- Formation of hydrates
- Formation of hemiacetals and hemiketals
- Acid and base catalyzed formation of hydrates and hemiacetals
- Formation of acetals
- Acetals as protecting groups and thioacetals
- Formation of imines and enamines
- Formation of oximes and hydrazones
- Addition of carbon nucleophiles to aldehydes and ketones
- Formation of alcohols using hydride reducing agents
- Oxidation of aldehydes using Tollens' reagent
Addition of carbon nucleophiles to aldehydes and ketones
How carbon-containing nucleophiles attack aldehydes and ketones to form alcohols. Created by Jay.
Want to join the conversation?
- I thought that Grignard reagents needed to react in absence of water, to avoid forming an R-H? Where does the H2O come into play?(5 votes)
- The H₂O is added after the Grignard adduct has been formed.
It is part of the "workup" procedure to isolate the alcohol.
R₂CHO⁻ ⁺MgBr + H₂O → R₂CHOH + Mg(OH)Br(11 votes)
- So just out of curiosity, this set of videos seems to have a lot of reactions. Does anyone know if this is an exhaustive list for the MCAT or are these just the nucleophilic reactions? Do we need to focus on the other kinds of reactions as well? Thanks!(2 votes)
- The way taught to me is that in acidic environments like the first problem, you first protonate, so creating the alcohol first is okay too? Or does the negatively charged nucleophile get preferred to add first?(1 vote)
- What are hydrins actually? I found that the structures of halohydrins had a halogen and OH on adjacent carbons, but in case of cyanohydrins, the OH and cyano groups are attached to the same carbon.(1 vote)
- At8:48what happens to the proton dissociated from the water molecule and the MgBr + ?(0 votes)
- The H from the water is the H in the alcohol, and the OH from the water ends up in the insoluble salt MgBr(OH).(3 votes)
- I thought (and read in the book written by Weid) you would need an acid enviroment for the second step in the addition with grignard reagent, because with water alone it is less likely to happen, or something like that. Is this true?(1 vote)
- its been a while since I took orgo, but at8:01what happens to the CH3 that it get converted to OH?(1 vote)
- Nothing happened to it, it’s still there, it didn’t get converted into OH. Look at the purple circles.(1 vote)
- How can we obtain phenylethanal from benzyl alcohol?(1 vote)
- how cyclopentanone form by wittig reaction?(0 votes)
- won't the Cl in the HCl also react with some of those aldehydes or ketones thus making there be a cyanohydrate and a carbon bonded to 2 R groups, an OH, and a Cl? What would you call this alcohol that also is bonded to at least 1 halogen?(0 votes)
Voiceover: We've seen a lot of nucleophilic addition reactions to aldehydes and ketones and in this video we're going to look at the addition of carbon nucleophiles. If I started with an aldehyde or ketone and I add something like potassium cyanide and a source of protons, I'm going to form a cyanohydrate over here. This is a cyano group and notice we are forming a carbon-carbon bond, so that can be useful for synthesis reactions. Let's look in more detail at our carbon nucleophile which is from the cyanide anion here. Let's go ahead and draw the cyanide anions, so that carbon triple bonded to nitrogen. I had the lone pair of electrons on this carbon like that, and so that gives this carbon a negative one formal charge which makes it a nucelophile. It's attracted to positively charged things and over here on our carbonyl, we know our oxgyen is partially negative and we know our carbonyl carbon is partially positive. The carbonyl carbon is our electrophile and so our nucleophile is going to attack our electrophiles, these opposite charges attract here and these electrons are going to attack right here at these carbon, pushing these electrons off onto the oxygen. Let's draw the intermediates. We would now form a carbon-carbon bond and then this carbon is triple bonded to a nitrogen like that. Following our electrons, these electrons right here are forming this carbon-carbon bond and then on the left side, we would have our oxygen. This time with three lone pairs of electrons on it which gives it a negative one formal charge and if we started with an aldehyde, we would have an R and an H here. In the final step, all we have to do is protonate our oxygen, so a lone pair of electrons picks up this proton to form this O-H and we formed our cyanohydrin as our product. Let's look at an example of cyanohydrin formation. If we started with something like acetone, so a ketone, and we added potassium cyanide and hydrochloric acid, we would form our cyanohydrins. We would have our carbon triple bonded to our nitrogen over here and then we would have our O-H over here. Since we started with a ketone this time, we would have these two methyl groups right here like that. Once you form a cyanogroup here, you can convert this into other functional groups, so that is another reason why it is useful for a synthesis. Let's look at another reaction where we have a carbon nucleophile and these would be the organometallics and specifically, we have a Grignard reagent here. We have an alkyl group bonded to magnesium and then a halogen, so an organomagnesium compound. We could have just as easily done something like R prime and then a bond to lithium, so organolithium compound. These are just organometallics. Now, the thing about organometallics is the electronegativity difference between the carbon that's bonded in this case to the magnesium, where the carbon bonded to lithium. Carbon is more electronegative than those atoms. If you think about the bond, I'm going to go ahead and highlight here, this bond between the carbon and our magnesium, carbon's more electronegative. Those electrons are polarized towards the carbon. You could think about these bonds as being very polar covalent bonds or you could even think about those electrons in blue being on your carbons. Let's go ahead and draw it that way, so another way to think about it would be it's so polarized that those electrons are on your carbon which gives you a negative one formal charge and you form a carbanion. Either way of thinking about it is fine with me, maybe some professors might care, but once you think about this being a carbanion, it makes these reactions a little bit easier because once again we have a polarized carbonyl situation, partial negative, partial positive, and so your nucleophile can attack right here, pushing these electrons off onto your oxygen. You could have just as easily thought about this way with a very polar covalent bond right here and attacking here or you could think about it as being ionic with these electrons on the carbon forming a carbanion. You'll see either mechanism, depending on which textbook that you look in. Once the nucleophile attacks, we would now have a carbon bonded to our R-double prime group and then over here on the left, we would have our negatively charged oxygen like that, so an R group. Then a hydrogen if we started with an aldehyde. Those electrons in blue, let's go ahead and change them, make then red here. These electrons right here, so these electrons. That formed this bond, so we formed a carbon-carbon bond now. In the next step of the mechanism, which needs to be kept separate, we're going to add a proton source or something like water. Once again, we're going to protonate our alkoxide intermediate. We pick up a proton here and then that gives us our final product, which is an alcohol and we can see we've added on our R-double prime group and formed a carbon-carbon bond. Once again, a very useful reaction for synthesis. Now, you have to do this reaction in two steps because if you try to do it in one step, if you tried to add on your proton at the same time you added your organometallic, these are good nucleophiles, they're also strong bases. If you think about a carbanion here reacting with water, this is going to pick up the proton from water to form an alkane. It's just going to go like that and form an alkane. You don't want that to happen which is why you need to separate these steps. This is an oversimplified mechanism of what's happening, but it's just the best way to think about it for an undergraduate student. Let's look at some examples of an organometallic reaction here. This time we're starting with an aldehyde and we have methylmagnesium bromide here in our first step, in our second step we need to add a proton source. Once again, we could think about the bonds between the carbon and the magnesium as being very polar covalent or you could think about that bond as being ionic and those two electrons in magenta on the carbon to form your carbanions. Like this, so a negative one formal charge in that carbon and so that's just a simple way of thinking about it here for trying to figure out the products. You would attack right here and push these electrons off onto your oxygen. If we showed the intermediate, we would have our carbon right here is now bonded to a methyl group. Still have a hydrogen and then we would have an alkoxide, an ion, as our intermediate like that. A negative one formal charge. The bond that's formed is between this carbon and this carbon like that. In the second step, we would protonate our alkoxide to form our alcohol. Let me go ahead and draw out our alcohol here, so we would have an O-H, and then you can see that we have increased the number of carbons. If we look at how many carbons we started with, we had one, two, three, four carbons, so we started with butanal and then when we were done, we ended with, let's see, one, two, three, four, five carbons and so we finished with two pentanol and we added one carbon on. This methyl group is the one we added on in our organometallic reaction for our Grignard reagent. This is what we started with. Notice when you start with an aldehyde, let me just change colors again. When you start with an aldehyde, you're going to end up with a secondary alcohol. This carbon that's bonded to your O-H is bonded to two other carbons, so an aldehyde goes to a secondary alcohol. Let's look at one more example. This time we're going to start with a ketone, so same reagent, but this time we're starting with a ketone and so again thinking about what happens. This time I can just show these electrons right here attacking this carbon, pushing these electrons off onto the oxygen and then in the second step, we know that we would protonate the alkoxide anion. We can just go ahead and draw the final product. We'd have our ring and we're adding on, again, a methyl group so we can go ahead and put a C-H three right here and then, of course, we would form an O-H like that. Let's, again, think about electrons. These electrons right here formed this bond and then we just added on C-H three too. That's one way to think about how to draw your final products. This time, if you think about our starting material, we started with a ketone and then we finished with a tertiary alcohol. If I look at this carbon, the one that's bonded to the O-H right here, this carbon is bonded to one, two, three other carbons. This is how to make, one way to make a tertiary alcohol.