- Aldehydes and ketones questions
- Nomenclature of aldehydes and ketones
- Physical properties of aldehydes and ketones
- Reactivity of aldehydes and ketones
- 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
- Cyclic hemiacetals and hemiketals
Oxidation of aldehydes using Tollens' reagent
Oxidation of aldehydes to carboxylic acids using Tollens' reagent. Created by Jay.
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- Very cool, one quick question.
So you oxidized aldehyde to produce a carboxylic acid. If you oxidized a ketone I'm assuming it would also produce a carboxylic acid?(3 votes)
- No, because there is no H. Ketones cannot be oxidized hence they cannot produce carboxylic acids. Thats why the Tollen's test is used to detect presence of aldehydes.(26 votes)
- Anybody is having any chart sheet regarding all these reactions?(3 votes)
- Ketones don't get oxidised using tollens reagent. However, fructose, which is a ketose, (a sugar containing a ketone group) does get oxidised using tollens reagent. How does that happen?(2 votes)
- In basic solution, fructose undergoes tautomeric isomerization to form a small amount of aldehyde.
If we use R to represent the last four carbons of fructose, we have
R-C(=O)CH₂OH ⇌ R-C(OH)=CHOH ⇌ R-CH(OH)CH(=O) or
fructose ⇌ an enediol ⇌ an aldehyde
The Tollens reagent oxidizes the aldehyde as fast as it is formed, so the equilibrium keeps getting pulled to the right (Le Châtelier's Principle) until eventually all the fructose is oxidized.(8 votes)
- How can aldehyde be carboxylate anion? I want to know the movement of electron.😢(2 votes)
- Remember the OH~ ion used while preparing the Tollen's reagent. It's that O atom from OH~ which get attached with aldehyde to form carboxalate anion.(2 votes)
- At6:05, wouldn't it be a hydrate instead of hemi-acetal? There is a OH group, H, and R group but there is no OR group. Infact instead of the OR group, there is another OH group so that makes it a hydrate no? OH, OH, OR, and H groups on a carbon make a hydrate right?(2 votes)
- He is talking about the anomeric carbon in the ring structure of glucose.
It has an R, an H, an OH, and an OR. That makes it a hemiacetal.(2 votes)
- I wonder why ketones are very difficult to oxidize.(3 votes)
- You cannot replace a C-C bond with a C-O bond, so there is no way to add additional Oxygen atoms to the alpha carbon(0 votes)
- does alpha hydroxy ketone tollen's reagent give silver mirror test?if yes then why?(2 votes)
- Why does ag+ and oh- do not form a unsoluble salt AGOH and precipitate(2 votes)
- Do carboxylic acids give a positive result (silver mirror) with Tollen's reagent? What about Fehling's solution?(2 votes)
- Why silver reaction of glucose is better than aldehyde?(2 votes)
Voiceover: There are several ways to oxidize aldehydes but perhaps the most fun way is to use Tollen's Reagent. And so we start with an aldehyde here and we add Tollen's reagent which was created by Bernhard Tollens, a German chemist. Need a source of silver ions, so silver nitrate works well. Need some hydroxide anHines, so sodium hydroxide, and some ammonia. And the order in which you add these and the concentrations will depend on which procedure you're using, but eventually you're gonna form this diamine silver cation here. And that's going to oxidize your aldehyde to a carboxylate anions, let's go ahead and show the formation of a carboxylate anion here. So let me go ahead and put in my electrons. So now the carbon has bonded to an oxygen over here on the right, so no longer bonded to a hydrogen, and that's an oxidation, that's an oxidation here. Let's go ahead and draw out some electrons, and let's assign some oxidation states so we know that this is indeed an oxidation reaction here. So let's put in the electrons in these bonds. So these electrons, right, we know that each bond consists of two electrons, so I'm putting in those electrons in here like that. One way to assign oxidation states is to think about differences in electronegativity. Oxygen is more electronegative than carbon, so we're gonna give all four of those electrons to oxygen, carbon and carbon have the exact same values, so one carbon gets one electron, one carbon gets the other electron. Between carbon and hydrogen, carbon is more electronegative than hydrogen so it takes those two electrons. Carbon has four valence electrons and here we have it surrounded by three electrons, so four minus three gives us an oxidation state of plus one. Over here for our carboxylate anion, let's go ahead and do the same thing here. Let's go ahead and draw everything out, and let's go ahead and use the same color, so we have carbon double bonded to our oxygen, and put in our electrons here, and then we have a carbon bonded to a carbon, and then carbon now bonded to an oxygen over here on the right. So let's put in our electrons, so we have our electrons in here like that. And once again, think about differences in electronegativity. Oxygen beats carbon, so oxygen gets all four of those electrons, we have a tie between these two carbons. And then over here now oxygen takes both of those electrons. So, carbon has four valence electrons, here we have it surrounded by one. So four minus three gives us an oxidation state of, four minus one I should say, gives us an oxidation state of plus three. And so you can see that our oxidation state has increased: we go from an oxidation state of plus one for the carbinol carbon or aldehyde, to an oxidation state of plus three for this [carbinol carbon for carboxylate anion,] so we've oxidized that carbon. And if you oxidize something you are reducing something else, and that something else is going to be your silver cation here. So we have a silver cation Ag Plus, so an oxidation state of plus one, and since we oxidize our aldehyde we're going to reduce our silver ions, so we're going to reduce it to solid silver with an oxidation state of zero. So it gains an electron, so this silver cation is reduced to solid silver, and this forms a silver mirror on your glassware if you do everything properly. So, the formation of a silver mirror indicates the presence of an aldehyde, and so this is a rather cool way, a rather cool diagnostic test for the presence of an aldehyde. And this will help distinguish an aldehyde from a ketone because in general only aldehydes are going to react with a Tollen's reagent, and you get a really cool silver mirror out of it. If you want to form the carboxylic acid you would need to then protonate your carboxylate anion, and so that would give you your carboxylic acid here. So oxidation of aldehydes and a reduction of your silver to form a silver mirror. Let's look at a reaction here. We're starting off with this compound, and the first reagent we're gonna use sodium dichromate, sulphuric acid and water, we know that's going to oxidize, different functional groups. And it's going to oxidize both our aldehydes and our alcohol. So let's go ahead and draw the final product here. So if we oxidize our aldehydes, we're going to form a carboxylic acid. And if we oxidize our alcohol, we're gonna form a ketone. So you oxidize a secondary alcohol you're going to form a ketone here. If you do this reaction using Tollen's reagent, and then go ahead and protonate, you're going to oxidize only the aldehyde. So the aldehyde gets oxidized to a carboxylic acid, and the OH remains untouched. So once again, in the work up when you protonate, you'd form your carboxylic acids, so the Tollen's reagent is selective for your aldehyde. So at leaves the alcohol untouched. This is because Tollen's reagent is a mild oxidizing agent, and it's pretty easy to oxidize aldehydes. Much harder to oxidize something like a ketone. Alright, let's look at another example where we're using Tollen's reagent, and this is a pretty cool experiment that's done in most undergraduate organic labs. And this is using glucose as your aldehyde. So right here you can see the open chain, the form of glucose, and you can see the aldehyde of functional group right here. And this is worth looking at because we've talked about some of this chemistry before. So lone pair of electrons on this oxygen can attack this carbonyl, push these electrons off onto this oxygen, and once you protonate, deprotonate and protonate, you're going to form a cyclic hemiacetal, so over here on the left let's say this is the cyclic hemiacetal that you form, so let's analyze those carbons there. So this carbonyl carbon right here becomes the anomeric carbon for our cyclic hemiacetal, so this is one of the cyclic forms of glucose. And you can see this is the one with the OH down relative to the plain of the ring, and if you're thinking about which anomer this is you can compare this OH to this CH two OH group, which is up relative to a flat plain of the ring. And since they're on opposite sides, this would be the trans, they're trans to each other. This is the alpha anomers, so this is the alpha, alpha glucose form in the cyclic hemiacetal form. So the other possibility, of course, would be to add that OH up relative to the plain of the ring. So since we have a plainer aldehyde here, we could have added the OH up, and then they'd be on the same side as the CH two OH, so they're sis to each other. So this would be the Beta anomer, so that would be Beta glucose, the cyclic form of it. And the cyclic hemiacetal is actually favored, so we talked about that in the video on hemiacetal formation. So glucose spends most of it's time in the cyclic forms, in the Beta and the Alpha form. But it's in equilibrium with the open chain form, with the open chain form containing our aldehyde functional group. This is what we need to react with Tollen's reagent. So when we oxidize the aldehydes, we're going to form a carboxylate anion, and when we oxidize the aldehyde we're going to reduce the silver. So the silver ion's go from Ag Plus to Ag and forming our silver mirror. And the reason that glucose is used is because this is a highly water soluble, so it just makes this reaction a lot easier. And because of all these OH groups on the glucose molecule. So you can form your silver mirror this way. You can make some really cool silver mirrors using glucose. And let me show you pictures of a couple of these things. So over here on the left my students actually made me an ornament. So you can do this reaction using a glass ornament. And then you can put your ornament on a chemist's tree. So let me go ahead and write on here Chemist Tree if it's the holiday season. Here is actually my wife holding the ornament that my students made for us. And then you can actually make flat mirrors out of this too. So over here on the right, here I am reflected in the flat silver mirror that I made, and I did this by starting off with a microscope slide which is glass, and then I put that at the bottom of a 600ml beaker. And then I silvered the top portion of it, and then I was able to make sure the other side was just glass. And then if you put some varnish on there you'll have a really nice silver mirror. And so making silver mirrors using chemistry is something that was perfected by the German chemist Justus von Liebig in the 1830's. So even though Bernhard Tollens is the creater of the Tollen's reagent, Liebig is the one who perfected the formation of the silver mirrors. And it's a lot of fun to make one and a highly encourage you to do so in your undergraduate organic chemistry lab.