in this lecture we're going to cover the confirmations of mono and disubstituted cycle hexanes we've already discussed the fact that a cyclohexane ring is going to adopt a chair confirmation so that it can have perfect tetrahedral carbons with 109.5 u Bond angles we also talked about the fact that the cyclohexane ring can flip and when the ring flips equatorial substituents become axial and a ax substituents become equatorial so let's look at an example of a monos substituted cyclohexane ring we're going to look at methyl cyclohexane when we draw methyl cyclohexene in the chair confirmation we can put the methyl substituent either in the axial or the equatorial position so we will start with that methyl substituent in the axial position if the methyl on this carbon is in the axial position the hydrogen then will be in the equatorial position if we ring flip this cyclohexane ring the axal substituents become equatorial and vice versa so what what I'm going to do first is number my carbons so I can keep track of them in the ring flipped version of this molecule I'll start uh with my methyl group carbon as number one and now when I ring flip this I'm going to keep track of all my carbons using these numbers remember that the head of my chair which in this case is carbon one will become the foot and the foot of my chair carbon 4 will become the head so here's my ring flipped version and I'll put my numbers in so we can keep track of those carbons again the easiest to identify will be the head and the foot which are now reversed and then I can fill in my numbers remembering that uh I'm numbering in this case going counterclockwise so now I've numbered my ring flipped cyclohexane and now I can add my substituents remember that in this case my methyl group is pointing up on the left as axial it's going to be up on the right but but now the opposition on carbon one is is equatorial my hydrogen on the left is down so on the right structure it is also going to be down but now on carbon one my down position is axial so this is the correct structure of the Ring flip version of my cyclohexane drawn and now my methyl group is equatorial I would strongly recommend making a model of this cyclohexane ring and Performing the ring flip so that you can convince yourself that that methyl group that starts out axial will become equatorial and when you reverse the ring flip uh it will go from equatorial to axial now that we've drawn these two confirmers of methyl cyclohexene we need to determine which is more stable we can determine which is more stable using Newman projection drawings to allow us to look at the type of interactions that are involved in each one of these confirmers of methyl cyclohexene when we're choosing how to draw a new Mo project ction you can choose uh two different views you need to make sure though that you're choosing uh one of the bonds that includes uh the carbon with the methyl substituent so just to simplify things I am going to choose to look at my cyclohexene ring down the carbon 6 to carbon one bond in this direction and remember that with a cyclohexane ring there are always two bonds that are parallel so we'll also be looking down the bond betwe carbon between carbons four and carbon three so let's draw this numer projection I've drawn the back carbons which are carbons one and three and carbons 1 and three are going to have a bond to Carbon 2 and that bond is pointing down my front carbons which are carbon six and carbon 4 are going to have a bond to carbon five which is pointing up and now I need to draw in the rest of my substituents which are mostly hydrogens except for on carbon one remember that these are all going to be spaced around the central carbon at about 120° they're going to be evenly spaced um because they're all staggered and because these bonds are 109.5 and so this is what all my substituents are going to look like and now I have to fill them in properly I'm going to start by numbering each of my carbons the two carbons in the front will be carbons 4 and carbon 6 six the two carbons in the back will be carbons three and carbon one and the two in the middle will be two and five five is between four and six and two is between carbons 1 and three and so except for carbon one all of the substituents on the rest of the carbons will be hydrogen so I'm going to draw those in and now I have to determine where to put my methyl substituent on carbon one and if you make this model and you look down car uh the bonds between carbons 4 and three and carbon 6 and one it'll be very easy to decide where to place this methyl group the other way you can think about it is um if you imagine looking down that Bond your methyl group is pointing straight up and it's very obvious when we look at our Newman projection which of the two Bonds on carbon one is pointing straight up it's going to be this one so this is where our methyl group goes leaving the last substituent spot for hydrogen if you're not convinced that that is how that numer projection would look I strongly recommend making this model and convincing yourself of that now let's draw the Newman projection for the equatorial methyl cyclohexane again I've drawn my two back carbons I'm going to number my cyclohexane ring so my two back carbons are drawn which in this case are carbons two and carbons four because again I'm going to look down a bond containing my methyl group and I'm going to choose just for uh to simplify things carbons 5 to 4 and carbons 1 to two and so the back carbons I've drawn in my numeral projection are 2 and four they're going to have a bond to carbon 3 which is going to be down my front carbons are carbons 5 and carbons 1 and they are going to have a bond to carbon 6 which is pointing up then I'm going to draw in the rest of my substituents at 120° from the ones I've drawn so far and then I'm going to number my carbons just like I did before remember the carbons in the front are carbons five and carbons one the carbons in the back are two and four and then it's fairly easy to figure out which is three and which is six now that I've numbered my carbons I can fill in my substituents all of them will be hydrogen except for one of the ones on carbon one and as I tried to determine where to put my methyl on carbon one remember that that methyl is pointing out to the side the hydrogen is pointing down and so hopefully that will help you figure out that the methyl group should go here and the hydrogen should go here if you're not convinced of that again make the model look at this numer projection on that model and make sure that you can convince yourself that this is the proper placement of the substituents on carbon one now remember the point of this exercise exercise was to determine whether that methyl substituent is more stable if it's in the axial or the equatorial position and to do that we can look at the confirmations on these numer projections and compare them to what we learned about numer projections for linear alkanes we know that the most stable confirmation for any alkane is going to be anti the least stable is going to be totally eclipsed now we have all staggered confirmations here and so what that means is the least stable confirmation we can have is go remembering what that means is um when the two largest groups are staggered but as close together as possible which is the interaction we have in the axial methyl group The methyl group and carbon 5 are go to one another in the equatorial methyl group that methyl group is anti to the closest large group on Carbon 2 what that tells us then is that the the equatorial methyl group is more stable than the axial methyl group because placing that substituent that methyl group in the equatorial position allows it to be anti from the carbons in the ring rather than go as it is when it's axial and this is a pattern we're going to see all the time in cyclohexane rings putting substituents in the equatorial position puts them anti to the carbons in the ring making them more stable the equatorial position is always going to be the more stable position for the larger substituents another name for this go interaction is a 13 di axial interaction it's called a 13 di axial interaction because it is it involves two axial substituents that are on carbons with a one three relationship to one another so for example we're talking about the interaction between the methyl group on carbon one and the hydrogen on carbon 5 both of which are in a positions and are three carbons away from one another there is a similar interaction between the hydrogen on carbon 3 and the methyl group on carbon one and you can actually draw the numer projection down carbons 3:2 and 5 to six and be able to see this in the newer projection as well and so you should start to be able to recognize these 1 three di axial interactions that I've shown here on cyclohexane rings and you can see that that one three di axial interaction is not possible for the equatorial methyl because it's not axial these go interactions or these 13 diaxial interactions become even more important when the groups are both larger than hydrogen such as in this C13 dimethyl cyclohexane which looks like this and if we draw out the chair confirmation you'll see that you can put those two methyl groups either both axial or both equatorial so here they're both axial if I flip the ring all of my axial substituents become equatorial and hopefully you can see that on the left structure we have a very unfavorable 1 three diaxial interaction between the two methyls that is not present in the structure where both of the methyls are equatorial and so the die equatorial substituted cyclohexane ring flip confirmer is more stable than the confirmer in which both of those methyl groups are ax what if we look at the trans isomer when we draw the trans isomer in the chair confirmation one of the methyls will be axial and one will be equatorial again if you're not convinced of this make the model of trans 13 dimethyl cyclohexane put it in the chair confirmation and see that no matter how you flip the ring one is always going to be equatorial and one will be axial so one methyl group will always be axial and one will always be equatorial this is in contrast to the Cy isomer where one conformer has both methyl groups axial and one has both equatorial because CIS 13 dimethyl cyclohexane has access to the most stable confer it is the most stable isomer as well because it can access the most stable confirmer so C13 dimethylcyclohexane is more stable than the trans 13 dimethyl cyclohexane because that CIS 13 dimethyl cyc hexane can access this confirmation where both methyl groups are equatorial at the same time now let's consider the one two dimethylhexane example this is isomer will look like this when we draw it in a chair confirmation both methyl groups will be up one way to draw that is like this where both of my methyl groups are in the utmost position on carbons that are right next to one another in a one two orientation and in this case one of them is axial and one of them is equatorial when we draw the ring flip the one that was axial becomes equatorial and the one that was equatorial becomes axial in the cis1 two dimethyl cyclohexane isomer there's always going to be one methyl that is axial and one that is equatorial in the trans isomer one methyl group will be up and one methyl group will be down when we draw that as a chair one methyl group will be up in this case axial and one will be down also axial when we draw the rim flip since what both were axial they will now both be equatorial and so in the trans one2 dimethyl cyclohexane isomer we have a less stable and a more stable isomer the isomer with both of the methyl groups equatorial is the most stable of any of the four I've drawn and therefore trans1 two dimethylcyclohexane has access to the most stable confirmer and therefore is the most stable of the two isomers so all of our examples so far of disubstituted cyclohexane rings have had two of the same substituent what happens when we look at larger substituents the larger the group the higher the preference it has for being in an equatorial position if there are two substituents and they are different sizes and both cannot be equ the more stable confirmer will have the larger group in the equatorial position let's look at an example of this CIS one ethyl four methyl cyclohexane which looks like this now let's draw it in a chair this is one confirmer of the chair confirmation of this molecule now we'll draw the ring flip where our methyl group that was axial has now become equatorial and our eth group that was equatorial has now become axial the confirmer that has the largest group equatorial the one on the left is going to be the most stable confirmer because the larger the group is the more preference it has to be in the equatorial position because those 1 three diate interactions only get worse as the group gets larger