Class 12 Chemistry (India)
- Formation of hydrates
- Factors determining the stability 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
- Hemiacetals and Acetals
- Formation of imines and enamines
- Formation of oximes and hydrazones
- Addition of ammonia derivatives to the carbonyl compounds
- Addition of carbon nucleophiles to aldehydes and ketones
- Addition of carbon nucleophiles to aldehydes and ketones.
- Formation of alcohols using hydride reducing agents
- Oxidation of aldehydes using Tollens' reagent
How ketones and aldehydes react with water to form hydrates. Created by Jay.
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- Could you do an intramolecular hydrogen transfer to form the hydrate? Instead of adding two more molecules of water.(17 votes)
- My understanding is that an intramolecular hydrogen transfer is possible, but unlikely for a couple of reasons. First, there are many solvent molecules for each molecule of the reactant, so the reaction with water is more probable. Second, the intermediate for the intramolecular hydrogen transfer (see2:24) would have bond angles of around 90°, which is highly strained and therefore energetically unfavorable.(3 votes)
- 1:17is it true that when we have a weak nuclephile we have to first protonate the oxygen to get a plus formal charge on carbon (so that it is more electrophilic)(5 votes)
- which is the most reactive carbonyl compound.ketones or carboxylic acids ?.if the methyl group is further would'nt it favour the formation of the product because of the steric effect? please answer(2 votes)
- Reactivity of carbonyl compounds is as follows from highest to lowest:
acyl halides > acid anhydrides > aldehydes > ketones > ester/carboxcylic acid >amides(5 votes)
- isnt 2 '-OH' gps on the same carbon unstable?(2 votes)
- Often, but not always. The stability can be increased by electronegative atoms in the molecule.(4 votes)
- At1:12, how does Jay know that those e- goes back to O are pi e- ? Is there such thing as "sigma e-"? Are e- that are being pushed in these mechanisms, all pi e- ?(2 votes)
- Sigma e- would be the electrons forming a sigma bond.
We know that π bonds are weaker than sigma bonds (and less tightly bound to the bonded nuclei)§, so it seems reasonable to conclude that they are easier to move around.
According to this logic, if π electrons are able to move that will be more favorable than moving sigma electrons ...
As you may see from how I've phrased this answer, I'm less than 100% certain about this ...
§note: I believe the explanation for this is that the sigma electrons are concentrated between the two bonded nuclei and so closest and most strongly attracted, while the π electrons are mostly out to the sides and thus further away from and less strongly bound to the nuclei.(3 votes)
- At4:05, would the product not also be an acetal? Or are hydrates a form of acetals?(2 votes)
- No, the product at4:05is not an Acetal. Acetals have two -OR groups that branch off the same carbon, at4:05the carbon has two -OH groups. To make an acetal, you need to use alcohols. Check out the video on acetals for more info!(2 votes)
- Hello! I've been studying carbohydrates structure. So, when they are in aqueous solutions, they suffer cyclization (if they are long enough). That's because one of their OH reacts with their own carbonyl. According to what I've read, they become hemiacetals, just like when they react with alcohol. My question is: do they form hydrates first, as soon as they are put in the water, and then hemiacetals? If not, how does the water contribute to the cyclization, ie, why does it only hapen in aquoes solutions? Thank you, hope you can help me!(2 votes)
- why acetaldehyde doesnot show cannizaro's reaction?(1 vote)
- The Cannizzaro reaction involves the base-induced disproportionation of an aldehyde lacking a hydrogen atom in the α position.
Acetaldehyde, CH₃CHO, has 3 α hydrogens. It undergoes the aldol condensation instead.(2 votes)
- Could you do an intramolecular hydrogen transfer to form the hydrate? Instead of adding two more molecules of water.
ps: this question is already posted but I couldn't find a proper answer to it. It's a common doubt. Thanks :-)(1 vote)
- Yes, you could, but the bonds have to be in the right position for this to happen.
It is more likely that a water molecule from the solution will be able to attack first.(2 votes)
- how is the first molecule in4:09an aldehyde? there is no R group, carbon is only bonded to two H and an O.(1 vote)
- R can mean a H atom too
First google result: http://www.chem.ucla.edu/~harding/IGOC/R/r_group.html
“R group: An abbreviation for any group in which a carbon or hydrogen atom is attached to the rest of the molecule”(2 votes)
Voiceover: Here's an example of a nucleophilic addition reaction to an aldehyde or a ketone. So over here on the left, it could be an aldehyde, or we could change that to form a ketone. And if you add water to an aldehyde or ketone, you form this product over here on the right, which is called a hydrate, or also called a gem-diol, or geminal diol because these two OHs here are on the same carbon, so like they're twins. And this reaction is at equilibrium. So let's think about the aldehyde, or the ketone. We know the carbonyl on the aldehyde or ketone is polarized, so we know that the oxygen has more electronegatives than carbons, so it withdraws some electron densities. So this oxygen here is partially negative, and this carbonyl carbon is partially positive, like that. And therefore the carbonyl carbon, since it's partially positive, is electrophillic, so it wants electrons. And it can get electrons from water. So let's go ahead and draw the water molecule right here. Water can function as a nucleophile. It has two lone pairs of electrons, this oxygen here is partially negative, and so we're going to get a nucleophile attacking our electrophile. So a lone pair of electrons on the oxygen is going to attack our carbonyl carbon like that, So the nucleophile attacks the electrophillic portion of the molecule, and these pi electrons here kick off onto the oxygen. So let's go ahead and draw the results of our nucleophilic attack here, and so we now have our oxygen bonded to this carbon, and this oxygen still has two hydrogens bonded to it, so I'm gonna go ahead and draw in those two hydrogens. There's still a lone pair of electrons on that oxygen, which gives that oxygen a +1 formal charge. And then this carbon here is bonded to another oxygen, which had two lone pairs of electrons around it, and now it picked up another one, so a -1 formal charge on this oxygen, and there's still an R group bonded to it, and a hydrogen over here, like that. And so let's try to follow some electrons here. So one of the lone pairs of electrons on the oxygen formed a bond with our carbon, so I'm saying that's these electrons right here like that. And then we can think about our pi electrons. So, our pi electrons in here, as kicking off onto our oxygen, so it doesn't really matter which one of these three it is, right, let's say it's that one, and we get this as our intermediate. And so next, we can think about an acid-base reaction. So, another water molecule comes along right here, and so we know water can function as an acid or a base, and so this lone pair of electrons could take, let's say it takes this proton right here, and leaves these electrons behind on our oxygen. So let's go ahead and draw the result of that acid-base reaction, and so we would have our oxygen here, would now be bonded to only one hydrogen, and, now let's see, we still have our negatively charged oxygen over here on the right, and then we have our R group, and our hydrogen like that, and we still have this lone pair of electrons, and we just picked up another lone pair of electrons, so let me go ahead and show those, so let me go ahead and use blue for those electrons, and these electrons in here, ended up on this oxygen, and so we're almost to our final product, we've almost formed our hydrate. We just need to do one more acid-base reaction. And so of course, another water molecule could come along. So water could function as an acid or a base, and a lone pair of electrons on this oxygen could take this proton, and leave these electrons behind, and then that of course would give us our final product. So that would give us our hydrate over here. And you could of course change this hydrogen to an R prime, change this hydrogen to an R prime if you wanted to, and you would get this over here, so your hydrate or your gem diol. And so that's the general mechanism for uncatalyzed reaction here. Let's look at some examples, and it's important to remember that this reaction is at equilibrium, and so let's look at three examples. Our first example is formaldehyde. So, if you react formaldehyde with water, you form your hydrate over here on the right, and in this case, the equilibrium is to the right, favors the formation of the hydrate, and that's because aldehydes are very reactive. And we talked about why in the video on the reactivity of aldehydes and ketones, so we've got very polarized carbonyl situation here, so this carbon right here is partially positive, and we have these hydrogens here, which don't take up a whole lot of space, and so we have greater polarization than ketones and also decreased steric hindrance, and so because of the high reactivity of aldehydes, then this product is favored. If we look at the next reaction, we no longer have an aldehyde. This is a ketone, this is acetone. And so adding water to acetone gives us this as our product for our hydrate. Except, we know that ketones are not as reactive as aldehydes, and so this time the equilibrium is to the left. It favors the formation of the ketone. Let's look at another one. So this is acetaldehyde. And so if we add water to acetaldehyde over here, we form this as our hydrate product. And once again, we know that these reactions occur because of our carbonyl carbon right here being partially positive, so the oxygen withdraws some electron density like that. So, we could make aldehydes or ketones more reactive by adding something else that withdraws electron density from that carbonyl carbon, and one thing you could do, is add an electronegative atom like halogen, so let's go ahead and do that. Let's add three halogens here. Let's add three chlorines to this carbon, the one adjacent to our carbonyl carbon. And those electron withdrawing groups, those very electronegative atoms withdraw some electron density, so they're gonna withdraw electron density this way, once again, away from our carbonyl carbon, and so this carbonyl carbon gets even more partially positive by the addition of these electronegative atoms. And the more positive you make that carbonyl carbon, the more electrophilic you make it, and therefore, the more the nucleophile, which is water, is going to attack, and so you make it even more reactive by adding these, and so you can push the equilibrium even more to the right. You can form more of your product. And so let's go ahead and draw the product of that reaction, so we would put three chlorines on here like that, and so you could also do this with ketones, and you could make ketones much more reactive by doing that. And so this particular reaction is a little bit famous. Over here on the left is trichloroacetaldehyde, and then, once you've formed the hydrate of that, you've formed chlorol hydrate, and this is famous for being knock-out drops. And so some of the old references to it are, "slip someone a Mickey Finn." So you could slip chloral hydrate into someone's drink, and it was a knock-out drop situation. And so just a little bit of an interesting reaction here for formation of hydrates.