Which has a higher percentage of the diequatorial-substituted ...

Question

Which has a higher percentage of the diequatorial-substituted conformer compared with the diaxialsubstituted conformer: trans-1,4-dimethylcyclohexane or cis-1-tert-butyl-3-methylcyclohexane?

Accepted Answer

Hello everyone. Let's do this problem. It says each of the compounds given below has a di equatorial substituted conformer and a di axial substituted conformer, which compound has a higher percentage of the di equatorial substituted conformer compound, one trans one ethyl, four methyl cyclohexane or compound two CS one TERT butty three ethyl cyclohexane. So what does it mean when the problem asks which compound has a higher percentage of the equatorial substituting conformer? Well, that means that the compound really prefers to be in that die equatorial substituted conformer. And that'll be the more stable conformer, right? Well, what makes a chair conformer want to be in the die equatorial substituted position when it has a larger substituent group? Right. So we want to look at the equatorial chair conformers for both of these compounds and look at which one has the bulkiest group. So which one has the bulkiest substituent group? Because that will be more desperate to be in the di equatorial substituted conformer, which means it will exist in that conformer at a higher percentage. Ok. So I'm going to start by drawing the two D structure of these compounds and I'll draw the substitute groups as wedge or dash. So we'll start with compound one trans one ethyl, four methyl cyclohexane draw cyclohexane ring numbered one through six, clockwise from the 12 o'clock position on the ring. And on carbon one, we have an ethyl group, I'll make that wedge. And on carbon four, we have a methyl group. And since this compound is trans, I'll make the methyl group dashed. All right. So now turning this into a chair confirmation, I'm going to draw two parallel lines going upwards from the left cap, the left side off with an upwards corner cap, the right side off with a downwards corner. And I'll number clockwise one through six, starting with the left most upper corner carbon. So now adding our substituent groups, carbon number one, the corner is pointed up in our ethyl group is wedge or up on the ring. So that group will be up in the Axio position. And we know the problem told us that these compounds have one equatorial substituted conformer and one di axial substituted conformer. So that means the substituent group on carbon number four should also be axial. So let's check the methyl group is dashed. So that means it is down and carbon number four in the chair is pointing down so that methyl group will be down in the axial position. So this is our di axial conformer. So let's do a chair flip and we should get the di equatorial substituted conformer So drawing two lines parallel down from the left capping, the left corner down, capping the right corner up, numbering one through six at the same positions as the first chair. So now corner number one is down, but our group, our ethyl group is up on the ring. So that means it will be up in the equatorial position. In carbon number four, the corner is pointing up but our methyl group is down on the ring. So that will be down in the equatorial position. So this is the di equatorial substituted conformer. OK. And what is the bulkiest group on this ring? The ethyl or the methyl, the ethel. So bulky group equals ethel. OK. Let's look at compound two. I'm going to draw the two D structure. So drawing cyclo heine numbering one through six, clockwise from the top. And on carbon one, we have a turtle group. So that is four carbons with a tertiary carbon. And I'll make that wedge. And then on carbon three, we have an ethyl group and this compound is CS. So this substituent group will also be wedged. OK. So now let's convert this to a chair. I'm going to draw two parallel lines upwards from the left cap, the left corner up, cap the right corner down, numbering it the same way as our first chair in our first compound. So corner number one is pointing up, our substituent group is up because it's wedged. So that will be in the axial position. Corner. # 3 is up and our substituent group is up because it is wedge. So that ethyl group will be axial as well. So we have our diao conformer here. Let me actually write that up here. And now let's do a chair flip. So flip the opposite direction. Numbering 1 through 6, that's an ugly chair. Let me redraw that. So numbering 1 through clockwise. And let's see, carbon number one is pointed down here, but our group is wedged. So it's up. So that will be up in the equatorial position. Carbon number three is down, but our ethyl group is up on the ring. So that needs to be up in the equatorial position. So that is our die equatorial conformer. So what is the bulky group in this conformer, the trt beel group or the group, the turtle group? Right? So, t but so now comparing compounds one and two, which one is going to be more desperate to be in the Dior position, which one has a bulkier substituent group? Compound two, right? That turtle group is more bulky than the ethyl group. So that means compound two is going to exist in the dye equatorial substituted conformer at a higher percentage than compound one. That's it for this problem. I hope this video was helpful and thank you for watching.

Which has a higher percentage of the diequatorial-substituted conformer compared with the diaxialsubstituted conformer: trans-1,4-dimethylcyclohexane or cis-1-tert-butyl-3-methylcyclohexane?

Key Concepts

Conformational Analysis

Steric Hindrance

Substituent Effects in Cyclohexane

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