Working backwards from a particular alcohol and determining what Grignard reagent you would need to synthesize it. Created by Jay.
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- I understand that the H3O+ is needed to protonate the Oxygen to make the alcohol... but I dont understand the need for the Et2O? what does it do in the mechanism?(12 votes)
- He touches on the concept in other videos, but Et2O acts as a solvent to conduct the reaction in. This can't (initially) take place in water because the Grignard reagent will react with it.(14 votes)
- At8:51Jay says that the lone pair of electrons from the oxygen will reform a double bond with carbon in order to kick off the other oxygen, but why is that a more likely scenario than the oxygen with a negative charge pulling in a hydrogen from the H3O group and continuing the reaction from that knowing that the Carbon will be partially positive?(5 votes)
- Usually hemiacetals (an OH and an OR on the same carbon atom) like that in the drawing are rather unstable. One example of stable hemiacetals is glucose in its cyclic form.(2 votes)
- I don't get the point of having grignards reagent once , on the arrow and once on the left hand side of the arrow ? Does it make any real difference ?(3 votes)
- No it doesn't make a difference! The grignard reagent will be in solution anyways and that's what it means when it's on the arrow and when it's on the left side of the arrow(2 votes)
- At9:20why does the double bond reform? Why can't the hydronium molecule protonate the molecule and yield the product with the ether group still present?(2 votes)
- At this stage of the reaction, the mixture is still basic. There is little hydronium present.
Even if hydronium ion were present, the product would be a hemiketal, which would also decompose to form the ketone.(4 votes)
- Won't the phenyl group be sterically hindered to attack the carbonyl group?(3 votes)
- if you look at the carbonyl carbon , it's sp2 hybridised and have planer geometry , so there won't be the case for steric hinderence.(1 vote)
- Why is preparation of alcohol from alkyl halides considered the best method for preparation
of primary alcohols?(2 votes)
- At2:09, where we suppose that propriophenone is the initial reactant, I'm having a hard time accepting that the "partial negative" on the carbon will not be stabilized by resonance of the pi system formed with the phenyl group. Would this factor strongly reduce the likelihood of that reaction?(2 votes)
- In propiophenone, the carbonyl carbon has a partial positive charge and, yes, it is partially stabilized by resonance with the phenyl group.
However, it still has a partial positive charge.
That’s why it is attacked by the partially negative-charged methyl group of the CH₃MgBr.(2 votes)
- in the reaction with ester will the alkoxide remain as it is? if so why?(1 vote)
- Why do Grignard reagents not react with ether solvents? Thinking about pKa values, Grignard reagents are way more basic than alkoxide anions, so I'd expect a substitution like this to happen: https://www.hiveworkshop.com/attachments/ethercleavage-gif.312364/
Why does this not occur? I know it occurs with epoxides due to steric strain, but I don't see why it doesn't occur (pardon the double negative) with any ether, strained cyclic system or not.(2 votes)
- I think The leaving group - O-R, it will get H form H3O+. So it will be HO-R. So we will get two type of alcohols. right?(1 vote)
- If you are referring to the reaction with an ester, yes. The reaction of RMgBr with R'COOR" gives RR'CHOH and R"OH. Often, R" is ethyl, so R"OH is ethanol, which is soluble in water. The other alcohol, RR'CHOH, usually has enough carbon atoms to make it insoluble in water, so the ethanol is easily removed by extraction with water.(3 votes)
In the last video, we saw how to make Grignard reagents and how to use Grignard reagents to make primary, secondary, or tertiary alcohols. In this video, we'll take a look at how to retrosynthesize alcohols, how to think backwards and figure out what Grignard reagent you would need to make a certain alcohol. And then, we'll take a look at the reaction of Grignard reagents with esters as well. So here is our alcohol. And we're going to think backwards. We're going to think using retrosynthesis and try to figure out how to make this alcohol using a Grignard reagent. And there are actually several different ways to do it. So if you understand the mechanism really well, you know that you add an alkyl group onto what used to be your carbonyl carbon. And your carbonyl carbon was the one connected to your oxygen. So if you're thinking backwards, you can think to yourself-- this must be my carbonyl carbon. All right. This must be the one. And I added an alkyl group onto that carbonyl carbon. So if you're taking backwards, you can disconnect a bond. So I could say-- oh, I'm going to make my alkyl group this group right here. So I'm going to say it's that methyl group. So I'm going to disconnect that bond. And that will allow me to figure out my starting materials. So if that's the bond that isn't going to exist over here when I'm drawing my starting materials-- I have my benzene ring right here. And the benzene ring is directly attached to what was my carbonyl carbon. So I'm going to go ahead and draw my carbonyl carbon in here. Like that. And I'm saying that this ethyl group over here in the right is still there. And what I added on in the mechanism was a methyl group. So the methyl group must have come from my Grignard reagent. So if I'm going to go ahead and write my Grignard reagent, it must be a methyl group. And so I'll make methyl magnesium bromide-- which is what we used in the previous video for our examples. So these two things in the right would be my starting materials to make the alcohol on the left. And if you're unsure of that, just think about working backwards. So you could think to yourself-- well, once again, from the last video, we know that this methyl group is actually negatively charged and is a carbanion. And it's going to attack our carbonyl carbon. Like that. And then, those electrons are going to kick off onto the oxygen. So when we add in our reagents-- in the first step, when you add your Grignard reagent and your ketone up here-- you would need to use ether as your solvent. And the second step-- you need to protonate your alkoxides to form your alcohol over here on the left. So you need H3O+. So that's how to think backwards and come up with your starting materials used to synthesize the alcohol on the left. This isn't the only way to do it. Right? Let's redraw this alcohol, and let's look at another way that we can synthesize the same alcohol here. So let's go ahead and redraw the alcohol on the left. And think about how else could I make it. OK. So here's my alcohol. And before, we first identified our carbonyl carbon. It's going to be the one attached to the oxygen right here. And I have all these alkyl groups attached to it. So I'm just going to disconnect another alkyl group. This time I'll disconnect the alkyl group on the right. And so I'm thinking retrosynthesis. I'm thinking about working backwards. All right. So if I'm going to go ahead and draw the carbonyl compounds, that has my benzene ring untouched. And the carbon in red is going to become my carbonyl carbon. Like that. And what's attached to that carbonyl? It's going to be a methyl group. Right? And this time, the methyl group was left. So when I'm thinking about what should be my alkyl group here, it would be the methyl group that I disconnected on the right. So my Grignard reagent is going to be the source of my ethyl group. So I'm going to go ahead and draw an ethyl group attached to the magnesium. So we have ethyl magnesium, a bromide. Like that. So this would give us the alcohol on the left. Again, same concept. Same mechanism. This carbon right here would end up attacking my carbonyl. Kick these electrons off. And once again, we need to specify that the first step of this synthesis is done in diethyl ether. And the second step-- we protonate to form our alcohol. Like that. All right. So let's look at yet another way to form this alcohol. We're kind of ignoring stereochemistry here. So if some of you were thinking, that carbon is a chirality center. We're just trying to think about how to make these alcohols here and not worry about what we're getting in terms of stereochemistry for our products. So if I start here with my benzene ring, and once again, we're going to make the exact same alcohol. And we're going to think backwards again. And once again, we know this is going to be my carbonyl carbon. This time, we're going to disconnect this bond. So that's the alkyl group that we're going to add on as our Grignard reagent. So there's a third way of doing it. So now, this time, I've disconnected my benzene ring. And this is going to be my Grignard agent this time. So this carbon on my benzene ring is going to be ionically bonded to magnesium. So MgBr. Like that. And so now it's called a "phenyl group." So you would term this Grignard reagent "phenylmagnesium bromide." And I can see the skeleton for the ketone that I'm going to use on the right there. So I have a carbonyl. And let's see. There's a methyl group on the left side and an ethyl group on the right side. So once again, if you mix those two together in ether and then protonate it, you will end up with the alcohol on the left. So three different ways to synthesize the same alcohol. And it's just thinking about what's my alkyl group and how can I add it on? All right. Let's look at the general reaction for a Grignard reagent with esters. OK. So there is my Grignard reagent, and I'm going to react it with an ester this time. So I'll make this R prime. And I'll make this one R double prime to distinguish it from the R group on our Grignard reagent. And the second step, once again, we add H3O+. In this situation, the Grignard reagent is going to add on twice to our carbonyl carbon. So the carbon attached to the OH for our product, this is another way to synthesize alcohols. It's going to have the R prime group still attached to it. And to the R group from the Grignard reagent is going to add on twice. So if you look at this, you can see that the R prime group and this carbon came from our ester. So did this oxygen right here. So this R group here from our Grignard reagent is going to add on twice in the mechanism. So let's take a look at what happens in our mechanism. We start with our ester. All right. So R prime, C double bond O, O, R double prime. And we'll go ahead and put in our lone pairs of electrons. We saw-- in the last video-- that that carbonyl is a polarized bond. Right? The oxygen ends up being partially negative, and the carbon ends up being partially positive. So the carbon is partially positive, making it an electrophile. The source of the electrons come from the Grignard reagent because we know that in that Grignard reagent we're treating that R group as a carbanion. All right. So this R group up here is actually negatively charged. So we saw that in the last video. So it is R with 2 electrons around it giving it a negative 1 formal charge, making it a carbanion, which is a good nucleophile. So the nucleophile's going to attack our electrophile. Opposite charges attract. All right. So nucleophilic attack. We'll kick these electrons off onto your oxygen in the first step of the mechanism. So now we have R prime, carbon with an oxygen up here, Three lone pairs of electrons around it-- giving it a negative 1 formal charge. And over here in the right, we have oxygen with an R prime. And then, our lone pairs of electrons. Like that. Our double prime, I should say. And then, down here, we attached our R group. OK. So in the next step of the mechanism, a lone pair electrons are going to move in here to reform our pi bond, to reform our double bond. And that's going to kick the electrons in this bond off onto the oxygen, since carbon cannot have five bonds. All right. So let's go ahead and draw the product of that. So we're going to reform our carbonyl. So we still have an R prime group attached to our carbonyl. Like that. We've just added an R group onto our carbonyl. And the leaving group will be this alkoxide over here. So R double prime with a negative 1 formal charge on the oxygen. So now we have a ketone. And if you have a molar excess of your Grignard reagent, the Grignard reagent is going to attack your ketone. So once again, another molar equivalent of our carbanion from our Grignard reagent is going to attack our carbonyl carbon-- just like before. Kick these electrons off onto the oxygen. So let's get some space here to show what happens. So our next intermediate is going to have our R prime attached to our carbon. This top oxygen here has three lone pairs of electrons and a negative 1 formal charge. We already added on one R group. We're going to add on another R group from our Grignard reagent. And in the last step, it's just acid-base chemistry. Right? We're going to add on H3O+. And this will allow us to protonate our alkoxide. Right? So a lone pair of electrons on here. Pick up a proton. Like that. Kick those electrons off. And then, we are done with our product. R prime. Carbon bonded to an oxygen bonded to a hydrogen. 2 lone pairs like that. And then, two R groups, which came from our Grignard reagent-- like that-- to form our alcohol. OK. So let's look at a reaction. So let's react an ester with excess Grignard reagent and see what our product will be. So we'll do a nice simple ester over here on the left. So here is our ester. And in the first step, we add excess methyl magnesium bromides. And in the second step, we add our H3O+. So we need to think to ourself-- what's going to happen? Well, I know that in my mechanism, this negatively charged carbanion methyl group is going to add on to this carbon. This is going to leave. It's going to be our leaving group. And then, the reaction happens again. All right. And the reaction happens again adding on another methyl group, and then, finally, protonated to form an alcohol. So when we draw the product, we know that this portion is going to come from our ester. We know we're going to form an alcohol. And we know that we're going to add our methyl group on twice. So all we have to do is put two methyl groups on there. Like that. So that, of course, would form a tertiary alcohol. So this is tertiary right here. This carbon bonded to three other carbons. So that does it for Grignard reagents and different ways to make alcohols.