Skip to main content
Ch. 24 - Amino Acids, Peptides, and Proteins
Wade - Organic Chemistry 9th Edition
Wade9th EditionOrganic ChemistryISBN: 9780135213728Not the one you use?Change textbook
All textbooksWade 9th EditionCh. 24 - Amino Acids, Peptides, and ProteinsProblem 41
Chapter 24, Problem 41

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.

Verified step by step guidance
1
Step 1: Analyze the structure of histidine's heterocyclic ring (imidazole). The imidazole ring contains two nitrogen atoms: one is part of a double bond (sp2 hybridized) and the other is bonded to a hydrogen atom (sp2 hybridized but capable of donating a lone pair). The nitrogen bonded to hydrogen is basic because it can donate its lone pair to accept a proton.
Step 2: To determine why the protonated form of histidine is stable, draw resonance structures for the protonated imidazole ring. Show how the positive charge can delocalize across the ring, stabilizing the cation through resonance.
Step 3: When histidine accepts a proton on the basic nitrogen, draw the resulting structure. Then, show the deprotonation of the other nitrogen atom in the heterocycle. This results in a neutral imidazole ring with a proton transferred from one nitrogen to the other.
Step 4: Explain how histidine functions as a proton pipeline. The ability of histidine to accept and donate protons through its imidazole ring allows it to transfer protons between the enzyme active site and the substrate. This is facilitated by the resonance stabilization of the intermediate forms.
Step 5: Relate the structure of urocanic acid (shown in the image) to histidine. Urocanic acid is a derivative of histidine where the amino group is replaced by a double bond to a carbon chain. This structure highlights the importance of the imidazole ring in proton transfer and catalytic activity.

Verified video answer for a similar problem:

This video solution was recommended by our tutors as helpful for the problem above.
Video duration:
8m
Was this helpful?

Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Basicity of Nitrogen Atoms

In histidine, the basicity of nitrogen atoms is determined by their ability to accept protons. The nitrogen in the imidazole ring can be protonated, making it a basic site, while the other nitrogen is part of a double bond and is less likely to accept a proton due to resonance stabilization. Understanding which nitrogen is basic is crucial for analyzing histidine's role in enzymatic reactions.
Recommended video:
1:45
Nitrogenous Bases Concept 3

Resonance Stabilization

Resonance stabilization occurs when a molecule can be represented by multiple valid Lewis structures, distributing electron density across the molecule. In the case of protonated histidine, resonance forms show that the positive charge can be delocalized over the nitrogen atoms and the adjacent carbon atoms, enhancing the stability of the cation. This concept is essential for understanding the stability of charged species in biochemical reactions.
Recommended video:
Guided course
03:43
The radical stability trend.

Proton Transfer Mechanism

Histidine can act as a proton shuttle in enzymatic reactions, facilitating the transfer of protons between different sites. When histidine accepts a proton on its basic nitrogen, it can later donate it from the other nitrogen, effectively transferring protons within the enzyme's active site. This mechanism is vital for many enzymatic processes, as it allows for the regulation of pH and the stabilization of transition states.
Recommended video:
3:05
Polyurethane Mechanism Example 1
Related Practice
Textbook Question

There are many methods for activating a carboxylic acid in preparation for coupling with an amine. The following method converts the acid to an N-hydroxysuccinimide (NHS) ester.

(b) Propose a mechanism for the reaction shown.

462
views
Textbook Question

Lipoic acid is often found near the active sites of enzymes, usually bound to the peptide by a long, flexible amide linkage with a lysine residue.


(a) Is lipoic acid a mild oxidizing agent or a mild reducing agent? Draw it in both its oxidized and reduced forms.

(b) Show how lipoic acid might react with two Cys residues to form a disulfide bridge.

(c) Give a balanced equation for the hypothetical oxidation or reduction, as you predicted in part (a), of an aldehyde by lipoic acid.

1358
views
Textbook Question

Metabolism of arginine produces urea and the rare amino acid ornithine. Ornithine has an isoelectric point close to 10. Propose a structure for ornithine.

1499
views
Textbook Question

Aspartame (Nutrasweet®) is a remarkably sweet-tasting dipeptide ester. Complete hydrolysis of aspartame gives phenyl alanine, aspartic acid, and methanol. Mild incubation with carboxypeptidase has no effect on aspartame. Treatment of aspartame with phenyl isothiocyanate, followed by mild hydrolysis, gives the phenylthiohydantoin of aspartic acid. Propose a structure for aspartame.

1420
views
Textbook Question

There are many methods for activating a carboxylic acid in preparation for coupling with an amine. The following method converts the acid to an N-hydroxysuccinimide (NHS) ester.

(a) Explain why an NHS ester is much more reactive than a simple alkyl ester.

649
views
Textbook Question

Show how you would use the Strecker synthesis to make tryptophan. What stereochemistry would you expect in your synthetic product?

1379
views