00:06
Part a of this problem is to assign eoc configuration to each double bond.
00:13
This compound, legimalade b, has nine double bonds.
00:18
Let's assign eoc configuration to each one.
00:23
Do you remember how to assign eoc configuration? so if you have a double bond and three four different groups, first we had to divide the double bond.
00:42
One in the two parts then for each side we have to assign the priorities to each group using khan in gold pre -legged rules so if this side has higher priority then we put this number one then the other one is number two if this side also as number one this is number two then higher priority now then number one sign the same side therefore higher product groups in the same side therefore this is seek configuration on the other way if here number one number two and this side number one is here and number two is here now higher priority groups are on the opposite side then this is e configuration okay let's use this to assign eoc configuration to each double bone.
01:48
Let's start here.
01:50
Here the double bond and there is a hydrogen atom here.
01:56
So let's divide the double bone in half here.
02:02
So this side, this is hydrogen, the directly attached, this side, the directly attached atom is the hydrogen.
02:15
This is carbon, therefore, side has higher priority so here this is number one this is number two and this side both are the the carbon is attached to the this side also carbon this one also carbon so both attached to the carbon because it has the same priority then we have to go to the second point different so do you remember so in this side here is a carbon here also carbon but this side this carbon attached to three hydrogen this carbon is attached to two hydrogens and one nitrogen so here it hydrogen hydrogen hydrogen now you see in the second point difference this is three hydrogen this is two hydrogen and one nitrogen so nitrogen as higher atomic mass therefore it has this side has a higher priority that means this side has number one this side number two so here if you aside therefore this is number one this is number two now to one opposite side therefore this is e configuration okay let's do the next one so in this double bond again here has a hydrogen atom therefore this side has double this side of the double bond has here number one this side number two how about the other side here this carbon is related to carbon he also carbon then we had to go second point there was like before this carbon attached to tea three hydrogens but this carbon attached to one hydrogen one carbon and one oxygen so therefore this side has higher priority therefore this one number one here number two again two ones are on the opposite side therefore this is again e configuration let's go to this double bone here it has a hydrogen atom here it's assign the priorities this side is this carbon this side is a one hydrogen as a lowest atom on the other side again this is attached to the carbon this is side also carbon but this carbonated to three hydrogen this attached to oxygen so in the case so in the case like this so you have double bond here so this carbon is attached to three hydrogens and this side this carbon is attached to here one oxygen and here double bond with oxygen so it's also it also we consider has attached to two oxygens so therefore this side as higher priority this is this side is one this side is two again to one sound the opposite side therefore this is again e -configuration okay let's to the next one in this double bond it has a two hydrogen one hydrogen here and one hydrogen here so definitely the other groups has higher priority so this is one hydrogen is two again this side is one this is hydrogen is two therefore two one sound opposite side therefore this is again e configuration then this one has hydrogen here so it's assigned the priority so this side this is number one this is number two and if you decide so uh this carbon is attached to the carbon this side is also carbon so we have to go to second point different so let's go to say it's real so you do double one so double bond is like this so this side it has a carbon and it it attached to three hydrogen but this side it has a this carbon is attached to one hydrogen and this side one hydrogen it has a carbon carbon double bond so it has one carbon and double bond also we consider it's attached to another carbon so this side two carbon and one hydrogen this is three hydrogen therefore this side has higher priority therefore let's the priority so this side is number one this is number two again two once on the opposite side therefore this is again e configuration so this is let's go to next double bond so on hydrogen here one hydrogen here so it is too easy higher priority groups are on the opposite side so this is again e okay next one the hydrogen is here so unison priorities this side definitely has high priority to this group and hydrogen has low on two how about this side so this carbon is attached to this carbon on this again carbon on this so we had to go to second point difference this side carbon attached to three hydrogen but this side one hydrogen and with the double bond is to two carbon therefore this side has higher one two now to one sound the same side therefore this double bond has c configuration okay let's do the other two so this is it has hydrogen hydrogen so definitely higher product groups are on the opposite side therefore this is e the last one one hydrogen here one hydrogen here so definitely two high product groups are on the opposite side therefore this is also e configuration okay so the eight double bonds are e and only one double bond is c configuration okay let's do part b in part b problem is to assign the rr is configuration to each tetrahedral stereogenic centers so first we have to find a tetrahedl steregidless erogenic centers or chiral centers okay let's let's start from this side so this is the this is the first one why this carbon is a two different group this is knight here there is a hydrogen so it has had four different groups therefore this is a first chiral set okay what are the others so this is the second one this is the third one and here it has a hydrogen here therefore this is the fourth one this is the fifth one and this is the six one six two so so this is six chiral centers okay do you remember how to assign the r &s configuration so when you assign r &s configuration to the chiral carbon so if you have a chiral carbon like this it is two bond and it's a tetrahedral carbon one one is out of the plane one is the into the plane so when you assign the rns configuration always we move the lowest part of the group into the plane number four then we assign priorities to the other three groups so if the priorities comes one two three that means then we start for number one and go to number three so if this go like clockwise then we call this is our configuration that on the on the other way so so lowest priority group always into the plane so this is one two three now one two three go in the other way it's a counterclockwise so we assign this as its configuration so let's use this to assign r and s configuration to each chiral centers okay let's start from here so here the lowest priority group is hydrogen it's already is into the plane so now directly assigned to priorities to the other groups because this is this hydrogen the lowest priority group number four is is already is into the plane so let's assign priorities to the other groups here the carbon is attached to the nitrogen here to the carbon here the third one also carbon therefore nitrogen has higher atomic mass therefore this one has number one this one is number one so now we have to assign the number two so if you go to this side both carbon is related to the carbon this carbon so then we have to go second point difference so so if if i write here this is carbon is attached to here then this side this carbon is attached to one hydrogen one hydrogen and here oh it on this side this is carbon and this side this carbon is this carbon is attached to here nitrogen and here oxygen and the double bond also we consider as a set it attached to the oxygen therefore this side this carbon is the second this carbon is right to nitrogen oxygen oxygen but this side hydrogen hydrogen oxygen therefore this side has higher priority it goes to oxygen so therefore this side there's number one this is number two so let's write here so this is number two because you already decide number one so here this side this is number two this size is number three now write the arrow so now two three if you go in the order one two three it's going the counterclockwise therefore this one is its configuration okay let's move to the other one just let's move to this one first so it has one hydrogen into the plane so the lowest product group is already is into the into the plane so we no need to worry about that so let's assign priorities to other tree groups so the in this side this is directly attached to carbon this also carbon this also carbon okay so then we have to go to the second point difference okay so let's go let's write so if i draw here so this one is attached to see h h h so this carbon on this side it's attached to carbon and there's a hydrogen here and there's double bond so double bond also we consider another another carbon attach to another carbon so to take to one hydrogen here one carbon and double bond also we consider as a carbon so this side is carbon carbon hydrogen how about the other side so this carbon it has a one hydrogen and it also has a double board so that side carbon one hydrogen and one carbon and double another one is carbon because it's a double board now let's see so this side carbon carbon hydrogen this side carbon carbon hydrogen decide carbon hydrogen decide three hydrogen therefore this has the next next lower priority so this this group is this one is number three okay let's aside let's check which one is number two so second point difference also same two carbons and two carbon hydrogen two carbon and one hydrogen next we have to go to third point difference so we have to go this way if we go this way this carbon is attached to the one carbon and this side carbon and this side two carbons and here also carbon so it's attached to carbon this carbon and this carbon and it's one the double -borne carbon then carbon and the next one also carbon is three carbon but if you go to this side it's attached to carbon and double bond carbon another carbon one hydrogen therefore this side has carbon carbon double one then there's another double bone here this is hydrogen carbon now this is three now this side has three carbon this side two carbon and one hydrogen therefore definitely this side has has the higher priority so this side there's number two number two sorry number one this is number two therefore it's assigned the priorities here so this side number one this is this side number two and this is number three now let's draw the arrow so it's go one two three this side so what this has said clockwise the arrows going the clockwise therefore this is our configuration okay let's do quickly the other one let's go let's go to this one so it has hydrogen here always it's already the lowest priority group is into the plane it's number four then let's assign priorities to others so uh this one is this carbon is another carbon with the tree hydrogen so this has the carbon with the double moon and this carbon is attached to the oxygen so the second carbon is attached to the oxygen therefore it has a highest priority so this side has highest priority so this is so this is one and this has higher part because this is three hydrogen this is two carbon and one hydrogen therefore number two three therefore the if you draw the arrows it's going this way that means it's a counterclockwise therefore this is its configuration okay let's do the other tree let's do this one now in this one the lowest product group is out of the plane so when we do this so we have to move the lowest product group into the plane but there is another way to do we can do like this way okay let's do the this one so we need so we need to do this one so we need to this one so here the lowest part group is out of the plane so let's assume this way assume assume lowest part of group is into the plane and into the so then we can assign priorities to this one.
25:28
So definitely this is the higher priority one because the carbon is attached to the oxygen.
25:37
And this side, this side the carbon is attached to another carbon, but that carbon is attached to three carbons.
25:53
But this side it has two carbons and one hydrogen.
25:58
Therefore this side has higher priority.
26:00
So this one number number two this one is this one is number three so in the in this one so this is number one number two number three so therefore it has one two three are configuration but the but the real one the higher product group is into the plane so therefore the real one is there the we have to switch the configuration that means the real one is the other way therefore this one should be its configuration so did you see how how we did that we assume the lowest product group is into the plane then we assign the configuration so it's came with r but the real molecule the lowest product group is on is out of the plane therefore we have to switch the configuration so therefore this is s let's do this one so it also has a hydrogen atom out of out of the plane so again the lowest product group is out of the plane so we uh when we do this uh we we should have the lowest product group is into the plane so let's assume so here to one let's assume the hydrogen is sorry let's assume the hydrogen is into the plane and och3 is out of the plane now assign priorities to each side so obviously the carbon attached to oxygen has higher priority and if you go to sides this side carbon is attached to the one hydrogen and two carbons this carbon is attached to two hydrogen and one carbon therefore definitely this side has but higher priority this is number three so in this figure so this is one two three so it is clockwise therefore our configuration but the in the real molecule the oxygen is into the plane and hydrogen is out of the plane so the real molecule so we have to switch the configuration so therefore real one is s configuration okay, let's do the last one.
29:58
Here it also has the hydrogen atom here.
30:06
So the same thing, the lowest priority group is out of the plane...