O chem question from DAT Achiever

[kponenation]

From 3rd test # 72

Select only heterocyclic aromatic compounds where the unshared electron pair of nitrogen occupies an sp2 orbital in the plane of the
Select only heterocyclic aromatic compounds where the unshared electron pair of nitrogen occupies an sp2 orbital in the plane of the ring.

I.

II.

III.

How do you attack this problem?
I've read o chem books but I can't seem to grasp this.
Can anyone help me nail this down?


Thanks.

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[UCB05]

From 3rd test # 72

Select only heterocyclic aromatic compounds where the unshared electron pair of nitrogen occupies an sp2 orbital in the plane of the
Select only heterocyclic aromatic compounds where the unshared electron pair of nitrogen occupies an sp2 orbital in the plane of the ring.

I.

II.

III.

How do you attack this problem?
I've read o chem books but I can't seem to grasp this.
Can anyone help me nail this down?


Thanks.

I generally just count the number of sigma bonds and free electron pairs to come up with the hybridization. I has 2 bonds and a pair, so sp2. II has 2 bonds and a pair, so sp2. III has 3 bonds and a pair, so sp3. sp2 centers have a trigonal planar shape, so i would say I and II have sp2 nitrogens with their e- pairs in the plane of the ring. III has an sp3 nitrogen with its e- pair sticking out of the page.

Is this correct or am i completely off in my logic?

 

[sixkiller]

I generally just count the number of sigma bonds and free electron pairs to come up with the hybridization. I has 2 bonds and a pair, so sp2. II has 2 bonds and a pair, so sp2. III has 3 bonds and a pair, so sp3. sp2 centers have a trigonal planar shape, so i would say I and II have sp2 nitrogens with their e- pairs in the plane of the ring. III has an sp3 nitrogen with its e- pair sticking out of the page.

Is this correct or am i completely off in my logic?

Actually, pyrrole (III) nitrogen is sp2 or spends more of it's "time" in the sp2 state than sp3. If you were to determine the bond lengths experimentally, the bond lengths would be shorter than that of an sp3 overlap....(this points to an orbital with greater s character therefore, shorter bond length).

The nitrogen participates in resonance and thus the structure is aromatic. If you draw the resonance structures, the nitrogen will bear a positive formal charge and a lone pair is distributed to one of the carbon giving it a negative charge.

It's just important to know that the structure is aromatic and that there are resonance structures invovled.

Same goes for furan....

 

[trentsquawkbox]

Compounds I and II have sp2 nitrogen atoms whose electrons are orthogonal to the aromatic system (in the plane of the ring.) Compound III also has an sp2 hybridized nitrogen even though it has 3 sigma bonds and a lone pair. We know pyrrole is aromatic and if the nitrogen was sp3 hybridized its electrons wouldn't be in the right place to be conjugated with the double bonds. The nitrogen in pyrrole rehybridizes to sp2 and the cost of rehybridization is bought back by the stability of aromaticity. In the case of compound III the hydrogen on the nitrogen atom is orthogonal to the aromatic system.

 

[kponenation]

Thanks guys..
The answer is I and II.
Btw, aren't these all aromatic compounds?

 

[sixkiller]

That's why it's a poorly written question....

Again, the nitrogen lone pair does occupy an sp2 orbital part of the time. It's not cut and dry. Just remember, standardized tests don't like to ask questions that aren't cut and dry. They are more likely to ask questions where a given choice MIGHT be true, but the other choice is ALWAYS true type questions....

That's like asking is the lone pair of an amide occupies a p-orbital. The answer is yes and no (if resonance theory is used), and the "real" answer is even more complicated.

I noticed several poorly written questions in achiever (the author does not seem to fully understand certain things, etc).

Wouldn't worry about achiever. Only glanced at the SNS (a friend of mine had the program) in other words basically didn't even use the program, and got a perfect score on both general and organic chemistry.

 

[Sublimation]

The way our chem professor taught us to tackle these was if the lone pairs show up on the inside of the ring then they are conjugated. If they are on the outside of the ring then they are not. They are basic Nitrogens when the lone pair is on the outside.

 

[loveoforganic]

If the Nitrogen is doubly bonded in the ring, then the lone pair cannot be in a p-orbital