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.