At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
I was not thouht this deep. SO I actually cnt answr.
I wish there was a way for me to draw all this out lol
to show how this gets from the reactant to product
I.E. drawing mechanism of how this get s to reactant to product.
Oooh this is fun, what's that say at the top left, it's blocked out. I can show a plausible mechanism or should I try to give you hints?
@empty Wanted to see how this was actually done I'm gonna try to draw this out haha..
that problem came from the book i told you about
Ok :) I think numbering the product and the reactant and trying to see where things connect up would be the best route to helping redraw it, since that'll at least tell you precisely what carbons line up and what has to leave. Cool, I'll try to find a pdf of it online or something
For step one I thought the acid causes removal of water to form a carbonation. Then what happens is the double bond from the heterocyclic compound attacks that carbon intramolecularly. I thought that the double bond on the heterocyclic compound wouldn't react w acid at first because it's aromatic... This stuff kind of confuses me b/c it's hard to keep track of the carbons.
It is aromatic, but we can still add to aromatic compounds by reacting. If you recall, toluene (methylbenzene) is an ortho-para director when thinking about benzene, which is due to the fact that carbons are electron donating groups, so they're what's known as activators since they increase reaction rates at locations next to them. Now if we specifically draw out furan's resonance structures, we get: |dw:1437255124435:dw| So it will actually be more likely to occur there after all! :) I have to go with my brother to get lunch, I'll be back a little later to help out if you still need it.
@empty so you're saying that the heterocycle would serve as the nucleophile? I have yet to post that other question I had for you but I'll do that. hmm the lone pair of electrons in the oxygen would make that serve as a better nucleophile..
Kind of, the electrons are sort of everywhere in that molecular orbital. Well right now it's not the lone pair of electrons in the oxygen, really it's just sort of a way to show resonance of how electron density moves around. Really since furan is aromatic it will be planar with all the p-orbitals lined up like this: |dw:1437447547279:dw| I've tried to draw the p-orbitals all overlapping above and below the ring as a giant donut. So one pair of electrons is going to participate in the conjugation with the other two pairs on the carbon, but they're all delocalized, while the other lone pair on oxygen will be pointing straight out just like the hydrogen atoms attached to the carbon atoms. Actually instead of worrying about drawing, here are some better pictures: https://upload.wikimedia.org/wikipedia/commons/thumb/9/90/Benzene_Orbitals.svg/2000px-Benzene_Orbitals.svg.png So that's basically what I'm drawing, they're all combined, the "split up into separate orbitals" like this picture isn't quite what's happening: https://teachthemechanism.files.wordpress.com/2013/03/karty16_imgb.jpg But you can kind of see how they will overlap this way easier.
I see one of the lone pair of elections in that furan ring doesn't participate in the pi system because it's not in the same plane as the other orbitals
That picture really does explain it well, and also explains why the lone pair of electrons in say that oxygen don't participate in that pi system. I guess it's like if you draw out the orbitals for carbon, it has 2 electrons in 1s^2-2s^2-2p^2 configuration, and like only three of the orbitals are available to participate in bonding and that picture doesn't really tell you much about how carbon bonds in a molecule. so looking at the picture with that furan, found your links to be extremely helpful by the way!, the carbons have to be sp^2 hybridized, as well as the oxygen? meaning that we have one s orbital combining with two p orbitals to create new molecular orbitals that can participate in the bonding described in that figure. now i'm guessing if you have 3 electrons in those new sp^2 orbitals and the last one is in a p orbital that doesn't participate in the pi system. so one of those bonds will be a pi, the other will be a sigma. that last p orbital will be used to bond with hydrogen.
with all that said; I think i'll close this out now
Ok you said a lot of good stuff, you're right about oxygen being in an \(sp^2\) orbital so I'll try to explain that in a reasonable sorta way. So it's a result that I can show you how to derive using Schrodinger's equation that Huckel's numbers are what is the most favorable and lowest energy way for a molecule to "snap" into. So I guess for the moment you'll just have to accept that oxygen will spontaneously go into the \(sp^2\) orbital because it's a much lower energy state since it would be between having 4 \(\pi\) electrons or 6 \(\pi\) electrons, so at least you already know that aromatic compounds are stable probably so at least it's reasonable to assume it will happen when in doubt like this for future problems. I just say that so you don't feel like you have to "just memorize" that oxygen will randomly turn into \(sp^2\) sometimes. Now you mentioned something about 3 electrons, I didn't fully know what you meant by that so instead of me trying to figure it out I'll just go ahead and explain something and if it's what you're saying, then awesome we agree so nothing to worry about, at the very least I'll confirm what you know! So Oxygen has 6 valence electrons and carbon has 4, so when we draw furan with all of the \(sp^2\) orbitals on them (excluding the hydrogen atoms) we can draw it out as: |dw:1437626065513:dw| So you can sort of see in this drawing that I only included the valence electrons of oxygen and two of the carbon atoms, it's sort of difficult to see, but you can tell now that there are exactly 4 \(\pi\) electrons, 2 electrons on a lone pair, and the rest of the electrons are in \(\sigma\) bonds (some with Hydrogen atoms not shown). So make sure you can find all 14 electrons properly placed. Cool I hope this helps, I think that's basically everything for this part of the molecule but did we actually finish the synthesis? :P