Synthesis of alcohols using Grignard reagents I
Synthesis of primary, secondary, and tertiary alcohols from aldehydes and ketones using Grignard reagents. Created by Jay.
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- So, my question is, what exactly happens to the magnesium-halogen compound? Does it simply leave or does it form a compound with something in the air/the ether.
Thanks a lot for the video, it truly explains a lot, really well!(5 votes)
- When the ketone or aldehyde (R = O) reacts with CH3MgBr, it is formed CH3 - R - O - MgBr. Then, when you add H2O (H - OH), you will get CH3 - R - OH (formed with the H from water) and HO - MgBr (formed with the OH from the water).(9 votes)
- At2:40, a carbon "anion radical" is formed, and it looks like there are 9 electrons around carbon. Since carbon's electrons can't expand into d and f orbitals, how is this possible?(6 votes)
- The Octet rule does not always hold! What you have described is a classical example of a violation of the rule.......(2 votes)
- What's the solvent for?(3 votes)
- Almost every reaction must occur in a solution. Because the grignard reagent acts as a carbanion (have two lone pair of electrons on a carbon), it is highly reactive and a strong nucleophile. Because of this, if we were to use water as our solvent, which contain hydrogens that are easy to steal by the carbanion, we would lose our grignard reagent and can no longer complete the reaction. By using ether (CH3-O-CH3) as our solvent, it is polar APROTIC, meaning that there aren't hydrogens for the carbanion to readily steal, and so we can keep our grignard reagent.(5 votes)
- at3:00Jay did the single electron addition and that halogen leaving in two different steps, so the intermediate is a carbon that has 9 electron and so breaking the octet rule, what is the reason that this could happen?(4 votes)
- I may be wrong, but I thought it was one step. The halogen's electrons are not really around the Carbon (electronegativity differences), so the single electron does have some room to come in(1 vote)
- can you explain to us how to read a chemistry given without get confused with all the information(2 votes)
- Just take it step by step and you can do it!(2 votes)
- Hi. Would you mind explaining to me why the CH3 anion chooses to bind to the carbonyl carbon instead of maintaining it ionic bond to the charged bromide? I understand that carbon is more electronegative than magnesium, however the magnesium is carrying a charge, and I thought a charged particle would always be more attractive than a non-charged one. Thanks(2 votes)
- Interesting perspective. However, it is faulty.
Remember that the methyl anion (Grignard) is extremely strong as a nucleophile and base.
You need to forget about charge and consider what atoms and molecular intermediates can form covalent bonds. Mg cannot ever do that. Therefore, the carbanion must be the nucleophile (MgX is just a spectator).(2 votes)
- With the cyclohexanone molecule, jay draws partial -ve charge on the O and partial +ve charge on the C.
My question is:
The carbonyl carbon is connected to three other carbons, isn't the inductive effect of the three carbons enough to counteract the polarisation of the -C=O ??(2 votes)
- The carbonyl carbon is connected to two other carbons. The inductive effect is real, but it is not enough to completely counteract the polarity of the C=O bond.(2 votes)
- In the first example, around05:46, we see that Grignard's reagent is our starting material (left of arrow) instead of the carbonyl containing compound. Is there a reason for this? Would it be wrong to write the carbonyl on the left side of the arrow and Grignard's reagent as being added/reacted in the first step?(2 votes)
- It makes not one iota of difference. You can switch the starting material and the reagent at any time. There is no etiquette/SOP (std of procedure) for doing this. No order is implied when doing either. It's the exact same thing.(2 votes)
- At2:04, what is driving the magnesium to donate its electrons to the carbon atom? The carbon atom already seems very stable in its octet so why would it want another electron?(2 votes)
- As Lithium and Magnesium shows diagonal relationship, why not Li is used to form Grignard reagent?(2 votes)
In this video we'll see how to synthesize alcohols using the Grignard reagents. So first, we have to learn how to make a Grignard reagent. So you start with an alkyl halide, so over here on the left. And you add a magnesium metal. And you need to add something like diethyl ether as your solvent. You can't have any water present because water will react with the Grignard reagent. And so this is what you make, over here on the right. You end up with a carbon atom bonded to a metal. Right? So carbon is bonded to magnesium. This is called an organometallic bond. And you can do this with other metals. You can do this with lithium, for example. But Grignard reagents are one of those things that's always talked about in undergraduate organic chemistry classes. And you can see that these two electrons here, these red ones, the ones in red. I've drawn it like a covalent bond. The bond between carbon and magnesium. But in reality, it's more ionic than covalent. So it's equivalent to the second structure down here. Now, in terms of electronegativities, carbon is actually more electronegative than magnesium. So the two electrons in red are actually going to be closer to the carbon atom, itself, giving the carbon a negative charge, and forming a carbanion. And so this is a carbanion that is formed. And this is unique because this carbanion can now act as a nucleophile in your mechanism to make alcohols. So this is the preparation of a Grignard reagent, it's proved to be a very, very useful thing in organic synthesis, so much so that Victor Grignard won the Nobel Prize for his research into this chemistry. Let's take a look at the mechanism to form a Grignard reagent. So I'm going to start with my alkyl halide. And this time I'll draw in all of my lone pairs on my halogen, like that. And we're going to add magnesium, which we know, being in Group 2, magnesium has two valence electrons. I'm going to draw magnesium's two valence electrons like that. In the first step of the mechanism, magnesium is going to donate one of its electrons. All right? So we're going to show the movement of one of its electrons over here to this carbon. We're going to use a half-headed arrow like that. So, we're going to make a new anion here, because this carbon actually picks up an electron. So it actually picks up a negative charge. And we call this an anion radical, OK? So this intermediate here is called an anion radical. So I'll go ahead and write that. So it's an anion radical. It's an anion because it picked up an electron, giving it a negative charge. And it's radical because that electron is unpaired. So this anion radical is unstable and it's going to fragment. So these two electrons right here are going to come off, onto the halogen. Right? So let's go ahead and draw what that would give us. We now have this carbon with one electron around it on the right side, like that. And now, our halogen over here, it had three lone pairs. It just picked up another. So it is now negatively charged, like that. And the magnesium that we started with donated an electron. Right? So this magnesium has one electron left around it. One valence electron. And it donated an electron, which gives it a plus 1 charge. Right? So this magnesium is now positively charged because it donated an electron in the first step. And in the next step of the mechanism, magnesium can donate its second valence electron. And its stable for it to do so, because then it'll have an electron configuration like a noble gas. So magnesium is going to go ahead and donate its second electron over here to the carbon, like that. And let's go ahead and draw what would result. So we're going to now have our carbon, right here, with three bonds. With now, with now two electrons around it. And that's what gives it its negative 1 formal charge to make a carbanion. Magnesium has donated both of its electrons. So this magnesium is now a plus 2 charged cation. So Mg2 plus, like that. And then the halogen is going to form an ionic bond with the magnesium on the right side here. So we have our halogen, which is negatively charged. Right? So we have a magnesium with two positive charges. And then, we have two things with negative charges around it, forming ionic bonds, right? So this is our carbanion. And, so I could redraw this stuff on the left with this carbon here. I can make it an R group with the lone pair of electrons, a negative 1 formal charge. And all the stuff on the right, I could just write it like this. There are several ways that you'll see this written. I could just say this is MgX with a plus 1 charge. So this just allows us to focus in on this carbanion here with a negative 1 charge. Or I could just pretend like everything's covalent and just save myself some time, right? I could go like that. And organic chemists understand what this organometallic compound means, that the R group is negatively charged as a carbanion. So, you'll see several different ways to write a Grignard reagent, just as long as you understand what's going on. That's the most important thing. So let's now take the Grignard reagent we just formed, and let's make an alcohol with it. OK, so I'll go ahead and write it the last way I did. It doesn't really matter how you do it. All right, so this is our generic reaction. We're going to introduce a carbonyl compound. So for this generic reaction I'm just going to say, it's some generic carbonyl. So I'm not going to show what's attached to either side of my carbonyl carbon here. And diethyl ether once again is our solvent. You have to exclude water from this reaction again, because the Grignard reagent will react with it. So in the first step, you want it to react with your carbonyl. And the second step, once it's reacted with the carbonyl, it's OK to add water in the form of H3O plus. And this is going to form our alcohol. So we're going to form an alcohol as our product. And this carbon here came from our carbonyl, and this R group is going to attach to that carbonyl carbon. That comes from our Grignard reagent. So it's a very useful reaction because it's a carbon, carbon, bond forming reaction. Right? So this R group had a carbon on the end. And then this carbon right here. So forming carbon, carbon bonds is very important when you're trying to build large, organic molecules using synthesis. So this is a very useful way to form either a primary, secondary, or a tertiary alcohol. It all depends on what sort of carbonyl compound that you're starting with. So let's show the mechanism for what's happening. All right. So I'm going to say the Grignard reagent is a source of carbanions, right? So I'm just going to draw my carbanion here like that. And then my carbonyl, right? I have carbon double bonded to an oxygen. And I'm going to make this, you know, anything could be attached the carbon for right now. Once again, when you're looking at carbonyl chemistry, all you have to do is think about electronegativity differences between carbon and oxygen, right? So go back and watch the electronegativity video. We know that oxygen, being more electronegative, will draw these electrons in the double bond closer to it, giving it a partial negative charge, leaving our carbon partially positive. Right, so carbon, being partially positive, carbon wants electrons, right? Carbon is an electrophile. And from our Grignard reagent, we have a nucleophile. The carbanion is going to act as a nucleophile. The negative charge is attracted to the positive charge. So the carbanion attacks the carbonyl carbon like that, which would kick these electrons off onto our oxygen. And I'll go ahead and draw the intermediate here. So we now have, we now have our R group directly attached to our, what used to be our carbonyl carbon. And now, our oxygen has three lone pairs of electrons around it, which give our oxygen a negative 1 formal charge, like that. So that's the first step of our reaction. In the second step, we have hydronium ions floating around, right? So H3O plus. So I'll go ahead and draw H3O plus here, like that. And the second step, of course, will be acid based chemistry. So a lone pair of electrons on our oxygen takes a proton from H3O plus, leaving these electrons behind to form water. And that is how we get our alcohol. OK? So that's going to give us our alcohol as our product. So, after our acid based reaction. All right, let's look at three different examples of synthesis of alcohols. OK, and let's show the different types of alcohols that can be produced. So we'll start with formaldehyde here. So we'll start with a very simple molecule like that. And we'll go ahead and already make our Grignard reagent. And we're going to make methyl magnesium bromide, so methyl magnesium bromide we're going to add in our first step. We're going to use her as our solvent. And in our second step, we're going to add H3O plus. Now, when you're analyzing a Grignard reagent, you pretty much have to think, where's my carbanion? Right? So this carbon right here is negatively charged. That's the carbon that's going to attack my carbonyl. So if this carbon attacks my carbonyl, right? And the electrons kick off onto here, right? As our intermediate, we would have hydrogens on either side of our carbon. A negatively charged oxygen up here. And the R group, this time, is a methyl group like that. So after you protonate it, right, in the second step. So after these, so we'll just say, these lone pair of electrons are going to pick up a proton from H3O plus, right? We would form this carbon with two hydrogens. We're going to protonate our alc oxide to form an alcohol up here for our product. And if you look at that molecule closely, you'll notice it is ethanol. Right? So you make a primary alcohols if you use formaldihide. So this is a primary alcohol. Its primary because the carbon attached to the OH is attached to one other carbon. Let's see how we can make secondary alcohols. OK, so this time you need to start with an aldehide. So instead of two hydrogens on either side of your carbon, as we did before, this time you have to have an R group on one side. So this will be our aldehide, like that. All right, and let's go ahead and use the same Grignard reagent. We use methyl magnesium bromide again. Like that, and second step, H3O plus. So once again, think about what is your nucleophile. Right? So the negatively charged carbon is going to be my nucleophile. Is going to attack my carbonyl, kick these electrons off. So once again, when we draw the intermediate, all right, up at the top here. We have our alc oxide anion, negatively charged. The hydrogen is still there. And what we did was, we added a methyl group on. So this CH3 at the bottom of our intermediate came from our Grignard reagent. And once again, acid based chemistry, to protonate the alc oxide, we'll form our secondary alcohol like that. So that would be the secondary alcohol that is produced from this reaction. Once again, this carbon is attached to two other carbons, making this a secondary alcohol. All right, so one more example here. This time we will react our Grignard reagent with a ketone. So we'll start with our ketone over here, on the left. So here's our ketone. So it's cyclohexanone. And once again, let's stick with methyl magnesium bromide. And so we have methyl magnesium bromide that we add. And again, our solvent is ether, excluding water. And second step, we're going to add a source of proton. So H3O plus. So once again, the exact same mechanism, exact same thinking involved. Right? Our negatively charged carbanion attacks, our carbonyl carbon, kicking these electrons off onto our oxygen. All right, so we form our intermediate. Right, so this top oxygen here now has three lone pairs of electrons, negatively charged. And actually, let me go ahead and take that off there so we can better show the atoms attached to that carbonyl carbon, right? So, if I'm showing that methyl group attacking that carbonyl, I'm going to push that alc oxide over here a little bit to the left. And that's going to get my negative charge. And that way, that just gives me some space to put my methyl group right here, like that. And that just looks a little bit closer to what my final product will look like. Right? So this lone pair of electrons, right? As usual, this lone pair, one of these one pairs is going to pick up a proton right here. And that's going to form our product. Right, so we now protonate our alc oxide to form our alcohol, like that. So our alcohol is going to form right here. And then, our methyl group adds on right here. And, if you look closely, you can see this is a tertiary alcohol that we just made. All right, so this carbon is connected to one, two, three other carbons. So, Grignard reagents are very useful for making alcohols. And you can make either primary, secondary, or tertiary alcohols from them. So it's a very versatile reagent to use. In the next video, we'll take a look at more about Grignard reagents, and we'll talk a little bit about how to work backwards and think about synthesis problems.