Is this valid:
Good try but that's not it. Could you tell me what resonance means first?
Why is that not valid?
Do you remember your book showing you something like this? Can you tell me what's going on in this picture? It's not a trick question. Honestly. heh..
is mine just not valid because it violates the octet rule since there are H's involved as well?
Lets not worry about your problem yet. Lets first see if you know what a resonance structure means. Do you know what it means?
Not to be rude, but I would love it if you could answer my question even briefly. I know what resonance is, I can assure you of that. :)
I don't believe you. At least try and meet me half way. :grumpy:
It's a way of showing more than one lewis structure where they are sort of moving in and out of each
Okay but that definition is a bit complicated. In laymen terms, it means a double bond can be in more then one place. Look at the picture I showed you earlier, it's clearly showing how the double bound can move around the ring. That's what resonance is. Now compare that with what you drew. It doesn't show any resonance. It doesn't show all the places a double bound can be in that ring.
After we figure out where the double bonds can go in your problem, we can then draw the arrows. Is this a question you pulled from your book? Can you post the question?
I know it doesn't show all the places, I'm asking if that is ONE of them. I don't think it is, since wouldn't that mean the carbon would have 5 bonds? (another carbon, the double and 2 hydrogens) ?
This is correct. When learning to draw resonance structures, it is sometimes helpful to explicitly draw all of the hydrogens in order to prevent mistakes like this.
Physically, a resonance structure like this does not make sense because resonance is interpreted as the sharing of electrons among unhybridized p-orbitals. In the cyclohexyl carbanion you drew, only one carbon atom could potentially have an unhybridized p-orbital (the carbon with the lone pair). However, none of the neighboring atoms have p-orbitals capable of overlapping with the lone pair on the carbanion, so resonance is not possible.
Thanks. My question was because of this:
I was wondering if you made the C have a negative formal charge, if that was the same thing as pulling off a hydrogen/proton. But I guess the negative charge is because of another electron around the atom, not the loss of a hydrogen, which would mean it would still have 2 attached to it.
is this correct?
The carbanion (carbon with a negative charge) shown would be produced by a loss of a proton (H+ ion) from cyclohexane. In place of the hydrogen that gets lost is a lone pair formed by one electron originally from the carbon and one electron that originally came from the hydrogen.
In practice you would rarely see a compound like this and it would be very unstable. You would need an extremely strong base to remove a proton from an alkene like cyclohexene.
So even though we remove a H, for purposes of bonds we pretend it is still there? Is what I said above correct?
No. You remove all of the hydrogen except for one of its electrons, which stays attached to the carbon atom. They hydrogen is completely gone and all that remains is an electron. In effect you are replacing a chemical bond to hydrogen with a lone pair.
This is the same thing that happens when you deprotonate an acid like in the case of
HCl --> H+ + Cl-
So why is the 2nd step of my resonance structure invalid, then?
Perhaps pictures would help here (see attachment).
Figure A shows the formation of the cyclohexyl anion. As I mentioned on my first post, a very strong nucleophile would remove the hydrogen from cyclohexene. The two electrons in the C-H bond that is broken are left on the carbon atom as a lone pair.
Figure B shows the resonance structure that you proposed (with hydrogen atoms and lone pairs explicitly shown). As mentioned previously, the resonance structure involves a carbon atom with five chemical bonds, a situation which is impossible.
So my 2nd structure isn't valid because that would mean the carbon would have 5 bonds? (another carbon, the double and 2 hydrogens) ? If that's right it seems the hydrogen is still there?
Again, refer to the diagrams in my previous post. To form the anion, you remove a hydogen from the bottom-most carbon (if you imagine that the cyclohexane ring is a clock, this is the 6-o-clock carbon). As you can see the 6-o-clock hydrogen has only one hydrogen, while the other carbons have two.
In the resonance structure, the 6-o-clock carbon is fine. It has a single bond to the 7-o-clock carbon, a single bond to a hydrogen, and a double bond to the 5-o-clock carbon. Since this carbon has four bonds and no lone pairs (giving a total of eight electrons), there is no problem.
However, there is a problem with the 5-o-clock carbon. This carbon has a single bond to the 2-o-clock carbon, two single bonds to hydrogens, and a double bond with the 6-o-clock carbon. This is a total of five bonds, totaling ten electrons and violating the octet rule.
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