so in this video we're going to talk about the interconversion of cycle accidents i've confirmed my cycle accident conformations so let's get started we know the cyclohexane ring it has a six carbon atom so six carbon-carbon single bonds and usually cyclics and rings are stable ring and it because it can add up it can be in sheer conformations if the cyclohexane ring appear in chair conformation there is no torsional strain so the ring is very stable the thing is right here since all this carbon here uh and also all the carbons here connected by carbon carbon single bonds and we know the carbon-carbon single bonds they can freely rotate and because of the result of simultaneous rotations about all carbon carbon sigma bond usually the chair conformation of cyclohexanes can interconvert to another chair conformations by ring flip so you can see here if these shear conformations if they're rotated about all carbon carbon bonds they will take the shape of another chair conformations shown here on the right the way that we usually take a look these ones if you just push this carbon this down carbon up and just half carbon down then you will have a ring flav chair conformation so these two are the chair conformations and since this is obtained by ring flipping of these chair conformations we call ring flip chair conformations again uh usually what we do we can we put we push this down carbon up and then we pull the down carb we pull this carbon down so that's why we get the ring filled flip chair conformations so in this process what actually happened you also can see here while the equatorial bond they turn into an axial bond an axial bond that they turn into an equatorial bonds so in this process when when there is a ring flip in chair conformation equatorial bonds become axials and axial bonds become equatorial now how to draw a ring flip chair confirmations so usually there are three rules and you must need to remember these three rules in order to write the ring flip chair confirmations first is draw the mirror image of your original ring so this is your chair conformations let's see you have a substituent r on this up carbon and pointing axially so you have excel r group um so if according to the step one it says draw the chair from draw the mirror image of your ringfield ring original ring so you know to do that what you have to do is just kind of pull this carbon down and then push this carbon up that way you will have the ring fl ring flip chair confirmations of the original one so you know so when you do that so we have now the mirror image of your original ring the second rule you can you can see here the up carbon they turn into now down carbon and down carbon turn into up carbons in the in the ring flip chair conformations now step two says on your knee ring move every substitution one position clockwise or counterclockwise it doesn't matter which way so if you number these two carbons one and two and now you just kind of on your new ring as it as it says move every substitute in one position clockwise or counterclockwise so we move every substrate by one positions this way so now i have my up carbon turn into a down carbon and down carbon turn into an up carbon in the new ring that's what it says here now what we see step 3 says draw each substituents on a new position it was on carbon number one now is the carbon number one if it was equatorial before then make it axial on the new ring if it was exhaled before then make it equatorial on the new ring so we have here the axial so that means this is going to be equatorial so axial become equatorial equatorial axials and another thing up so this is up carbon and up substituent will remain as the up substituent if you have it down then it will remain as a down right so these three steps you need to follow in order to draw the ring flip chair conformations again the first is draw the mirror image of your original ring by pulling this carbon down and pushing this carbon up and then on your new ring move every substituents one position clockwise or or counter clockwise it doesn't matter anyway and then draw its substituents on its new positions only on your new ring and remember if it was axial then that will turn into equatorial if it was equatorial it will turn into axial and up stage always up down the ways down now let's solve a problem together so that way you should be able to uh you should be able to draw a ribbing flip chair confirmations so here is that we want to draw a ring flip chair confirmations of these molecules again if you would like you can pause the video and try to draw the the ring flexion confirmation on your own then you can check your answer attention i'm going to show you in a second so first release it says i have to draw the mirror image of this original ring you know to do that what i'm going to do and pull this i'm going to pull this one down and this one up that way i will have my mirror image of this original ring now if i number this one this is one and four then my new carbon the carbon number one is with this one and then carbon number four is that one usually up carbon turn into down carbon down carbon turn into and half carbon now i have to have these two substituents on carbon number one and carbon number four i don't know what would be the position of those two however i can see the rule number step number three says if it if it was equatorial here as you see if it if it is axial here then it will turn into equatorial if it is axial here it will turn into equator so i will have two equatorial so this substitute in here is going to be equatorial this substrate is going to be here equatorial so then i completed my drawing the ring flip chair confirmations so what i did if i summarize here i have the mirror image of this original ring but i pull it this one down push this one up so then that way i will have my new mirror image of the ring then i move one move the substitutions clock move the substituents counterclockwise this direction so i have one and then this is four new position of carbon number one and four and i have drawn the r and this um substituents uh as equatorial because they are uh they were they are here axial position so axial turn into negligible positions so that's the final structure next here is a draw the chair and ring fleece structure for the following molecules so here is the the chair conformation of these molecules as you see these this just they're trends to each other so this is actually up this is actually down so this is basically a chair conformations of these molecules now i would like to draw the ring flips chair conformation of these original ring you know to do that i want to pull this one down and push this one up that will give me my the mirror image of this original rig then i want to number this carbon one and two then i want to move this one counterclockwise um this way then i can see this is turning into carbon number one down carbon this down carbon turned into half carbon now i want to put the position now uh draw the substituents of methyl group two methyl group carbon number one and two you can see here if i put it like this so i put it but now i have to indicate whether they would be equator or axial i see axial here axial here the rule says axial will turn into equatorial so that means this was axial up that should be equatorial up this is actually down this is equatorial down the down will remain stays down upwards stays up that's what another thing that you also remember here so this transport to dimethyl cyclohexane has two chair conformations where in one case we have two methyl group on the exhale position another case we see the two methyl group on equatorial positions we later on we'll talk about which one which ring would be more stable this one or that one so when there is one oxygen on the cyclohexane ring there there are two chair conformation possible but those two chair conformations are no longer equivalent are no longer say um probably no longer probably will no longer have the same stability for example if you look at these cyclics and shear conformations where we have an excel methyl group if you uh kind of pull this carbon down push this carbon up then that will give you the mirror image of that origin ring and if you want to put that methyl group right there that would be in the equatorial positions you can see here that the methyl group here on equatorial up this was initially axial lapse of upstairs up but axial turned into equatorial so this is the ring flipped chair confirmation of this original ring now the question is you have here the methyl group on axial position you have here equatorial um the methylcarbonyl position so methyl gluconaxial positions this year confirmation usually the less stable shear confirmation and metal and equatorial positions are more stable uh chia confirmations and the question is why why is actual methylcyclization is less stable compared to the the chair conformation where the equator where the equatorial method is on equational position the reason is when you have a substitute in an axial position there are closed approaches from the hydrogens or other substances located at the axial positions two carbon away usually we call those as one three dioxide interactions if you look at this chair conformation we can see here if you have let's say r is the methyl group so this methyl group can see these two hydrogen axial hydrogens pretty close to each other this is type of straining interactions and if you look at the ball stick model you can see even pretty close the methyl group hydrogen is very close to these two axial hydrogens and that interactions again we know when the atoms are getting too close to each other they rebuild each other and they interact with clustering interactions and because of this straining interaction it makes this ring unstable and how ministering interaction can you see this methyl group can interact with one of this hydrogen this is one axial interactions and here you have another axial interactions we call one three diaxial interactions so those are kind of two carbons away you can see it has a one two carbonyl these hydrogens axial hydrogen one two again this is also two carbon number excel hydrogen so this type of steric interaction with one three diaxial interactions and each repulsions each of these repulsion is 3.6 kilojoules if r is the methyl group if r is the methyl group then it is 3.6 kilojoules if r is an ethyl group then it would be around 4.0 if r is a tertiary butyl group then it would be even bigger number it will be 11 point something kilojoules per mole that means if you have r larger the r group and if it is on the actual position the ring will become more and more unstable so if you look at here in this table what you can see here you have a y group if y is fluorine then the diox one diaxial interaction is 0.5 if you have chlorine is 1.0 bromine 1.0 if you have methyl group 3.8 kilojoules or 3.6 sometimes people consider that if an ethyl group is a 4.0 so if you have ethyl group again one diaxial interaction is 4.0 if you have tertiary butyl isopropyl group this is 4.6 so if you have tertiary butyl group that is 11.4 kilojoules per mole so if you have a larger and larger substituents here on carbon number one and axial opposition so then the one three di dioxin interactions um cost more energy cause more strain energy so here is one of the problem that we're going to calculate how much energy involved when you are considering one two diaxial interactions and other type of interactions so if you look at here one two dimethyl cyclohexane we want to count the total strain energy for this system uh one two dimethylcycloxin then we'll compare with the say uh transport to diametrocycloxin and then we will see which one is most stable right which chair conformation would be most stable so let's first begin with the says one to dimethyl cycle hexane and we want to see how much each of the chair confirmation um that the total strain energy how much uh total strain energy for for each of the chair conformations so here we have one two diameter cycle accent and this is the chair confirmations of um the first chair confirmations you can see this is a methyl up and this is equatorial up so both are up means they are um they're ceased to each other and then if you just kind of draw the mirror image of these then it will be another chair conformation this is a ring flip chair conformations so this methyl group this methyl group now is turning into an equatorial up and this this was equatorial up this should be actually up so again these are again the ring flip chair conformation of this one now if you want to count the total energy involved here we see one methyl group here so one methyl group can see two diaxial interactions there are two carbon airways so one two so one hydrogen is right here and here's one two another hydrogen right there so these two hydrogens would be close to this methyl group and each of those interactions cost 3.6 the total of 7.3 kilojoules that's the diaxial interaction so 2 times 3.6 is around 7.2 kilojoules per mole we call wanted diaxial interactions another type of interactions one can see here this methyl group this is actually up and incubatory and this methyl group is equatorial up so when you have axial liquidual combinations and they can have a gauche interactions same way here if you look at this chair ring flip chair conformation you can see this methyl group is pointing axially then this methyl group can have one three diaxial interactions such as one two so you can see these hydrogens close right here on this carbon and on this carbon there's another hydrogens here so you can see this methyl group will have seven point two kilojoules uh which is coming from one three diaxial interactions it also have axial and equatorial methyl group so they there will be gauss interaction so that is 3.6 kilojoules per mole and in order to look at it whether they have a gauche interaction or not one can draw the neumann projections if you look at along this axis you can draw the newman projection which would look like this where you said this methyl group and this methyl group they're 60 degree apart from each other so dietary angle is close to 60 degree this is the gauss interactions when you have a gauss interaction this is around 3.6 kilojoules per mole same way if you look at this way then you will have this metal this metal uh 60 degree apart from each other uh this is required to gauss metal so this is a gauss interaction again 3.6 kilojoules so if you calculate the total amount here for this year confirmation this is 7.3 plus 3.6 which is add up to 10.9 kilojoules similarly if you calculate the total energy involved for this case we have 7.3 that's uh um that's the one to diesel interaction then one gauss interaction which is 3.6 so total is 10.9 kilojoules so in this case these two chair conformations they have the equal energy so it means these two will appear 50 50 in the equilibrium equilibrium mixture how about for the trans so trans 1 to dimethyl case so this is one of the chair confirmation as you see this is equitable down and this is equatorial up up and down this is kind of relating to the trans confirmations so in this case if you have a ring flip one if you do that so this will move on to this side so that's a equatorial up will turn into axial up and this is equatorial down to turn into axial down so this is the second ring uh the re so this is this ring flip chair confirmation of this one now if you count the total strain energy involved for each of this confirmation then you will see here that this is this is these two are equatorial it means there is no one to diaxial interactions this is zero kilojoules uh per mole that's at so zero kilojoules per mole that's a one to the iso interaction because there's no axial so only thing that you can see here these two methyl grouping neutral positions so those can be uh consider that those are gauss metals so you can have 3.6 kilojoule per mole in this case you can see one methyl group here axial positions one methyl group this for this methyl group you'll have two this way one and two which is seven point three kilojoules diet diet one through dioxin interactions for this methyl group you can see this down methyl the hydrogen this down hydrogen that also gives 7.3 kilojoules and again you can look at here you can see in the neumann position these two methyl groups are close to each other this methyl group they're entered to each other so there's no gauss interaction for this case so if you count the total amount of energy strain energy involved for this ring so we have zero kilojoules coming from one two diaxial interactions and three point six coming from gauss's interactions the total is 3.6 and for this chair confirmation we have 14.6 because of this two the four diaxial interaction you can see two for this methyl group two for this fourteen point six plus zero kilojoules so no gauss interaction so total of fourteen point six kilojoules so we can see here between these two uh chair confirmation the first shared confirmation is most stable check confirmation which is 3.6 kilojoule and if you see here these two methyl group there in equatorial position so whenever the sub the substituents are in equatorial position usually that type of conformations are stable conformations compared to the axial substituents all right just remember that so now if you see here we have the wiener here which is this one 3.6 kilojoules and if we go back right here for the cis case we have 10.9 for both of them that means if i say one three dimensional cycle x in which chair conformation would be the stable one then you need to tell you need to show this one would be the most stable one of one three dimethylcycloxene the bottom line is here whenever you are given this type of problem if i say okay which one would be or which ear conformations would be more stable for one two disubstituted cyclohexane are one three disubstituted cyclohexanes were one for disubstituted cyclohexane so you first start with the c's then you start start with the trans and then you calculate the amount of energy involved and you should be able to tell which one would be the most stable chair conformation at the end so let's another thing that i would also like to point out here when to consider that you have gauche interactions when you have the two substituents or axial and equatorial positions on the adjacent carbon remember that if it is on carbon number one and two on the adjacent carbon if the substituents are axial and equatorial you will have gauss interactions and that is 63.6 kilojoules per mole if you have two equatorial position on again on the adjacent carbon then you will have 3.6 kilojoules per mole um that's the gaussian interactions if you have two substituents those are the points those are an axial position this is actually up and actually down then there shouldn't be any gauss interaction so you won't see a gauss interactions when you have two substituents or adjacent carbons and they are on axial positions all right so now here is another practice problem this says below are the tissue conformations of one two four diet trimethylcyclics and estimate the amount of one three dyson strain in each conformer and predict which controller is most stable so here is the problem where we would like to calculate the total strain energy for each of the chain chair conformations so first of all we want to look at the conformer a we see we have this methyl group on carbon number one which is axial methyl group carbon number two has a methyl group which is also an axial methyl group and carbon number four which is the methyl grouping on equatorial position we know when we have one methyl as an axial substituent then it gives two diaxial interact interactions and each of the dioxin interaction is 3.6 kilojoules so 2 times 3.6 kilojoules around 7.3 kilojoules or 7.2 kilojoules so here is it should be 7.2 if we have for this methyl group again 2 times 3.6 this is around 7.3 kilojoules and for carbon number 4 this is an equatorial position so there shouldn't be any diagonal interaction involved so total amount of energy 7.3 7.3 plus 0 which is 14.6 kilojoules per mole next we want to calculate the total energy for this one so if you look at here we see we have carbon number one which is the equatorial carbon number two which is also an equatorial state means there is no diesel interaction for this to methyl group but we have carbon number four which is here is the axial position and that is having two times 3.6 kilojoules per mole and that is 7.3 but if you look at these two carbon number one and two those two are adjacent carbon and those adjacent carbon has two equatorial methyl group if there is two equatorial methyl carbon adjacent carbon then there would be a gauss interaction which is 3.6 so 3.6 plus 7.3 that is 10.9 remember here for the conformer a we have one and two this these two methyl groupers are just in carbon but they are on axial positions when they are on axial position then there shouldn't be any gauss interaction this is why uh we haven't seen any gauss interaction involved here so between these two ring so confirmer a has 14.6 kilojoule per mole conformer b has 10.9 kilojoules per mole so converter b is the winner and this would be the most stable ring now when you have a die substitute recycle alkanes then if you would like to find out which one which chair confirmation would be the most double chair conformation then you should start with the cease disubstituted compound then draw the chair conformations and ring flip share confirmation of that and then figured out which one would be more stable then compare with the trendshare confirmations and finally should be able to tell which which uh chair confirmation will be the winner so two dice dye substituted cycle accents could be like one two disubstituted cyclohexane hexane or could be one three disubstituted cycle accident or could be one for disubstitute to cycle accident so let's see let's see uh confirmations of one for disubstituted cycle accents which one which confirmation share confirmation of one for disaster cycle xn would be more stable the c is one or trans one if it disease which particular system confirmation would be more stable so what we are going to do to solve this type of problems we will first begin with uh the says conformations so here we can see this is equatorial down and actually downs both are down means they are equatorial they are cease so this is the c dimethyl uh one forces seize one for diameter cyclohexane and this is going down this is going up it means this is trans one for disubstituted cyclohexane so now what we're going to do we're going to compare the confirmation for this one and then we're going to compare that to three flip confirmations uh with this one and then we'll finally see which one is the winner so first let's figure it out here so in this case what we see we have one methyl group in equatorial position and this is the method that's axial position this is why we call cs and this is two are opposite this is why we call uh trans one two one four dimethyl uh cyclohexane now first we will start with the caesar isomers so seize isomers is this is our original ring as you see here this methyl group in a quantum position this methyl group in axial positions if you if you push this one up pull this one down then you will have the mirror image of that original ring chair confirmations and then you can see this is actually down that should be equatorial down and this was initially equitable down that should return into axial methyl group down now what can you see here you see here without even calculating you can tell he has one methyl group axial one methyl of a critical one methyl grad cell one methyl could be neutral so the both will have the same amount of energy all right then if you compare with the trans isomers so here is the chair conformation of the trends right here you can see where these two methylcarbonic visual positions and here we have two methyl group on axial position so what we know you can if you want you can calculate the total strain energy for this one so this one would have would be zero but this one you can see each methyl group will have two dioxygen interaction which is 7.3 and this one is the 7.3 so total would be 14.3 right and this would be zero so this is most stable share confirmation so this is zero but if you compare with the says says you still have one methyl group excellent position still have one methyl carboxyl position so it means you will have 7.3 kilojoules per mole that would be the energy strain energy for both of these 7.3 7.3 but the for trends we have this is zero and this is 14.6 so if i just say which confirmations of one for disubstituted cycle x and would be most stable then the answer would be this one because this is this was the clear winner for those among four share confirmations needed zero energy involved so this is that take us for our next lecture problem here this is the practice problem is to draw the most stable chair conformations of one target b12 three methylcyclonic sand again uh this is a disubstituted cyclonic sin and this is one three so uh it means you need to start with c's then you check the trends and then you compare which one has the highest energy and which one has the lowest and least energy so the least energy will be the winner if you want you can pause the video and then you can try to answer these questions on your own then you can then you can check the answer that i'm going to show you in a second so so first of all as i said we'll start with the says and it's a 1-3 die substituted case so what you can see here this is we this is where we put the the start button neutral position and this is a methyl group or neutral position both are equatorially down it means they are says if you do the ring flip what you can see in the ring flick position this carbon this equatorial on so we moved the ring one carbon clockwise these directions so this is turning into an axially down it was equatorial down the equatorial will turn into axial so excel down and this was equitable down this is turning into the axial down so now we can see here so when you have these two methyl group in your on negative position usually there is no diaxial interactions and they're not adjacent to each other so there is no gauss interaction so the energy for this ring will be zero but for this one when you have a turbutrial group on the axial position it cost more energy it is 11.6 or 11.5 kilojoules per mole uh for each of the multi-directional interactions so you have two of them um so you have like one with these hydrogens another with this methyl group which is much much uh more energy so you have methyl group on the axial position it can also give you one three dioxin interaction so this is very light stable ring so for the between these two this is clearly a winner case now if we consider for the trans one so for the trans case we can see this is equator up and this is methyl is actually down and in this case what we see this is the ring flip one where is the metal tart b12 group turning into actually up and this is turned into metal down so now we can see here between this two chair conformation the larger group the bulky group tertiary beautiful gibranicular position and that is a stable ring because then you will have a lace strain energy here this methyl group is smaller than tert-butyl group so it only give you one two diaxial interactions uh 3.6 kilojoules per mole but when you have the third butyl group on excel position it gives 11.5 closures per mole that that's a kind of one three diaxial interaction energy so this ring is going to be less stable again in future remember when you have a bulky group always try to put it on a neutral position because that caused less energy so between this two ring what we can see here so we have a methyl group on axial position so at least you will have 7.3 kilojoules per mole this is the winner ring but if you compare with the cis we saw here this is this has zero energy involved because those two are in one equatorial position so if you compare those four share confirmation this would be the winner more stable share confirmations of one three disubstituted cyclohexane so this is the most stable chair conformations so that's all for this video you