[ipromiseidomyhw]

I've been studying lately for my test, and I ran into some problems I just can't figure out, so I provided pictures.



In this picture, the answer in the back of the book is that this molecule is aromatic. But doesn't the O atom inside the ring have an sp3 hybridization, and therefore not planar? Shouldn't this molecule be nonaromatic?




For this problem, I  know a positive charge means there has been a loss of electrons, so Oxygen is missing a lone pair, therefore shouldn't there be 4 pi electrons, which doesn't satisfy Huckels rule, so it's antiaromatic? This book said this was aromatic as well.



For this last picture, I provided the answer in the back of the book of the product, and I don't understand why the Bromine was placed where it was on the ring.
In red I put where I thought ortho, meta, para locations were, and since it is an activating group over a deactivating group, I wanted to place it on the para position, but the book put it elsewhere. Why is the Br in that location?



Thanks for the help I stink at chemistry lol, hopefully my thinking is somewhat okay

[mjc123]

1. No, it is sp²; one lone pair is in a p orbital wich overlaps with the π bonds to give an aromatic system. In an aliphatic ether it would be sp³, but if changing to sp² allows the molecule to be stabilised by resonance, then this will happen.

2. There is no loss of electrons; one of the lone pairs (the one not involved in the aromatic system) bonds to a proton to give a molecule with overall positive charge. The aromatic sextet is not affected.

3. You have an activating, o,p-directing alkyl group and a deactivating, m-directing carbonyl group. Therefore the bromine will go where it is shown, p to the alkyl group and m to the carbonyl (o to the alkyl is sterically hindered).

[spirochete]

Somebody on the forum recently asked a question almost identical to yours regarding aromaticity. I gave a simplified answer using VSPR theory and Corribus gave a more in depth answer talking about molecular orbital theory: http://www.chemicalforums.com/index.php?topic=75811.msg275550#msg275550.

Regarding the last question: the bromine in the product is meta to the ketone and indeed also para to the alkyl group. So it is not a surprising result. The other reasonable answer would be to put the bromine ortho to the alkyl group and meta to the alkyl group (on "the bottom" of the ring) but this must be a minor product for steric reasons.

[caters]

1. No, it is sp²; one lone pair is in a p orbital wich overlaps with the π bonds to give an aromatic system. In an aliphatic ether it would be sp³, but if changing to sp² allows the molecule to be stabilised by resonance, then this will happen.

2. There is no loss of electrons; one of the lone pairs (the one not involved in the aromatic system) bonds to a proton to give a molecule with overall positive charge. The aromatic sextet is not affected.

3. You have an activating, o,p-directing alkyl group and a deactivating, m-directing carbonyl group. Therefore the bromine will go where it is shown, p to the alkyl group and m to the carbonyl (o to the alkyl is sterically hindered).

If oxygen has 2 lone pairs and 2 bonds, none of which are double than O is sp3 hybridized just like how 0 lone pairs and 4 single bonds on carbon = sp3 hybridized. The O has 2 lone pairs and 2 single bonds(thus 2 sigma bonds and 0 pi bonds) in the first molecule and is thus sp3 hybridized and so the first one is not aromatic for 2 reasons:

1) there are 4 pi electrons which violates Huckels rule
and
2) the O is sp3 hybridized and thus the 5 membered ring will be bent at the O

The second one isn't aromatic either because again Huckels rule is violated. It is also a very unstable molecule since the O is positive and O HATES being positive which is why H3O+ is in a complex of ions making H9O3+ which itself is in a complex of ions and so on. It will likely pick up a proton and leave as water making you have cyclo-1,3-butadiene and acyclic 1,3-butadiene of which the cyclic form is not aromatic.

The bromine is meta to the carbonyl and para to the alkyl group which is the least sterically hindered position because if it was meta or ortho to the alkyl group or if it was ortho to the carbonyl it would have quite a bit of steric hinderance, especially if it was ortho to that dimethyl group.

[kriggy]

The second one isn't aromatic either because again Huckels rule is violated. It is also a very unstable molecule since the O is positive and O HATES being positive which is why H3O+ is in a complex of ions making H9O3+ which itself is in a complex of ions and so on. It will likely pick up a proton and leave as water making you have cyclo-1,3-butadiene and acyclic 1,3-butadiene of which the cyclic form is not aromatic.

While I agree with you that its not aromatic, I dont agree with the rest of your reasoning.  You have oxygen with positive charge and it will pick up a proton with same positive charge?

[caters]

 you have O with 1 hydrogen bonded to it. It is positive. Water(H2O) is more stable than positively charged hydroxide(OH+) At the same time the water leaves a single C-C bond will form.

Another possible thing the oxygen could do to get rid of the positive charge(and this is probably more likely since a +2 charge on O is even more unstable than a +1 charge on O.) is to have a hydroxide anion pick up the proton leaving the 2 electrons on the oxygen and you get a cyclic ether(also know as epoxide) as the result.

[mjc123]

In the face of your denials, I reassert my statements.

If oxygen has 2 lone pairs and 2 bonds, none of which are double than O is sp3 hybridized just like how 0 lone pairs and 4 single bonds on carbon = sp3 hybridized.

Not necessarily. A fundamental difference between sp³ carbon and oxygen (or nitrogen) is that a C with bonds to 4 other atoms or groups is constrained to be tetrahedral, because of the repulsion between the attached groups. Electrons can move around more freely, and it only needs a small adjustment of atomic positions for an sp³ O or N to adopt an sp² configuration, with a lone pair in a p orbital available to overlap with a pi system. sp³ may be more stable in isolation, but if sp² allows resonance stabilisation, then that will happen. (This also applies to C with 3 bonds; thus cyclopentadiene is not aromatic, but Cp^- is.)

The O has 2 lone pairs and 2 single bonds(thus 2 sigma bonds and 0 pi bonds) in the first molecule and is thus sp3 hybridized and so the first one is not aromatic

You can draw resonance structures (try it) in which one or other of the C-O bonds is double, so they have partial double-bond character.

1) there are 4 pi electrons which violates Huckels rule
and
2) the O is sp3 hybridized and thus the 5 membered ring will be bent at the O

There are 6 pi electrons (2 from the oxygen p orbital) and the molecule is aromatic. The oxygen is sp² hybridised and the ring is planar (an experimental fact).

The second one isn't aromatic either because again Huckels rule is violated.

This also is aromatic; there are again 6 pi electrons, two from the oxygen.

O HATES being positive

Let's not bring the pathetic fallacy into it.

It will likely pick up a proton and leave as water making you have cyclo-1,3-butadiene and acyclic 1,3-butadiene of which the cyclic form is not aromatic.

This is just nonsense; it makes no chemical sense.

Another possible thing the oxygen could do to get rid of the positive charge(and this is probably more likely since a +2 charge on O is even more unstable than a +1 charge on O.) is to have a hydroxide anion pick up the proton leaving the 2 electrons on the oxygen

That's more reasonable. But who said there were any hydroxides around? The question just asked whether the molecule was aromatic, not about what reactions it would undergo.

you get a cyclic ether(also know as epoxide) as the result.

An epoxide is a cyclic ether with a 3-membered ring (so called because it can be considered as the result of adding an O atom across a C=C bond). Other cyclic ethers are not called epoxides.

The bromine is meta to the carbonyl and para to the alkyl group which is the least sterically hindered position because if it was meta or ortho to the alkyl group or if it was ortho to the carbonyl it would have quite a bit of steric hinderance, especially if it was ortho to that dimethyl group.

True as far as it goes, but it's not all about steric hindrance (what about the position para to the carbonyl?) You have to consider the directing influence of substituents too.

[caters]

In the face of your denials, I reassert my statements.

If oxygen has 2 lone pairs and 2 bonds, none of which are double than O is sp3 hybridized just like how 0 lone pairs and 4 single bonds on carbon = sp3 hybridized.

Not necessarily. A fundamental difference between sp³ carbon and oxygen (or nitrogen) is that a C with bonds to 4 other atoms or groups is constrained to be tetrahedral, because of the repulsion between the attached groups. Electrons can move around more freely, and it only needs a small adjustment of atomic positions for an sp³ O or N to adopt an sp² configuration, with a lone pair in a p orbital available to overlap with a pi system. sp³ may be more stable in isolation, but if sp² allows resonance stabilisation, then that will happen. (This also applies to C with 3 bonds; thus cyclopentadiene is not aromatic, but Cp^- is.)

The O has 2 lone pairs and 2 single bonds(thus 2 sigma bonds and 0 pi bonds) in the first molecule and is thus sp3 hybridized and so the first one is not aromatic

You can draw resonance structures (try it) in which one or other of the C-O bonds is double, so they have partial double-bond character.

1) there are 4 pi electrons which violates Huckels rule
and
2) the O is sp3 hybridized and thus the 5 membered ring will be bent at the O

There are 6 pi electrons (2 from the oxygen p orbital) and the molecule is aromatic. The oxygen is sp² hybridised and the ring is planar (an experimental fact).

The second one isn't aromatic either because again Huckels rule is violated.

This also is aromatic; there are again 6 pi electrons, two from the oxygen.

O HATES being positive

Let's not bring the pathetic fallacy into it.

It will likely pick up a proton and leave as water making you have cyclo-1,3-butadiene and acyclic 1,3-butadiene of which the cyclic form is not aromatic.

This is just nonsense; it makes no chemical sense.

Another possible thing the oxygen could do to get rid of the positive charge(and this is probably more likely since a +2 charge on O is even more unstable than a +1 charge on O.) is to have a hydroxide anion pick up the proton leaving the 2 electrons on the oxygen

That's more reasonable. But who said there were any hydroxides around? The question just asked whether the molecule was aromatic, not about what reactions it would undergo.

you get a cyclic ether(also know as epoxide) as the result.

An epoxide is a cyclic ether with a 3-membered ring (so called because it can be considered as the result of adding an O atom across a C=C bond). Other cyclic ethers are not called epoxides.

The bromine is meta to the carbonyl and para to the alkyl group which is the least sterically hindered position because if it was meta or ortho to the alkyl group or if it was ortho to the carbonyl it would have quite a bit of steric hinderance, especially if it was ortho to that dimethyl group.

True as far as it goes, but it's not all about steric hindrance (what about the position para to the carbonyl?) You have to consider the directing influence of substituents too.

It isn't false that O hates being positive. I mean H3O+ isn't really H3O+ but an individual ion in a cation complex(to spread the positive charge so that it isn't localized to 1 oxygen atom). In fact if oxygen is charged it prefers negative since it has the second greatest electronegativity. This is how come the positively charged OH is so unstable as is and is very likely going to deprotonate due to hydroxide anions in solution and form a cyclic ether.

I don't consider 1 of the lone pairs on O to be pi electrons in the structure with C=C bonds because they aren't in a double bond(which would more likely happen if the O wasn't there at first and O is introduced and form a carbonyl than be in the plane of the ring) and even in the resonance structures with a C=O double bond in the ring and not sticking out of the ring(which as I said is very unlikely since that makes oxygen positive) there are still only 2 double bonds, 1 C=C and 1 C=O and there is also a carbanion in this case. Thus my saying that there are 4 pi electrons, not 6 in the first one.

In the one with the positive OH there is 1 lone pair on the oxygen and again in any resonance structure 2 pi bonds which means 4 pi electrons which means once again that it is not aromatic.

[Dan]

Caters: you are out of your depth, posting nonsense and causing confusion.

Both are aromatic. I will post diagrams later to explain this - I'm away from my computer right now.

[caters]

I am not posting nonsense, especially about positively charged oxygen. 2 pi bonds = 4 pi electrons and because of how the first 2 have 2 pi bonds per resonance structure and because of the fact that R'-C=O-C-R is a very unlikely resonance structure because oxygen is unstable with a pi bond and 2 sigma bonds making it be O^1+ and the fact that there is a carbanion when the oxygen could just as easily become an oxyanion(which couterintuitively is more stable(I mean that oxyanions are extremely reactive and so they are very unstable(and I thought that it was even more unstable than positively charged oxygen)) through a ring opening reaction, we can safely say that neither of the 2 are aromatic, even the one where oxygen isn't positive at first.

[Corribus]

You are making erroneous conclusions because you are taking snapshots of the problem without considering the context.

Furan is aromatic despite the transient positive charge such resonance structures place on the oxygen. Yes, removing electron density from an electronegative element like oxygen requires an energy sacrifice, but that is more than made up from the energy gains deriving from resonance stabilization across the entire molecule.  (You're not placing a whole unit of charge on the oxygen anyway. It's delocalized and so the charge is "spread out".)

[Dan]

Here:

[orgopete]

I thought this was sufficiently elementary that I was surprised that it should evoke controversy. Although this was not part of the question, I thought everyone might understand that pyridine and a pyridinium cation are aromatic. It has the same elements to support aromaticity as pointed out in Dan's most recent post. If you can understand why pyridine is aromatic, then the others should follow.