Grignard reagents add to carbonate esters as they add to other...

Question

Grignard reagents add to carbonate esters as they add to other esters.

(a) Predict the major product of the following reaction.

(b) Show how you would synthesize 3-ethylpentan-3-ol using diethyl carbonate and ethyl bromide as your only organic reagents

Accepted Answer

All right. Hi, everyone. So this question says that green red reactions react with carbonate ethers in a similar fashion as with other ethers. Part one determine the identity of the major product of the reaction shown below. And number two show how to synthesize two methyl propan tool from dimethyl carbonate and Methylbromide as the starting material. So dimethyl carbonate is the carbonate ester that we have on the screen right now. And for part one of the reaction we're reacting our dimethyl carbonate here with ethyl magnesium bromide in excess. So here, recalled greener reagents produce new carbon carbon bonds as a result, right, as a product of the reaction itself. So here, recalled carbonate ester like the one we have on the screen can react with greener reagent in excess to produce alcohols with identical substituent on the carbonyl carbon. So here, let's go ahead and illustrate that point, right. So let's go ahead and take our here and react with our first equivalent of ethyl magnesium bromide. So recall that the magnesium bromide group itself is going to have a partial positive charge associated to it. And as a result, the carbon that it's connected to is going to have a partial negative charge and be nucleic as a result. So because the carbon is nucleic, the carbon is going to go ahead and attack the carbon. And that is what creates the new carbon carbon bond. Now, as a result of this nucleic attack, the pi electrons of my carbonyl are going to shift upwards towards oxygen. And as a result, we're going to have a tetrahedral intermediate in which we have four groups connecting to the carbonyl carbon. No, because this tetrahedral intermediate is not the most stable, the carbona is going to reform after a pair of electrons on oxygen moves downward between itself and carbon. Now that as a consequence is going to, I realized that I drew that meth oxy group wrong because the oxygen is supposed to connect to the carbon, not the methyl group. But anyways back to the topic at hand, right, as a consequence of this carbon or reforming the tetrahedral intermediate is going to fall apart. And one of my Mathy groups of my carbonate ester here, our dimethyl carbonate is going to go ahead and be displaced, it's going to be kicked out, so to speak. In fact, that is going to form the following. So after our first reaction right here, we have a regular and so when this process repeats itself again, in the second step with the second molecule of ethyl magnesium bromide, then what's going to happen is that both metox groups are going to be removed and we are left with a key to instead. So here, right, my third equivalent of eyl magnesium bromide is going to attack my carbon once again. And that is going to go ahead and produce an alcohol after protonation because at this point, right, there are no more leaving groups. And so when the pi electrons of the carbon move upwards towards oxygen, that oxygen must be pronated in a separate step to yield our final alcohol, right? Because here, here's my third equivalent or f for magnesium bromide. So once the nucleic carbon attacks my carbonyl carbon, my pie bo is going to break and now I have three ethyl groups on the same carpet. So now, right now I have an oxide ion with magnesium bromite as a counter ais. So here, right, recall that the very last step of agreeing your reagent is acid, work up in order to proton that al oxide and yield are final alcohol at the end. So my product is an alcohol with three identical ethyl substitutes. So for part one, that's what I'm going to draw right here is my alcohol and here is my major product for part one. So now, for part two, right, for part two, we're trying to figure out how to synthesize two methyl propan two or dimethyl carbonate. So if I scroll down for part two, right, we have dimethyl carbonate once again and we are reacting to form, we are reacting to form two methyl propan 20. OK. So we have a three carbon chain with a hydroxy group on carbon two as well as another methyl substituent on carbon two. No notice here how our target product is a tertiary alcohol with three identical substitutes, right? We have three identical methyl substituent on our tertiary alcohol. So because of this right, our carbonate ester can react with excess green yard reagent to produce the alcohol we need with three identical substitutes. So notice how the question is asking us to use dimethyl carbonate and methyl bromide. So methyl bromide is what's going to provide, right, our green reagent. But first, we have to go ahead and convert Methylbromide into a greener reagent and recall that we can do that by reacting with magnesium in the presence of ether because that is going to give methyl magnesium Promide, which is my green reagent. So from here, oops, from here, I can take my carbonate ester again, I can take it again and just like in, in part one, I can go ahead and react my carbonate ester with an excess of greener reagent because as a result of my excess reagent, I'm going to result in an alcide ion with three identical methyl substitutes. So from there, in my second step, I have to introduce acid to perate. It's a pro my oxide ion and yield my final tertiary alcohol. And here we've arrived at the answer right. So the bottom line here is that carbonate ester can react with multiple equivalents of greener reagents to yield tertiary alcohols with identical substitutes. So as a result of treating dimethyl carbonate with multiple equivalents of ethyl magnesium bromite, in part one, we ended up with a tertiary alcohol in which there were three identical ethyl substitutes. Now, in part two, in order to showcase how two methyl propan tool can be synthesized. First, we had to go ahead and convert Methylbromide into Meho masi bromide using magnesium in the presence of ether. And then from there, we reacted dimethyl carbonate with an excess of methyl magnesium bromide to yield a tertiary alcohol with three identical method substitutes. So if you stuck around to the end, thank you so much for watching. And I hope you found this helpful.

Grignard reagents add to carbonate esters as they add to other esters.
(a) Predict the major product of the following reaction.

(b) Show how you would synthesize 3-ethylpentan-3-ol using diethyl carbonate and ethyl bromide as your only organic reagents

Key Concepts

Grignard Reagents

Carbonate Esters

Synthesis of Alcohols

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