Hydrolysis of Nucleosides - Video Tutorials & Practice Problems

Hydrolysis of Nucleosides Concept 1

Hey, everyone. So in this video, let's talk about the hydrolysis of nucleosides. Now we're going to say that nucleosides are not easily hydrolyzed. However, given a strong enough acid and time, it's possible. Now, when we do the hydrolysis of a nucleoside, we produce as a result of this our sugar, our free sugar, and a base, our nitrogenous base. Now, here when we're looking at the acid-catalyzed hydrolysis mechanism, it can be broken down into 4 steps. Step 1 deals with a proton transfer. Step 2, we have the loss of our leaving group. Step 3, we have our nucleophilic attack. And then finally, step 4, we have yet another proton transfer. So, just remember these individual steps help to make the overall mechanism for the acidic hydrolysis of our nucleoside. And now that we've talked about it, click on the next video and let's take a look at an example question.

Hydrolysis of Nucleosides Example 1

Hey everyone. So in this example question, it says, provide the mechanism for the acid catalyzed hydrolysis of Caedidine. Step 1, we're going to use the hydronium ion, and we're going to use it to protonate the carbonyl oxygen. Here is our carbonyl oxygen. We're going to introduce our hydronium ion, which is H₃O⁺. It's water mixed with acid here that we would do this. The HCl here is just helping provide an acidic enough environment that we can protonate the water. As a result, we create H₃O⁺. So we're going to protonate this carbon and oxygen. Doing that gives us this structure initially. We still have this CH₂OH portion here. We have our two OH groups. We have our nitrogenous base. And the nitrogenous base has its lone pair. Here goes our oxygen now, which is positively charged because it gained that H⁺.

Now, moving to step 2, we create a double bond between the anomeric carbon and the anomeric oxygen to expel the base. Remember, our anomeric oxygen is this oxygen right here, and the anomeric carbon is this carbon here. They're going to form a double bond with each other. Carbon can only make four bonds at one time, so that's going to expel this entire base here. Now we have this temporary structure. Here's our double bond between the anomeric carbon and the anomeric oxygen. We still have our two OH groups. We have our CH₂OH group, plus our expelled base. The H from the bond now belongs to it. Here is our free sugar and our base.

For our nucleophilic attack, we're going to use water as a nucleophile to attack the anomeric carbon. Water is going to come in and hit this anomeric carbon, kicking this bond back to the oxygen. We still have our base just sitting there. Now, at this point, here's our sugar. We have our OH groups. We have our CH₂OH. And now we have our oxygen, which is attached. This water, which detaches, can attach from the top or bottom, so either stereochemistry of a wedge or dash bond is possible. To show this, we show this squiggly line. Now, oxygen is making three bonds, so it's positively charged. And we still have this base here in its same form.

Now we use another water to deprotonate our positively charged oxygen. We can't just leave it like this; we can't have a charge on our final product. A second water molecule comes in. It's going to deprotonate, so it's going to remove this H. Oxygen holds on to the electrons. Here is our sugar again. The OH groups are still present. We have our CH₂OH. Again, we don't know the stereochemistry of this OH group. It could align itself top or bottom and give us alpha or beta of the sugar. We're just showing the squiggly line to show both are possible. Here is our nitrogenous base. Technically, this nitrogenous base could be in this form, its enol form, or it could be in its keto form, where you have the nitrogens here. N and N. So we're showing the tautomers of this particular base. Either one is a possibility. But here we have our free sugar, and we have our base that has been formed. And here I've shown both tautomers of that particular base.

This will represent the acid catalyzed mechanism when it comes to breaking down this nucleoside yet again into its free sugar and its base.

Propose a mechanism for acid-catalyzed hydrolysis of adenosine.

Hey, everyone. Let's take a look at the following practice question. Here it says, propose a mechanism for acid catalyzed hydrolysis of adenosine. Alright. So here we have our structure. And remember, the first step that we're going to do is we're going to protonate this nucleoside. What we're going to do here is HCl is within an aqueous environment. Its presence helps to create H3O+. Now, for this particular nucleoside, what happens is we're going to protonate this nitrogen here. It's going to come in. It's going to grab this H+H3O+. We protonate that one. Here our sugar is still intact. And here we have to draw this out. So this nitrogen now is positively charged. So there goes our structure, and we're going to do this. Next, what we're going to do is we're going to have a loss of leaving group. This nitrogen decide well, this anomeric oxygen decides to make a double bond with the anomeric carbon, causing this to break off and fly away. The base is going to be removed. So we come down here. Here's our free sugar now. These two are double bonded to each other, so now this is positively charged here. We have our base hanging around still. Actually, let's draw that base. This is positive. Okay. And what's going to happen next is we're going to do a nucleophilic attack. So the water molecule comes and hits the anomeric carbon, kicking this double bond to the oxygen. Now remember, that water molecule can come in from the top or bottom, so it could potentially have a wedged or dashed bond. To illustrate this possibility of both, we show this squiggly line. Here goes the water molecule. We're going to use a second water molecule to deprotonate it so that we get a neutral thing – a neutral sugar at the end. So here goes our second water molecule. It comes in deprotonates. So here goes our free sugar. Here goes the Oh group. Plus, we have our nitrogenous base here. So this nitrogenous base could also be in this form. Where this nitrogen doesn't have that H+ on it. And that H+ can gravitate down here to this nitrogen here so that both of them make three bonds and are neutral. Now here, I'm sure I'm not showing the stuff where we use another water molecule to deprotonate this nitrogen, and then this bottom nitrogen can then remove the H+. We're not doing all of that. This nitrogen down here, technically. This, so this nitrogen down here would be negatively charged since it's making two bonds. This is the better version because in this one, they're both neutral. They're both making the ideal three bonds. Here are our two free sugars. So these will be our hydrolysis products when we hydrolyze that initial nucleoside.