Orgoman Question

[Ibraiz]

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Why is answer choice B is horrible and choice C is anti aromatic but choice D is fair? Why is the answer choice E the best answer? I am confused here. Using 2n + 2 = # of Pi electrons I am getting multiple answer.



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[UndergradGuy7]

http://forums.studentdoctor.net/showthread.php?p=9724623#post9724623

 

[RCT PC CRN]

Why is answer choice B is horrible and choice C is anti aromatic but choice D is fair? Why is the answer choice E the best answer? I am confused here. Using 2n + 2 = # of Pi electrons I am getting multiple answer.



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I didn't look at your question in detail but one thing I know is it's 4n+2 not 2n+2...may be that's why you are confused..just my guess. I can't see the image you have attached in my phone well.

 

[Ibraiz]

Basically I think the question wants you to identify between aromatic vs nonaromatic vs anti-aromatic.

So you should know how to identify each.
Nonaromatic = the norm, Aromatic like benzene = more stable, anti-aromatic = less stable.

Look at the carbocations...

So A is a normal ring secondary carbocation.

B is a benzene. A benzene wiht a + charge = bad. Remember benzene does not even react with Br2 it needs a catalyst like FeBr3.

C is antiaromatic because you have 4n pi electrons (4 pi electrons). But you still satisfy all the other rules of aromaticity.

D is non-aromatic because the bottom carbon is sp3 hybridized. But it has resonance so it is more stable than A (I think).

E is aromatic. It has 4n+2 pi electrons (6 here). All rings atoms are sp2 hybridized. It is planar and cyclic.

So E is most stable.


Google the rules of determining aromaticity vs non vs anti. Or watch chad's videos.

Thanks, I knew I asked this question before. Answer choice E when Cl leaves, has a SP3 hybridization because it has a lone pair of electron + two Carbons + an H attached to it which makes it SP3 hybridized then how is it stable? or am I looking at it wrong?

 

[UCB05]

Thanks, I knew I asked this question before. Answer choice E when Cl leaves, has a SP3 hybridization because it has a lone pair of electron + two Carbons + an H attached to it which makes it SP3 hybridized then how is it stable? or am I looking at it wrong?

In an Sn1 mechanism, the molecule loses Cl-, not a Cl+, so what's left is a sp2 hybridized C with two C-C bonds and a C-H bond with no lone pair of electrons.

 

[Troyvdg]

Thanks, I knew I asked this question before. Answer choice E when Cl leaves, has a SP3 hybridization because it has a lone pair of electron + two Carbons + an H attached to it which makes it SP3 hybridized then how is it stable? or am I looking at it wrong?

your looking at it wrong, when Cl leaves it takes its electrons with it, leaving the carbon sp2 hybridized with an empty p orbital that is contributing to the pi system. Because the cycloheptatriene forms an aromatic compound (4n+2=6) as its carbocation intermediate, it is the most stable.

hope this helps

 

[Ibraiz]

Thanks, it helps, my last confusion is about the answer choice B. Why is it not a good option. Destroyer calls it a horrible choice, why?

Using 4n+2 = 6 it gives us 1 which means it is an aromatic compound, correct? or am I missing something here.

 

[FredDurst]

Thanks, it helps, my last confusion is about the answer choice B. Why is it not a good option. Destroyer calls it a horrible choice, why?

Using 4n+2 = 6 it gives us 1 which means it is an aromatic compound, correct? or am I missing something here.

When looking at SN1, you want to focus on how stabilized the carbocation is. In the case of B, the system is aromatic, but this aromaticity is doing nothing to stabilize the positive charge. The empty orbital can't overlap with the aromatic pi system because it is directed out from the ring. Where you'll see the aromatic ring like this help is when the carbocation is in the benzylic position; the aromatic ring then would be able to interact with the empty orbital.