@YusufHasan can't we find gi like we do in oxime

?

@Yuvraj r u talking bout johan's question?

@YusufHasan yes

I mean like we assign cis/Trans in oxime

With respect to oh position

Yes you can, that's what I am doing ultimately if u carefully read what I have done above, but ultimately the issue is that, if you start assigning all possible cis trans combinations, and then eliminate the redundant possibilities, you will end up making 16 cases like Guru Vishnu said

And that will be a long method. So any method which allows us to reduce the amount of structures you have to make/list out is surely better. The way I have stated, you will get all 4 distinct stereoisomers of the molecule in one go, without having to list repetitions,and then eliminate

@YusufHasan if you do not mine can we count using my own method?

I mean can I share my thought?

yeah sure, go ahead

In an exam where time matters it's good to follow the shortest route. I really enjoyed making 16 structures and removing the repeating and mirror image structures. I think there's a game similar to this activity. But I don't know its name.

OK so we have group R and OH at each oxime

So OH can be above or below the plane

Wait a minute...how are you sending things above or below the plane?

R is in plane

And oh is above and below the plane?

And lone pair is same

Above and below?

I guess you and I have a different plane in mind...I am thinking about the molecular plane, as in the plane containing the double bond, R, OH and lone pair. Is this the plane you are referring to?

from your statement, it seems that you are referring to the plane perpendicular to the molecular plane, then OH above and below makes sense. Am I following you correctly?

@Yuvraj did you get what I was saying?

wait a seccond

yes imean the same

@YusufHasan

ok let's continue using your convention; what next?

so R N are in a same plane and it is the oh and lone pair which arrange up and down right?

yeah

so each N has two isomer cis and trans

yep

total 8 isomer of all the other N

and now I am confused how do I eliminate the one which are similar

Why? if each N has 2 configs associated with it(cis and trans) like you said, then logically total isomers should come out to be 2*2*2*2=16

Fundamental principles of counting

And you have circled back to the argument that @GuruVishnu made yesterday XD

oops sorry yes total 16

And now, you have to look at symmetry elements(Plane and centre of symmetry) and enantiomeric pairs to eliminate the repetitive cases by making out the structures

@YusufHasan I think now, I'm using a hybrid version of your method and mine. I got used to drawing structures. So, here afterwards, I'm going to fix one and vary others. Added advantage, sometimes, I can figure out optical isomers. Do you use circular permutation for such cases too?

it must be 8 then ?@YusufHasan

like with 4 centre of symmetry)

@Yuvraj No, the answer is 4.. If you are going to follow thus long path, then first draw all 16 possibilities, then, try reflecting them against a mirror plane, and see which compounds are actually the same...Also look out for meso compounds(if they exist)...I guess Guru Vishnu can give a detailed solution on this method since he has already done it this way yesterday

@YusufHasan i think you misunderstood me

@GuruVishnu No I don't really do that, altho it seems like a nice idea to probe..Wait, I will send a detailed solution of what I actually do rather than providing hints so u can judge for urself

i mean we can eliminate total 8 isomers like this

and then two plane of symmetry and meso compoud

sorry if i irritated you with silly doubts i should tr by my own!

@YusufHasan

@Yuvraj Ohh that way..Yes it maybe right, so what answer are you getting?

four

Well that's fine then...so you managed to visualize all this without drawing, right?

yes

i read your and vishu chat two time and i got the idea what you were thinking

@YusufHasan @GuruVishnu a beauty for u

well, that's great if you can do that..Although I do try to avoid this elimination process, because especially in an exam scenario, there's a high possibility that you manage to miss eliminating a repetition..instead, the forward approach of directly making the distinct isomers is more easy for me

I will now clearly send the way I would do things

Don't get me wrong, ur method is absolutely fine @Yuvraj , it's just that I want you to appreciate the risk this method poses, that's all

@YusufHasan no i am insane, and it is better to have faster exam method

@Yuvraj Sorry for the over writing

Also,please ignore the 2nd OH that you see marked on N2 in the diagram