- Carboxylic acid questions
- Carboxylic acid reactions overview
- Carboxylic acid nomenclature and properties
- Reduction of carboxylic acids
- Preparation of esters via Fischer esterification
- Preparation of acyl (acid) chlorides
- Preparation of acid anhydrides
- Preparation of amides using DCC
- Alpha-substitution of carboxylic acids
The decarboxylation (loss of carbon dioxide) of malonic acid and a beta-keto acid. Created by Jay.
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- what is cannizarro's reaction ? please explain.(3 votes)
- The Cannizzaro reaction is the base catalyzed disproportionation of an aldehyde with no α-hydrogens.
Disproportionation means that the aldehyde is simultaneously oxidized *to a carboxylic acid (salt) and reduced (to an alcohol).
Aromatic aldehydes are the most common aldehydes with no α-hydrogens. For example,
2Ar-CHO + KOH → Ar-CH₂OH + Ar-COO⁻K⁺(4 votes)
- Is the salt of carboxylic acid reduced or oxidised in kolbe's electrolysis?(3 votes)
- It is oxidized.
In the reaction 2CH₃-COO⁻ → CH₃-CH₃ + O=C=O, the oxidation number of the carboxylate carbons increases from +3 to +4,
This is an oxidation.(2 votes)
- At4:17, instead of removing CO2, can the acid undergo dehydration and form an anhydride? If yes then which is the major product, the anhydride or ketone?(2 votes)
- You could form an anhydride, but the driving force for the reaction is the exceptional stability of CO₂. The anhydride never gets a chance to form.(3 votes)
- What is the role of Sodium hydroxide in the decarboxylation reaction and where from the Hydrogen attaches itself to the carbanion?(2 votes)
- What is orithine ? How does it forms ?(2 votes)
- why methane cannot be prepared by Kolbe's electrolytic method??(1 vote)
- The Kolbe electrolysis involves the dimerization of free radicals, so you can make only hydrocarbons with an even number of carbon atoms.
If you tried to make methane by the electrolysis of sodium acetate, you would get ethane, not methane.
2CH₃COO⁻ → 2CH₃COO·
2CH₃COO· → 2CH₃· + 2CO₂
2CH₃· → CH₃-CH₃(3 votes)
- what does 2 propanate on oxidation gives? is it ketones or carbon dioxide....and to which category does it belong? confused! pls help(2 votes)
- My professor has been doing Decarboxylation without heat. Instead she has been using H+ and H2O, is that correct? cause in the textbook it uses heat just like in this video.(2 votes)
- What happens if the alpha carbon is instead a tertiary carbon? Does the reaction not take place?(2 votes)
- Why decarboxylation is not easy in CH3CH2COOH ?
please help soon .
thank you .(1 vote)
- You need a carbonyl group on a beta carbon for the reaction to work. CH3CH2COOH has only C and H on the beta carbon.(3 votes)
Voiceover: Here's the dot structure for propanedioic acid, or malonic acid and if you heat it up it's going to undergo a decarboxylation reaction, so if we show free rotation about this bond, and it's a sigma bond, so we can show a different confirmation. Let me go ahead and draw in this carboxylic acid on the left, and then we're going to have our carboxylic acid on the right too, this time the carbonyl is going to be going to the right, so let me go ahead and put in those electrons, and the OH will be going to the left, so there we have it. So in this mechanism we're actually going to form a bond between this oxygen and this proton, and it's a cyclic mechanism, so if these electrons in here move in to here that's going to push these electrons in to here, and then these electrons are going to form the bond between the oxygen and the hydrogen, so let's go ahead and show the result of our cyclic mechanism. We would have a carbon bonded to an oxygen, bonded to hydrogen and then we have an OH over here, and then we'd have a double bond between this carbon and another carbon, on the right we could actually form CO2. So let me go ahead and put in lone pairs of electrons on that oxygen so we can see that we would form our carbon dioxide molecule here. So let me go ahead and draw those in. And let's follow some of those electrons, so the electrons in magenta, right in here, are going to move in to form this bond, to form our double bond for CO2. And then these electrons in here in blue, so between this carbon and this carbon are going to move over here to form this double bond between this carbon, and then there's a carbon right here, and then there's also two hydrogens bonded to this carbon. Let me go ahead and draw those in so we can see it a little bit better. And then finally, I want to make these electrons in here red, so these electrons are the ones that are going to form this bond between the oxygen and the hydrogen, so that we formed our CO2 and we've also formed an acid enol. So this right here is called an "acid enol". And we saw in earlier videos how the enol is in equilibrium with the keto form, with keto–enol tautomerization. And so this is actually the enol form of acetic acid, and so that's actually going to be our product. So let me go ahead and draw acetic acid up here, and here we have the OH on the left side, and then we would have our carbon over here with three hydrogens bonded to it, so this would be if we're thinking about the keto-type form. So the difference between the enol and the keto form are the movement of one proton, so there's a proton here and the oxygen in here, it's one of these on the carbon, and then the double bond. Here we have the double bond between the two carbons, and here we've moved the double bond between the carbon and the oxygen. So once again, we've seen how to do that in earlier videos. And then we also produce CO2, so carbon dioxide as the other product for this reaction. So the key to a decarboxylation reaction is having a cabonyl beta to a carboxylic acid. So, for example, here's our carboxylic acid, and we know the carbon next to a carboxylic acid is the alpha carbon, and the carbon next to that is the beta carbon, and we saw how this carbonyl was necessary in the mechanism. And so the fact that there's an OH here isn't really necessary, and what we really need is a cardonyl that's beta to our carboxylic acid in order for a decarboxylation reaction to occur. So let's look at another example where we don't have a dioic acid anymore, we have a carboxylic acid on the right and then over here on the left we have a ketone. But again the key point is here's the alpha carbon and here's the beta carbon, we have a carbonyl that's beta to our carboxylic acid, and so therefore a decarboxylation reaction can take place. So if we heat up this molecule, once again thinking about the mechanism rotating about that single bond. Let's go ahead and redraw this. So we have our benzene ring, and then we have our carbonyl right here, and then we would have once again our carbonyl going off to the right this time, and then our OH over here on the left. So thinking about our mechanism once again, we know it's a cyclic mechanism, we know this oxygen is going to bond to this proton, and so these electrons are going to move in to here, and these electrons move in to here, and these electrons move in to here, so that's our cyclic mechanism and we draw what happens, moving all those electrons around, we have our benzene ring, we would have it bonded to an oxygen, this oxygen was bonded to this proton now, and we have a double bond right in here, and then we also form CO2. So once again let's run through those electrons and try to follow some of those electrons here, so let me draw in these lone pairs of electrons on our oxygens so we can see where the CO2 comes from. So once again, these electrons in here in magenta are going to move in here to form the double bond on CO2. At the same time these electrons in here are going to move in to here to form this double bond, and then finally these electrons in here are going to move out to form the bond between the oxygen and the hydrogen so we've made our CO2. So the oxygen-carbon-oxygen comes from this oxygen-carbon-oxygen right here on the molecule on the left, so hopefully that's a little bit easier to see now. And once again we have an enol, so we form our CO2, we also create our enol, and so we can think about keto-enol tautomerization for our product, so the enol is going to be in equilibrium with the keto form. So let's go ahead and draw the keto form, which we know is moving one proton and moving the double bond. So we move the double bond between the carbon and the oxygen and we move that proton to this carbon right here, so this carbon right here picks up a proton. So down here that carbon had two hyrdogens, and it doesn't have to be this one right here, but we are going to add a hydrogen to that carbon and that gives us our final product, our ketone. So this decarboxylation reaction produces a ketone, and once again it doesn't really matter what this R group is here, here we have a benzene ring instead of the OH in the previous example. And then of course we're also going to make CO2. So decarboxylation reactions are going to be important in future videos where we're synthesizing some complicated molecules, and so this is the cyclic mechanism for it.