You do not know how much this has helped me for my test next tuesday. I was doing so well undersatnding addition reactions to alkenes and alkynes, and all I neede to get back to was understanding the stereochemistry of them! Stereochemistry isnt my strong suit, but you seemed to have made the topic very managable.
here's a tip if you are adding syn to a double bong then for fast result just assume it on 2D and add 2 group on end and remain the configuration like H CH3 / | | ------ + KMnO4/OH- cold (B.R.) --> OH------------------OH [ optically active ] ------ | | / CH3 H { or } H H | | ------ + KMnO4/OH- cold (B.R.) --> OH------------------OH [ meso ] ------ | | / \ CH3 CH3 and on anti addition just switch one like CH3 CH3 / | | ____ + Br2/CS2 --> Br-------------------------Br [meso] ------ | | / H H { and these are cis or trans butene }
At 8:40 the wedge-dash structures of the molecule seem incorrect as after anti addition to trans, we obtain an erythro product. After converting the wedge-dash to fischer, it indicates a meso compound. The one in the video doesn't give one, it instead gives an enantiomeric pair. The correct structure must have one wedge and one dash for bromine unlike in the video where it is given as both wedges or dashes. Please confirm my doubt or correct me if i am wrong. Thanks :) Man! You're the best. Cheers.
I had the same doubt. Isn't it a bromination reaction which means that if there is a bridge intermediate and both bonds to the Br at the bridge intermediate are on wedges then the other Br does a back-side attack and attaches to the molecule on the dash. It's trans-addition.
U r right but even u missed at 15:15 there pdt. Shud b meso cuz Trans + syn = EN but both CH3 grup r same n syn addition is there Hence it will be meso not EN .
No guys actually, you're all wrong. He didn't explain it there but what happens is that any molecule is able to rotate about its sigma bond (that is a single bond) so if you draw a wedge and a dash then you can rotate one side of molecule which will include one Br either the wedge one or the dash one to match the other one. In doing so, the methyl group will also symmetrically align with the other methyl group about our line of symmetry. Also, saying that these molecules are symmetric about the line which he's talking about without rotating the molecule to match the methyl groups would be incorrect as the molecule is not mirrored about the line of symmetry. So we need to rotate the molecule to 1). Match the wedge and dash bromines. 2). To make the molecule mirrored about our line of symmetry. Only then we can call it a meso compound.
Also that depends on steric factors, such as in halogenation(addition of halogen)of alkenes where after the formation of cyclonium ion in the substrate the other X- ion has to take the other way and thus anti addition takes place
You have a misconception in differentiating between meso and enantiomers in some of the examples. For example, Trans and syn addition will eventually get a meso compound, you said it would give a pair of enantiomers. Don’t take it with the bond shape, instead get the configuration for each asymmetric center if they are different with the same groups so they are meso compound Again in this particular example, you will get one R and the Other S so it will be only one stereoisomer
In several of the examples you explain that in a trans-anti addition of Br2 to an alkene that each Br may be drawn on a solid wedge of the alkane. However this seems to provide the incorrect R/S assignment. For example, you draw a molecule of 2,3-bromopentane (@10:40) with each Br atom being added on a solid wedge adjacent to each other (2,3 positions). This would provide the 2R,3R configuration. However if an anti-trans addition to an alkene forms an erythro product then the actual configuration would be 2S,3R. Please correct me if I am mistaken or missing something. Just looking for some clarification. Thanks!
The starting material is a TRANS alkene (pent-2-ene 2-pentene). The addition of molecular bromine in DCM solvent to an alkene is an ANTI addition. An ANTI addition to a TRANS alkene produces erythro enantiomers or, potentially, a meso compound. In this case the erythro enantiomers are formed because the product does not have a plane of symmetry, in which case it would be a meso compound. You can remember this by the acronym "ATE, standing for Anti addition to a Trans alkene forms Erythro enantiomers (Anti-Trans-Erythro). Anti addition to a Cis alkene forms Threo enantiomers. You can similarly remember this by the acronym "ACT". Hopefully, this helps a bit.
Thank you for sharing. I have a question!! How I know this is a sun or anti addition to start with? Is there any strategy or I just have to memorize which one will give a syne an which one gives an anti? Thanks in advance.
I don't know about your country, but here in India most of us have memorized the fact that KMnO4 gives syn addition, Br2 gives anti addition reaction(same can be said for Cl2) etc. From this video I learnt that D2,OSO4 also give syn addition reactions. As for the products, u can check them using the R/S nomenclature.
D2 addition is like H2 reduction where Pd gives its surface where D2 is adsorbed so when the MCTS(Membered Cyclic Transition State) happen, so it attached from one side and in BR2/CCl4 or CS2 cyclic bromonium ion forms as intermediate then Br- attack from opposite side one on carbon where carbocation in the transition state is stable, hence anti addition for more stability
if you assume it to be in the plane of paper then if syn addition is there both should be attached on dash or wedge and if anti then one on the dash and other on wedge. then check chirality of compound and if chiral then make its mirror image. (if a compound have one chiral carbon then the compound is also chiral and if not have then check others criterias)
The lowest priority group should be on the dash. here in the example, H is on dash so no problem < if not then do 2 switches on the carbon bond and make it on the dash. now the priority is decided by atomic number, the more the atomic number the more we give priority and if the first group is same then we go towards the second atom attached to it. in case of isotope mass dominate or like -T>-D>-H. now in this carbon OH will be marked as 1 (since O is there) then the 2 carbon is attached so we go towards bond attached to the two carbon one has (Br, C, and H) and another one has (C and 2 H) and so we give priority to C with Br, marked it out 2. and then other C as 3 and H as 4. finally, make a circle going from 1 to 4 if come clockwise then R and anticlockwise then S.
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This lecture gave me more interest in organic chemistry
U r a great teacher
Thank you
You do not know how much this has helped me for my test next tuesday. I was doing so well undersatnding addition reactions to alkenes and alkynes, and all I neede to get back to was understanding the stereochemistry of them! Stereochemistry isnt my strong suit, but you seemed to have made the topic very managable.
You have provided so many examples for better understanding, thank you!
You're very welcome
If I don’t find your videos related to my subject I always get depressed
Only you can make me understand things
Thanks a ton...
You would be very famous if you ever come to India
here's a tip if you are adding syn to a double bong then for fast result just assume it on 2D and add 2 group on end and remain the configuration like
H CH3
/ | |
------ + KMnO4/OH- cold (B.R.) --> OH------------------OH [ optically active ]
------ | |
/ CH3 H
{ or }
H H
| |
------ + KMnO4/OH- cold (B.R.) --> OH------------------OH [ meso ]
------ | |
/ \ CH3 CH3
and on anti addition just switch one like
CH3 CH3
/ | |
____ + Br2/CS2 --> Br-------------------------Br [meso]
------ | |
/ H H
{ and these are cis or trans butene }
How do you know student will struggle through this topic and you make them one 🙃
At 8:40 the wedge-dash structures of the molecule seem incorrect as after anti addition to trans, we obtain an erythro product. After converting the wedge-dash to fischer, it indicates a meso compound. The one in the video doesn't give one, it instead gives an enantiomeric pair.
The correct structure must have one wedge and one dash for bromine unlike in the video where it is given as both wedges or dashes.
Please confirm my doubt or correct me if i am wrong. Thanks :) Man! You're the best. Cheers.
I had the same doubt. Isn't it a bromination reaction which means that if there is a bridge intermediate and both bonds to the Br at the bridge intermediate are on wedges then the other Br does a back-side attack and attaches to the molecule on the dash. It's trans-addition.
You are right
U r right but even u missed at 15:15 there pdt. Shud b meso cuz
Trans + syn = EN but both CH3 grup r same n syn addition is there
Hence it will be meso not EN .
Yes your are thinking right. There is a mistake that one br has to go to wedge and another dash. But he draw both in dash / wedge.
No guys actually, you're all wrong. He didn't explain it there but what happens is that any molecule is able to rotate about its sigma bond (that is a single bond) so if you draw a wedge and a dash then you can rotate one side of molecule which will include one Br either the wedge one or the dash one to match the other one. In doing so, the methyl group will also symmetrically align with the other methyl group about our line of symmetry. Also, saying that these molecules are symmetric about the line which he's talking about without rotating the molecule to match the methyl groups would be incorrect as the molecule is not mirrored about the line of symmetry. So we need to rotate the molecule to 1). Match the wedge and dash bromines. 2). To make the molecule mirrored about our line of symmetry. Only then we can call it a meso compound.
Absolutely fantastic broooo
Really well explained. Thanks a lot.
Can someone explain to me .......How are we suppose to know the reagent's gonna added the syn way or the anti way?
we need to remember those
Also that depends on steric factors, such as in halogenation(addition of halogen)of alkenes where after the formation of cyclonium ion in the substrate the other X- ion has to take the other way and thus anti addition takes place
Thank you so much for this great video 🙏🏻🙏🏻🙏🏻
This is very nice explanation
Thank you so much for such an effective explaination .
super sir i have become as your fan sir
thank you for wonderful content!
You have a misconception in differentiating between meso and enantiomers in some of the examples.
For example, Trans and syn addition will eventually get a meso compound, you said it would give a pair of enantiomers.
Don’t take it with the bond shape, instead get the configuration for each asymmetric center if they are different with the same groups so they are meso compound
Again in this particular example, you will get one R and the Other S so it will be only one stereoisomer
how to know when it is syn or anti?
Depends on type of reagents used.
@@vinayakdixit721 So Hydrogen is both Syn & Anti ?
In several of the examples you explain that in a trans-anti addition of Br2 to an alkene that each Br may be drawn on a solid wedge of the alkane. However this seems to provide the incorrect R/S assignment. For example, you draw a molecule of 2,3-bromopentane (@10:40) with each Br atom being added on a solid wedge adjacent to each other (2,3 positions). This would provide the 2R,3R configuration. However if an anti-trans addition to an alkene forms an erythro product then the actual configuration would be 2S,3R. Please correct me if I am mistaken or missing something. Just looking for some clarification. Thanks!
I have the same question. Struggling with the meso/en thing when doing alkene synthesis
The starting material is a TRANS alkene (pent-2-ene 2-pentene). The addition of molecular bromine in DCM solvent to an alkene is an ANTI addition. An ANTI addition to a TRANS alkene produces erythro enantiomers or, potentially, a meso compound. In this case the erythro enantiomers are formed because the product does not have a plane of symmetry, in which case it would be a meso compound.
You can remember this by the acronym "ATE, standing for Anti addition to a Trans alkene forms Erythro enantiomers (Anti-Trans-Erythro). Anti addition to a Cis alkene forms Threo enantiomers. You can similarly remember this by the acronym "ACT".
Hopefully, this helps a bit.
A M thank you
Thank you for sharing. I have a question!! How I know this is a sun or anti addition to start with? Is there any strategy or I just have to memorize which one will give a syne an which one gives an anti?
Thanks in advance.
I don't know about your country, but here in India most of us have memorized the fact that KMnO4 gives syn addition, Br2 gives anti addition reaction(same can be said for Cl2) etc. From this video I learnt that D2,OSO4 also give syn addition reactions. As for the products, u can check them using the R/S nomenclature.
@@shivamsrivastava5762 are u preparing for jee??
@@shivamsrivastava5762 Br2 is anti, but due to achiral center you just draw them sunstituents without stereochem.
rex like addition of H2,OH,BH3 gives syn addition and halogenation gives anti addition
Thank you this much examples
Thanks a lot Sir
How do you know for which reagent are you going to use syn or anti ?
U shud check at 8:40 n 15:15
There is a bit an error
The symmetry should be centre of symmetry and not plane of symmetry....I guess
Thank youuuuu
CIS + syn can give ERYTRO also
Wouldn't there be a carbocation shift on to the 3° carbon for the question on 26:17 ....doesn't the OH prefer to be more stable.
Thx sir
Is could be helpful for me but my O Chen class wants me to use the EZ system instead of cis and trans
How do you know many product(s) will formed?
Hasan Kuliyev 2^n (n= number of chiral centers) that’s tells you how many stereoisomers you’ll have
among us thank uou
Why does D2 do a syn addition in 12:03 b/c I thought things like Br2 or Cl2 were doing anti additions
D2 addition is like H2 reduction where Pd gives its surface where D2 is adsorbed so when the MCTS(Membered Cyclic Transition State) happen, so it attached from one side and in BR2/CCl4 or CS2 cyclic bromonium ion forms as intermediate then Br- attack from opposite side one on carbon where carbocation in the transition state is stable, hence anti addition for more stability
How would I know to add a wedge or a dash ???
if you assume it to be in the plane of paper then if syn addition is there both should be attached on dash or wedge and if anti then one on the dash and other on wedge. then check chirality of compound and if chiral then make its mirror image.
(if a compound have one chiral carbon then the compound is also chiral and if not have then check others criterias)
Anyone have youtube red and get ads on these videos still?
What does he mean when he says "chiral centers change"?
gaurav shinde because chiral center has 4 different group
this video was really confusing :')
a lot of these are wrong lol, or at least I am pretty sure, came here for practice and its just messing me up
Most of the subscribers are from india😅
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.👇
help me pls how it was r isomer !? 29:09
The lowest priority group should be on the dash. here in the example, H is on dash so no problem < if not then do 2 switches on the carbon bond and make it on the dash.
now the priority is decided by atomic number, the more the atomic number the more we give priority and if the first group is same then we go towards the second atom attached to it. in case of isotope mass dominate or like -T>-D>-H.
now in this carbon OH will be marked as 1 (since O is there) then the 2 carbon is attached so we go towards bond attached to the two carbon one has (Br, C, and H) and another one has (C and 2 H) and so we give priority to C with Br, marked it out 2. and then other C as 3 and H as 4.
finally, make a circle going from 1 to 4 if come clockwise then R and anticlockwise then S.
@@farhankh_17 thank you so much sir !!!
You should learn to use the solid-dash notation properly. You are confusing a lot of people here.
You need to correct this video. It’s incorrect.
im fucked