OC aromatic question (from achiever)

[Amos2014]

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Achiever Test #3, OC #72
Q: select heterocyclic compounds where the unshared electron pair of nitrogen occupies an sp2 orbital in the plane of the ring.

I. pyrimidine
II. Isoquinoline
III. pyrrole

My answer was I, II & III, and the correct answer is I & II.

I do not know why III is not. Sorry for not having figures. Thanks in advance.

 

[MowgliR]

My answer would have been just I. I dont know why II would be the answer, since there are no N.

Can some one clarify the question and correct me?

 

[Parle G]

I agree with Randy822 naphthalene is not even a hetrocyclic compound; so, I think the answer is a typo. I'd go with I & III.

 

[nainaaz]

i would select (I) too
NAPHTALENE-- has no nitrogen in the ring
PYRROLE--nitrogen hybridization is sp3 as it has 3 bonds bonded to carbons and a hydrogen plus a lone pair.

 

[Parle G]

whoops, i made a boo-boo, yeah nainaaz and randy822 are right. Only I is correct

 

[loveoforganic]

It's asking in which case nitrogen's lone pair lies in an sp2 orbital not a p orbital. When nitrogen's lone pair participates in resonance, as in pyrrole, it lies in the p orbital. When nitrogen uses one of its bonding electrons to form a pi bond, as in pyrimidine, the nonbonding lone pair is in an sp2 orbital. This orbital is in the plane of the ring (normal to the pi bond's p orbital).

So the answer is I only.

 

[Amos2014]

My answer would have been just I. I dont know why II would be the answer, since there are no N.

Can some one clarify the question and correct me?

Woops, my bad! II has one N atom on the ring.

 

[Amos2014]

So, does aromaticity not require sp2 hybridization?

 

[UndergradGuy7]

The answer is I.

pyrimidine is sp2 since it is bonded to two atoms and has 1 lone pair, and there are 6 pi electrons. So it is aromatic.

pyrrole nitrogen is also sp2. If you count the number of bonds and lone pairs and say it is sp3, you are wrong. The lone pair on nitrogen can do resonance and becomes part of the ring. This makes the nitrogen sp2 and pyrrole aromatic. Since pyrrole is aromatic is has to be planar and all of the atoms have to be sp2. The lone pair from nitrogen goes to a p orbital and is part of resonance, so there are no lone pairs left in the sp2.

The wikipedia drawing of II shows no nitrogens?

 

[MowgliR]

Woops, my bad! II has one N atom on the ring.

so then its not napthalene...

The answer is I and III.

pyrimidine is sp2 since it is bonded to two atoms and has 1 lone pair, and there are 6 pi electrons. So it is aromatic.

pyrrole is also sp2. If you count the number of bonds and lone pairs and say it is sp3, you are wrong. The lone pair on nitrogen can do resonance and becomes part of the ring. This makes the nitrogen sp2 and pyrrole aromatic. Since pyrrole is aromatic is has to be planar and all of the atoms have to be sp2.

True.



Btw, I thought the question is referring to an unshared electron pair of N. Does pyrrole have an unshared electron pair. Is the N-H bond considered shared?

 

[UndergradGuy7]

so then its not napthalene...





True.



Btw, I thought the question is referring to an unshared electron pair of N. Does pyrrole have an unshared electron pair. Is the N-H bond considered shared?

Check post again. I thought it was asking about hybridization, not the location of electrons. The right answer is up now.

 

[MowgliR]

got it. totally makes sense now.

 

[Amos2014]

Now I got it. Thanks for all your inputs!

 

[2ndFn]

The second structure is called is called an isoquinoline, not napthalene. Here's its structure from wiki:

http://en.wikipedia.org/wiki/Isoquinoline

Just follow what the solution tells you. When you see odd number of double bonds in the ring, nitrogen's lone pair would lie in an sp2 orbital not a p orbital like loveoforganic said.

 

[CANgnome]

so from what i gather: resonance electrons go into p orbitals and not sp2 orbitals. pyrimidine has a lone pair on top of its p orbital resonance electrons so the lone pair is not in p but in sp2.

pyrrole lone pair electrons are in resonance so, they are in the p orbital?

 

[Amos2014]

The second structure is called is called an isoquinoline, not napthalene. Here's its structure from wiki:

http://en.wikipedia.org/wiki/Isoquinoline

Just follow what the solution tells you. When you see odd number of double bonds in the ring, nitrogen's lone pair would lie in an sp2 orbital not a p orbital like loveoforganic said.

Thanks, made a correction!

 

[UndergradGuy7]

Cangnome...

If you have say N with a lone pair and it is single bonded to a carbon and double bonded to another carbon in a ring like in I above. Then the lone pair can't do resonance because N already has 1 double bond. So its hybridization is double bond + single bond + lone pair = 3 electron domains = sp2
So the lone pair is in a sp2.

In II above, Isoquinoline, same thing.

In III, pyrrole the nitrogen only has single bonds. So its lone pair can be used to make a double bond. After we make this double bond, the nitrogen does not have any more lone pairs. So there is no lone pair in sp2. The lone pair was moved to a p orbital to make a double bond. The nitrogen also becomes sp2.