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6:44 I'm confused with this example that doesnt have chiral centers but has an internal line of symmetry. I thought meso compounds need to have two or more chiral centers?
THERE IS AN ERROR: the 1-3-cyclobutanediol is not a meso compound because it does not have any chiral centers: the C connected to the hydroxyl groups are not chiral centres because they both attached to two same group of atoms CH2-CH-0H. We can however say that it is a cis compound (both OH groups point on the same direction).
@@dysprosiumdead5078 calm down bro, professors aren’t teachers. you learn on your own, professors are just there to clean up the edges and answer any questions you have. welcome to college
@@101booie101 i just learned this so im not an expert so if what i say ends up being wrong im sorry i watched the khan academy video on meso compounds, but in fischer compounds horizontal lines = a wedge (forward) and vertical lines = dashes (away) and the 4 group is at a horizontal line so you reverse the rotation to make it away/towards the back.
At 7:15, the cyclobutane with the OHs, isn't that not a meso compoud since it doesn't have chiral centers? carbons are attached to the same thing on both sides/symmetrical. Wouldn't it just be achiral and not meso?
I'm convinced it's achiral, since it has no chiral centers and has line of symmetry. I'm very frustrated because a similar compound like this was on my exam today, where it was achiral and symmetrical. However, I could only use the terms "not meso (identical)" to describe it, but the other comparable compound was a diastereomer of it.
Awww man. I get you. These are super confusing. I had a test yesterday and i was so glad my professor didnt tell us to identify meso, constitutional, conformers, diastereomers or enantiomers. I think i could easily tell if something is meso now. As long as you have at least 2 chiral carbon, and they re symmetrical then they re meso
The one on the bottom left of the of the worked examples is not a chiral molecule. Both carbons connected to the carbons with the OHs have the exact same connectivity rendering the those carbons achrial.
Noticed that also. To be meso the molecule must be achiral but still have at least one chiral center, which there are no chiral centers in this molecule. While this molecule is achiral/ has a sigma plane, it has no chiral centers. It is not meso. I could very well be wrong, but I don’t see any chiral centers on the molecule.
He was doing something called the flip rule. The horizontal lines are wedges coming out at you, so if the fourth priority attachment is a horizontal line, you flip the R/S.
For Fischer projections, the horizontal lines are assumed to be wedges (coming out of the page towards you), and the 2 vertical lines at the top and bottom are assumed to be dashed lines (going into the page away from you). So when figuring out R vs S, when you order the priority of the substituent groups and H is NOT on a dashed line (meaning away from you), you flip the stereochemistry from R to S (or vice versa).
And for these specific examples in the video, since H was always horizontal (and therefore a wedge coming out of the page towards you; NOT on a dashed line), he had to flip the stereochemistry for all of them.
At 3:45, couldn't the bond going through the dotted red line rotate 180 degrees, causing the compound to have symmetry? EDIT: Nevermind, I think I understand now. The way I learned these projections are *with* the wedges, but after drawing the projections with the assumed wedges and rotating the bond 180 degrees, it makes sense.
At 5:15 the first compound, 2,4-Dibromopentane can have three stereoisomers, which means it can have (2S, 4S)-, (2R, 4R)-, and (2R, 4S)- configuration. Only (2R, 4S) configuration is meso compound but others are enantiomers because they are not superimporsable in mirror image each other. If I am wrong with this, please let me know what I am wrong.
Sarah Gloria if you think so try numbering the substituents 1 through 4. You are gunna get stuck at which one is the second and which is the third substituent.
The cyclobutane ring with 2 OH is not a meso compound. Because the requirement for meso compound is to have at least 2 chiral center. Where chiral center means 4 different group attached. Here the 2 Carbons with OH only have 3 groups attached.
Nah they have 4 groups. Both carbons attach to 2 carbons (each other and the carbon on the opposite corner) the OH group but also a H which is not shown. So that’s four different groups.
David MacPherson if you look at the connection of carbon-carbon to the left and to the right. They are the same group. C-C-C-OH connection when you go to the left and also to the right. How are they 2 different group then?
Alexandra Long you need to look at more than just the carbons next to each. You’ve got COH, an OH and CH2 group connected to each chiral centre carbon. Plus the H not shown.
David MacPherson OH is 1 group, H is 2nd group, 3rd group is C-C-OH. Please find me the 4th group please! If you are saying the CCOH group from the right is different from the left then you need to get count them again. They are exactly the same. Meso compounds have 2 chiral carbons and 4 DIFFERENT groups. FIND ME 4 EXACTLY DIFFERENT GROUP OKAY?
Alexandra Long OMG SORRY!!!! I was looking at the wrong molecule, I thought you meant the one middle left, not bottom left. Yes you are absolutely correct, definitely not a chiral centre. There’s no way it can be. He’s made a mistake. Sorry for the confusion!
Some of these seem like they’re superimposable if you were to flip it completely. For lack of a better term, are just not allowed to flip the whole molecule over?
As far as I know from other examples Ive done, it might be that when the fourth substituent ( in this case H) is on a wedge, then the orientation is flipped. Not sure if that might help.
I remember it by VCR-Vertical Clockwise R, which means that if hydrogen is vertical and if the rotation is clockwise then its R. HCS- horizontal clockwise S.
@@kartikarora3521isn't molecular dissymmetry a criterion for optical activity? bruh... I am sharing a drive link just observe these images and tell me can't they have pos along the plane of paper drive.google.com/drive/folders/1_NlA_3y4J7CKfER3DckMj5AeichUE1be?usp=sharing and ignore the chair shape as we take benzene as planar at this level?
@@kartikarora3521 to be optically active a molecule must have molecular dissymmetry !!! and this molecule posses a pos on the plane of paper so it is not dissymmetric...so it is optically inactive??
Hi! thanks for the video! I have a question please about the last Example [ the (2R,3S)-(-)-2,3-pentanediol ] : I see there is no line of symmetry but the chiral centers are still R and S, like the butanediol you showed at the beginning of the video, so why is one optical active and the other isn't? Let's say I have 2 compounds: (2R)-bromobutane and (2R)-bromo pentane; if I measure the optical activity of the two, would I see any difference?
Hi there In the first example of butanediol, the 2 chiral centres 'cancel' each other since they are exact mirror images of each other, giving a net optically inactive compound. However, in the last example, one of the chiral centres has -CH3 and the other has -C2H5. Because of this the chirality of both the chiral carbons are different, hence giving a net optically active compound.
Thanksss. I have a question, for 2,5-dibromohexane, with one Bromo out of the page and one in, why is is this the meso compound although there isn’t a plane of symmetry?
It is a racemic mixture because one chiral centre is in S configuration and the other is in R configuration. If you search the enantiomer of this molecule, you will have the same molecule but flipped.
Can I say if both chiral centeres has same direction configuration like R-R and S-S . they are not meso compounds. if R-S is possible, it is always meso compound?
bro canu make a vid on how to convert newman projections to fischer and to bondline like that means can u make a vid on how to interconvert a structure to different projections
hey i ve got a question. Shouldnt the inversion centre be also enough of a symmetry element for the compounds to be meso? Dont really understand why in the 2nd example you said the two molecules to be enantiomers (?) thanks for the help in advance :)
The reason why the compounds in the second examples are enantiomers is because they both show asymmetry and all the chiral centres change which makes them non superimposable.But for the first question both the compounds are symmetrical making it Meso .Hope you got it 🙂
If number 4 substituent is on horizontal or dash posting the absolute configuration of that chiral carbon is reversed. Its just a trick that my elder taught me
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Y'all keep in mind these are Fischer projections. The horizontal lines are assumed to be wedges, coming out of the page towards you.
Omg thank you I was so confused as to why we had to reverse it
Thank you so much for your input. Wasn't sure why he kept saying we gotta reverse it
glad i wasnt the only one who was like "wtf" i was about to click off because i did not understand it. tysm.
And because H isn’t behind(on dashed line) we must reverse the S/R to get the correct answer. Thank you for clearing it up
omg...thanks a lott bro, it was really helpful! most teachers usually dont tell these things and just tell the process...without telling the reason
Thank you! I was about to cry over this! Can’t believe how simple it is!
Thank you very much for this ... imagine sitting in class for 40 minutes without anything but here only ten minutes
I understand completely.
Seriously, I don't know how I'd survive lectures without this
6:44 I'm confused with this example that doesnt have chiral centers but has an internal line of symmetry. I thought meso compounds need to have two or more chiral centers?
Exactly my thoughts
this guy gets me every time
THERE IS AN ERROR: the 1-3-cyclobutanediol is not a meso compound because it does not have any chiral centers: the C connected to the hydroxyl groups are not chiral centres because they both attached to two same group of atoms CH2-CH-0H. We can however say that it is a cis compound (both OH groups point on the same direction).
i was actually just about to type this😅
bro im going insane over the fact that im in a university and i have to learn through yt and constantly look out for these. fvck education
@@dysprosiumdead5078 calm down bro, professors aren’t teachers. you learn on your own, professors are just there to clean up the edges and answer any questions you have. welcome to college
@@OreoWaffles44 in low fields like yours for sure.
@@dysprosiumdead5078 biochem’s a low field?
It is really awesome sir...u humbly simplified the complexity of meso compounds..thanks a lot 🥰🥰🤩
1:29 why do we need to reverse it?
Cuz the H must be always away from us but the H is towards us in the fischer diagram. Hence, we need to reverse it
@@vennaiz2198 How do you know the H is towards us? He didn't use the thick triangle or line bonds?
@@101booie101 i just learned this so im not an expert so if what i say ends up being wrong im sorry i watched the khan academy video on meso compounds, but in fischer compounds horizontal lines = a wedge (forward) and vertical lines = dashes (away) and the 4 group is at a horizontal line so you reverse the rotation to make it away/towards the back.
@@katherinewolinski9245 Cause the H is on a horizontal
These are Fischer Projections. It is assumed in this format that the HORIZONTAL lines are WEDGES, coming "out of the page".
At 7:15, the cyclobutane with the OHs, isn't that not a meso compoud since it doesn't have chiral centers? carbons are attached to the same thing on both sides/symmetrical. Wouldn't it just be achiral and not meso?
Peter Cohn omg thank you! Someone agrees with me. Please help me convince the other dude. Go to my comments and you'll see my frustrations :)
I'm convinced it's achiral, since it has no chiral centers and has line of symmetry. I'm very frustrated because a similar compound like this was on my exam today, where it was achiral and symmetrical. However, I could only use the terms "not meso (identical)" to describe it, but the other comparable compound was a diastereomer of it.
Awww man. I get you. These are super confusing. I had a test yesterday and i was so glad my professor didnt tell us to identify meso, constitutional, conformers, diastereomers or enantiomers. I think i could easily tell if something is meso now. As long as you have at least 2 chiral carbon, and they re symmetrical then they re meso
This video explained to me what several 2 hour videos couldn't properly.
The one on the bottom left of the of the worked examples is not a chiral molecule. Both carbons connected to the carbons with the OHs have the exact same connectivity rendering the those carbons achrial.
Noticed that also. To be meso the molecule must be achiral but still have at least one chiral center, which there are no chiral centers in this molecule. While this molecule is achiral/ has a sigma plane, it has no chiral centers. It is not meso. I could very well be wrong, but I don’t see any chiral centers on the molecule.
Was looking for this comment!!
Thank you !
Thanks!
Why did he reverse R/S in the beginning with the lefthand compound?
He was doing something called the flip rule. The horizontal lines are wedges coming out at you, so if the fourth priority attachment is a horizontal line, you flip the R/S.
@@GPStan2157 And this is because we typically want the lowest priority atom facing away from us (on a dashed line) not towards us.
How we can compound by assigning 2R, 3S configuration ?
Thanks!
why did you flipped the R/ S group at about 1:24 ? You didn't draw any wedge/dashed group but you still flipped it ? Can you please explain that ?
For Fischer projections, the horizontal lines are assumed to be wedges (coming out of the page towards you), and the 2 vertical lines at the top and bottom are assumed to be dashed lines (going into the page away from you). So when figuring out R vs S, when you order the priority of the substituent groups and H is NOT on a dashed line (meaning away from you), you flip the stereochemistry from R to S (or vice versa).
And for these specific examples in the video, since H was always horizontal (and therefore a wedge coming out of the page towards you; NOT on a dashed line), he had to flip the stereochemistry for all of them.
Thank you so much for these videos ❤ they actually helped me get good grades 😍❤😘
At 3:45, couldn't the bond going through the dotted red line rotate 180 degrees, causing the compound to have symmetry?
EDIT: Nevermind, I think I understand now. The way I learned these projections are *with* the wedges, but after drawing the projections with the assumed wedges and rotating the bond 180 degrees, it makes sense.
There is something wrong with the R and S configuration.R is for clockwise direction band s for anti clockwise direction
You constantly save my life every finals
At the 3:07 molecule, there's an inversion center. It shouldn't be chiral then, right?
You are right
At 5:15 the first compound, 2,4-Dibromopentane can have three stereoisomers, which means it can have (2S, 4S)-, (2R, 4R)-, and (2R, 4S)- configuration.
Only (2R, 4S) configuration is meso compound but others are enantiomers because they are not superimporsable in mirror image each other. If I am wrong with this, please let me know what I am wrong.
Yea even I have the same doubt. I think when he drew the bond, he was supposed to draw both coming in our out of the plane.
Thank you so much tomorrow is my Medical Entrance
2:45 but they dont have restricted rotation.. so HO-H can be written as H-OH right?
Isn't the last example achiral because it can have a vertical plane of symmetry?
I'm too wondering the same!
7 minutes of my life well spent
In short. If there is a line of simmetry and chiral centers it is a mesocompound.
1:49 Why do we have to reverse it?
Rule 😊
how is the 1,3-cyclobutanol a meso compound when it doesnt seem to have chiral carbons?
you're right, its not meso because it has no stereocenters to begin with
Sarah Gloria if you think so try numbering the substituents 1 through 4. You are gunna get stuck at which one is the second and which is the third substituent.
j e sorry i was looking at the 1,2 cyclobutane diol you are right .
Actually you have to look at its trans and cis forms in order to see that it actually has one and two planes of symmetry respectively.
However i would say it is correct when drawn without stereochemistry that it has no chiral centers.
@ 1:17 why do we reverse it? why isn't it "R"?
hello there, tell me why did we reverse it i'm struggling T-T
@@aalaanassar4332 I forgot how to do all this lmao I ended up getting an A tho lol
@@doinitlive3015 I see, well done mate T-T
Bro H is above the plane ,4th group should be placed away from plane so when H is on the plane or above the plane R becomes S and S becomes R
That was helpful I used to find it confusing first but this video has disproven my point🙏🏻😁
The cyclobutane ring with 2 OH is not a meso compound. Because the requirement for meso compound is to have at least 2 chiral center. Where chiral center means 4 different group attached. Here the 2 Carbons with OH only have 3 groups attached.
Nah they have 4 groups. Both carbons attach to 2 carbons (each other and the carbon on the opposite corner) the OH group but also a H which is not shown. So that’s four different groups.
David MacPherson if you look at the connection of carbon-carbon to the left and to the right. They are the same group. C-C-C-OH connection when you go to the left and also to the right. How are they 2 different group then?
Alexandra Long you need to look at more than just the carbons next to each. You’ve got COH, an OH and CH2 group connected to each chiral centre carbon. Plus the H not shown.
David MacPherson OH is 1 group, H is 2nd group, 3rd group is C-C-OH. Please find me the 4th group please! If you are saying the CCOH group from the right is different from the left then you need to get count them again. They are exactly the same. Meso compounds have 2 chiral carbons and 4 DIFFERENT groups. FIND ME 4 EXACTLY DIFFERENT GROUP OKAY?
Alexandra Long OMG SORRY!!!! I was looking at the wrong molecule, I thought you meant the one middle left, not bottom left. Yes you are absolutely correct, definitely not a chiral centre. There’s no way it can be. He’s made a mistake. Sorry for the confusion!
Bravo sir🥰....loved it.. clearly mentioned
Some of these seem like they’re superimposable if you were to flip it completely. For lack of a better term, are just not allowed to flip the whole molecule over?
Doesn't the 2-hexanol have symmetry inline with the paper?
Cyclobutan-1,3-diol doesn't have any chiral centers, hence no internal compensation. So, I don't think it can possibly be meso.
Yup me too ...resemblin with you stickin with the same question......
That just really cleared my head, thanks brother:)
Great video! But why do you reverse it from S to R in 1.13? I dont understand why it apears to be the other first en then you switch it?
As far as I know from other examples Ive done, it might be that when the fourth substituent ( in this case H) is on a wedge, then the orientation is flipped. Not sure if that might help.
In an easy language, whenever the 4th priority group lies on a horizontal direction/line, the config gets reversed
I remember it by VCR-Vertical Clockwise R, which means that if hydrogen is vertical and if the rotation is clockwise then its R. HCS- horizontal clockwise S.
Thanks a lot mate, was really helpful - best wishes from Denmark :)
the cis cyclobutane with the two alcohols, even though it has a plane of symmetry wouldn't it NOT be mso because it has no chiral centers?
thanks for the examples to clarify.
at 6:12 this compound has POS present then why is it optically active?
@@kartikarora3521 cutting from the plane of paper will result in two identical molecules!??
@@kartikarora3521 then just flip one of the molecules created after cutting and then you will be able to make these two co-inside with each other ?
@@kartikarora3521isn't molecular dissymmetry a criterion for optical activity? bruh... I am sharing a drive link just observe these images and tell me can't they have pos along the plane of paper drive.google.com/drive/folders/1_NlA_3y4J7CKfER3DckMj5AeichUE1be?usp=sharing and ignore the chair shape as we take benzene as planar at this level?
@@kartikarora3521 bruh... A molecule not having pos and cos is termed to have molecular dissymetry it's that simple!!!
@@kartikarora3521 to be optically active a molecule must have molecular dissymmetry !!! and this molecule posses a pos on the plane of paper so it is not dissymmetric...so it is optically inactive??
I need the structure of Meso compound, please, and where we can use this drug?
Eternally grateful, thanks a lot!
Fantastic videos as usual thank you!
Great Vid Btw!!!!!!!
Does the relation between two compound is meso or we refer to a particular compound as meso?
A particular compound
do we have to reverse them because of they are coming out of the page?
Hi! thanks for the video!
I have a question please about the last Example [ the (2R,3S)-(-)-2,3-pentanediol ] :
I see there is no line of symmetry but the chiral centers are still R and S, like the butanediol you showed at the beginning of the video, so why is one optical active and the other isn't?
Let's say I have 2 compounds: (2R)-bromobutane and (2R)-bromo pentane; if I measure the optical activity of the two, would I see any difference?
Hi there
In the first example of butanediol, the 2 chiral centres 'cancel' each other since they are exact mirror images of each other, giving a net optically inactive compound. However, in the last example, one of the chiral centres has -CH3 and the other has -C2H5. Because of this the chirality of both the chiral carbons are different, hence giving a net optically active compound.
ur r and s config too. tysm it helped me understand better :)
Is an erythro isomer same thing as a meso compound??
Thanksss.
I have a question, for 2,5-dibromohexane, with one Bromo out of the page and one in, why is is this the meso compound although there isn’t a plane of symmetry?
It is a racemic mixture because one chiral centre is in S configuration and the other is in R configuration. If you search the enantiomer of this molecule, you will have the same molecule but flipped.
Is this racemic mixture also?
thank u so much omgomg u saved my life on meso compounds xD
thanks a lot for the vid, helped a great lot
Thank you!! I'm really in love with you thanks again and again 😘♥️♥️
great explanation
Can I say if both chiral centeres has same direction configuration like R-R and S-S . they are not meso compounds. if R-S is possible, it is always meso compound?
When I apply this method, I got always true answers.
Why do you have to reverse it?
Is the last compound in last example achiral tho its not meso
bro canu make a vid on how to convert newman projections to fischer and to bondline like that means can u make a vid on how to interconvert a structure to different projections
can someone please tell me the R S config. of chiral carbons of cyclobutane-1,2-diol and how did you find it
Thank u vm. Really helpful
ilysm bbg🥰🥰🥰
thank you for your help!
thank you so much ,it was rally helpful
wait how do we know which group has a higher priority?
Thanks my bro❤
I got it .. thank you!!
which software do you use and which device is that a tablet (I am not going to start making videos)
thanks for the video
hey i ve got a question.
Shouldnt the inversion centre be also enough of a symmetry element for the compounds to be meso? Dont really understand why in the 2nd example you said the two molecules to be enantiomers (?)
thanks for the help in advance :)
The reason why the compounds in the second examples are enantiomers is because they both show asymmetry and all the chiral centres change which makes them non superimposable.But for the first question both the compounds are symmetrical making it Meso .Hope you got it 🙂
Which app you used to draw?
1; 4 Cyclohexane diol is not a mesoo compound because it has not Assymetric center which is compulsory for a meso compound
Welldone prof
Why do reverse the R and S
If number 4 substituent is on horizontal or dash posting the absolute configuration of that chiral carbon is reversed. Its just a trick that my elder taught me
I love you thx 💕💕💕💕💕
and miso is fermented soybean paste from which you can make delicious soup
So easy until test time
Thank u so much
Thanks 🙏
THANK YOU!
Thankyou!
Thanks
Sir can make a video on stability of me so compound
That’s really helpful thanks bro.
Is there a way to be in touch for future ?
I think the R and S configuration you told was wrong
amazing
Hey man you don't know what you do to me ... You are a nice guy n god sees the nice guy
Ty
What is symmetry?
ablility of the molecule to show similarity in its structure
This is in my 12th class syllabus not in college
Thank you. My chem teacher sucks
❤❤❤❤❤🥺
Why does he not explain why the R and S are reversed?
Tnaks.
Why r tutor donot teach like you
It has open up me
Puriyiramaari pesuda onnume puriyala
"chiral center" does not exist... a center can't be chiral.