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Organic Chemistry: Stabillity of Nitrogen Containing Cyclic Compounds

  • Thread starter rgk13
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  • #1
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Homework Statement


Pyridine and pyrrolidine react rapidly with dilute aqueous HCl to form the corresponding hydrochloride salts which are easily purified, isolated and stored in a charge. However, pyrrole, which is another nitrogen-containing heterocycle, does not form a hydrochloride salt under these conditions, explain.

Homework Equations


Pyridine:
80px-Pyridine-2D-Skeletal.png


Pyrrolidine:
Pyrrolidine.png


Pyrrole:
100px-Pyrrole-2D-numbered.svg.png


The Attempt at a Solution



I am pretty sure that resonance plays a major role in the answer. Is it because by protonating the pyrrole with the HCl you create an aromatic compound that prefers to stay in that form?
 

Answers and Replies

  • #2
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You're right, resonance does have a role, but its role is a lot simpler than you're thinking.

Draw the hypothetical protonated form of pyrrole. Now ask yourself if it is aromatic. Remember: breaking aromaticity comes at a cost of ~35 kcal/mol, which is a lot of energy and often unfavorable!

What I'll also mention: the lone pair in pyrridine is not involved in the conjugated pi system (that p orbital is orthogonal to the other p orbitals, meaning it has zero interaction). In pyrrole, the lone pair/p orbital is involved in the conjugation.
 
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  • #3
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so by protonating the pyrrole you would lose its stability because it would eliminate all its possible resonance structures by inducing the + charge on the nitrogen while the others still are able to resonate?
 
  • #4
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so by protonating the pyrrole you would lose its stability because it would eliminate all its possible resonance structures by inducing the + charge on the nitrogen while the others still are able to resonate?
Eep, watch your terminology because that will be your pitfall!! By protonating the nitrogen in pyrrole, its lone pair electrons are no longer delocalized across the ring in the pi molecular orbital--they would be confined to a sigma bond between the N and the new H. Note that in pyrrole, every atom is sp2 hybridized. If any of them are changed to sp3 hybridization, as protonating the N would do, then the aromaticity is broken (i.e., 4n + 2 pi electrons in the ring--Huckel rule!).

There is no charge on the nitrogen when it is protonated! It is electrically neutral (count the protons and electrons around it--neutral). It is only a formal charge, which has no relation to actual electrostatic charges.

Also, be careful with the word "resonating." The structures/bonds are not resonating between each other in any classical sense. It is both structures *at the same time.* Weird eh? Basically, the "true" structure, representing the real bond strengths, distances, and charge densities cannot be represented by these simple drawings. So, we draw a series of "contributing structures" to describe its actual chemical behavior. What it comes down to, especially in regard to conjugated pi molecular orbitals, is that the electrons are not confined between any two atoms of a bond, but are distributed across the entire molecular orbital!

Yes, I realize I'm using a hybrid of valence bond theory and MO theory here, but I'm doing so because *it works.* There are some people on this forum who only want to discuss things in terms of MO theory, but that's not my style :)
 
  • #5
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Ahhh so essentially the HCl can react with the pyridine because the lone pair of electrons is available on the nitrogen and it can essentially interact with those electrons without effecting the aromaticity while in the second compound the lone pair is there and aromaticity is not a factor while the third structure, using the lone pair on the nitrogen would cause the "destruction" of the pi orbital system that makes it aromatic and that would be too energetically unfavorable.
 
  • #6
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Bingo.

Forming or breaking aromaticity has a *huge* impact in the reactivity of molecules!

On a similar note: forming or breaking anti-aromaticity (i.e., particularly unstable cyclic compounds with 4n pi electrons) also has a huge impact on reactivity.
 
  • #7
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Thank you very much!
 

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