Histidine is an important catalytic residue found at the activ...

Question

Histidine is an important catalytic residue found at the active sites of many enzymes. In many cases, histidine appears to remove protons or to transfer protons from one location to another.

(a) Show which nitrogen atom of the histidine heterocycle is basic and which is not.

(b) Use resonance forms to show why the protonated form of histidine is a particularly stable cation.

(c) Show the structure that results when histidine accepts a proton on the basic nitrogen of the heterocycle and then is deprotonated on the other heterocyclic nitrogen. Explain how histidine might function as a pipeline to transfer protons between sites within an enzyme and its substrate.

Accepted Answer

Hello everyone. Let's do this problem. It says UIC acid is an intermediate in the metabolism of histidine. We have the structure of UK acid given to us here part A says determine which nitrogen of UIC acid is basic and which one is not. Part B using resonance forms show how the proton form of UCA acid is a particularly stable Canion. And part C draw the structures of UCA acid showing protonation at the basic nitrogen and then de protonation of the other nitrogen. So, UK acid is a five membered or it has a five membered heterocyclic ring that has two nitrogen atoms and this ring is called imidazole and this is an aromatic ring. So remember rings have to be aromatic if they follow the four N plus two huckle rule, right? And that relates to the number of pi electrons. So notice how there are only two double bonds. So that's four pi electrons in order to be aromatic, we're going to need two more electrons to satisfy huckle rule. So those will come from one of the nitrogens that has these lone pairs, right? So notice how both nitrogens are going to have a lone pair. But why does only one nitrogen contribute to the aromatic? Well, the nitrogen that has no hydrogen is going to be the basic nitrogen because it has an SP- two orbital that is not involved in the aromatic conjugation. It already has this pi bond that is involved in the conjugation. So its long pair is going to stay out of it and that's going to be basic. Whereas the lone pair on the other nitrogen that has the hydrogen is not involved in aromatic, right? There are no pi bonds touching this nitrogen. So in order to keep that full conjugation of the ring, going, the lone pair on this nitrogen is going to have to participate in the aromatic conjugation. And since it's participating in the aromatic, it is going to be not basic as it will not be able to accept a hydrogen or else that will disrupt the conjugation. OK. So that is The answer for part one, the nitrogen with a hydrogen is not basic and the nitrogen without a hydrogen is basic. So part B says using resonance forms show how the proton form of UK acid is a particularly stable ca so let me start by drawing UK acid again and it says to show the proton form of UCA acid. So we know our nitrogen that is basic is the one that will accept a proton, right? And that will give a positive charge on that nitrogen. So the Canion formed by this protonation can be stabilized by shifting around the pi electrons. So the pi electrons that are also involved with that nitrogen can go on to the nitrogen as a lone pair to shift that positive charge. So now we have nitrogen with a lone pair and a hydrogen and that neighboring carbon now has the positive charge the carbon between the two nitrogens. So now the lone pair on the proin in nitrogen, which I forgot to draw on. The first structure can swing down to shift that those pi electrons and form a double bond. Now between this nitrogen and that positive carbon oops and that leaves our pronated nitrogen now with a positive charge and involved in the pi bond. And our originally basic nitrogen also still has that proton. So these are three resonance structures for pronated form of UIC acid. And since there are three resonance structures that is going to make this a particularly stable cat ion like the problem mentions. All right, moving on to the final part, part c drawing the structures of UK acid showing protonation at the basic nitrogen and then de protonation of the other nitrogen. Well, why would this occur? Well, once the imidazole is pronated, let's show that form. First, I'll show the original structure once more. And then we want to perate this at our basic nitrogen protonation of basic nitrogen, which gives us a positive charge on that nitrogen. So once this aminosol is proton, the two nitrogens are now chemically equivalent. So either nitrogen can lose a proton and return to the Amitiza molecule. So the problem is specified, showing the de protonation of the other nitrogen. So that will occur if these pi electrons shift over towards that other nitrogen, the originally nonbasic nitrogen. So now we still have our two pi bonds and one of them is being shared with the originally nonbasic nitrogen. And now the originally basic nitrogen is the one with the hydrogen and the available lone pair that can contribute to conjugation. So see how this structure is chemically equivalent after a deep protonation of the other nitrogen. That is it for this problem. I hope this video was helpful and thank you for watching.

Key Concepts

Basicity of Nitrogen Atoms

Resonance Stabilization

Proton Transfer Mechanism

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