In this video, we're going to focus on reactions associated with amines. So let's start with butyl bromide, or 1-bromobutane. And let's react it with excess ammonia.
So what's going to be the product of this reaction? Ammonia is going to act as a nucleophile in this reaction, attacking the carbon from the back, expelling the leaving group. And so this is going to be an SN2 reaction.
And so right now, the nitrogen atom has a positive charge since it has four bonds. Now, another NH3 molecule is going to remove a hydrogen. And so right now we have butylamine.
And so that's the simple way in which we can make an amine. Now there are some problems with that reaction, particularly exhaustive methylation. For example, let's say if we were to react ammonia with methyl bromide. The ammonia molecule can attack the methyl group, kicking out the leaving group.
And so we're going to get CH3NH3. Now, in the next step, another NH3 molecule is going to basically remove a hydrogen. And so now what we have is methyl amine.
Now this will react with methyl bromide again, and so we can add another methyl group to the nitrogen atom. And so right now the nitrogen atom has two methyl groups. and two hydrogen atoms. So now it carries a positive charge.
And this process can repeat until the nitrogen atom has four methyl groups. And so this is called exhaustive methylation. So we need to find better ways to produce an amine because the nitrogen could attack an alkyl halide multiple times. Another way in which we could synthesize an amine is by reacting an alkyl halide with sodium azide, NaN3. So the N3-ion is the nucleophile in this reaction and so in an SN2 reaction is going to displace the bromine group and so we're going to get this.
Now the N3 ion looks like this. In the middle we have a nitrogen with a positive charge and at the end each nitrogen has two lone pairs so they both carry a negative charge. So one of these two nitrogens attacks the carbon atom.
And so this structure, this compound, looks like this. So here is the nitrogen with a positive charge. On the left, the nitrogen is neutral because it has three bonds.
On the right, it has a negative charge. So once we have this compound, we can reduce it using lithium aluminum hydride followed by H2O. And we're going to lose two of the three nitrogen atoms, but the last one that remains will be reduced to an amine.
And so that's another way in which we can make an amine from an alkyl halide. Now the next thing that we could do, starting with an alkyl halide, is we can react it with sodium cyanide. And so the cyanide ion is going to be the nucleophile in this reaction.
The carbon has the negative charge, and so it attacks the carbon with the halogen, kicking out the leaving group. And so we're going to get a nitrile. Now we can reduce the nitrile into an amine.
And so we could use hydrogen gas with a metal catalyst such as nickel. We could also use lithium aluminum hydride in the first step, followed by... H2O. And so what we're going to have is a total of five carbons and an NH2 group. So this is another way to convert an alkyl halide into an amine by the addition of a carbon atom.
So before we had four carbon atoms but now we have a total of five. Another way in which we can make a primary amine is by using the Gabriel synthesis reaction. And so in this reaction we're going to start with this compound which is called thalamide. And in the first step we're going to react it with...
potassium hydroxide and so this is to deprotonate the hydrogen put in a negative charge on the nitrogen which becomes a nucleophile for the next step and then we're going to react it with an alcohol halide attaching the nitrogen to those four carbons and finally we're going to use hydrazine you can write it as n2h4 or h2n-nh2 with heat and so what's going to happen is we're going to replace the bromine atom with an NH2 group. So this is going to be the main product. So we have butylamine and a side product that we're going to get will look like this. So the hydrogen molecule will replace this NH group.
And so this five atom ring is going to expand into a more stable six atom ring. And so we're going to get something that looks like this. So that's the other side product for this reaction. So this is called a thalamide hydroxide.
But the purpose of the Gabriel synthesis reaction is to make primary amines. So the first step in the mechanism of this reaction is that the hydroxide ion takes off this hydrogen. And so we're going to have a negative charge. on the nitrogen atom, but the reason why that hydrogen is relatively acidic is because the conjugate base is stabilized by resonance.
So now that the nitrogen atom has two lone pairs, we can draw a resonance structure with the carbonyl group, and so that's why this hydrogen is relatively acidic compared to other NH bonds. So now that we have a nucleophilic nitrogen atom, we can now react it with butyl bromide. And so it's going to attack the carbon, kicking out the bromine group.
And so right now, we have this structure. So we have an N with a butyl group. And once we react it with hydrazine, this will give us... the primary amine plus thalamide hydroxide, which we drew already.
But let's say instead of reacting it with hydrazine and heat, what if we did something else? What if we reacted with H2O plus and heat? What's going to happen?
Well, if you have, let's say, an amide and you react it with H2O plus with heat, you can hydrolyze the amide bond. So you're going to get a carboxylic acid and an ammonium ion. So something similar will happen here. So this bond will be hydrolyzed, turning this into a carboxylic acid. And the same is going to happen to this part.
So if we react this with H2O plus and heat in the final step, we're going to get a side product which basically contains a benzene ring. with two carboxylic acid functional groups and we're going to get our product but in its protonated form because the solution is acidic so we're going to get an ammonium ion instead but we can add some base to this let's say hydroxide just enough to basically remove the hydrogen and turn it to an amine So if you use H2O plus instead of hydrazine, just know that you're going to get two carbosilic acids on a benzene ring as your side product. But you could still get the same amine in the end if you basically increase the pH of the solution.
Now let's say if we have a nitro group on a carbon chain. If we reduce it with iron metal, hydrochloric acid this will turn into an NH3 plus group. And then if we add a base to the solution, we could turn it into an amine.
So that's another way in which we can make an amine, is by converting a nitro group to an NH2 group by reduction. So here's another example. So let's say if the NO2 group was on a secondary carbon.
Another way in which we could reduce the NO2 group is by using catalytic hydrogen gas. So hydrogen gas with a metal catalyst. And that's going to turn the NO2 group into an NH2 group.
So that's how you can make an amine from a nitro group. Next up, we have a ketone and NH3. What do you think is going to happen if we mix a ketone with ammonia? So what we're going to lose is water.
We need to take off two hydrogens and an oxygen. So we'll get water as our side product. And then the main product will be an imine, which will look like this. So one of the hydrogens will remain.
Now once we have an imine, we can reduce it using sodium cyanoborohydride. And so it's going to reduce the double bond into a single bond. converting the imine into a primary amine which looks like this or we can draw it this way so that's how we can convert a ketone into a primary amine this process is called reductive amination Now let's say if we have an aldehyde or a ketone, and we chose to react it with NH3, we can get this product.
We're still going to get a primary amine, but first we're going to get the imine, just like before. and then we're going to reduce it using the same reagent, sodium cyanoborohydride. And so this time, the amine is on the end.
It's still a primary amine, but it's not on the middle of the chain, but rather, it's on the end of the chain. now we can also make secondary and tertiary amines so let's start with a ketone and we're going to react it with a primary amine So we're going to lose water if you take off the O and two hydrogen atoms. We'll get water as a side product.
And so this amine has an R group attached to it. In this case, an ethyl group. And now, once we reduce it with sodium cyanoborohydride, this is going to turn into a secondary amine. The double bond will be reduced into a single bond.
and we're still going to have the ethyl group attached to it, but now we're going to have a hydrogen and a lone pair. So that's how we can make a secondary amine by using reductive amination. So to make the secondary amine, react the aldehyde or ketone with a primary amine. Now, if we want to make a tertiary amine, starting from a ketone, we need to react it with a secondary amine.
So let's use diethylamine. And initially we're going to get the iminium ion, which looks like this. And this is in equilibrium with the enamine. So you may want to keep that in mind.
So this is the enamine. But regardless of which form it's in, once we add sodium cyanoborohydride, these two forms will be reduced into a tertiary amine. So the double bond, regardless of where it is, is going to be reduced to a single bond.
I'm going to need more space. So our final product will look like this. So this is how we could synthesize a tertiary amine.
It's by reacting an aldehyde or ketone with a secondary amine. For the next reaction, let's react an acetyl chloride with ammonia. But instead of writing NH3, I'm going to write it as HNH2.
And so when you react an acid chloride with ammonia, you're going to get hydrochloric acid as a side product. So this will leave as a gas. And then for the other product, you need to pair up these two together.
And so an acid chloride with NH3 will produce an amide. If you want to see the mechanism for this reaction, check out the description section of this video. You can find my new organic chemistry playlist. And in that playlist, look for carboxylic acid derivatives. Those reactions, they have the mechanism where you can react an acid derivative with an amine or an alcohol or something else.
But the mechanisms for those reactions are the same. So in this video, I'm just going to focus on the major product. Now, once we have the amide, we can reduce this into an amine. And so what we need to use at this point is lithium aluminum hydride followed by H2O. And so what's going to happen is the carbonyl group is going to be reduced to a CH2 group, converting the amide into an amine.
So that's another way in which we can make amines. So this process is called acylation reduction. Now notice the pattern.
in these reactions. In the last example, we reacted an acyl chloride with ammonia, and we got a primary amine. So let's see what's going to happen if we react a primary amine with an acyl chloride.
So this is going to be H, N with an R group, and here's the other hydrogen. So let's take out HCl as our side product, and so this will give us an amide. with an ethyl group on it and then we're going to reduce it.
Sometimes you might see lithium aluminum hydride written like that but if you do see that it's the same and so this carbonyl group is going to convert into a CH2 group and so our product will look like this. So the end result is that we have a secondary amine if we react the acetyl chloride with a primary amine. And the reason being is, we started with one R group, but we're going to add this R group to the nitrogen. So that's why it's going to be a secondary amine. Now, if we react the acyl chloride with a secondary amine, you can guess what's going to happen.
Eventually, we're going to get a tertiary amine. But first... Let's take away hydrochloric acid and let's produce the amide.
And then let's reduce it using lithium aluminum hydride followed by H2O. And so that's going to be reduced to a CH2. And so this product is triethylamine.
So you can make a tertiary amine if you react an acid chloride with... a secondary amine, followed by reduction. Now the next reaction we're going to talk about is something called the Hoffman rearrangement reaction. And so what we're going to do is we're going to react with a halogen, it could be Br2 or Cl2, under basic conditions in water. And so what's going to happen is we're going to lose the carbonyl group.
So we started with a total of four carbons, but now we're going to have three carbons. And so this is going to be the end result of this reaction. If you want to see the mechanism for this reaction, I have a separate video on YouTube.
So if you just type in Hoffman rearrangement and then organic chemistry tutor, it should come up. Now the next reaction, which I also have a mechanism for on YouTube, is called the Curtis rearrangement. And so this reaction starts with an acyl chloride, but instead of using these reagents, you're going to use sodium azide, followed by H2O, with heat.
Now the effect will be very similar. So we started with four carbons, and we're going to have three. So we're going to lose the carbonyl group. The intermediate for this reaction is an isocyanate intermediate. It looks something like this.
It's an R group with an N double bonded to a carbon, double bonded to an oxygen. But in that other video on YouTube that I have, you'll see how you can get that isocyanate intermediate. So once we lose this carbonyl group, we're going to have three carbons.
And then the chlorine group we're also going to lose, but that's going to be replaced with an NH2 group. which comes from the N3 minus azide ion. So the end result is that we're going to replace this entire acid chloride group with an NH2 group.
So we should have a total of three carbons instead of four. So let me give you some example problems. So go ahead and draw the major product of these two reactions. So here we have Cl2 with hydroxide and water and for the second reaction, here we have an acyl chloride and we're going to react it with sodium azide followed by H2O and heat.
So the first reaction is the Hoffman rearrangement. All we need to do is get rid of the carbonyl group. And so we're going to have one less carbon.
Everything else is going to be the same. So just to make sure, let's count the carbons. So we have the six-membered cyclohexane ring, and then these three additional carbons.
So we started with nine carbons, and now we have six plus two. So we have a total of eight carbons now. Now the next one is the Curtius rearrangement.
So basically what we're going to do is take away this whole entire acid chloride group and just replace it with an NH2 group. And so that's going to be the end result for that reaction. Now the next reaction that we're going to go over is something called the Hoffman elimination reaction. And so the overall process of this reaction is that it converts an amine into an alkene.
So it helps us to get rid of the NH2 group that's on a carbon chain or on a ring. So now the first step in this reaction, in the Hoffman elimination reaction, is to react the amine with methyl iodide in excess. So we have to use excess methyl iodide.
And so in a process called exhaustive methylation, we're going to add as many methyl groups as we can. to the nitrogen atom. So nitrogen can only form four bonds and when it has four bonds it's going to have a positive charge.
Now we're also going to have an iodide ion in the solution. Now the next step is to react it with silver oxide and water. So what's going to happen is iodide has a strong affinity for the silver cation. It forms an insoluble product called silver iodide.
Now the iodide ion will displace the oxygen and in water the oxide will form hydroxide. So the other product will be this product. We're still going to have the quaternary ammonium ion, but instead of having iodide as a free ion, we now have hydroxide in the solution. The purpose of this transformation is to convert a bad leaving group.
into a better leaving group. So once we have a better leaving group, we can now perform elimination. Whenever the leaving group has a positive charge, it's a better leaving group than if it was neutral. Now we have hydroxide in the solution, so hydroxide will act as a strong base. It can remove hydrogen or hydrogen B.
Now this reaction is called the Hoffman elimination reaction. And guess what? It's going to give us the Hoffman product, the less stable alkene.
So hydroxide is going to go for hydrogen A, forming a double bond, kicking out the leaving group. And so the major product will be 1-butene. Now You can still get the other products, but this is going to be the major product. The other products are cis and trans-2-butene. But these two, they constitute the minor product, which is the Z-tiff product in this case.
But here, the major product is going to be the elimination product. Now, let me give you another example. So let's say we have...
and amine, how can we convert it to cyclohexane? So how can we completely get rid of the NH2 group? What reagents do we need here? So the first thing we need to do is perform the Hoffman elimination.
So we need to use excess methyl iodide. followed by silver oxide in water, followed by heat. And so this is going to turn the NH2 group into an alkene. So we're going to get this. Now the last thing we need to do is basically hydrogenation.
So we can react this with hydrogen gas. and a palladium catalyst. And so that's how we can completely get rid of the NH2 group from that ring. Now consider this example.
Let's say if we have a nitrogen that's inside a ring. What's going to happen if we react it with excess methyl iodide followed by silver oxide in water. Followed by heat. What's going to happen in this reaction? So we know that the first two steps, before you add heat, the first two steps involve methylation.
Well, the first step involves methylation. So after step one, we're going to add as many methyl groups as we can to the nitrogen atom until it has a positive charge. So we can add two methyl groups right now.
And right now we have iodide in the solution. But once we react it with silver oxide and water, the iodide will be replaced with a hydroxide ion. Now what's going to happen here once we heat it?
Now once we heat it, we're going to form an alkene. So let me redraw the hydroxide ion. And so the base is going to grab a proton, form a double bond, and expel the nitrogen leaving group. And so the end result is that the ring is going to open.
And so we have a carbon, another carbon, another carbon, and then a carbon-carbon double bond. And we still have two methyl groups on the nitrogen atom. So we can get a mixture of products here, but this is one of the products that we can get.
Because technically, we can also take off this hydrogen, form a double bond here, break in this bond. So that's possible too, but I'm just going to focus on one of these products. The end result is that the ring opens, and so we get a product that looks like this.
But right now, the nitrogen shouldn't have a positive charge. Because it has three bonds now, it has a lone pair instead. So this is our product, or at least one of the products that we can get in this reaction.
Now there's another similar reaction that you need to be familiar with, and it's called the Cope elimination reaction. Now instead of having four bonds or three methyl groups on this nitrogen, we're only going to have two methyl groups on it. So what we have is a tertiary amine as opposed to a quaternary ammonium ion.
And now we're going to oxidize the tertiary amine using hydrogen peroxide. And so this is going to give us an amine oxide, which looks like this. So we're going to have an oxygen atom on the nitrogen.
So at this point, the nitrogen now has... four bonds, so it has a positive charge. Now once we heat the solution, the cope elimination will occur. So the oxygen with a negative charge is going to go for the hydrogen that has the same direction as the nitrogen.
So if this is on the wedge, this must be on the wedge too. it occurs with syn stereochemistry. So it's going to go for the hydrogen, the carbon-hydrogen bond is going to break, forming a double bond, kicking out the leaving group.
And so this reaction will proceed by means of a five-membered cyclic transition state. And so we're going to get an alkene, one butene, and then the nitrogen now has two methyl groups. and an OH group with a lone pair.
So the end result is that we still get the Hoffman product over the Zetov product as the major product. So the Culp elimination and the Hoffman elimination, they have the same regional chemistry. They favor the less substituted alkene over the more substituted alkene. Now before concluding this video, there's one more topic that I would like to mention. So here we have cyclohexanone, and here we have an alpha-beta unsaturated cyclic ketone.
So this is the alpha carbon, this is the beta carbon. What's going to happen if we react each of these with ammonia? Now we know that if we react a ketone with ammonia, we're going to get an imine.
So we're going to lose water, and we're going to get this product. So the nitrogen atom will attack the carbonyl group. Now if we have an alpha-beta unsaturated ketone, the nitrogen atom will attack the beta carbon, as opposed to the carbonyl carbon. And so it's going to favor conjugate addition.
as opposed to direct addition. It's important to understand that weak bases, they attack at the beta carbon, as opposed to the carbonyl carbon. And so we're going to get this product instead, if we have an alpha-beta unsaturated ketone. So the NH2 group is going to go here.
And so that's it for this video. Those are the reactions of amines that I wanted to discuss. So thanks again for watching.