Disulfide Bonds - Online Tutor, Practice Problems & Exam Prep

Disulfide bonds, formed through the oxidation of thiols, play a crucial role in stabilizing the tertiary structure of proteins. The process involves three steps: thiol ionization, where a strong base deprotonates thiols to create thiolate anions; an SN2 reaction with a dihalide to form halogenated thiols; and a second SN2 reaction to yield disulfides. These bonds, also known as disulfide bridges, are essential for maintaining protein shape, particularly through interactions between cysteine residues. Disulfides can be reverted to thiols using reducing agents like hydrochloric acid and zinc.

Disulfide Bonds Concept 1

Hey everyone. Let's take a look at disulfide bond formation. So here we're going to say that thiols can be oxidized into a disulfide in the presence of the mild oxidizing agent in the forms of either Br₂ or I₂ within a sodium hydroxide solution. Now, here we're going to say the disulfide bond itself, represents 2 sulfur atoms covalently bonded. And we're going to say it takes 3 steps to create our disulfide.

In step 1, we have thiolionization. Here a strong base deprotonates the thiol to create a thiolate anion. So if we take a look here, we have our hydroxide ion which is our strong base. It comes in and deprotonates this thiol. The bond breaks and sulfur keeps the electrons.

This creates our thiolate anion. In step 2, we have an SN₂ reaction. Here, the thiolate anion reacts with a dihalide molecule, so this molecule here. Sulfur will use its lone pairs to hit this bromine, and then the bromine leaves. This

Provide the name of the product created when the following reaction takes place. 

• A. Isopropyl thiol
• B. 2-methyl-1-propanethiol
• C. 3-methyl-1-propanethiol
• D. 2-methyl-3-propanethiol

Using 3-methyl-1-propene as a starting material, predict the final product based on the list of reagents given.

In this practice question, using 3 Methyl 1 Propene as a starting material, predict the final product based on the list of reagents given. So here, we're starting out with H2S, H+. This should seem kind of familiar to us because we have H2O, H+. Sulfur and oxygen are in the same group, so they should react similarly to one another because they have an equal number of valence electrons. We remember that H2O with H+ represents acid-catalyzed hydration. Right? So this is in essence similar to that. Remember, in this reaction, we would have this double bond grabbing this H+. And by Markovnikov's rule, we would protonate the less substituted alkene carbon, which would be this one, making this carbocation. But that carbocation is secondary, so it would undergo a hydride shift to make a more stable tertiary carbocation. So this here would attack and attach here, and we know that it would get deprot

Beginning benzene, determine the chemical steps needed to prepare the following compound. 

Hey, everyone. So in this practice question, we're beginning with Benzene, and we have to find the chemical steps needed to create this final product. Now, what is this product? Well, this product, initially, we see that we added 2 carbons to Benzene. And at the end, what we've created is a disulfide. Right. So let's tackle the easy part first. How do we add 2 carbons directly to benzene? That's right. We're going to do Friedel-Crafts alkylation. Because it's only 2 carbons that we're adding, we don't have to worry about any type of rearrangement. So we're just going to use Ethyl Chloride over our Lewis acid catalyst. So we created our ethylbenzene. Now we need to functionalize this alkyl chain. So remember, alkanes, alkyl groups, they're not really reactive. They only do one real reaction that's important, and that is our radical catalyzed halogenation. So we'd use NBS over UV light. So that will replace one of these benzylic hydrogens with a bromine. So now that we have that bromine here, we can use a strong base so that we can do E2 elimination. So here goes our hydrogens. This base would come in and deprotonate, forming the double bond here, kicking out the Br. So we've just created an alkene. Alright. So we've just created an alkene. What else can we do here? Well, what we could do next is we could do hydroboration-oxidation. So let's see what that's going to do. We have BH3 THF over H2O2 and NaOH. Remember, this creates an alcohol by anti-Markovnikov's rule. And we did this because now we have a terminal alcohol, primary alcohol. And because it's primary, I could then change it into a good leaving group. So SOCl2 changes it into a chlorine. And this is a primary alkyl halide. Remember, primary alkyl halides, we could do SN2 reactions with them. So I'm going to bring in Na+, SH-. And by SN2, this

Disulfide Bonds Concept 2

So with disulfides, we can now apply them to the idea of tertiary structures of proteins. We're going to say here, recall that the tertiary structure of a protein is its overall three-dimensional shape. This comes from the bending and twisting of the peptide chain itself. One of the interactions responsible for this structure are the disulfide bonds, which you can also call disulfide bridges. Disulfide bridges form when our thiol groups, or SH groups, from 2 cysteine residues reacting.

Now if we take a look here at cysteine and disulfide bridges, here we have one cysteine amino acid in another one. We do oxidation which means we're going to have the loss of hydrogen. So this sulfur would lose a hydrogen and this one as well. To basically make up for that loss of a bond, they would bond to one another. This will create our disulfide bond.

Here are 2 sulfurs and they bond to each other. If we want to look at this in terms of a three-dimensional structure, here is our peptide chain, and one of the important types of interactions that help to hold these tertiary structures is disulfide bridges. Here, just imagine that these two blue spheres represent cysteine and their thiol groups are what are connecting together, kind of forming this shape here from their interactions. Just remember, we know how to create disulfide bonds. These disulfide bonds or bridges are an important interaction that helps to maintain the tertiary structure of a typical protein.

Disulfide Bonds Example 2

Hey, everyone. So in this example question, it says, illustrate the pentapeptide formed when a molecule of cysteine connected to glycine or dipeptide connects with cysteine, valine, and phenylalanine via a disulfide bridge. Right. So here we have a dipeptide, and here we have our tripeptide, and we need to connect them together. We know that the cysteines through their thiol groups are the ones that are going to connect to make the connection.

So if we're looking at it in just a simplistic way, we know that these two would be the ones connecting. Now, how exactly would we show this? Alright. So if we scroll down, we're going to draw here our remember, when we're drawing this peptide, we want to do the N-terminus end on the left and the C-terminus end on the right. Alright.

So, glycine is the simplest amino acid with its R group just being a hydrogen. So we're going to have here our protonated amino group. So here, I'm just doing the protonated form. So we have that. It's connected to our alpha carbon.

Our alpha carbon has its hydrogen and its R group is also another hydrogen. And here's our carbonyl. That carbonyl now is connected to the nitrogen group of our cysteine. So we're going to say here our cysteine we're drawing this connected to the alpha carbon. And it's connected to our alpha carbon.

That alpha carbon is connected to CH₂ group connected to an SH. And then here we'd have our carbonyl. Now, here's the thing. We're going to say at this point, this SH is going to connect to the other SH of the other thiol. They're connecting together to make this disulfide connection right here.

We're going to remove hydrogens from them. So here's our disulfide bridge. Now we just have to draw the rest of it. So again, we're going to have here, this connected to actually we're going to draw this in red to show the distinction. We're going to be connected to our CH₂, which is connected to our alpha carbon.

That alpha carbon there, it's going to have its amino group out here, then we're connected to this carbonyl, then we're connected so this is our cysteine, now we're going to do valine. Now we have our NH of valine. Valine's R group is an isopropyl. Then we have our carbonyl, then it's connected to the nitrogen of phenylalanine, and then that's connected to our alpha carbon, which is connected to a CH₂ and then a benzene ring. And then we have our carboxyl end here.

So this will represent our pentapeptide that we created and here is our disulfide bridge that connects the dipeptide to the tripeptide in order to make our pentapeptide.

Determine which of the following peptide chains could be further stabilized by intramolecular a disulfide bridge(s). 

• A. Trp–Cys–Met–Ile–Val–Gln–Arg–Ala–Arg–Gly–Glu–Met–Pro–Leu–Phe
• B. Ala–Phe–Met–Ser–Trp–Ser–Leu–His–Arg–Arg–Ala–Pro–Thr–His–Val
• C. Gly–Leu–Ala–Glu–Gln–Cys–Ser–Met–Gln–Leu–Asn–Cys–Trp–Ile–Val–Ala
• D. Phe–Ile–Leu–Cys–His–Arg–Tyr–Thr–Phe–Met–Arg–Ala–Asn–Ile–Phe–Lys
• E. Gly–Asn–Val–Val–Ille–His–Leu–Thr–Pro–Met–Tyr–Ser–Cys–Trp–Leu–Ala