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To simplify the question, they are basically asking which molecule has most conjugated double bonds. How do you do that?😕
Answer is C but why
conjugation
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To simplify the question, they are basically asking which molecule has most conjugated double bonds. How do you do that?😕
Answer is C but why
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Conjugation is a path through atoms that alternates single --> double bonds.
The longest chain in A is two single/doubles.
In B the longest chain is one. The doubles are too far apart.
With C, you could start on an oxygen and get a chain with 4 doubles and 3 singles.
And D has two groups with 3 single / double patterns. But these two groups are not connected by just one single bond so they can't be combined into a pattern with 6 doubles.
Molecule C, with 4 doubles, wins. And now I have a hankering for In n Out burgers for some reason.
The longest chain in A is two single/doubles.
In B the longest chain is one. The doubles are too far apart.
With C, you could start on an oxygen and get a chain with 4 doubles and 3 singles.
And D has two groups with 3 single / double patterns. But these two groups are not connected by just one single bond so they can't be combined into a pattern with 6 doubles.
Molecule C, with 4 doubles, wins. And now I have a hankering for In n Out burgers for some reason.
Isn't C, the only one with a fully conjugated system (only because the C-O bond provides available p orbitals)?
For the other molecules, it doesn't really matter how many conjugated bonds are present but rather the fact that their system is disrupted --so they don't have a fully conjugated system. Therefore it's choice C.
(I hope this is a passage-based question and that we weren't supposed to know the relationship b/w color and fully conjugated systems out of the blue...)
For the other molecules, it doesn't really matter how many conjugated bonds are present but rather the fact that their system is disrupted --so they don't have a fully conjugated system. Therefore it's choice C.
(I hope this is a passage-based question and that we weren't supposed to know the relationship b/w color and fully conjugated systems out of the blue...)
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😕 I'm not sure what you are getting at here. You are pushing the concept of "fully" conjugated and I don't know what "fully" is supposed to mean, or why that would have any bearing on the original question.
Molecule C has that weird amine tail so it isn't like the whole molecule is participating in the conjugation. Since we are therefore only talking about the conjugation of parts of a molecule, molecule A certainly has the classic pattern you also see in 1,3-pentadiene, which is kind of the poster-child for conjugation.
I think they are asking you to select the best conjugated system because it is capable of absorbing not just uV but visible spectra.
On the first round B is out because its not conjugated the sp2 Carbons are double bonds that are isolated from eachother. D is then out because it too has a disrupted system of conjugation. Leaving A and C. A is a conjugated system (4 sp2 Cs- when you do the resonance structures you get 3 with two that are uncomfortable with charges). When you look at C you have six sp2 Cs plus the 2 oxygen also sp2 hybridised. There are a lot more resonance structures.proving that has pretty much all its atoms communicating with eachother. I started doing them there are more than 3 but for the most part the -ve charge is on the O and the + on the C so they are more comfortable. Its a complete roustabout too all of the sp2 carbons and O participate in the conjugation system without interuption. But the real story is how the molecular orbitals are arranged and they fill up lowest energy to highest energy. C structure has the lowest energy state at any one time and it is easy then for visible light to boot one of those lower energy electrons fom the HOMO (highest occupied molecular orbital) crowd into the LUMO (lowest unoccupied molecular orbital) crowd giving you the colour- somebody in that structure is always ready to go HOMO to LUMO.
On the first round B is out because its not conjugated the sp2 Carbons are double bonds that are isolated from eachother. D is then out because it too has a disrupted system of conjugation. Leaving A and C. A is a conjugated system (4 sp2 Cs- when you do the resonance structures you get 3 with two that are uncomfortable with charges). When you look at C you have six sp2 Cs plus the 2 oxygen also sp2 hybridised. There are a lot more resonance structures.proving that has pretty much all its atoms communicating with eachother. I started doing them there are more than 3 but for the most part the -ve charge is on the O and the + on the C so they are more comfortable. Its a complete roustabout too all of the sp2 carbons and O participate in the conjugation system without interuption. But the real story is how the molecular orbitals are arranged and they fill up lowest energy to highest energy. C structure has the lowest energy state at any one time and it is easy then for visible light to boot one of those lower energy electrons fom the HOMO (highest occupied molecular orbital) crowd into the LUMO (lowest unoccupied molecular orbital) crowd giving you the colour- somebody in that structure is always ready to go HOMO to LUMO.
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There are no such thing as "conjugated bonds." Only conjugated systems. These are compounds, or parts of compounds like you said MT Headed, that contain 2 or more double or triple bonds separated by a single bond.
C is the only fully conjugated molecule! (Note that Nitrogen has a lone pair--empty p orbital--that will contribute to this conjugated system)
For a great, short review, this link is awesome. You probably already know it, but if not...
http://masterorganicchemistry.com/2011/03/08/are-these-alkenes-conjugated/
C is the only fully conjugated molecule! (Note that Nitrogen has a lone pair--empty p orbital--that will contribute to this conjugated system)
For a great, short review, this link is awesome. You probably already know it, but if not...
http://masterorganicchemistry.com/2011/03/08/are-these-alkenes-conjugated/
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But your link says you block conjugation if you are connected to four atoms (i.e. sp3 hybridized), and that conjugation happens because of resonance. That amine tail is attached to a carbon with four atoms, which is sp3 hybridized, which is not going to participate in resonance. I still disagree that the nitrogen can participate in the conjugation. I think he is stuck in a corner by himself.
The amine group isn't the issue its the occupation of molecular orbitals that is the issue and which molecule is the most stable. Everybody should revise MO theory the MCAT sneaks that in where ever they can.This q is not like colour coming from electron jumping to a lower energy orbital this question is about electrons capturing energy to go to the higher unoccupied orbital and what they don't absorb is the colour you see.
"A" is a conjugated system but it is not the best one there.
"A" is a conjugated system but it is not the best one there.
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There most certainly are "conjugated double bonds." In a conjugated system, you can have isolated double bonds, so what would you call the alternating double bonds in a conjugated system?
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So far the only thing I've found related to those that is on the MCAT is the fact that UV spectroscopy requires pi bonds, and conjugated pi bonds show up more.. or something.
I guess I need to review that too lol.
