kamepan
Registered: 10/09/09
Posts: 13
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Reply with quote | #1 |
Can someone explain to me as to why the second structure isn't aromatic? So even if the other ring isn't in the same plane, you have to take its bonds into account as well? Even so, wouldn't that be 10? (4n+ 2=10 n=2?) And if you did only the larger ring, wouldn't that still fit the rule? How do you determine if something is aromatic or not if both of the rings aren't in the same plane? |
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RA
Registered: 09/29/09
Posts: 22
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Reply with quote | #2 |
In that larger ring, there will always be a carbon atom that is going to hinder aromaticity. This is because it will have two single bonds on either side of it. Even if you move the double bonds around, there will always be two single bonds next to each other regardless of which carbon it is. Thus this means that that ring does not have a ring of closed p orbitals and therefore cannot be aromatic.
So in answer to your question, it's not the matter of the two rings being in the same plane but the continuity of the p orbitals.
Hope that helped!
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richardn90
Registered: 10/20/09
Posts: 18
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Reply with quote | #3 |
Hey guys, I was wondering along the same lines.. the fourth molecule is not aromatic. So even if a part of a molecule IS aromatic, if the entire thing isn't aromatic, then it isn't considered aromatic?
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DrH
Moderator
Registered: 09/22/08
Posts: 431
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Reply with quote | #4 |
The double bond on the right hand side cannot be ignored or excluded from the closed p orbital loop, even if this exclusion suggests the molecule is aromatic. The molecule has eight pi electrons in the p orbital loop, so it is not aromatic. |
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dhlee624
Registered: 10/08/09
Posts: 19
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Reply with quote | #5 |
It also said circle the molecules that are not aromatic, so if the molecule as a whole was not aromatic then it would have to be circled
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