Propose a mechanism for methylation of any one of the hydroxy ...

Question

Propose a mechanism for methylation of any one of the hydroxy groups of methyl α-D-glucopyranoside, using NaOH and dimethyl sulfate.

Accepted Answer

All right. Hi, everyone. So for this question, I'd suggest a suitable mechanism to show the formation of methyl ether at any one of the hydroxy groups of methyl alpha D galacto Perano side by treatment with dimethyl sulfate and sodium hydroxide. And up on the screen here is the structure of alpha D galacto Peran. Aside on the screen now, because were trying to synthesize an ether specifically recall that the Williamson Ether synthesis is essentially the most common method for making ethers, right? But the problem with the Williamson Ether synthesis specifically for this question is that the Williamson Ether synthesis involves the use of a very strong base such as an oxy. And the reason why I am referring to this as a problem for this question is because our starting material is in fact a sugar molecule. So if I was to attempt to proceed with a regular Williamson ether synthesis with this simple sugar and an oxide ion, the simple sugar would not only isomerize but it would also decompose. So I can't quite do that in this case, right? Because I'm trying to keep this structure intact. So instead, right, there is such a thing as a slightly modified Williamson ether synthesis because instead of using or oxide, I can go ahead and use hydroxide instead. But before he can do this, right, the simple sugar has to first be converted into a glycoside because the glycoside here is what contains an ace to over on position one on carbon, one of the ring and that stabilizes the compound overall and prevents decomposition. So for this scenario, our base is going to be hydroxide, not oxide. And for the purposes of my illustration, the hydroxy group that's going to react is going to be the one on carbon two over here. So that's the one that we're going to react. So to do that right, let's go ahead and bring in our hydroxide, which I'm drawing in blue. Now hydroxide being a base is going to go ahead and remove the hydroxy proton of the hydroxy group on carbon two, which is therefore going to result in a negative charge on our oxygen. So here I'm actually going to scroll down to draw the second step. And it's worth noting that this first step earlier, I said second step, but I meant first step, this step is reversible. So that's why I'm drawing the equilibrium arrows. So here here is my ring structure and for this scenario, just to make sure that I'm not missing anything during my, during my drawing here, I'm going to go ahead and make sure I include everything that hasn't reacted before. I actually showcase what has changed. So here I'm just going around and making sure that I'm adding everything that I'm that's not reacting in case I miss it. Right. So here are the hydroxy groups and the hydrogens of other carbons. Carbon one on the bottom right corner is an acea because it has a metox group instead of a hydroxy group. And now on carbon too, we're going to have the same hydrogen in the axial position, but we're now going to have an oxide ion. Excuse me, there was a hydrogen in the axial position and now in the equatorial position, there is an oxide instead of a hydroxy group. So at this point, let's recall here that the text of the question mentioned specifically dimethyl sulfate in addition to sodium hydroxide. And the reason that matters is because it is dimethyl sulfate that we're going to use in the next part of our synthesis. So I'm going to go ahead and draw out dimethyl sulfate in green. And I am going to make sure that I kind of separate one methyl group from the rest of the structure because it is this one methyl group that will attach itself to the oxide oxygen on my alpha de galacto Perano site. So in this case, right, they are cox side as a nuclear file and the methyl carbon of dimethyl sulfate is the electro considering it's connecting to the more electron oxygen atom, meaning it is electronic. So now oxygen is going to attack the methyl carbon and that is going to break the bond between carbon and oxygen and that is going to generate our final product and this process is actually irreversible. So I'm drawing an arrow pointing only in one direction. And now we can go ahead and draw the final product. And again, I'm going to make sure I add everything that hasn't reacted just in case I miss it, I can fix that. Give me one second. So now, right, we've gone from a hydroxy group to an oxide and now we have a meth boxy group at the very end of this process because now the hydroxy oxygen of carbon too of alpha D galacto piran aside has since been converted into a meth group to form an ether. So instead of a hydrogen, we have a methyl group and such is our final product. But in addition to this new ether, right, recall it, you also have to include your side products which in this case is going to be dimethyl sulfate or excuse me, not dimethyl sulfate, but the product of a methyl group being removed from dimethyl sulfate. That's what I meant. But there you have it. So this problem right here exported an example of a modified Williamson ether synthesis involving a simple sugar and hydroxide as the base. So in the first step, right, the hydroxy proton was detonated by hydroxide to yield the corresponding oxide ion on carbon two of the simple sugar. And as a result, the resulting oxide ion received an additional methyl group or rather a methyl group from dimethyl sulfate. They yield the corresponding meth boxy group and creating an ester on carbon tube. So with that being said, if you stuck around to the end, thank you so very much for watching. And I hope you found this helpful.

Propose a mechanism for methylation of any one of the hydroxy groups of methyl α-D-glucopyranoside, using NaOH and dimethyl sulfate.

Key Concepts

Methylation

Nucleophilic Substitution

Role of Base (NaOH)

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