Spotting Chiral Centres
When given a large organic molecule, often with multiple carbon cycles and rings all bonded together, what sort of rules can you apply to find the chiral centres?
Take the molecules chlosterol and ergosterol as examples to which your rules would apply (you don't need to actually perform them, just should preferably make sure they actually do work on tougher molecules like these).
Link to structures of cholesterol and ergosterol: http://www.google.co.uk/imgres?imgurl=http://www.doctorfungus.org/thedrugs/images/cholesterol_ergosterol.jpg&imgrefurl=http://www.doctorfungus.org/thedrugs/antif_pharm.php&h=480&w=640&sz=10&tbnid=svvDw7gEgpJKrM:&tbnh=92&tbnw=122&prev=/search%3Fq%3Dergosterol%26tbm%3Disch%26tbo%3Du&zoom=1&q=ergosterol&usg=__P3n9QritzR-W3nktw--wpkgZe98=&docid=GChADYRyc6Ou9M&hl=en&sa=X&ei=V_2BUNiWKKqx0AWUqIC4BA&sqi=2&ved=0CDYQ9QEwBA&dur=122
Take the molecules chlosterol and ergosterol as examples to which your rules would apply (you don't need to actually perform them, just should preferably make sure they actually do work on tougher molecules like these).
Link to structures of cholesterol and ergosterol: http://www.google.co.uk/imgres?imgurl=http://www.doctorfungus.org/thedrugs/images/cholesterol_ergosterol.jpg&imgrefurl=http://www.doctorfungus.org/thedrugs/antif_pharm.php&h=480&w=640&sz=10&tbnid=svvDw7gEgpJKrM:&tbnh=92&tbnw=122&prev=/search%3Fq%3Dergosterol%26tbm%3Disch%26tbo%3Du&zoom=1&q=ergosterol&usg=__P3n9QritzR-W3nktw--wpkgZe98=&docid=GChADYRyc6Ou9M&hl=en&sa=X&ei=V_2BUNiWKKqx0AWUqIC4BA&sqi=2&ved=0CDYQ9QEwBA&dur=122
(edited 11 years ago)
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Original post by Big-Daddy
When given a large organic molecule, often with multiple carbon cycles and rings all bonded together, what sort of rules can you apply to find the chiral centres?
They are always bonded to FOUR DIFFERENT groups.
The 'groups' can be anything from H or Cl to things you wouldn't normally call groups like CH2CH2C-OHCH3.
Basically, just look for carbons with four different things around it.
Original post by SpiggyTopes
They are always bonded to FOUR DIFFERENT groups.
The 'groups' can be anything from H or Cl to things you wouldn't normally call groups like CH2CH2C-OHCH3.
Basically, just look for carbons with four different things around it.
The 'groups' can be anything from H or Cl to things you wouldn't normally call groups like CH2CH2C-OHCH3.
Basically, just look for carbons with four different things around it.
Original post by Revent
Just look for carbons with 4 single bonds and make sure that all molecules attached to the carbon are different. Takes practice but you'll get faster.
So carbons attached to double bonds can't be chiral centres?
I take it that a group can count as a molecule, so you have to look at the whole structure around it to decide whether each part of the structure attached to this carbon is different. But what counts as "different" in a carbon ring/cycle?
Also, do you have access to any practice questions you could give me?
(edited 11 years ago)
Original post by Big-Daddy
So carbons attached to double bonds can't be chiral centres?
I take it that a group can count as a molecule, so you have to look at the whole structure around it to decide whether each part of the structure attached to this carbon is different. But what counts as "different" in a carbon ring/cycle?
Also, do you have access to any practice questions you could give me?
I take it that a group can count as a molecule, so you have to look at the whole structure around it to decide whether each part of the structure attached to this carbon is different. But what counts as "different" in a carbon ring/cycle?
Also, do you have access to any practice questions you could give me?
Nope, the carbons can't have double bonds.
I don't know what you mean by 'a group can count as a molecule'. But yes, you do have to look at the whole structure around the carbons.
I don't know where there are any questions just on chiral molecules, but you can have a look through past papers on the AQA website and find some.
Original post by Big-Daddy
So carbons attached to double bonds can't be chiral centres?
no
I take it that a group can count as a molecule, so you have to look at the whole structure around it to decide whether each part of the structure attached to this carbon is different.
yes
But what counts as "different" in a carbon ring/cycle?
If you keep going bond to bond and find something different ...
So for example 2-methylcyclohexan-1-ol has 2 chiral centres
Original post by charco
no
How about carbons double bonded to atoms other than carbon, e.g. C=N or C=O double bonds? Does that also disqualify the centre from being chiral?
Original post by charco
If you keep going bond to bond and find something different ...
So for example 2-methylcyclohexan-1-ol has 2 chiral centres
So for example 2-methylcyclohexan-1-ol has 2 chiral centres
But in what direction do you go if you're inspecting a carbon that is part of a ring/cycle? Look at the image I posted linking to structures of cholesterol and ergosterol for an example of what I mean.
If your explanation is correct, then primarily it is carbon atoms attached to two hydrogens that are not chiral centres and most of the time the other carbon atoms would be (assuming the structure is only hydrogen and carbon primarily, with the odd oxygen or nitrogen atom as well). How accurate, generally, is the claim that a carbon has to be bonded to 4 different groups to be a chiral centre?
Also, I take it H3C is the same as CH3 in these problems and the shape of the molecules (e.g. whether a particular bond is protruding into the page, etc.) do not matter, so long as the molecules at the ends of the bonds themselves are 4 in number and each different from the other?
(edited 11 years ago)
Original post by Big-Daddy
How about carbons double bonded to atoms other than carbon, e.g. C=N or C=O double bonds? Does that also disqualify the centre from being chiral?
Carbon can only form four bonds, so it is irrelevant what element it is forming the double bond with. If it forms a double bond, then it only has two other bonds available. This means that it can only attach to three groups in total, and you need four substituents for it to be an asymmetric chiral carbon (chiral centre).
Original post by Big-Daddy
How accurate, generally, is the claim that a carbon has to be bonded to 4 different groups to be a chiral centre?
It is always a chiral centre unless there is some other form of symmetry that cancels out the non-superimposible mirror image.
This happens with meso-forms in (for example) tartaric acid.
Original post by charco
It is always a chiral centre unless there is some other form of symmetry that cancels out the non-superimposible mirror image.
This happens with meso-forms in (for example) tartaric acid.
This happens with meso-forms in (for example) tartaric acid.
Thanks. Having looked at the structure of tartaric acid shown by Wikipedia, should not the 2nd and 3rd carbons in the chain be chiral centres? (The 1st and 4th can be immediately disqualified by the fact that they are double-bonded to something, an O.)
I take it H3C is the same as CH3 in these problems and the shape of the molecules (e.g. whether a particular bond is protruding into the page, etc.) do not matter, so long as the molecules at the ends of the bonds themselves are 4 in number and each different from the other?
(edited 11 years ago)
Original post by Big-Daddy
Thanks. Having looked at the structure of tartaric acid shown by Wikipedia, should not the 2nd and 3rd carbons in the chain be chiral centres? (The 1st and 4th can be immediately disqualified by the fact that they are double-bonded to something, an O.)
I take it H3C is the same as CH3 in these problems and the shape of the molecules (e.g. whether a particular bond is protruding into the page, etc.) do not matter, so long as the molecules at the ends of the bonds themselves are 4 in number and each different from the other?
I take it H3C is the same as CH3 in these problems and the shape of the molecules (e.g. whether a particular bond is protruding into the page, etc.) do not matter, so long as the molecules at the ends of the bonds themselves are 4 in number and each different from the other?
Usually the total number of optical isomers is 2n, where 'n' is the number of optically active centres.
Tartaric acid, however, has two enantiomers and one meso form in which the two optically active carbons are mirror images of each other and cancel out the optical activity. This is called internal compensation.
Original post by charco
Usually the total number of optical isomers is 2n, where 'n' is the number of optically active centres.
Tartaric acid, however, has two enantiomers and one meso form in which the two optically active carbons are mirror images of each other and cancel out the optical activity. This is called internal compensation.
Tartaric acid, however, has two enantiomers and one meso form in which the two optically active carbons are mirror images of each other and cancel out the optical activity. This is called internal compensation.
Thanks.
Again:
I take it H3C is the same as CH3 in these problems and the shape of the molecules (e.g. whether a particular bond is protruding into the page, etc.) do not matter, so long as the molecules at the ends of the bonds themselves are 4 in number and each different from the other?
If you don't know (which is probably unlikely), then please at least tell me so I don't hang around waiting.
Original post by Big-Daddy
Thanks.
Again:
I take it H3C is the same as CH3 in these problems and the shape of the molecules (e.g. whether a particular bond is protruding into the page, etc.) do not matter, so long as the molecules at the ends of the bonds themselves are 4 in number and each different from the other?
If you don't know (which is probably unlikely), then please at least tell me so I don't hang around waiting.
Again:
I take it H3C is the same as CH3 in these problems and the shape of the molecules (e.g. whether a particular bond is protruding into the page, etc.) do not matter, so long as the molecules at the ends of the bonds themselves are 4 in number and each different from the other?
If you don't know (which is probably unlikely), then please at least tell me so I don't hang around waiting.
That's correct, couldn't let you wait any longer!
Original post by Big-Daddy
Thanks.
Again:
I take it H3C is the same as CH3 in these problems and the shape of the molecules (e.g. whether a particular bond is protruding into the page, etc.) do not matter, so long as the molecules at the ends of the bonds themselves are 4 in number and each different from the other?
If you don't know (which is probably unlikely), then please at least tell me so I don't hang around waiting.
Again:
I take it H3C is the same as CH3 in these problems and the shape of the molecules (e.g. whether a particular bond is protruding into the page, etc.) do not matter, so long as the molecules at the ends of the bonds themselves are 4 in number and each different from the other?
If you don't know (which is probably unlikely), then please at least tell me so I don't hang around waiting.
BD you have to understand that there is life outside of TSR - I have spent the last 2 hours in a bar watching football and quaffing some well earned cervezas. Sometimes you just have to wait!
But as the poster above says 'there is no difference between CH3 and H3C'.
Original post by charco
BD you have to understand that there is life outside of TSR - I have spent the last 2 hours in a bar watching football and quaffing some well earned cervezas. Sometimes you just have to wait!
But as the poster above says 'there is no difference between CH3 and H3C'.
But as the poster above says 'there is no difference between CH3 and H3C'.
I do know that.
It's just that people often don't bother mentioning when they don't know the answer to something (even if they've been helping me up till that point) so I'm often waiting for days without confirmation as to whether I should give up on them and look elsewhere or if they're just taking a break and will be back soon. This happens to me not only on TSR but on any internet medium.Thanks for the confirmation. What about the molecular geometry (e.g. if a bond is shown as going into the page or coming out of it, and another is shown as in the plane of the page or as the opposite to the last one, do these count as two different groups, making it a chiral center, or the same group, meaning it isn't a chiral center?)?
Original post by Big-Daddy
Thanks for the confirmation. What about the molecular geometry (e.g. if a bond is shown as going into the page or coming out of it, and another is shown as in the plane of the page or as the opposite to the last one, do these count as two different groups, making it a chiral center, or the same group, meaning it isn't a chiral center?)?
Thanks for the confirmation. What about the molecular geometry (e.g. if a bond is shown as going into the page or coming out of it, and another is shown as in the plane of the page or as the opposite to the last one, do these count as two different groups, making it a chiral center, or the same group, meaning it isn't a chiral center?)?
You would have to be more specific as I'm not 100% clear what you mean.
The key to chirality is asymmetry. A chiral centre must have a non-superimposible mirror image.
Be very careful with bond geometry as single bonds nearly always have free rotation... BUT not always. For example the 2,6 disubstituted biphenyls may have optically active isomers (depending on the substituents) as free rotation of the bond holding the two ring systems may be sterically hindered.
This is a special case that you are only likely to meet at university along with optically active systems such as hexihelicene
Original post by charco
You would have to be more specific as I'm not 100% clear what you mean.
The key to chirality is asymmetry. A chiral centre must have a non-superimposible mirror image.
Be very careful with bond geometry as single bonds nearly always have free rotation... BUT not always. For example the 2,6 disubstituted biphenyls may have optically active isomers (depending on the substituents) as free rotation of the bond holding the two ring systems may be sterically hindered.
This is a special case that you are only likely to meet at university along with optically active systems such as hexihelicene
The key to chirality is asymmetry. A chiral centre must have a non-superimposible mirror image.
Be very careful with bond geometry as single bonds nearly always have free rotation... BUT not always. For example the 2,6 disubstituted biphenyls may have optically active isomers (depending on the substituents) as free rotation of the bond holding the two ring systems may be sterically hindered.
This is a special case that you are only likely to meet at university along with optically active systems such as hexihelicene
I'll look into that. But am I to take it that, as a general rule, it makes no difference if a bond is coming out of the page or going into the page?
e.g. The C atom we're looking at is single-bonded to an OH group (in the plane of the page), an O (coming out of the page) and a CH2CH3 (going into the page) as well as to a CH2CH3 in the plane of the page. Do the two ethyl groups cause the C atom not to be a chiral center, despite the fact that one is angled completely differently to the other? Or do their angles and directions of the bonds with the C atom cause them to be considered different groups?
(edited 11 years ago)
Original post by Big-Daddy
I'll look into that. But am I to take it that, as a general rule, it makes no difference if a bond is coming out of the page or going into the page?
e.g. The C atom we're looking at is single-bonded to an OH group (in the plane of the page), an O (coming out of the page) and a CH2CH3 (going into the page) as well as to a CH2CH3 in the plane of the page. Do the two ethyl groups cause the C atom not to be a chiral center, despite the fact that one is angled completely differently to the other? Or do their angles and directions of the bonds with the C atom cause them to be considered different groups?
e.g. The C atom we're looking at is single-bonded to an OH group (in the plane of the page), an O (coming out of the page) and a CH2CH3 (going into the page) as well as to a CH2CH3 in the plane of the page. Do the two ethyl groups cause the C atom not to be a chiral center, despite the fact that one is angled completely differently to the other? Or do their angles and directions of the bonds with the C atom cause them to be considered different groups?
The two ethyl groups prevent any possibility of chirality. Remember what I said about free rotation in single bonds...
I'll try and make this as easy as possible
when theres an skeletal formula given, just look for 4 different groups on each side of the carbon atom, thats the essence of a chiral centre
when youre given a badass closed chain skeletal formula the key lies in the hydrogen atoms. First you draw the hydrogen atoms yourself on each carbon(since theyre not drawn by default) to make things simpler. if there are 2 hydrogens then it cant be chiral. Also, the carbon atoms with double bonds attached aren't chiral. Don't go too much in depth or try to break the whole chain down or you'll lose your mind and it'll take too much of your time.
when theres an skeletal formula given, just look for 4 different groups on each side of the carbon atom, thats the essence of a chiral centre
when youre given a badass closed chain skeletal formula the key lies in the hydrogen atoms. First you draw the hydrogen atoms yourself on each carbon(since theyre not drawn by default) to make things simpler. if there are 2 hydrogens then it cant be chiral. Also, the carbon atoms with double bonds attached aren't chiral. Don't go too much in depth or try to break the whole chain down or you'll lose your mind and it'll take too much of your time.
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