in this video we're going to talk about the chair conformation of cyclohexane as well as how to draw a ring flip but let's start with the bond line structure of cyclohexane it looks like that now let's convert it into its chair conformation so we're going to do is we're going to draw two parallel lines and then two more parallel lines with the same length and then another two parallel lines so that's a simple way of drawing the chair conformation of cyclohexane now this first carbon atom has two hydrogen atoms the same is true on this carbon atom it has two hydrogen atoms but the way we show it is going to be a little different there's two types of bonds that you need to be familiar with the axial bond and the equatorial bond the axial bond can go straight up or straight down the equatorial bond kind of veers off to the side so let's draw the axial bonds so this is axial up the next axial bond will be down and it's going to alternate this is going to be axial up axial down axial up axial down this equatorial bond is kind of slanted downward the next one will be slanted upward and then down up down up so that's how you can show all of the axial and equatorial bonds for the chair conformation of cyclohexane now let's work on an example problem draw the most stable chair conformation of methylcyclohexane so first let's draw the bond line structure so this is cyclohexane and we just have a methyl group now let's try the chair conformation so let's start with our two parallel lines and then we'll draw two additional parallel lines and so forth now we could put the method group in the axial position or we could put the method group in the equatorial position in both cases this represents methyl cyclohexane but which one is more stable than the other is it better to put the method group in the axial position or in the equator position which configuration is the lower energy configuration if we put it in the axial position there's something known as 1 3 diaxial strain this methyl group is going to interact with these hydrogens if we call this carbon one it's going to interact with the hydrogen on carbon three regardless of where you count it from the methyl group so because of this one three diaxial interaction the axial position is less stable than the equatorial position so this is going to be the answer the chair conformation will be more stable if we put the methyl group off to the side in the equatorial position because we will avoid this one three diaxial interaction now what about this example problem draw the most stable chair conformation of one ter-butyl for methyl cyclohexane so let's begin by drawing the chair conformation so let's put the tert-butyl group on carbon one so that's our tert-butyl group and let's put the methyl group on carbon four so this is one way in which we can draw one terbutal 4 methyl cyclohexane this is in the axial position and this is in the actual position so in terms of stabilities there's four different types of stabilities that we can have here it could be axioaxial axial equatorial equatorial axial or equatorial equatorial so let's draw four different combinations based on what we just saw so let's keep the tert-butyl group in the axial position but now let's move the method group to the equatorial position so this is axial down this is going to be equatorial up and let's try another one so this time we're going to put the terbuta group in the equatorial position and let's keep the method group in the axial position and let's draw one more confirmation so this time the terbutal group will still be in the equatorial position and the same will be instead of the methyl group now let's rank the stability of the four confirmations that we see here so we're going to say that number one is the most stable number four is the least stable so which one is going to be the most stable chair conformation out of what we have listed here the most stable is going to be the one where both groups are in the equatorial position so this is number one the least stable is going to be the one where both groups are in the axial position so that's this particular conformation now between these two which one is more stable will the stability increase if we put the tert-butyl group in the equatorial position or the methyl group in the equatorial position because in both cases one is accurate one is equatorial in order to determine which conformation will be more stable you want to look for the one where the bulkiest group is off to the side so we can reduce the potential energy of this molecule by putting the bulkiest group on the equatorial position so this is going to be more stable than this one so this is the most stable this is the second most stable that's the third this one is the least stable so the answer that we're looking for in this problem we just want to draw the most stable chair conformation and it's this one it's the one where we put the most bulkiest groups in the equatorial position now consider this problem identify the following chair conformations as cis or trans isomers so how can we do this for each of the conformations that we see here all we need to do is identify if it's up or down so looking at this methyl group it's in the axial position but it's up for this one it's axial but it's facing downward if one is going up and the other is going down it's going to be trans i'm going to put t for trans and looking at the next one this is axial up this is equatorial up because they're both facing upward this is going to be cis they're going in the same side now for the next one this is equatorial down this is axial down they're both going in the downward direction so the same direction this is going to be cis next this is equatorial down axial down same direction this will be the cis isomer for this one this is equatorial up equatorial down so if one is going up the other is going down this is going to be trans because they're going in opposite directions and for the last one this is axial down equatorial down so that's going to be the cis isomer so that's how you can determine if a particular chair conformation is a cis isomer or a trans isomer now let's focus on ring flips let's talk about how we can draw a ring flip so let's begin by drawing the chair conformation of cyclohexane and let's say we have a methyl group on carbon one so let's number this let's say this is one two three four five six now to draw the ring flip instead of drawing these parallel lines slanted to the right we're gonna draw the mirror image we're going to slant them towards the left and then we're going to draw two parallel lines and then two short parallel lines so let's call this form a and form b going from form a to form b you want to move these numbers in the clockwise direction so this carbon which was carbon 1 was there it's now going to be here and each of the number is going to move one position in the clockwise direction now the methyl group was in the axial position and it was in the upward direction anytime you draw a ring flip the groups the axial position they're going to go into the equatorial position and vice versa so if the method group was in the equatorial position it's going to go in the axial position now if a group is in the up position it's going to stay up if it's in a down position it's going to stay down so when doing a ring flip the only thing that changes is the axial and equatorial bonds the way they're directed up or down that part doesn't change so the methyl group it was in the axial position now it's going to be in the equatorial position so this part changed from a to e but it's not going down it's still going up so that's a quick and simple way of how you can draw a ring flip now just to recap if you're going from form a to form b just remember these numbers you need to move them in the clockwise direction and if you're going from let's say form b to form a then these numbers they get moved to or in the counterclockwise direction so be mindful of which form you're starting from now these two forms of methylcyclohexane these two chair conformations they're in a convertible one form can convert into the other form so what type of arrows should we put here should we put arrows that are equal in size or should one be significantly bigger than the other one which set of arrows should we use well notice that these two chair conformations are not equally stable here the methyl is in the axis position here it's in the equatorial position and we know putting a methyl group in the equatorial position reduces the 1 3 dioxial interaction between the methyl group and the hydrogen atoms so therefore this structure is more stable so because it's more stable this particular reaction the way it's written it's going to be product favored so we're going to use this set of arrows so we're going to have a bigger arrow towards the right a smaller arrow towards the left so the two arrows help us to see that it's reversible however the position of equilibrium is towards the right this particular reaction is product favored the bigger arrow shows you which conformation is more stable let's work on this example problem perform a ring flip on the chair conformations shown below so looking at the first example let's call this carbon one two three four five and six now what type of what form of the chair conformation do we have form a or form b notice that the lines are slanted to the right with a positive slope so therefore this is for me to go to form b we need to move the numbers in the clockwise direction so first let's draw two parallel lines with uh going down with a negative slope and then let's draw two additional parallel lines and then connect them so this bromine atom well first let's move the numbers so this was carbon one this is now going to be carbon 1. so this will be 2 3 4 5 and 6. this bromine atom was in the axial position now it's going to be in the equatorial position it was axial up but now it's going to be equatorial up now this bromine atom is equatorial up but on carbon 2 here it's going to be axial up so up remains up but axial change to equatorial and vice versa so this is the ring flip of this structure well that's the confirmation that is produced when we flip this ring now which structure is more stable notice that the stability of these two molecules is the same this is in the axial position that is in the equator position so we have one axial one equatorial and the same is true here we have an axial bromine and an equatorial bromine so because they're equally stable we can draw a reversible reaction with two arrows of the same length so 50 will be in form a the other 50 will be in form b so they can easily interconvert from one form to the other now let's focus on the second example which form of the chair conformation do we have is it form a or form b so notice the lines the slope of those two lines are negative so therefore we have form b and we need to convert it to form a so let's draw two parallel lines with a positive slope and then everything else will follow now in order to do the ring flip because we have form b we need to go to form a this time we're going to move the numbers in this direction so this is carbon 1 2 three so this is going to be carbon one this will be two and this is three so now the tert-butyl group was in carbon one and it was in the actual position in the upward direction so axial is going to become equatorial but it's still going to be in the upward direction now on carbon 3 we have a methyl group in the equatorial position going in the downward direction so carbon 3 we're still going to have a methyl group but it's going to change from equatorial to axial but it's going to remain in the downward direction so this is down that's down this is up and that's up so that's how we could perform the ring flip in that case so we just need to move the numbers in the clockwise direction going from b to a but going from a to b we would move the numbers in a clockwise direction so wait let me just just in case i said that wrong i'm just going to repeat that so going from b to a we need to move the numbers in the counterclockwise direction but from a to b we move it in the clockwise direction so now that we've drawn the chair conformation that results from performing a ring flip on form b to get form a now let's talk about stability which of these two structures or two chair conformations is more stable is it form b or form a in each case we have one group in the axial position one group in the equatorial position but we do have a bulky terbuta group and we can greatly increase the stability of the whole chair conformation by putting the bulky buta group in the equator position therefore the structure on the right will be more stable than the structure on the left this structure is less stable so the position of equilibrium is going to favor the product or the reactant that is more stable so we're going to have a bigger arrow that points to the right and a smaller arrow that points towards the left now let's work on this example convert the following structures into their corresponding chair conformations so for the first one we have the trans isomer now let's draw a chair conformation that corresponds to it so let's put the method group on the first carbon at the top now notice that the methyl group is on the wedge which means it's coming out of the page therefore we're going to put it in the up position now the bromine atom is on the dashed line it's on the hatch wedge so it's going it's going into the page this is coming out of the page that's going into the page so for into the page we're going to put the bromine atom going down so if we were to number the carbons this is carbon 1 2 3 1 2 3 so on carbon 3 bromine has to be going down it's not going to be axial up but it has to be equatorial down so that's one possible chair conformation that we can draw for uh this structure it's not the only one there are other ways you can draw it as well but this is just one possibility so for instance let me give an example of another possibility so instead of putting the method group on the axial up position what we could have done we could put the methyl group in the equatorial opposition as long as in the it's in the up position it's going to work because the method group is on the wedge so anything that's on a wedge you want to place it in the up position if if it's on a dash place it in the down position so now this will be carbon one this is going to be two and this is going to be three so this is going to be down but it's going to be axial down so these two structures they both correspond to the trans isomer as we can see this is up and this is down which makes it trans so now draw one possible chair conformation that corresponds to that structure so let's put the hydroxyl group on this carbon so this is on the solid wedge we want the oh group to be going up so that's axial up now the chlorine atom is also on a solid wedge which means that when we put it on the chair conformation it needs to be going up now let's number this so let's put oh on carbon 1. where should we put the chlorine atom should this be carbon 2 or should this be carbon 2 in a situation like this we need to make sure that we're counted in the same direction so here we're counted in the clockwise direction so we need to do the same thing we need to count in a clockwise direction therefore carbon 2 should be here and this is going to be equatorial up so that's it for that example so that's one way in which we can draw the chair conformation for it now let's focus on this one so we can make this carbon one or we can make that carbon one we're not trying to name it using iupac rules we're just putting the numbers in so we can draw the correct share confirmation so let's count this in the clockwise direction and let's put the method group on carbon one so it's on a solid wedge which means we need to put the method group in the up position and if this is carbon one counting it in the clockwise direction this is going to be carbon four so on carbon 3 the bromine is in the back it's on the hatch wedge so that means it needs to be facing down so on carbon 3 we can either put it in the axial position or the equatorial down position so we need to put it in the equatorial down position now on carbon 4 the oh group is on a solid wedge so we want it going up so we got to put it in the equatorial opposition and so that's it for these problems now you know how to convert a bond line structure into the corresponding chair conformation you