This video was very relieving not gonna lie. This question had gotten stuck in my head from the moment I saw it. And in that comment section, I guess I made a mistake of calling a 'centre' of symmetry to be a 'plane' of symmetry for some molecules. ('cause I inclusively called all of them to have a plane of symmetry) Thanks a lot making me think this much.
Thanks a lot, Sir. No one taught me this in my organic classes; it was there in the book, but I could not understand by reading it. You helped me a ton :)
Awesome! This is indeed one of those topics that a lot of instructors assume is “obvious” but my experience shows that it’s anything but, so I made a video with a few practice questions. I’m glad it helped you.
Hello ! In 0:54 , you merely just did it's mirror image as if it is a picture . Same goes for 1:02 . But at 1:19 , you made it like you overturned . One group was lying upside and another down , it changes . So while finding enentiomer of one optically active compound , how we will take the mirror image ?
All of those are mirror images. There are three ways how you can reflect your molecule. If the molecule is chiral and has a non-superimposable mirror image either works b/c all mirror images are the same between themselves. I also have more tutorials on my website on stereochemistry, you might wanna check those out.
Sir what actually is optical isomerism? Is it interaction with plane polarized light? Or is it just that a molecule [an optically active one] has a non superimposable mirror image? Is a chiral centre a necessary condition for optical isomerism? And can we just rotate the molecule in the plane of the paper?
Sir in my skol they never thought about rotation ,so I thought opt A ,, idk abt it we just take the mirror image and check if it is superimposable or not ,btw tqsm sirr !!
My doubt is dumb, but i want to know why an optically active compound has only one mirror image. Why cant there be more than one non superimposable mirror image to it? Regardless of where you place the mirror, you always have a Dextro and a corresponding leavo form, not more than the two possibilities...
@@VictortheOrganicChemistryTutorone, but whatif i place the mirror at some other angle? For a molecule, if i keep the mirror at A, and the next time I keep it at B, the images of the molecule in the mirror in the two cases, are identical. This , although dumb, is giving me a hard time. Especially for compounds with multiple rings it's hard to imagine..
That's the point, the *are* superimposable. Those dashes and wedges are meaningless in this example. You can only have non-superimposable mirror images for chiral objects, this molecule is *not* chiral. As the molecules are not stationary, you should consider possible conformations as well.
@@VictortheOrganicChemistryTutor Thank you for your answer. Should we then check if the molecule is not chiral first before creating a mirror image? I can't tell when we have to check for such conformations and when we don't have to consider them. Also, why are those dashes and wedges meaningless? I imagined the molecule as being made of sticks of the structure shown, and the orientation of D and its mirror image were definitely different.! If you considered the possibility that methyl groups and hydride groups are rotating, then those dashes and wedges would be meaningless... but I don't know why you suddenly consider such aspects in D. Thanks always for your help!
Unless you're specifically told to compare conformations, we have to take into consideration movement around the single bonds and all possible conformations. Using dashes and wedges on non-chiral atoms is a common trick. This topic is a bit more involved than what I can answer in the short comment though. Stereochemistry is a complex topic and questions like this keep popping up, so I think I should make a dedicated video on when dashes and wedges are meaningless or not or maybe talk about it during one of my office hours. The short answer here is since those carbons are not chiral *and* we have free 360° rotation around the single bonds, those dashes and wedges are meaningless. If your molecule has rotational restrictions (for instance, in cycles), dashes and wedges are going to be important even on achiral atoms.
This video was very relieving not gonna lie.
This question had gotten stuck in my head from the moment I saw it.
And in that comment section, I guess I made a mistake of calling a 'centre' of symmetry to be a 'plane' of symmetry for some molecules. ('cause I inclusively called all of them to have a plane of symmetry)
Thanks a lot making me think this much.
Center or an inversion point, yeah, plane, not so much, hence my original comment under that post in the community.
Thanks a lot, Sir. No one taught me this in my organic classes; it was there in the book, but I could not understand by reading it. You helped me a ton :)
Awesome! This is indeed one of those topics that a lot of instructors assume is “obvious” but my experience shows that it’s anything but, so I made a video with a few practice questions. I’m glad it helped you.
Hello ! In 0:54 , you merely just did it's mirror image as if it is a picture . Same goes for 1:02 . But at 1:19 , you made it like you overturned . One group was lying upside and another down , it changes . So while finding enentiomer of one optically active compound , how we will take the mirror image ?
All of those are mirror images. There are three ways how you can reflect your molecule. If the molecule is chiral and has a non-superimposable mirror image either works b/c all mirror images are the same between themselves. I also have more tutorials on my website on stereochemistry, you might wanna check those out.
Sir what actually is optical isomerism? Is it interaction with plane polarized light? Or is it just that a molecule [an optically active one] has a non superimposable mirror image? Is a chiral centre a necessary condition for optical isomerism? And can we just rotate the molecule in the plane of the paper?
Watch my other tutorials on stereochemistry.
Yay so i guess i was right answering this question yesterday in the community section.
Thanks a lot sir. It helps me a lot.
Most welcome!
Sir in my skol they never thought about rotation ,so I thought opt A ,, idk abt it we just take the mirror image and check if it is superimposable or not ,btw tqsm sirr !!
This is a very incorrect approach as molecules are not stationary or "stuck" in space, nor molecules care how you draw them.
@@VictortheOrganicChemistryTutor oh oke sir now I got it ,, tq for your amazing video
My doubt is dumb, but i want to know why an optically active compound has only one mirror image. Why cant there be more than one non superimposable mirror image to it? Regardless of where you place the mirror, you always have a Dextro and a corresponding leavo form, not more than the two possibilities...
Grab a mirror and look at yourself. How many mirror images of yourself do you see?
@@VictortheOrganicChemistryTutorone, but whatif i place the mirror at some other angle? For a molecule, if i keep the mirror at A, and the next time I keep it at B, the images of the molecule in the mirror in the two cases, are identical. This , although dumb, is giving me a hard time. Especially for compounds with multiple rings it's hard to imagine..
I didn't understand the part on D -- if you used the normal method of making a mirror image, it's not possible to superimpose them right?
That's the point, the *are* superimposable. Those dashes and wedges are meaningless in this example. You can only have non-superimposable mirror images for chiral objects, this molecule is *not* chiral. As the molecules are not stationary, you should consider possible conformations as well.
@@VictortheOrganicChemistryTutor Thank you for your answer. Should we then check if the molecule is not chiral first before creating a mirror image? I can't tell when we have to check for such conformations and when we don't have to consider them.
Also, why are those dashes and wedges meaningless? I imagined the molecule as being made of sticks of the structure shown, and the orientation of D and its mirror image were definitely different.! If you considered the possibility that methyl groups and hydride groups are rotating, then those dashes and wedges would be meaningless... but I don't know why you suddenly consider such aspects in D.
Thanks always for your help!
Unless you're specifically told to compare conformations, we have to take into consideration movement around the single bonds and all possible conformations. Using dashes and wedges on non-chiral atoms is a common trick.
This topic is a bit more involved than what I can answer in the short comment though. Stereochemistry is a complex topic and questions like this keep popping up, so I think I should make a dedicated video on when dashes and wedges are meaningless or not or maybe talk about it during one of my office hours.
The short answer here is since those carbons are not chiral *and* we have free 360° rotation around the single bonds, those dashes and wedges are meaningless. If your molecule has rotational restrictions (for instance, in cycles), dashes and wedges are going to be important even on achiral atoms.
@@VictortheOrganicChemistryTutor Why are the carbons on example D not chiral? Aren't there 4 different groups attached to it?
@@shad0wstrife what are the groups that those C’s are attached to?
Sir you are best.would u plz recommend me a nice freely available pdf book to delve into the details of sterochemistry.
I have a playlist and things are also organized in a slightly different way on my website. You can go with either sequence.
How do you make a mirror image out of the paper plane
Like if both molecules are standing in front of each other. Just like you *you* stood in front of a mirror on a wall.
are we listening to a robot or live person or voice changing app
Don't you think if I were using a voice app of some sort I would've chosen a more "classic sounding" voice? 😉