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Topic: Nitration of Pyridine  (Read 26290 times)

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Offline Hunt

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Nitration of Pyridine
« on: August 28, 2008, 12:42:40 PM »
In the nitration of pyridine which is activated by CH3O on one side and by NH2 on the other , is it correct to assume that nitration occurs on C3 and C5 with equal probability ? 


Offline spirochete

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Re: Nitration of Pyridine
« Reply #1 on: August 28, 2008, 02:49:39 PM »
In some cases you  have to consider that those two substituents don't have identical "strengths" in terms of directing effects, due mostly to their different electronegativities.  Both are O/P directing but one has more influence relative to the other.  I don't think you have to consider that in this case though, can you see why? 

EXCEPT this question seems to be ignoring the fact that acidic conditions will alter the directing effects of one of those substituents.  Everything changes if you take that into account.   

Offline Hunt

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Re: Nitration of Pyridine
« Reply #2 on: August 28, 2008, 03:27:27 PM »
In some cases you  have to consider that those two substituents don't have identical "strengths" in terms of directing effects, due mostly to their different electronegativities.  Both are O/P directing but one has more influence relative to the other.  I don't think you have to consider that in this case though, can you see why? 

Yes I did think about the difference in their strengths but frankly I do not know which is a stronger electron donating group. However, I fail to see why it makes no difference here.

EXCEPT this question seems to be ignoring the fact that acidic conditions will alter the directing effects of one of those substituents.  Everything changes if you take that into account.   

I really have no idea. How does an acidic medium change the directing effects ?

Offline spirochete

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Re: Nitration of Pyridine
« Reply #3 on: August 28, 2008, 03:39:53 PM »
First tell me what you know about the mechanism then I can try to clear up any confusion.  I can't just explain/draw out the whole mechanism there are textbooks for that.

Offline macman104

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Re: Nitration of Pyridine
« Reply #4 on: August 28, 2008, 03:42:20 PM »
EXCEPT this question seems to be ignoring the fact that acidic conditions will alter the directing effects of one of those substituents.  Everything changes if you take that into account.

I really have no idea. How does an acidic medium change the directing effects ?
Are either of the substiuents affected by an acidic medium?  What happens to this substiuent when it comes in contact with an acid?

Offline Hunt

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Re: Nitration of Pyridine
« Reply #5 on: August 29, 2008, 01:34:07 PM »
Well first things first the mechanism for electrophilic substitution for activated pyridine should be like this :




Now I think I understand why the ED strength of -OMe and -NH2 makes no difference. It's because the positive charge is not static . There's a delocalization of the Pi electrons due to resonance , correct ?





So I can safely assume that electrophilic attack occurs on C3 and C5 with equal probability.

Now if the medium is acidic , shouldn't N get protonated resulting in N+ -H ? This in effect changes the electrophilic attack to C2 or C6 , thereby the obtained product is not the one shown in my 1st post . Is this logical ?

Offline spirochete

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Re: Nitration of Pyridine
« Reply #6 on: August 29, 2008, 08:55:57 PM »
Well first things first the mechanism for electrophilic substitution for activated pyridine should be like this :




Now I think I understand why the ED strength of -OMe and -NH2 makes no difference. It's because the positive charge is not static . There's a delocalization of the Pi electrons due to resonance , correct ?


It sounds like you already understand this but just to clarify, the reason they have different electron donating effects is because nitrogen is less electronegative and therefore more "willing" to donate it's lone pair and take on a positive charge than oxygen.  So the intermediate with an amine group benefits more from that electron donation and has less carbocation character.

I'm not sure I completely understand the logic in your explanation but it's not what I had in mind.  My observation was that your molecule has a special kind of symmetry which gives exactly equal distribution of attack via electrophilic addition.  I'll try to explain without a picture:  Think of things in terms of probability.  Obviously there's a 100% probability of substitution with each successfull reaction.  Let's say for the sake of explanation that oxygen is worth 40% and the amine group is worth 60%.  And we'll also assume ortho and meta occur with equal probability (this is close to true).  Next cut each percent in half and assign it to each location on the ring.  You'll notice the probability of substitution at each location ends up being 50%.  This math will work with ANY set of values you assign to the two groups.

This would NOT be the case with a 1,4 substituted benzene ring.  In this case the para position is blocked and you'd see preferential substitution ortho to the amine group.

I hope that wasn't too convoluted but it's the best objective way I can think of to explain it.

Now if the medium is acidic , shouldn't N get protonated resulting in N+ -H ? This in effect changes the electrophilic attack to C2 or C6 , thereby the obtained product is not the one shown in my 1st post . Is this logical ?

I'm not familiar with the numbering on pyridine but you have the correct general idea.  The protonated nitrogen no longer has a lone pair to donate.  Now inductive effects dominate and it becomes meta directing.  Inductive meta direction is generally weaker than direction provided by lone pair donators like oxygen.  This means the sum of the products resulting from ortho/para substitution relative to the oxygen would be greater than the product resulting from substitution meta to the protonated amine.


Offline Hunt

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Re: Nitration of Pyridine
« Reply #7 on: August 30, 2008, 04:28:27 AM »
Quote from: spirochete
It sounds like you already understand this but just to clarify, the reason they have different electron donating effects is because nitrogen is less electronegative and therefore more "willing" to donate it's lone pair and take on a positive charge than oxygen.  So the intermediate with an amine group benefits more from that electron donation and has less carbocation character.

I understand that you concluded that based on electronegativity , but shouldn't we also take into account the -CH3 attached to O and the 2 H's attached to N to figure out which has a more electron donating strength ?

Quote from: spirochete
I'm not sure I completely understand the logic in your explanation but it's not what I had in mind.  My observation was that your molecule has a special kind of symmetry which gives exactly equal distribution of attack via electrophilic addition.  I'll try to explain without a picture:  Think of things in terms of probability.  Obviously there's a 100% probability of substitution with each successfull reaction.  Let's say for the sake of explanation that oxygen is worth 40% and the amine group is worth 60%.  And we'll also assume ortho and meta occur with equal probability (this is close to true).  Next cut each percent in half and assign it to each location on the ring.  You'll notice the probability of substitution at each location ends up being 50%.  This math will work with ANY set of values you assign to the two groups.

This would NOT be the case with a 1,4 substituted benzene ring.  In this case the para position is blocked and you'd see preferential substitution ortho to the amine group.

I hope that wasn't too convoluted but it's the best objective way I can think of to explain it.


I'm not familiar with the numbering on pyridine but you have the correct general idea.  The protonated nitrogen no longer has a lone pair to donate.  Now inductive effects dominate and it becomes meta directing.  Inductive meta direction is generally weaker than direction provided by lone pair donators like oxygen.  This means the sum of the products resulting from ortho/para substitution relative to the oxygen would be greater than the product resulting from substitution meta to the protonated amine.



I think I got it. Incase of trouble , I'll come back.

Thanks spirochete,

Offline spirochete

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Re: Nitration of Pyridine
« Reply #8 on: August 30, 2008, 12:04:39 PM »
I understand that you concluded that based on electronegativity , but shouldn't we also take into account the -CH3 attached to O and the 2 H's attached to N to figure out which has a more electron donating strength ?

There are a lot of examples in organic chem where you have major factors influencing something and minor factors.  In this case the electronegativity of the atom directly attatched to the ring is by far the major contributing factor. 

You're absolutely correct that what's connected to the O/N will affect it's electron donating ability, and probably also have other minor effects on the overall strenth of directing effects.  For example, alkyl groups are electron donating and thus will stabilize a positive charge.  If we were comparing "-OH" to "-O-CH3" we'd probably find "-O-CH3" would be stronger. I'd guess however in most cases the effect would be so minor it probably wouldn't have any practical implications for an experiment.

There are also frequent examples of major/minor effects when comparing relative acid/base strength.  For example R2N- is always going to be a stronger base then RO-, regardless of whether "R" is hydrogen or alkyl group/s.

Offline spirochete

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Re: Nitration of Pyridine
« Reply #9 on: August 30, 2008, 01:28:23 PM »
I should add that one take home lesson from this is that in that in general, effects which stem from resonance will usually be stronger than inductive effects. 

Offline Hunt

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Re: Nitration of Pyridine
« Reply #10 on: August 30, 2008, 03:18:31 PM »
Hey Spirochete , thanks for replying...

I have another question concerning EW/ED groups , something that always perplexes me in organic.

Suppose I react a nucleophile / base with the compound shown below , there are two products that could form . But which is more stable ? Oxygen is an EW group so it should stabalize a carboanion , but then sometimes we say oxygen is an ED group as it can give away its electrons and stabalize a carbocation. Doesnt this mean it should destabalize a carboanion ? This seems very confusing to me and even contradictory. What is the right way of thinking ?   



P.S. What do we name this compound ? I suppose it's a cyclic ether but what's its IUPAC nomenclature ?

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Re: Nitration of Pyridine
« Reply #11 on: August 30, 2008, 03:31:33 PM »
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info, pH-meter.info

Offline spirochete

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Re: Nitration of Pyridine
« Reply #12 on: August 30, 2008, 04:23:20 PM »
An atom can only be electron donating be resonance if there is a potential resonance structure to be drawn where that donation takes place.  If not, then inductive effects dominate. 

Offline Hunt

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Re: Nitration of Pyridine
« Reply #13 on: August 30, 2008, 04:42:48 PM »
Thanks borek ...

Spirochete , then why does O act as an ED group in the pinacole rearrangement ? I couldnt see any resonance there. Incase you need an image , I'll attach one.

Offline Hunt

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Re: Nitration of Pyridine
« Reply #14 on: August 30, 2008, 04:45:07 PM »
oh and btw A is more stable than B , right ?

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