Chemical Forums
Chemistry Forums for Students => Organic Chemistry Forum => Topic started by: kelaklub on May 22, 2006, 11:11:24 PM
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Due to all the help I received on this forum, I was able to answer most of the aromatic questions on my sample tests correctly. This one however has me a little stumped. It may be a typo, I don't know. The answer is stated as choice C.
But I thought that choice D also represents an Aromatic molecule, because
- It is cyclic
- All the carbons are sp2 hybridized.
- It satisfies Huckels rule, because I don't think the lone pairs of electrons will take part in the resonance because that would not make all the carbons sp2 hybridized.
Are these valid reasons? Thanks.
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As you said all the carbons are sp2 hybridrized, the lone pairs of that carbon can also be delocalized into the pi system (remember, there is a hydrogen attached to that carbon). Now the no. of e- in the pi system is 4 which cannot fulfill 4n + 2 rule. If the lone pair doesn't delocalize to pi system, this carbon will become sp3 hybridrized that blocks the delocalization of e- in pi system.
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I agree with kelaklub. If it's a singlet carbene, I would answer with E (ie. C and D).
The reason I would expect D to be aromatic in it's singlet state (that's with a lone pair on the C rather than a diradical) is because, when sp2 hybridised, you can put the lone pair in an sp2 orbital, leaving an empty p orbital => 2 electrons in the pi system => aromatic.
I don't think the lone pairs of electrons will take part in the resonance because that would not make all the carbons sp2 hybridized
I don't agree with that argument.
The only way anything on that c will interact with the pi system (take part in resonance) is if it is in a p orbital - so in the hypothetical reverse situation to what I described above, the lone pair is in a p orbital, and there is an empty sp2 hybrid orbital. All the carbons would still be still sp2! - but it is not aromatic (and so the cofiguratiion would be unstable) because the number of electrons is 4.
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(remember, there is a hydrogen attached to that carbon).
I don't think there is... I've never seen a trivalent carbene
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The problem probably meant to put a negative formal charge on the carbon atom with the lone pair making it a standard carbanion. In this case, because there is a hydrogen and two C-C bonds occupying the sp2[/sub] orbitals, the lone pair must be in the p-orbital creating a system with 4n electrons which is antiaromatic by Huckle's criteria.
However, as the problem is written with no formal charge on the carbon, I agree with Dan's answer. But I really doubt someone would mean to put a carbene into a problem like this.
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Due to all the help I received on this forum, I was able to answer most of the aromatic questions on my sample tests correctly. This one however has me a little stumped. It may be a typo, I don't know. The answer is stated as choice C.
But I thought that choice D also represents an Aromatic molecule, because
- It is cyclic
- All the carbons are sp2 hybridized.
- It satisfies Huckels rule, because I don't think the lone pairs of electrons will take part in the resonance because that would not make all the carbons sp2 hybridized.
Are these valid reasons? Thanks.
Only pyridine is aromatic (C)
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Only pyridine is aromatic (C)
What is your justification for D not being aromatic?
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I think that the authors of this question made a silly mistake, in that they did not include the formal charge. Let's assume for now that it is indeed a carbanion, and not a carbene. To have p orbitals available for resonance at every carbon, the carbanion must be sp2 hybridized. At the same time, doing this does not satisfy Hückel's rule. If we leave the carbon sp3 hyrbridized, then it is not a fully conjugated system. So, assuming it is indeed a carbanion, it is certainly not aromatic.
Taking the case of the carbene, well I don't know enough about carbenes to answer it, honestly. But from the context of the question, I highly doubt the authors would intend for you to consider a carbene.
So, my answer: C
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Only pyridine is aromatic (C)
What is your justification for D not being aromatic?
As Winga after Huckel said - 4n+2 rule
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But look at the structure of a singlet carbene, http://wngr343-pc1.science.oregonstate.edu/gablek/CH630/carbenorbs.gif
The empty p orbital (LUMO) overlaps with the pi system, giving 2 electrons in the pi system, ie n=0 for Huckel's rule => aromatic. The HOMO is orthogonal to the pi system.
If it was a carbanion, then it wouldn't be aromatic, but it isn't, it's a carbene.
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Will the structure D achieve maximum multipilicity?
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i think there is nothing too much in this question which worth such a lengthy debate. as suggested above, the system D doesnt obey the 4n+2 rule. hence it is not aromatic.
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the system D doesnt obey the 4n+2 rule. hence it is not aromatic.
My point is that D will obey the 4n + 2 rule in the singlet state (where n = 0). The LUMO of the singlet state is a p orbital, see link above.
Sure, the triplet state will not be aromatic, but I expect the singlet state is lower in energy than the triplet due to the aromatic stabilisation of the singlet state => the ground state the carbene D is the singlet state which is aromatic.
I have drawn a pretty picture http://users.ox.ac.uk/~some1599/Images/SingletCarbeneD.jpg
If there is a formal negative charge then, I agree, it will not be aromatic.
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Can the structure D undergo aromatic reactions? (e.g. electrophilic aromatic substitution)
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Well, I have never made it so I dont know, but I would have thought that it's reactivity resembles that of a carbene more than benzene. An example would be (rapid) insertion to C-H bonds.
I don't think it would be stable.
If the point you are getting at is "If it doesn't react like an aromatic compound, is it really aromatic?", then I can see exactly where you are coming from! Technically it is aromatic, in the sense that it obeys the 4n+2 rule, but in practice I would be suprised if this label is meaningful.
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I can't believe anyone has picked up on Dan's point! I think he is completely right.
The notation used means it must be a singlet carbene (if it was triplet the electrons would be one above the carbon and one below).
Effectively it is the same as cycloprop-2-en-1-ylium (cyclopropene with a positive charge) but with a pair of electrons in a sp2 orbital rather than a C-H bond. If that is aromatic then surely D is as well.
I think everything Dan has said is right, there is no formal negative charge etc. etc.
The only problem with this carbene is that it might react with itself, it would make a really cool addition polymer though- like a long chain of back-to-back cyclopropanes, or a chain of "bicyclo[1.1.0]butane"s as ChemDraw calls it!
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cyclopropene with a positive charge
Yes! It's just a deprotonated one, I'm glad someone can see what I'm getting at.
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Not to over-discuss this topic, but I wanted to throw my two cents in:
As Dan has said, if it is a singlet carbene, it's aromatic, if it's a carbanion, then it's anti-aromatic.
My guess, however, is that it's a typo in the problem and they intended for it to be a carbanion.
By the way, the triplet form of a carbene is about 10 kcal/mol lower in energy than the singlet form, according to calculations. There are plenty of examples of both types of carbenes, however; that energy difference is for the simplest example: H2C:
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the singlet form of a carbene is about 10 kcal/mol lower in energy than the singlet form
???
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He means the triplet form of a carbene is about 10 kcal/mol lower in energy than the singlet form.
That what I also found from a book.
Of course, I (everyone discussing in here) got what Dan said, and I also agree what he said.
About the triplet and singlet forms, they should be interchangable, so we can regard it as aromatic.
If under a magnetic field, it is not easy to tell!
I wonder where is the source of this question!
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Sorry, I fixed the mistake.
Whether or not the carbene has time to isomerize to whichever form is more stable will depend on how long it hangs around for before it reacts.
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Of course, I (everyone discussing in here) got what Dan said, and I also agree what he said.
About the triplet and singlet forms, they should be interchangable, so we can regard it as aromatic.
:-[ My bad! I just read what a couple of people wrote before I posted, dismissing the carbene as antiaromatic, and got a little bit worked up lol!
Another thing in favour of the singlet state in D, as well as the aromaticity, is the smaller bond angles in the molecule, which allow a greater space than normal for the electrons in the sp2 orbital, so the "incentive" for the electrons to be in the triplet state is reduced further. Also, the general idea of the triplet state being favoured (due to stability) is useful, sometimes, but useless when there are electron-rich substituents on the carbene molecule, like in :CCl2, where the electrons would be in the singlet state (but, by coincidence :CCl2 is formed in the singlet state!).
Sorry for going on about carbenes, I just find them really interesting! [Edit: especially the Arndt-Eistert synthesis, and the Hofmann rearrangement!]
I wonder where is the source of this question!
So do I! There seems to be so many mistakes, though, in chemistry books nowadays. I've even found at least 3 'chemistry' errors in my organic chem book (the Clayden, Greeves, Warren, Wothers one), and my current A-Level book has at least 6 'chemistry' errors in it, one being half a page devoted to a mechanism (acid catalysed halogenation of acetone), which is entirely wrong! When my teacher tried to write out the mechanism out for that reaction last year he wrote the one in the book, and without reading up about this reaction I could tell it looked wrong, and suggested a mechanism which happened to be right after checking it out afterwards. Good thing you don't need to know that mechanism for the A-Level, the book was just trying to demonstrate the rate equation and its relation to the RDS.
I think there seems to be a lack of chemistry book editors- I am yet to find a chemistry book without a mistake in it! ;)
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cyclopropene is also aromatis because the ring is polar.
4n+2=2,pi electrons(n=0).So the answer is E
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cyclopropene is also aromatis because the ring is polar.
4n+2=2,pi electrons(n=0).So the answer is E
Cyclopropene (C3H4) is not aromatic - cyclopropenyl cation (cycloperopenylium) C3H3+ is aromatic according to Huckel rule 4n+2