c. Draw Newman projections of the two conformers of the trans ...

Question

c. Draw Newman projections of the two conformers of the trans isomer.

d. Which of the conformers predominates at equilibrium?

Accepted Answer

Hello everyone. Let's do this problem. It says two conformers can be drawn for the trans isomer of dimethyl cyclohexane part one, draw Newman projections of the two conformers, part two, which conformer is present in greatest concentration at equilibrium. So why can we only have two conformers for this compound? Usually we can draw many if we're rotating about a carbon carbon, single bond. Well, this is a cyclohexane based compound. So we are talking about chair confirmations. OK. So first, before we draw chair confirmations, I think it's helpful to draw the two D structure of the compound. So just the regular bird's eye view structure, including wedges and dashes, then we'll draw our chair conformers. And lastly, we will draw the Newman projections from those chair conformers. OK. So starting with step one, draw the two D structure. So let's just draw 12 dimethyl cyclohexane. Cyclohexane is a six membered ring made of carbons And on carbons, one and two, we have methyl groups and this is the trans isomer. So those are going to have opposite orientation. So we can make the methyl group on carbon one wedge and will make the methyl group on carbon two dash. And let me number my ring actually. So carbon 123456, numbered clockwise from the top of the ring. OK. First step complete. Now let's turn this structure into chair conformers and we can draw two conformers. So we'll start by drawing our parallel lines slanted upwards to the right, cap those off, left side, up, right side down. And I'm going to number my ring, I'll start in the bottom left corner and number counterclockwise one through 6. So now let's add our methyl groups. I'm going to add them in red just so they stand out a little bit. So the methyl group on carbon one in our two D structure we see is wedge. So we can consider that up on the up face of the ring and the methyl group on carbon two will be on the bottom face or the down face of the ring. So on carbon one, we want our methyl group to be up, but the carbon, the corner is down. So it can't be axial. That group will have to be up in the equatorial position and the hydrogen that is also sharing that carbon will be Axio. Does that make sense? So now on carbon # two, the corner is up but our methyl group is down. So the hydrogen oops I should do the methyl group first, the methyl group will also be down in the equatorial position. And the hydrogen will be in the axial position. And I'm not going to draw the other hydrogens for now, since that isn't really important here. So we can refer to this as conformer. A both of our metal groups are on opposite faces of the ring and they are in the equatorial positions. So now let's perform a chair flip, that will be our other conformer. So we'll draw our parallel lines now diagonal upwards to the left capping that off with the left corner going down, right corner going up will number our ring in the same positions. The same carbons 1, 2, 3, 4, 56 going counterclockwise. And again, our methyl group on carbon one is on the top face pointing up. So that will be axial since this corner, carbon one is pointing upwards in the axial position follows the direction of the corner. So that means the hydrogen will be down in the equatorial position. And on Carbon two, that methyl group is down or on the bottom face, the corner in our ring is down. So that methyl group will be axial down and the hydrogen will be equatorial up. So this is conformer B. All right, we're almost to step one which is draw the Newman projections Or part one which says draw the newman projections of the two conformers. We just completed step two in our mini list. So now let's draw our newman projections from these conformers. OK. So to draw our Newman projection. We're going to place an eyeball on one side of our chair. I'm going to place it on the right side, facing the compound towards the left. So the eyeballs in front of carbon three, carbons two and one are on the left side of the eye, carbons four and five are on the right hand side of the eye and carbons three and six are right down the middle, one in front of the other. And we'll do the same position for conformer B. So converting this to Newman projections, We're going to have two circles for bonds 2-1 and 4-5 carbons, 2-1 and 4-5. And we're connecting Carbon three in the front and carbon six going up in the back. So now we're going to add our substituent groups or not substituent groups atoms on each carbon. So carbon two is the left front atom and we have on a conformer, a we have an axial hydrogen going up that equatorial metal going down. OK. Going back onto carbon one behind this atom behind this carbon carbon one has the hydrogen Axio pointing down and the methyl group equatorial pointing up to the left And moving over to the other side. Carbon four on the front of our newman projection, we'll have all hydrogen groups, right. So hydrogen axial hydrogen equatorial and same for carbon number five, right. That's why we didn't draw those groups because there are just hydrogens there. And again, this is carbon two On the left side, this is carbon four on the front, right side. So that is conformer a the newman projection for one of our trans isomers trans conformer I should say. And now let's do the same four conformer B except this time, we have our two circles for our carbons. And then carbon three is pointing up in the front. Carbon six is pointing down in the back. And let's see, Carbons four and oh, this is ugly. Let me redraw this. OK. Carbons four and 5 have just hydrogen. So we can draw those Adams. Now, carbons two and one on the left hand side, we have a hydrogen up in the equatorial position. The methyl is down in the axial position. And on carbon one the methyl is up in the axial position and the hydrogen is down left in the equatorial position. And let's connect carbons one in 5 via six down in the back. And that is our new in projection for conformer beat. So that is part one of this question drawing both of the Newman projections In part two asked which conformer is in the greatest concentration at equilibrium. So do you think that has something to do with stability? Yeah, the most stable conformer is the one that will be the most present at equilibrium. So we know that in chair confirmations, the largest group favors being in the equatorial position So conformer A has both methyl groups which are the largest groups on our ring in the equatorial position. So conformer A is going to be present in the greatest concentration at equilibrium, which makes b the correct answer choice for this problem. That's it for this one. I hope this video was helpful and thank you for watching.

c. Draw Newman projections of the two conformers of the trans isomer.
d. Which of the conformers predominates at equilibrium?

Key Concepts

Newman Projections

Conformers

Equilibrium and Stability

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