TBR Strongest nucleophile

[Jengreef]

I'm looking at one of the TBR passage questions where they give you this compound looks a little like histidine except with a phenyl group in between. It's hard to explain, but it looks something like this:

Imidazole
NH2-C=O-Phenyl-NH2

Eh...the formatting isn't carrying over, but the amide and amine are para each other relative to the phenyl group and the imidazole is connected directly above the phenyl group, using two of the phenyl carbons as its base. I'll whip out a quick paint file if need be.

The question asks which Nitrogen on the compound is most nucleophilic. The answer key states that the N that isn't protonated in the imidazole group is the only one that is nucleophilic and that the others are actually hardly nucleophilic if at all.
But I understand amide groups to be moderate electron donating groups, and amine to be strongly activating. Then I feel it's reasonable to expect either of those Nitrogens to also be relatively decent nucleophiles as well given the donated electron density.

So when you have more than one atom in a compound that seems to have nucloephilic factors supporting it, then how do you choose which is most nucleophilic? For those who have TBR, this is Chapter 7 Passage 1 Question 5 on the 52 question exam.

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[Big N Tasty]

Think of it like this, the lone pair on the amine group is actively donating in its electrons to the ring, thus if it were to act as a nucleophile, you would lose aromaticity and that would be unfavorable. That N is out.

The N of the amide group, though, is separated by another group and is not directly contributing to the resonance structure. The lone pair is a great electron source so that N is your nucleophile.

Hope that helps

 

[Jengreef]

Excellent stuff. I assume the reason you gave for the amide also applies to the amine then, so both are actively donating their electrons toward the ring instead of each other?

I have a second question, would it follow that molecules with electron withdrawing groups also make good nucleophiles? And even better nucleophiles if those deactivating groups have a corresponding electron donating group?

 

[Big N Tasty]

So, I realized I accidentally goofed with a spelling error in that first post, so I edited it to show that I meant the amine is donating while the amide isn't because it is too far from the conjugated ring. I was trying to show that the p orbitals all along the ring are in conjugation with the lone pair of the amine (also in a p orbital). Thus, this one is less nucleophilic because if that lone pair acted as a nucleophile, the ring loses some of its electron delocalization ability.

Electron withdrawing groups tend to make the ring nucleophilic at the meta position (in the presence of an electrophile) or electrophilic at ortho/para (in the presence of a nucleophile). Depending on the withdrawing group, consider an aromatic ring bonded to a carbon that is double bonded to an oxygen as well as a nitrogen. In this case, it is possible for the nitrogen to act as a nucleophile because it's resonance is only weakening the withdrawing power of the carbonyl as opposed to actually contributing to the ring's aromaticity.

So, to answer that second question, I would say generally donating groups make for better nucleophiles for the molecule as a whole because there is more electron density that is able to attack a local electrophile while withdrawing groups remove electron density as a whole, making it more prone to a nucleophilic attack. A donating group and a withdrawing group effectively cancel each other out depending on their relative strengths. As for the most nucleophilic/electrophilic atom, however, you need to look for lone pairs, formal charges and whether or not that there is stability via resonance.

Hope that cleared things up.

 

[Big N Tasty]

Also, you mentioned that your molecule looks like histidine?

If that's the case, the nitrogen with the double bond is the nucleophilic atom because it is donating into the ring via a double bond and its lone pair is in an sp2 hybridized orbital and thus is not in p conjugation with the ring. The nitrogen with the hydrogen and the lone pair, however, is the one in conjugation with the p orbitals of the ring and thus is less nucleophilic because of conjugation stability.

 

[Jengreef]

A little confused on a couple points. Let's first talk about the second question I put out.

First, to cover the ground base (please correct me anywhere I'm wrong), I understand a good nucleophile is characterized by a negative charge and seeks out a nucleus, a positive charge, to stabilize charges. And typically the nucleophile itself is a single atom in a molecule that contains a negative charge by excess lone pairs. So it stands that if there is an electron withdrawing group (ie a group that has an electronegative atom that is pulling in electrons towards itself), and an electron donating group on that same molecule that is giving away electrons toward the withdrawing group, you'd have an increased electron density at one end of that molecule, causing it to be more negative, thus more nucleophilic, right? I expected one to complement the other, so could you explain how they tend to cancel each other out?

Next, I thought I understood conjugation, but after reading TBR's answer and yours (both of you agree), I think I may need a refresher. Conjugation is a network of alternating single and multiple bonds right? How can I determine which lone pairs are in conjugation with the system and which ones aren't?

 

[Big N Tasty]

A little confused on a couple points. Let's first talk about the second question I put out.

First, to cover the ground base (please correct me anywhere I'm wrong), I understand a good nucleophile is characterized by a negative charge and seeks out a nucleus, a positive charge, to stabilize charges. And typically the nucleophile itself is a single atom in a molecule that contains a negative charge by excess lone pairs. So it stands that if there is an electron withdrawing group (ie a group that has an electronegative atom that is pulling in electrons towards itself), and an electron donating group on that same molecule that is giving away electrons toward the withdrawing group, you'd have an increased electron density at one end of that molecule, causing it to be more negative, thus more nucleophilic, right? I expected one to complement the other, so could you explain how they tend to cancel each other out?

Next, I thought I understood conjugation, but after reading TBR's answer and yours (both of you agree), I think I may need a refresher. Conjugation is a network of alternating single and multiple bonds right? How can I determine which lone pairs are in conjugation with the system and which ones aren't?

1) You are correct in thinking that. Also, negative charges aren't necessary to have a nucleophile because a nucleophile could have a neutral charge but be a relatively strong base such as NH3. The key is to understand whether or not it would donate that lone pair into making another bond. NH3 is nucleophilic because it could easily attack a proton and become NH4+. Thus that lone pair and N's ability to accept a + charge makes it nucleophilic. The same can be said of high electron density. I can see how I may have been a little unclear but you are correct about withdrawing groups. When I said they could cancel out, I meant in a situation where a ring has both a withdrawing and donating group, depending on their positions, they could cancel out the ring's nucleophilicity. BUT the nucleophilic atom then would lie on the withdrawing group itself. You were correct in the way you explained it.

2) The key is to either visualize the atoms or (the easier way) to draw resonance structures. An amine group has a lone pair and 3 bonds. It's normal hybridization suggests that it's lone pair is found in a p orbital (convenient because all of the double bonds throughout the ring are via pi type p interactions). It can easily donate a lone pair into the ring to form a new double bond. After a series of shifting the double bonds around the ring, you can eventually have that nitrogen accept a double bond shifting into it as a lone pair. Thus there is resonance throughout. However, the nitrogen that already has a double bond in the ring. Think of this molecule as SP2 hybridized (read trigonal planar). That lone pair is sort of jutting out away from the ring, perpendicular to where a p orbital would lie. Thus, it couldn't interact with a neighboring p orbital and could not form a pi bond. Because it does not contribute to the resonance structure, these lone pair electrons are free to go and attack an electrophile. But the key is to look for resonance. So for this nitrogen with the double bond, think of it as a withdrawing group. Break the double bond to the ring and give it to nitrogen. Nitrogen now has two lone pairs, two bonds and a negative charge. It is a great nucleophile.

 

[Jengreef]

First, thanks for keeping up with me on my perpetual questions. Just need to make sure I understand what you just said, and I think I'm good. Thanks again.

1) So by "canceling out" we're talking about the nucleophilicity of the molecule as a whole, whereas the nucleouphilicity of any given atom may increase/decrease based on that atom's electronegativity, and other electron donating groups. Based on your response, I also get the impression that if the atom itself is too electronegative, it would also make a poor nucleophile because it would be too content to seek out a positive charge and give away its lone pair. So would a good nucleophile be an atom that wants to give away its electrons (ie unstable as a negative atom)?

2) It looks like a simple way of looking at it is to see if the nitrogen is double-bonded or not as a neutral atom. Again, correct me if I'm wrong, but it looks like any nitrogen that is sp3 hybridized with 3 bonds have a lone pair vested into the conjugated system, and cannot easily pull their lone pair away from the ring. Whereas on the other hand, an sp2 hybridized nitrogen has a pair of electrons that isn't its normal lone pair that is already a part of the conjugated system, so its current lone pair doesn't need to be involved in the conjugation, thus it would be the best fit for nucleophilicity?

 

[Big N Tasty]

Yes and yes!

 

[Jengreef]

Excellent stuff. Thanks very much!

 

[Big N Tasty]

No problem, happy to help