Chiral carbon in carvone moleculesWhat is the perfect definition for chirality?Why was the definition of chiral carbon changed?Using CIP rules in a bicyclic compound with carbonyl subsituentAssigning Carbon NMRIs chirality a physical or a chemical property?Identifying chiral centersWhy was the definition of chiral carbon changed?Chiral Centers and StereochemistryAre bi-cycles chiral molecules?What is it meant by two different groups in stereoisomerism?Stereogenic Center QuestionWhy does a chiral carbon affect some hydrogen's hydrogen environment?Can two lone pairs in different orbitals make a chiral centre?
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Chiral carbon in carvone molecules
What is the perfect definition for chirality?Why was the definition of chiral carbon changed?Using CIP rules in a bicyclic compound with carbonyl subsituentAssigning Carbon NMRIs chirality a physical or a chemical property?Identifying chiral centersWhy was the definition of chiral carbon changed?Chiral Centers and StereochemistryAre bi-cycles chiral molecules?What is it meant by two different groups in stereoisomerism?Stereogenic Center QuestionWhy does a chiral carbon affect some hydrogen's hydrogen environment?Can two lone pairs in different orbitals make a chiral centre?
$begingroup$
This is the structure of carvone, with the chiral carbon noted by *.
I don't understand why this is a chiral carbon, to my understanding a chiral carbon is a carbon with 4 different chemical groups attached to it.
In the molecule, the chiral carbon is surrounded by $ceH$, $ceC(CH3)=CH2$ and two $ceCH2$ groups.
Why are both $ceCH2$ groups different and thus making that carbon chiral?
organic-chemistry molecular-structure chirality
edited May 7 at 11:26
andselisk
20.8k770136
asked May 7 at 10:32
Rafael FrancoRafael Franco
735
$endgroup$
|
show 2 more comments
$begingroup$
This is the structure of carvone, with the chiral carbon noted by *.
I don't understand why this is a chiral carbon, to my understanding a chiral carbon is a carbon with 4 different chemical groups attached to it.
In the molecule, the chiral carbon is surrounded by $ceH$, $ceC(CH3)=CH2$ and two $ceCH2$ groups.
Why are both $ceCH2$ groups different and thus making that carbon chiral?
organic-chemistry molecular-structure chirality
edited May 7 at 11:26
andselisk
20.8k770136
asked May 7 at 10:32
Rafael FrancoRafael Franco
735
$endgroup$
2
$begingroup$
Both CH (which are actually both CH2, but that doesn't matter) are different in what else is attached to them. Imagine a simpler molecule, with the following four substituents: H, Cl, CH3, and C2H5. You'll see right away that they are different. But wait, how are CH3 and C2H5 different if they both start with CH2? Well, just like that.
$endgroup$
– Ivan Neretin
May 7 at 11:03
1
$begingroup$
I edited the CH to CH2 (whoops), I think about benzene every time I see a cyclic molecule.
$endgroup$
– Rafael Franco
May 7 at 11:10
$begingroup$
So one of the CH2 groups have a CO group attached, and then the next carbon is just a C(after the CO), does this matter at all? or what matters is what is immediately next to that CH2 and everything else down the chain is not relevant
$endgroup$
– Rafael Franco
May 7 at 11:20
3
$begingroup$
Yes it does matter. Everything down the chain is relevant. You can imagine two chains longer than these, which are very similar and only differ at the very end, and yes, they will be different.
$endgroup$
– Ivan Neretin
May 7 at 11:34
1
$begingroup$
Related: chemistry.stackexchange.com/q/39536/7951
$endgroup$
– Loong♦
May 7 at 13:11
|
show 2 more comments
$begingroup$
This is the structure of carvone, with the chiral carbon noted by *.
I don't understand why this is a chiral carbon, to my understanding a chiral carbon is a carbon with 4 different chemical groups attached to it.
In the molecule, the chiral carbon is surrounded by $ceH$, $ceC(CH3)=CH2$ and two $ceCH2$ groups.
Why are both $ceCH2$ groups different and thus making that carbon chiral?
organic-chemistry molecular-structure chirality
edited May 7 at 11:26
andselisk
20.8k770136
asked May 7 at 10:32
Rafael FrancoRafael Franco
735
$endgroup$
This is the structure of carvone, with the chiral carbon noted by *.
I don't understand why this is a chiral carbon, to my understanding a chiral carbon is a carbon with 4 different chemical groups attached to it.
In the molecule, the chiral carbon is surrounded by $ceH$, $ceC(CH3)=CH2$ and two $ceCH2$ groups.
Why are both $ceCH2$ groups different and thus making that carbon chiral?
organic-chemistry molecular-structure chirality
organic-chemistry molecular-structure chirality
edited May 7 at 11:26
andselisk
20.8k770136
asked May 7 at 10:32
Rafael FrancoRafael Franco
735
edited May 7 at 11:26
andselisk
20.8k770136
asked May 7 at 10:32
Rafael FrancoRafael Franco
735
edited May 7 at 11:26
andselisk
20.8k770136
edited May 7 at 11:26
andselisk
20.8k770136
edited May 7 at 11:26
andselisk
20.8k770136
20.8k770136
asked May 7 at 10:32
Rafael FrancoRafael Franco
735
asked May 7 at 10:32
Rafael FrancoRafael Franco
735
asked May 7 at 10:32
Rafael FrancoRafael Franco
735
735
2
$begingroup$
Both CH (which are actually both CH2, but that doesn't matter) are different in what else is attached to them. Imagine a simpler molecule, with the following four substituents: H, Cl, CH3, and C2H5. You'll see right away that they are different. But wait, how are CH3 and C2H5 different if they both start with CH2? Well, just like that.
$endgroup$
– Ivan Neretin
May 7 at 11:03
1
$begingroup$
I edited the CH to CH2 (whoops), I think about benzene every time I see a cyclic molecule.
$endgroup$
– Rafael Franco
May 7 at 11:10
$begingroup$
So one of the CH2 groups have a CO group attached, and then the next carbon is just a C(after the CO), does this matter at all? or what matters is what is immediately next to that CH2 and everything else down the chain is not relevant
$endgroup$
– Rafael Franco
May 7 at 11:20
3
$begingroup$
Yes it does matter. Everything down the chain is relevant. You can imagine two chains longer than these, which are very similar and only differ at the very end, and yes, they will be different.
$endgroup$
– Ivan Neretin
May 7 at 11:34
1
$begingroup$
Related: chemistry.stackexchange.com/q/39536/7951
$endgroup$
– Loong♦
May 7 at 13:11
|
show 2 more comments
2
$begingroup$
Both CH (which are actually both CH2, but that doesn't matter) are different in what else is attached to them. Imagine a simpler molecule, with the following four substituents: H, Cl, CH3, and C2H5. You'll see right away that they are different. But wait, how are CH3 and C2H5 different if they both start with CH2? Well, just like that.
$endgroup$
– Ivan Neretin
May 7 at 11:03
1
$begingroup$
I edited the CH to CH2 (whoops), I think about benzene every time I see a cyclic molecule.
$endgroup$
– Rafael Franco
May 7 at 11:10
$begingroup$
So one of the CH2 groups have a CO group attached, and then the next carbon is just a C(after the CO), does this matter at all? or what matters is what is immediately next to that CH2 and everything else down the chain is not relevant
$endgroup$
– Rafael Franco
May 7 at 11:20
3
$begingroup$
Yes it does matter. Everything down the chain is relevant. You can imagine two chains longer than these, which are very similar and only differ at the very end, and yes, they will be different.
$endgroup$
– Ivan Neretin
May 7 at 11:34
1
$begingroup$
Related: chemistry.stackexchange.com/q/39536/7951
$endgroup$
– Loong♦
May 7 at 13:11
2
2
$begingroup$
Both CH (which are actually both CH2, but that doesn't matter) are different in what else is attached to them. Imagine a simpler molecule, with the following four substituents: H, Cl, CH3, and C2H5. You'll see right away that they are different. But wait, how are CH3 and C2H5 different if they both start with CH2? Well, just like that.
$endgroup$
– Ivan Neretin
May 7 at 11:03
$begingroup$
Both CH (which are actually both CH2, but that doesn't matter) are different in what else is attached to them. Imagine a simpler molecule, with the following four substituents: H, Cl, CH3, and C2H5. You'll see right away that they are different. But wait, how are CH3 and C2H5 different if they both start with CH2? Well, just like that.
$endgroup$
– Ivan Neretin
May 7 at 11:03
1
1
$begingroup$
I edited the CH to CH2 (whoops), I think about benzene every time I see a cyclic molecule.
$endgroup$
– Rafael Franco
May 7 at 11:10
$begingroup$
I edited the CH to CH2 (whoops), I think about benzene every time I see a cyclic molecule.
$endgroup$
– Rafael Franco
May 7 at 11:10
$begingroup$
So one of the CH2 groups have a CO group attached, and then the next carbon is just a C(after the CO), does this matter at all? or what matters is what is immediately next to that CH2 and everything else down the chain is not relevant
$endgroup$
– Rafael Franco
May 7 at 11:20
$begingroup$
So one of the CH2 groups have a CO group attached, and then the next carbon is just a C(after the CO), does this matter at all? or what matters is what is immediately next to that CH2 and everything else down the chain is not relevant
$endgroup$
– Rafael Franco
May 7 at 11:20
3
3
$begingroup$
Yes it does matter. Everything down the chain is relevant. You can imagine two chains longer than these, which are very similar and only differ at the very end, and yes, they will be different.
$endgroup$
– Ivan Neretin
May 7 at 11:34
$begingroup$
Yes it does matter. Everything down the chain is relevant. You can imagine two chains longer than these, which are very similar and only differ at the very end, and yes, they will be different.
$endgroup$
– Ivan Neretin
May 7 at 11:34
1
1
$begingroup$
Related: chemistry.stackexchange.com/q/39536/7951
$endgroup$
– Loong♦
May 7 at 13:11
$begingroup$
Related: chemistry.stackexchange.com/q/39536/7951
$endgroup$
– Loong♦
May 7 at 13:11
|
show 2 more comments
2 Answers
2
active
oldest
votes
$begingroup$
If you consider the carbon with as star (C*), it bears a C(CH3)(=CH2) group and an H group (not shown). But it is also part of a ring and in that case, we actually follow the exact same rules!
From C*, if we go to the left or the right, we get a CH2 group. So let's get further: on the left we get a CH group with a double bond to another carbon, while on the right, we get a C atom with a double bond to an oxygen and a single bond to a carbon atom.
Because you get different groups if you follow the ring from C* to the left or the right, it means that C* is chiral!
answered May 7 at 14:59
SteffXSteffX
2,239410
$endgroup$
add a comment |
$begingroup$
This is just a general note on handling this type of question. Many students struggle with line diagrams when they are first introduced. This is because a lot of important information is implied by the drawing, rather than showing it explicitly.
Take the time to redraw the molecule, clearly showing all of the missing atoms (especially the hydrogen atoms). This should make things clear for you.
It is well worth your time to redraw the molecule until you are very familiar with line drawings.
answered 2 days ago
Michael LautmanMichael Lautman
42629
$endgroup$
$begingroup$
I'm afraid that doesn't actually answer the question, I don't think OP has any problem identifying what surrounds the asymmetric carbon (next neighbour), but rather why it is asymmetric in the first place.
$endgroup$
– Martin - マーチン♦
2 days ago
$begingroup$
I had actually meant to post this as a comment rather than an answer.
$endgroup$
– Michael Lautman
2 days ago
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
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votes
active
oldest
votes
$begingroup$
If you consider the carbon with as star (C*), it bears a C(CH3)(=CH2) group and an H group (not shown). But it is also part of a ring and in that case, we actually follow the exact same rules!
From C*, if we go to the left or the right, we get a CH2 group. So let's get further: on the left we get a CH group with a double bond to another carbon, while on the right, we get a C atom with a double bond to an oxygen and a single bond to a carbon atom.
Because you get different groups if you follow the ring from C* to the left or the right, it means that C* is chiral!
answered May 7 at 14:59
SteffXSteffX
2,239410
$endgroup$
add a comment |
$begingroup$
If you consider the carbon with as star (C*), it bears a C(CH3)(=CH2) group and an H group (not shown). But it is also part of a ring and in that case, we actually follow the exact same rules!
From C*, if we go to the left or the right, we get a CH2 group. So let's get further: on the left we get a CH group with a double bond to another carbon, while on the right, we get a C atom with a double bond to an oxygen and a single bond to a carbon atom.
Because you get different groups if you follow the ring from C* to the left or the right, it means that C* is chiral!
answered May 7 at 14:59
SteffXSteffX
2,239410
$endgroup$
add a comment |
$begingroup$
If you consider the carbon with as star (C*), it bears a C(CH3)(=CH2) group and an H group (not shown). But it is also part of a ring and in that case, we actually follow the exact same rules!
From C*, if we go to the left or the right, we get a CH2 group. So let's get further: on the left we get a CH group with a double bond to another carbon, while on the right, we get a C atom with a double bond to an oxygen and a single bond to a carbon atom.
Because you get different groups if you follow the ring from C* to the left or the right, it means that C* is chiral!
answered May 7 at 14:59
SteffXSteffX
2,239410
$endgroup$
If you consider the carbon with as star (C*), it bears a C(CH3)(=CH2) group and an H group (not shown). But it is also part of a ring and in that case, we actually follow the exact same rules!
From C*, if we go to the left or the right, we get a CH2 group. So let's get further: on the left we get a CH group with a double bond to another carbon, while on the right, we get a C atom with a double bond to an oxygen and a single bond to a carbon atom.
Because you get different groups if you follow the ring from C* to the left or the right, it means that C* is chiral!
answered May 7 at 14:59
SteffXSteffX
2,239410
answered May 7 at 14:59
SteffXSteffX
2,239410
answered May 7 at 14:59
SteffXSteffX
2,239410
answered May 7 at 14:59
SteffXSteffX
2,239410
2,239410
add a comment |
add a comment |
$begingroup$
This is just a general note on handling this type of question. Many students struggle with line diagrams when they are first introduced. This is because a lot of important information is implied by the drawing, rather than showing it explicitly.
Take the time to redraw the molecule, clearly showing all of the missing atoms (especially the hydrogen atoms). This should make things clear for you.
It is well worth your time to redraw the molecule until you are very familiar with line drawings.
answered 2 days ago
Michael LautmanMichael Lautman
42629
$endgroup$
$begingroup$
I'm afraid that doesn't actually answer the question, I don't think OP has any problem identifying what surrounds the asymmetric carbon (next neighbour), but rather why it is asymmetric in the first place.
$endgroup$
– Martin - マーチン♦
2 days ago
$begingroup$
I had actually meant to post this as a comment rather than an answer.
$endgroup$
– Michael Lautman
2 days ago
add a comment |
$begingroup$
This is just a general note on handling this type of question. Many students struggle with line diagrams when they are first introduced. This is because a lot of important information is implied by the drawing, rather than showing it explicitly.
Take the time to redraw the molecule, clearly showing all of the missing atoms (especially the hydrogen atoms). This should make things clear for you.
It is well worth your time to redraw the molecule until you are very familiar with line drawings.
answered 2 days ago
Michael LautmanMichael Lautman
42629
$endgroup$
$begingroup$
I'm afraid that doesn't actually answer the question, I don't think OP has any problem identifying what surrounds the asymmetric carbon (next neighbour), but rather why it is asymmetric in the first place.
$endgroup$
– Martin - マーチン♦
2 days ago
$begingroup$
I had actually meant to post this as a comment rather than an answer.
$endgroup$
– Michael Lautman
2 days ago
add a comment |
$begingroup$
This is just a general note on handling this type of question. Many students struggle with line diagrams when they are first introduced. This is because a lot of important information is implied by the drawing, rather than showing it explicitly.
Take the time to redraw the molecule, clearly showing all of the missing atoms (especially the hydrogen atoms). This should make things clear for you.
It is well worth your time to redraw the molecule until you are very familiar with line drawings.
answered 2 days ago
Michael LautmanMichael Lautman
42629
$endgroup$
This is just a general note on handling this type of question. Many students struggle with line diagrams when they are first introduced. This is because a lot of important information is implied by the drawing, rather than showing it explicitly.
Take the time to redraw the molecule, clearly showing all of the missing atoms (especially the hydrogen atoms). This should make things clear for you.
It is well worth your time to redraw the molecule until you are very familiar with line drawings.
answered 2 days ago
Michael LautmanMichael Lautman
42629
answered 2 days ago
Michael LautmanMichael Lautman
42629
answered 2 days ago
Michael LautmanMichael Lautman
42629
answered 2 days ago
Michael LautmanMichael Lautman
42629
42629
$begingroup$
I'm afraid that doesn't actually answer the question, I don't think OP has any problem identifying what surrounds the asymmetric carbon (next neighbour), but rather why it is asymmetric in the first place.
$endgroup$
– Martin - マーチン♦
2 days ago
$begingroup$
I had actually meant to post this as a comment rather than an answer.
$endgroup$
– Michael Lautman
2 days ago
add a comment |
$begingroup$
I'm afraid that doesn't actually answer the question, I don't think OP has any problem identifying what surrounds the asymmetric carbon (next neighbour), but rather why it is asymmetric in the first place.
$endgroup$
– Martin - マーチン♦
2 days ago
$begingroup$
I had actually meant to post this as a comment rather than an answer.
$endgroup$
– Michael Lautman
2 days ago
$begingroup$
I'm afraid that doesn't actually answer the question, I don't think OP has any problem identifying what surrounds the asymmetric carbon (next neighbour), but rather why it is asymmetric in the first place.
$endgroup$
– Martin - マーチン♦
2 days ago
$begingroup$
I'm afraid that doesn't actually answer the question, I don't think OP has any problem identifying what surrounds the asymmetric carbon (next neighbour), but rather why it is asymmetric in the first place.
$endgroup$
– Martin - マーチン♦
2 days ago
$begingroup$
I had actually meant to post this as a comment rather than an answer.
$endgroup$
– Michael Lautman
2 days ago
$begingroup$
I had actually meant to post this as a comment rather than an answer.
$endgroup$
– Michael Lautman
2 days ago
add a comment |
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2
$begingroup$
Both CH (which are actually both CH2, but that doesn't matter) are different in what else is attached to them. Imagine a simpler molecule, with the following four substituents: H, Cl, CH3, and C2H5. You'll see right away that they are different. But wait, how are CH3 and C2H5 different if they both start with CH2? Well, just like that.
$endgroup$
– Ivan Neretin
May 7 at 11:03
1
$begingroup$
I edited the CH to CH2 (whoops), I think about benzene every time I see a cyclic molecule.
$endgroup$
– Rafael Franco
May 7 at 11:10
$begingroup$
So one of the CH2 groups have a CO group attached, and then the next carbon is just a C(after the CO), does this matter at all? or what matters is what is immediately next to that CH2 and everything else down the chain is not relevant
$endgroup$
– Rafael Franco
May 7 at 11:20
3
$begingroup$
Yes it does matter. Everything down the chain is relevant. You can imagine two chains longer than these, which are very similar and only differ at the very end, and yes, they will be different.
$endgroup$
– Ivan Neretin
May 7 at 11:34
1
$begingroup$
Related: chemistry.stackexchange.com/q/39536/7951
$endgroup$
– Loong♦
May 7 at 13:11