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spectroscopy quiz: on aromaticity (huckel's rule)

Posted: Wed Jan 15, 2020 4:40 pm
by lotus0618
So i came across this question on the quiz, can you please show me how to identify which electrons are pi electrons (aside from the obvious pi electrons in the pi bond of a conjugated ring). For instance, how am I supposed to know if the lone pairs on this ring count towards aromaticity or not. Ex: the picture (attached with this question)-why are there 6 electrons in this? but pyridine's lone pair is considered to be pi electrons? thank you!

Re: spectroscopy quiz: on aromaticity (huckel's rule)

Posted: Thu Jan 16, 2020 12:43 am
by NS_Tutor_Mathias
Right, lone pairs can participate in pi bonds, and the oxygen on furan is therefore going to be sp2 hybridized and its lone pair will be participating in resonance. They'll pretty much always count >if< they can be used to satisfy Hueckel's rule. I don't think the MCAT features a lot of cases where you'll need to overthink this - just keep an eye out mostly for lone pairs in rings, and if they can turn it aromatic, they will.

The most common exception would be cases where the atom whose lone pair you care about is already sp2 hybridized. But that is honestly too rare of a case to worry about, so don't.

Re: spectroscopy quiz: on aromaticity (huckel's rule)

Posted: Thu Jan 16, 2020 4:21 pm
by lotus0618
Wait, I still don't understand why there aren't 8 electrons instead of 6 electrons in a furan...
So long as the lone pair can participate in the resonance, we will consider them to be pi electrons? If this is the case, why don't we consider the lone pair on N of Pyridine to be pi electrons and thus Pyridine would satisfy the Huckel's rule?

Re: spectroscopy quiz: on aromaticity (huckel's rule)

Posted: Thu Jan 16, 2020 7:11 pm
by NS_Tutor_Mathias
The system just trends towards low energy states. So you don't have to worry about 'extra' lone pairs gumming up the works. It just grabs the ones that are actually helpful and has those participate in pi bonds.

So for example, purine is an aromatic heterocycle:
https://en.wikipedia.org/wiki/Purine

But it has waaaaaaaaaaay too many lone pairs by your reasoning. The answer to that is easy: Take only the ones you need to fulfill Hueckel's rule. On the flipside, you can of course not decide to count or not-count already existing pi bonds. Those must be counted.

Re: spectroscopy quiz: on aromaticity (huckel's rule)

Posted: Thu Jan 16, 2020 8:25 pm
by lotus0618
I know this concept is supposed to be easy but it's harder for me to understand your explanation via texts here. Based on what you said, then all conjugate rings could satisfy the Huckel's rule then cause I can decide if i want to count the electron pairs or not. With that furan molecule, we decide to consider only on extra lone pair to make a total of 6 instead of 8 cause that would give us an integer for n. Is this what you meant?

Re: spectroscopy quiz: on aromaticity (huckel's rule)

Posted: Thu Jan 16, 2020 11:00 pm
by NS_Tutor_Mathias
lotus0618 wrote:
Thu Jan 16, 2020 8:25 pm
I know this concept is supposed to be easy but it's harder for me to understand your explanation via texts here. Based on what you said, then all conjugate rings could satisfy the Huckel's rule then cause I can decide if i want to count the electron pairs or not. With that furan molecule, we decide to consider only on extra lone pair to make a total of 6 instead of 8 cause that would give us an integer for n. Is this what you meant?
Pretty much. Not sure what you mean with 'conjugate rings' (a conjugated ring system is the definition of aromaticity, yes), but you are totally on the right track. Aromatic situations are more energically favorable, so they'll happen whenever they're easily possible. But having excess lone pairs doesn't preclude aromaticity. Those extra lone pairs just don't participate in the conjugated system, that is all.

If visual problems help you work through this, look at hypoxanthine below - the aromatic polyheterocycle. Notice how many excess lone pairs it has, versus just the 10 electrons that must be participating in resonance. Edit: Looking at it for an extra second, the lone pairs from the extra sp2 hybridized nitrogens cannot participate in resonance anyway. That is entirely my bad. Hypoxanthine turns out to be fairly straight-forward, since each nitrogen with only a lone pair (no pi bonds) will be contributing it's lone pair, and the others with already-existing pi bonds will contribute those.

Also take a look at the two states of imidazole below. What does that suggest about the aromaticity of histidine at low and at high ph?

Re: spectroscopy quiz: on aromaticity (huckel's rule)

Posted: Thu Jan 16, 2020 11:12 pm
by NS_Tutor_Mathias
As an added bonus, check these out for an idea of under which conditions a ring system with pi bonds might NOT be aromatic.

Re: spectroscopy quiz: on aromaticity (huckel's rule)

Posted: Wed Jan 22, 2020 10:56 pm
by lotus0618
I spent some time on this and sitll don't understand how you know that the lone pairs form the extra sp2 hybridized nitrogens cannot participate in resonance. I thought you could spread the lone pairs of sp2 nitrogen around. 2) I just discovered an interesting thing in my note to satisfy the Huckle's rule, a compound must also have continuous ring of p orbitals with NO sp3 carbons interupting. In this hypoxanthine molecule, it has sp3 so isn't it not aromatic? 3) From the previous post, you explained that we can choose which lone pairs to contribute to resonance, but if that's the case then all compounds can be aromatic then cause we can manipulate in such a way that total of electrons will produce an integer.

4) Regarding the 2 states of imidazole below, my answer to your question is that at low pH, the ring of histidine wouldn't be aromatic. But at high pH, it would be aromatic.

5) A ring system with pi bonds might not be aromatic under low pH cause the third molecule (from left to right) would gain a H. Is this correct?

Re: spectroscopy quiz: on aromaticity (huckel's rule)

Posted: Thu Jan 23, 2020 1:11 am
by NS_Tutor_Mathias
1) You can only either contribute what is in an existing pi bond OR a lone pair, not both.

2) Your notes are correct. I have been dancing around stating this outright a bit: An sp3 atom with a free lone pair can instead hybridize to sp2, in which case it will again have a free p orbital. This is why it can contribute to the conjugated system. This is also why it can only contribute one lone pair.

3) Not at all. Most compounds don't have lone pairs whatsoever. Where will you find any lone pairs in cyclopentane?

4) Correct. when the lone pair is present, it can participate in resonance.

5) What you said is true, but I wanted to point out that cyclobutene itself is not aromatic, despite looking like it should be (two alternating pi bonds).



New challenge question on related material: Consider the peptide bond. Knowing what you know now about lone pairs and unexpected sp2 hybridization, does this influence the geometry of peptide bonds?

Study this image: https://en.wikipedia.org/wiki/Peptide_s ... upling.svg

And maybe download it, circle in red where you expect sp2 character and in blue where sp3, and where you would therefore expect free rotation.

Re: spectroscopy quiz: on aromaticity (huckel's rule)

Posted: Wed Jan 29, 2020 12:20 am
by lotus0618
5) the cyclobutene with only one double bond-shouldn't this be aromatic cause the interger is 0

Challenge: the lone pair and unexpected sp2 hybiridzation shouldn't affect the geometry of peptide bonds cause they contribute to resonance

Based on the picture you sent to me. The carbonyl (in blue print) is sp2, C (connected to R2 in red print) is sp3. There's restricted rotation around the bond of carbonyl connected to an amine cause that's where the resonance takes place (it's stable and thus we cannot rotate the bond around there)