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Why is cyclopentadiene anion is aromatic but cycloheptatrienyl anion is not?
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Miguel
Why is cyclopentadiene anion is aromatic but cycloheptatrienyl anion is not?
It's quite simple, really. Aromatic rings have all their relatively stable electronic orbitals filled and unstable ones empty. The 4n+2 rule just says that a conjugated ring will have an odd number of stable pi orbitals to be filled, that is 2n+1 orbitals to be occupied by two electrons apiece for some whole number n.
When you try to put 4n (or 4n+4) electrons into the ring as in cycloheptatrienyl anion, you have filled and empty states at the same energy level; that's what the Frost circle tells you. Sometimes molecules do that, but they generally are not as stable as what you get with all the filled states well below all the empty ones. The latter is what the proper 4n+2 electron count does.
It's quite simple, really. Aromatic rings have all their relatively stable electronic orbitals filled and unstable ones empty. The 4n+2 rule just says that a conjugated ring will have an odd number of stable pi orbitals to be filled, that is 2n+1 orbitals to be occupied by two electrons apiece for some whole number n.
When you try to put 4n (or 4n+4) electrons into the ring as in cycloheptatrienyl anion, you have filled and empty states at the same energy level; that's what the Frost circle tells you. Sometimes molecules do that, but they generally are not as stable as what you get with all the filled states well below all the empty ones. The latter is what the proper 4n+2 electron count does.
Cycloheptatrienyl anion(tropylium anion) has 8 pi electron system, therefore it must be antiaromatic but the extra lone pair on the one carbon would cause that carbon to be become sp3 hybridized and put those extra elctrons in one of the sp3 orbitals. This would make it non-planar and non-aromatic.
Cycloheptatrienyl anion(tropylium anion) has 8 pi electron system, therefore it must be antiaromatic but the extra lone pair on the one carbon would cause that carbon to be become sp3 hybridized and put those extra elctrons in one of the sp3 orbitals. This would make it non-planar and non-aromatic.
That happens because of Huckel's rule of aromaticity. The cycloheptatrienyl anion has 8 electrons which translates as 4n electrons, not the 4n+2 as implied by Huckel. If you are about to construct the MOs for both cycles, you'll find that putting 4n electrons will rise in a diradical molecule, not stable as it already sounds. A quick trick for doing that is using Frost's circle. All you need to do is inscribe the correspondent polygon in a circle and make sure that you have a vertex touching the circle as low as possible. The vertices will give you the qualitative energy of the MOs.
That happens because of Huckel's rule of aromaticity. The cycloheptatrienyl anion has 8 electrons which translates as 4n electrons, not the 4n+2 as implied by Huckel. If you are about to construct the MOs for both cycles, you'll find that putting 4n electrons will rise in a diradical molecule, not stable as it already sounds. A quick trick for doing that is using Frost's circle. All you need to do is inscribe the correspondent polygon in a circle and make sure that you have a vertex touching the circle as low as possible. The vertices will give you the qualitative energy of the MOs.
Ground state O2 is a diradical and is pretty stable. Id say its just a form of a Jahn-Teller distortion - a nonlinear molecule with incompletely filled degenerate orbitals.More
Agree. Indeed pseudo-Jahn-Teller effect is present in molecules like cyclobutadiene and distorts the geometry from square to a rectangular. It may also happens in cycloheptatrienyl anion.More
"That happens because of Huckels rule of aromaticity. "No, it doesnt happen because of this rule. The rule is a tool to predict aromaticity/antiaromaticity in very simple systems.More
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Aromaticity exists when there are $4n+2$ (i.e. 2, 6, 10, 14...) electrons in a planar, cyclic pi system. Planar, cyclic pi systems that have $4n$ (4, 8, 12, 16...) electrons are antiaromatic.
In the case of the cyclopentadienyl anion, there are 6 electrons in the pi system. This makes it aromatic. The cycloheptatrienyl anion has 8 electrons in its pi system. This makes it antiaromatic and highly unstable. The cycloheptatrienyl (tropylium) cation is aromatic because it also has 6 electronics in its pi system.
Aromaticity exists when there are $4n+2$ (i.e. 2, 6, 10, 14...) electrons in a planar, cyclic pi system. Planar, cyclic pi systems that have $4n$ (4, 8, 12, 16...) electrons are antiaromatic.
In the case of the cyclopentadienyl anion, there are 6 electrons in the pi system. This makes it aromatic. The cycloheptatrienyl anion has 8 electrons in its pi system. This makes it antiaromatic and highly unstable. The cycloheptatrienyl (tropylium) cation is aromatic because it also has 6 electronics in its pi system.
![What is non-hückel double bond of Aromaticity? [duplicate] image](https://file.echemi.com/fileManage/upload/community/category/category_picture_0416.jpg)
It's quite simple, really. Aromatic rings have all their relatively stable electronic orbitals filled and unstable ones empty. The 4n+2 rule just says that a conjugated ring will have an odd number of stable pi orbitals to be filled, that is 2n+1 orbitals to be occupied by two electrons apiece for some whole number n.
When you try to put 4n (or 4n+4) electrons into the ring as in cycloheptatrienyl anion, you have filled and empty states at the same energy level; that's what the Frost circle tells you. Sometimes molecules do that, but they generally are not as stable as what you get with all the filled states well below all the empty ones. The latter is what the proper 4n+2 electron count does.
It's quite simple, really. Aromatic rings have all their relatively stable electronic orbitals filled and unstable ones empty. The 4n+2 rule just says that a conjugated ring will have an odd number of stable pi orbitals to be filled, that is 2n+1 orbitals to be occupied by two electrons apiece for some whole number n.
When you try to put 4n (or 4n+4) electrons into the ring as in cycloheptatrienyl anion, you have filled and empty states at the same energy level; that's what the Frost circle tells you. Sometimes molecules do that, but they generally are not as stable as what you get with all the filled states well below all the empty ones. The latter is what the proper 4n+2 electron count does.
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Cycloheptatrienyl anion(tropylium anion) has 8 pi electron system, therefore it must be antiaromatic but the extra lone pair on the one carbon would cause that carbon to be become sp3 hybridized and put those extra elctrons in one of the sp3 orbitals. This would make it non-planar and non-aromatic.
Cycloheptatrienyl anion(tropylium anion) has 8 pi electron system, therefore it must be antiaromatic but the extra lone pair on the one carbon would cause that carbon to be become sp3 hybridized and put those extra elctrons in one of the sp3 orbitals. This would make it non-planar and non-aromatic.
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That happens because of Huckel's rule of aromaticity. The cycloheptatrienyl anion has 8 electrons which translates as 4n electrons, not the 4n+2 as implied by Huckel. If you are about to construct the MOs for both cycles, you'll find that putting 4n electrons will rise in a diradical molecule, not stable as it already sounds. A quick trick for doing that is using Frost's circle. All you need to do is inscribe the correspondent polygon in a circle and make sure that you have a vertex touching the circle as low as possible. The vertices will give you the qualitative energy of the MOs.
That happens because of Huckel's rule of aromaticity. The cycloheptatrienyl anion has 8 electrons which translates as 4n electrons, not the 4n+2 as implied by Huckel. If you are about to construct the MOs for both cycles, you'll find that putting 4n electrons will rise in a diradical molecule, not stable as it already sounds. A quick trick for doing that is using Frost's circle. All you need to do is inscribe the correspondent polygon in a circle and make sure that you have a vertex touching the circle as low as possible. The vertices will give you the qualitative energy of the MOs.
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Aromaticity exists when there are $4n+2$ (i.e. 2, 6, 10, 14...) electrons in a planar, cyclic pi system. Planar, cyclic pi systems that have $4n$ (4, 8, 12, 16...) electrons are antiaromatic.
In the case of the cyclopentadienyl anion, there are 6 electrons in the pi system. This makes it aromatic. The cycloheptatrienyl anion has 8 electrons in its pi system. This makes it antiaromatic and highly unstable. The cycloheptatrienyl (tropylium) cation is aromatic because it also has 6 electronics in its pi system.
Aromaticity exists when there are $4n+2$ (i.e. 2, 6, 10, 14...) electrons in a planar, cyclic pi system. Planar, cyclic pi systems that have $4n$ (4, 8, 12, 16...) electrons are antiaromatic.
In the case of the cyclopentadienyl anion, there are 6 electrons in the pi system. This makes it aromatic. The cycloheptatrienyl anion has 8 electrons in its pi system. This makes it antiaromatic and highly unstable. The cycloheptatrienyl (tropylium) cation is aromatic because it also has 6 electronics in its pi system.
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