TPR ICC Orgo Passage 11 Q4 (question included)

[Codaster]

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The question asks:

Which of the following is the major resonance contributor to the anion formed after diazepam is treated with the strong base lithium diisoproyl amide?

I thought it'd be A since the alpha-carbon is the nucleophile after extraction of a hydrogen near a carbonyl group. Also, I think that makes the ring stabilized because the two double bonds + lone pair on nitrogen make it aromatic.

However, the answer is C. Isn't that compound anti-aromatic? I'm confused why that would be the major resonance contributor.

 

[sunflower18]

C is aromatic. The nitrogen in the ring that does not have a double bond has a lone pair that it can donate and contribute to the aromaticity of the ring. Besides, A has the same number of electrons as C -- that negative charge on the carbon comes with a lone pair.

C > A because carbanions are not stable. Oxygen is more electronegative than carbon, so it'd much rather have the lone pair and negative charge.

Hope that helps!

 

[Codaster]

C is aromatic. The nitrogen in the ring that does not have a double bond has a lone pair that it can donate and contribute to the aromaticity of the ring. Besides, A has the same number of electrons as C -- that negative charge on the carbon comes with a lone pair.

C > A because carbanions are not stable. Oxygen is more electronegative than carbon, so it'd much rather have the lone pair and negative charge.

Hope that helps!

I guess I get why the answer choice is C, but on a different point:

I agree that the nitrogen w/o the double bond can contribute to the ring. There are three double bonds contributing and the lone pair from the nitrogen (4*2 = 8 pi electrons). Any multiple of 4n is anti-aromatic where any (4n+2) is aromatic, by Huckel's rule.

Just a point of clarification to make sure I'm not missing something.

Thanks for your help!

 

[sunflower18]

I guess I get why the answer choice is C, but on a different point:

I agree that the nitrogen w/o the double bond can contribute to the ring. There are three double bonds contributing and the lone pair from the nitrogen (4*2 = 8 pi electrons). Any multiple of 4n is anti-aromatic where any (4n+2) is aromatic, by Huckel's rule.

Just a point of clarification to make sure I'm not missing something.

Thanks for your help!

I didn't look carefully enough! Yes indeed, it seems anti aromatic. However, all of the answer choices are anti aromatic, so I guess it's not really a factor because its a commonly shared feature.

 

[Codaster]

I didn't look carefully enough! Yes indeed, it seems anti aromatic. However, all of the answer choices are anti aromatic, so I guess it's not really a factor because its a commonly shared feature.

Indeed, I realized that after the fact. I wasn't thinking that the lone pair from the negative charge contributed to aromaticity for some reason.

 

[FeinMS]

Huckel's rule doesn't always apply to compounds with multiple rings.

Also, Nitrogen without a double bond can choose to hybridize in sp3 to avoid anti-aromaticity. Remember stability determines geometry and hybridization. What makes NH3 a sp3 hybridized and amide Nitrogen sp2 hybridized? It's all up to which hybridization leads to more stable compound.