The reaction of a nitrile with an alcohol in the presence of a...

Question

The reaction of a nitrile with an alcohol in the presence of a strong acid forms an N-substituted amide. This reaction, known as the Ritter reaction, does not work with primary alcohols.

Chemical equation illustrating the Ritter reaction: nitrile and alcohol react with HCl to form an N-substituted amide.

b. Why does the Ritter reaction not work with primary alcohols?

Accepted Answer

All right. Hi, everyone. So for this question, it says that the Ritter reaction occurs when a night trial and a secondary or tertiary alcohol reacts in the presence of an acid to form an N substituted. Am I explain why the Ritter reaction does not work with primary alcohols? All right. So, just to recap um what the question is telling us when we combine a nitrile with either a secondary or a tertiary alcohol, we get an un substituted a mind. And what that means is that our amide has a substituent attached to our nitrogen. So, in order to understand the answers to this question, let's go over the general mechanism of the rider reaction and recall that the first step is actually going to be acid catalyzed dehydration of an alcohol, right? Because we're starting off with an alcohol. These are uh two R groups, we're starting off with our alcohol and are strong acid. And for this example, we'll use H C L. So our first step in an acid catalyzed dehydration of an alcohol is of course the protonation of our O group using our acid. So that's going to generate a chloride ion here because I used H C O and then what that does is that creates a better leaving group, right? Because we want our O H group to detach itself from the al group of our alcohol. But O H by itself is a terrible leaving group. So now that we've protend it, right, what we've done is we've given the oxygen a positive charge and O H two plus is an excellent lead group, right? So our next step is going to be dehydration because a water molecule detaches itself from the group of an alcohol. And so that creates a carbo ion intermediate. So now here is our carbo cut ion, here's our carbo K ion and this is actually where our night trial comes in are night trial. Now can go ahead and attack this caramel ion intermediate. And I meant to do that in a different color. Great. So here's our night trial and like I mentioned before, our night trial can go ahead and attack are carbo ion intermediate. And so now what we've done is we've combined, we've essentially combined the alco group of our alcohol to the nitrogen of our, of our nitrile. Sorry. So here's the a the a group of our alcohol and that's attached to the nitrogen of our nitrile. So now nitrogen is going to have a positive charge. Nitrogen is going to have a positive charge. And so the pipe bombs in our night trial are actually going to shift around, not only are they going to shift around our water molecule here that detach itself to form our carbo ion is actually going to go ahead and attach itself to the S P carbon of our nitro. So in order to show that, let me go ahead and draw our molecule of water may use your water. And this water molecule is going to go ahead and attack the S P carbon of our night trap. So that causes one of the pi bonds in our night trial to shift towards the nitrogen to go ahead and make it neutral. So after that, we are going to have the following. So we're going to end up with the following intermediate and I used the wrong color initially, right. So now our night trial has an O H group attached to it, actually an O H two to use our oxygen and it's still associated to its two hydrogens, right? So it's going to have a positive charge while attached to this molecule to this intermediate. But now our nitrile carbon is only double bonded to our nitrogen. And then here's the group of our starting alcohol, right? So now oxygen has a positive charge, which of course is problematic. So now, recall, chloride formed in the very first step of the reaction after our O H group and our alcohol was proin. So that's actually going to make a comeback here. These are chloride and our chloride is going to go ahead and deproteinate our O H two plus to leave us with H C L once again and a neutral O H group. So when I scroll down to give myself some space, our intermediate remains the same except this time, our O group is an O H group, meaning it's neutral. All right. So now our final step is the actual formation of our AM, right? Because we're not quite there yet, but from here are Amy forms via acidic hydrolysis. So we're going to once again introduce our H C O that we had reformed in the previous step. So here's our H C L and here, the nitrogen of our amide or are am might intermediate, not quite an amide yet. It's going to go ahead and take the hydrogen of our Akea and so chloride forms once again. So now because nitrogen is attached to this hydrogen here from H C L, it's going to have a positive charge, it's going to have a positive charge. And right now, I'm just making sure that I'm drawing everything correctly, right, that I'm taking into account all of my carbons and all of my hetero atoms. So now because our nitrogen has a positive charge, our carbo is going to form, right? Because what will happen is that a pair of electrons on oxygen? Well, go ahead and move in between itself and the carbon it's attached to, to form that carbon. Because what that will do is that will allow a pi bond or rather a pair of pi electrons in our c double bond n here to move towards the nitrogen to make it neutral. So now we have a single bond in between carbon and nitrogen, which is great. But now our oxygen is going to have a positive charge. Our oxygen is going to have a positive charge because it's going to have a double bond to the carbon it's attached to while also being associated to that hydrogen that it was attached to, right? So there's a positive charge. And so now recall that once again, we have chloride in solution, right? So chloride just like before, is going to go ahead and deproteinate that hydrogen attached to our oxygen to make it neutral. And this time we end up with are a mind, right? Because here's chloride. And so that's going to go ahead and take the hydrogen to form our car bottle. And so now we have our am these are neutral carbona and here's the a group of our starting alcohol, right? So in addition to this, a, we're going to end up forming H C L once again. Mhm All right. So why did I do all this? Right? Well, the point is really, so that you guys can see the importance of having a secondary or a tertiary alcohol during this process, right? Because recall that our very first step was the formation of a carbo ion after acid catalyzed dehydration of an alcohol. So because a Carbocaine intermediate is formed during the reaction, we have to make sure that it's going to be stable throughout this process. Right. So our Carbocaine intermediate is formed and primary carbo CAD ions stemming from primary alcohols recall that they are not stable, right. So because they do not form stable Carbocaine and intermediates, the rider reaction cannot proceed with a primary alcohol. So our answer here is that a carbo ion intermediate is formed during the reaction and primary alcohols do not make stable Carbocaine intermediates, right? So because of this, this is going to be our final answer, right. So because the Ritter reaction relies on the formation of a Carbocaine intermediate, we know that a primary alcohol cannot be used because primary carbo cads are not stable. So if you stuck around to the end, thank you so much for watching. I appreciate it. And yeah, thank you very much for watching.

The reaction of a nitrile with an alcohol in the presence of a strong acid forms an N-substituted amide. This reaction, known as the Ritter reaction, does not work with primary alcohols.

b. Why does the Ritter reaction not work with primary alcohols?

Key Concepts

Ritter Reaction

Nucleophilicity

Steric Hindrance

Watch next