mole and 39.7 kilojoules per mole. Well, notice to get to our gauch conformers, we have two energy levels, right? Point D is the first major hump and then we have to go past point E or point F, right, same energy level to get to the other gauch conformer C. So let's draw a dotted line horizontally from the Y axis through point A in through point C. And we can label the distance from A to D as one energy barrier and the distance from this gauch level A or C to E and that will be the other energy barrier, right? So which one is larger the one from A to D? Right? So that'll be 39.7 kilojoules per mole and the one from A to E or C to F, right? It's the same amount is 22.6 kilojoules per mole. Now, they also gave us another amount. The anti conformer is 5.86 kilojoules more stable than a gauch conformer. So more stable is going to be less energy. So that means this difference below that ac line that we drew is going to be 5.86 kilojoules per mole. OK. So now we have our graph labeled with all of our information. Let's answer the question and the question says, determine the energy difference between the most stable eclipsed conformer and the most stable staggered conformer. So where are those points? Well, the most stable eclipsed conformer is going to be E and F, right? We said that conformer D is the highest energy and our most stable staggered conformer is our anti conformer B. So let's look at the difference between those on our diagram. We have E and F and B. So we see that the difference between these is the sum of 22.6 and 5.86. So you see why I think it's easier to draw these on the diagram, we can see exactly what amounts they are looking for. So let's add together 22.6 kilojoules per mole And 5.86 kilojoules per mole. And that gives us 28.5 kilojoules per mole. So that is the energy difference between the most stable eclipse conformer and the most stable staggered conformer for 12 Di Bromo Ethane, which makes c the correct answer for this problem. That's it for this video. I hope this was helpful and thank you for watching.

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

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