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Topic: stability, orbitals, specific questions  (Read 2998 times)

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zephy

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stability, orbitals, specific questions
« on: October 11, 2004, 10:49:29 PM »
I have a few questions from a sample midterm that I am going over for my midterm this week.

This one asks for which anion is more stable based on one of four factors (the atom the charge is on, resonance, inductance, or orbitals), and I want to make sure my reasoning is correct. An image is available here: http://www.sharemation.com/flare/orgo01.jpg

For a, I don't know the reason and need help to figure out which one. b) I decided on inductance since the electrons left behind are on the nitrogen and since the nitrogen is so close, it will cause instability. So I picked the first one was more stable. c) I said the first was more stable because all the chlorine atoms will produce inductance and make it more unstable. d) I said the second one because of orbitals; the sp bond there will stabilize it more than the first one.

http://www.sharemation.com/flare/orgo02.jpg

I chose D because the resonace structure in A pushes a lone pair from oxygen to create a double bond and pushes a bond to create a corner. B does something similar, too, and C pushes bonds around the carbon ring and one into the corner. I just want to make sure I'm reasoning correctly.

http://www.sharemation.com/flare/orgo03.jpg

I'm assuming the original structure has a double bond in the biddle, since it is 2-butene, but because the resonance structure creates charge separation, I reasoned that it's false because that structure would not be significant. Is this correct?

http://www.sharemation.com/flare/orgo04.gif
I thought D because the lone pair on nitrogen is close to two oxygens so they would be delocalized whearas the oxygen on the OH is only near the nitrogen so it is localized.

http://www.sharemation.com/flare/orgo05.gif
I don't really know how to approach this problem. Is it supposed to imply a stable base?

http://www.sharemation.com/flare/orgo06.gif
I saw 5 stereocenters - one at the OH, the one below OH, the one to the left of OH, one at Cl, one below Cl. Therefore, 2^5 means 32 stereoisomers. It also asks how many enantiomers there are and I said 64 because each stereoisomer can be mirrored and would not be superimposable anymore. Am I right?

That's all. Please help confirm my answers and help on the ones I could not figure out. Thank you.

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Re:stability, orbitals, specific questions
« Reply #1 on: October 12, 2004, 01:37:55 AM »
I'll answer some of these.  Someone else can answer the rest.

Here goes:

1) a) the second one is more stable because of resonance into the alkene adjacent to the carbanion.
b) I assume the - represents the lone pair on nitrogen.  Otherwise it doesn't make much sense.  Anyway, the second one is more stable because nitrogen is more electronegative than carbon.
c) This is where your inductive effect comes into play.  The Cls are inductively withdrawing, therefor stabilizing.  (CHCl3 is somewhat acidic!)
d) The second one is more stable because of the s-character in the hybridized orbitals.  The second one has more s-character, therefor is better.

5) The question is asking which carbocation is most stable.  Go over the carbocation stability in your textbook and this one should be pretty obvious.

6) You've got it right except for the number of enantiomers.  Remember that the term "stereoisomer" is more inclusive than "enantiomer."  There must be fewer than 32 enantiomers because all enantiomers are stereoisomers!

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