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Lecture Comments (10)

1 answer

Last reply by: Professor Starkey
Wed Dec 3, 2014 1:48 AM

Post by beatrice vaillant beatrice vaillant on December 2, 2014

what about the anomeric effect? I think it is missing here.

1 answer

Last reply by: Professor Starkey
Thu Nov 21, 2013 11:52 PM

Post by Amirali Aghili on November 18, 2013

Hi Dr.Starkey:

In your lecture around minute 21 you mentioned the thermodynamic product predominates which makes sense here since you form conjugated system. but in the Organic chemistry book by Klein it is mentioned with examples that the less-substituted double bond, or kinetic product is the major product of the reaction when there is unsymmetrical ketone. I am confused about this.


1 answer

Last reply by: Professor Starkey
Sat Aug 24, 2013 11:33 AM

Post by Christian Benz on August 23, 2013

Great that´s exactly what our learn-intensive 1 Semester Organic Chemistry Course also covered.

1 answer

Last reply by: Professor Starkey
Sun Aug 11, 2013 10:37 AM

Post by Parabjit Kaur on August 2, 2013

where do you cover gabriel synthesis of amines?

1 answer

Last reply by: Professor Starkey
Sun May 5, 2013 11:06 PM

Post by Nawaphan Jedjomnongkit on May 5, 2013

Why in reaction of secondary amine with ketone, when N have + charge is unstable and have to be deprotonated while in previous slide that talk about RNH2 is better nucleophile N is stable to have + charge in form of quaternary ammonium salt?


Rank the compound in order of increasing basicity:
  • CH3 is an electron-donating group, making the p-methylaniline more basic than the aniline.
  • NO2 is an electron-withdrawing group, making the p-nitroaniline less basic than aniline.
Rank the compound in order of increasing basicity:
Draw the product(s) for this reaction:
Draw the products for these 2 reactions:
  • These 2 reactions involve Hofmann Elimination
  • When constitutional isomers are possible in a Hofmann elimination, the major alkene product is the one with the less substituted double bond.
Draw the product(s) for this reaction:
  • Less substituted C=C bond is favored in Hofmann elimination.
Draw the product(s) for this reaction:

*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.



Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

  • Intro 0:00
  • Amines: Properties and Reactivity 0:04
    • Compare Amines to Alcohols
  • Amines: Lower Boiling Point than ROH 0:55
    • 1) RNH₂ Has Lower Boiling Point than ROH
  • Amines: Better Nu: Than ROH 2:22
    • 2) RNH₂ is a Better Nucleophile than ROH Example 1
    • RNH₂ is a Better Nucleophile than ROH Example 2
  • Amines: Better Nu: than ROH 3:47
    • Example
  • Amines are Good Bases 5:41
    • 3) RNH₂ is a Good Base
  • Amines are Good Bases 7:06
    • Example 1
    • Example 2: Amino Acid
  • Alkyl vs. Aryl Amines 9:56
    • Example: Which is Strongest Base?
  • Alkyl vs. Aryl Amines 14:55
    • Verify by Comparing Conjugate Acids
  • Reaction of Amines 17:42
    • Reaction with Ketone/Aldehyde: 1° Amine (RNH₂)
  • Reaction of Amines 18:48
    • Reaction with Ketone/Aldehyde: 2° Amine (R2NH)
  • Use of Enamine: Synthetic Equivalent of Enolate 20:08
    • Use of Enamine: Synthetic Equivalent of Enolate
  • Reaction of Amines 24:10
    • Hofmann Elimination
  • Hofmann Elimination 26:16
    • Kinetic Product
  • Structure Analysis Using Hofmann Elimination 28:22
    • Structure Analysis Using Hofmann Elimination
  • Biological Activity of Amines 30:30
    • Adrenaline
    • Mescaline (Peyote Alkaloid)
    • Amino Acids, Amide, and Protein
  • Biological Activity of Amines 32:50
    • Morphine (Opium Alkaloid)
    • Epibatidine (Poison Dart Frog)
    • Nicotine
    • Choline (Nerve Impulse)

Transcription: Amines

Hi, and welcome back to Educator.0000

Today, we are going to be talking about amines.0001

Most of the conversation we are going to have is going to be focused on comparing amines to a functional group we already know pretty well, and that is alcohols.0005

Now, the big difference between amines and alcohols, or nitrogen compared to oxygen, is that oxygen is more electronegative than nitrogen.0013

Or, you could say that nitrogen is more electric positive than oxygen; remember, oxygen is the second most electronegative atom in the periodic table.0029

Any time oxygen has lone pairs of electrons, or has a covalent bond, it is always pulling those electrons toward itself; that is what it means to be electronegative.0036

Nitrogen does not do that as strongly, so its lone pairs are more available; and we are going to see several consequences of that.0046

OK, one such consequence is that amines have lower boiling points than alcohols.0056

When we look at the polarity of the N-H bond, we know that nitrogen is more electronegative than hydrogen, so this is a polar bond.0061

And it is so polar that it can undergo hydrogen bonding, but because oxygen is more electronegative, that is an even bigger dipole, and so this hydrogen bonding is stronger.0071

If we look at some boiling points, we have an example here of pentane: pentane is a nonpolar molecule, so you expect to have a fairly low boiling point.0085

An amine--this butanamine: butanamine has hydrogen bonding; it can form hydrogen bonds, so we see a big jump in the boiling point, not just because the nitrogen is more polar and we have some dipole-dipole attractions, but we can actually have some hydrogen bonding between molecules.0097

But, when you compare that to the oxygen--again, a big jump again, and that is because it has stronger hydrogen bonding than is capable between alcohol molecules.0117

Alcohols we expect to be one of the highest-boiling functional groups there are, and amines are also high-boiling, but not as high, because it has weaker hydrogen bonding capabilities.0130

OK, amines are better nucleophiles than alcohols: if we have a neutral oxygen, such as an alcohol molecule, we describe that as being a weak nucleophile.0145

It is something that wouldn't want to do an SN2 mechanism, for example.0156

If we had it react with a neutral carbonyl like this, if it attacked, we would end up with an O+ and an O-, and we wouldn't want to have that kind of intermediate in a mechanism.0161

There is going to be no O- and O+ in the same mechanism.0174

So, if we wanted to get an alcohol to react with a carbonyl, we would need an acid catalyst to get the reaction going, to protonate the carbonyl first.0179

OK, on the other hand, an amine is a very good nucleophile, even when neutral; because that nitrogen is less electronegative, you could imagine those lone pairs being not held so tightly, more available, more able to go out and attack an electrophile.0188

And so, a neutral amine can attack a neutral carbonyl and get this kind of an intermediate, and N+ and an O- are OK.0206

Those are compatible with each other; those are both possible to be in the same mechanism; so this is where we have seen some examples of the difference in nucleophilicity between an oxygen and a nitrogen.0216

And of course, neutral amines like ammonia love to do SN2 mechanisms; they are very good at doing that; so if we gave it a good nucleophile and a good electrophile, like methyl iodide--an unhindered electrophile--we would expect the SN2 to take place--no problem.0228

And that is after a deprotonation of the NH3 group.0249

But in fact, amines are so good at being nucleophiles, and the fact that alkyl groups actually add electron density--what we have here is still a great nucleophile, so this product can continue reacting with the methyl iodide, and we can end up with having two methyl groups on the nitrogen.0254

It's still a great nucleophile; in fact, nitrogen is such a great nucleophile that it will continue reacting as a nucleophile until it literally can no longer do that because it has no lone pairs of electrons left.0277

It will use its very last lone pair; it will use its lone pair continuously until it's all gone, and we would get this: it would be the iodide salt if we used methyl iodide--this is a quaternary ammonium salt--that is what it is called.0289

And these are very easy to form with amines, if it is given an excess of alkyl halide.0306

Even without an excess of alkyl halide, you can get over-alkylation, possibly.0314

So, if you are trying to just do a single alkylation, it is actually quite difficult to do that; you would have to have a huge excess of ammonia in order to get a good yield of methylamine without having a significant amount of dimethylamine in there, as well.0323

Amines are very good nucleophiles.0340

They are also very good bases; in fact, we use amines routinely in organic preparations, in organic reactions, as bases.0342

So, if we were to react an amine with a strong acid like HCl, it would most definitely and very much favor the forward reaction of having NH3+ and the Cl-.0350

So, we would get an amine salt here; and this is extremely stable--nitrogen is very happy with a positive charge; these salts are good.0365

In fact, you might see these sometimes, if you open up the little...even if you just read the name of some of the drugs that you might find--the medicines in your medicine cabinet.0379

Sometimes, you will see the name of a compound: will have HCl or HBr as part of the name of the drug, and that is because that drug has an amine in it, and it was reacted with an equivalent of HCl or HBr to protonate that amine, and that actually makes the amine much, much more stable.0390

It's much, much less likely to decompose or oxidize or do other reactions.0401

And so, that is where that comes from: so this should be recognized as a quite stable species.0419

Some examples we have seen of amines being very good bases can be found when we try to do an electrophilic aromatic substitution reaction: these are nitration conditions that could be used to add a nitro group to a ring.0428

And we would expect that this amine, this aniline derivative--because it is an electron donating group, we expect this to be an ortho/para director, so that the incoming electrophile will go to the ortho position or to the para position.0443

However, because aniline is very good base, and these reaction conditions (nitric acid, sulfuric acid) are clearly acidic, what is going to happen first is: we are going to protonate your aniline.0458

And it is going to be protonated to a very, very large extent, and now it turns that group into an electron withdrawing group, because it's positively charged, so it's pulling electron density now instead of adding electron density.0475

And that makes it a meta director; so in this reaction, what we find is that our major product is going to be aniline with a nitro group added in the meta position, instead of the ortho or para position that we were expecting initially.0488

This was one example; we saw an amine behaving as a good base.0504

Another example you might find is: when you look at an amino acid (here is a structure of an amino acid; you will see a lot of these if you move on to biochemistry after organic), it has both an amino group and a carboxylic acid group.0508

Now, this carboxylic acid group has a pKa of about 5; that is quite acidic, and that is certainly acidic enough to protonate the amine base.0520

So, in fact, the way it is...only an organic chemist, really, would draw an amino acid like this, because we like to see our carboxylic acid and our amino group on their own.0530

Because, in reality, in a neutral pH, it actually exists in the zwitterionic form; it's called a zwitterion when we have a + and a -, because this is the more stable form.0542

This ammonium group has a pKa somewhere around 10, so it's a much, much weaker acid than the carboxylic acid; so the proton would rather be on the nitrogen, giving it a positive charge, and having this oxygen with a negative charge.0559

That is more stable than having this neutral molecule, where the hydrogen is on the carboxylic acid functional group; so very interesting...0575

Of course, this charge is resonance-stabilized, and the nitrogen is quite happy having a positive charge, so you should recognize this zwitterion as also something that is reasonable and something that is stable.0584

Now, let's take a look at alkyl amines versus aryl amines and their basicity.0598

Let's ask which is the strongest base of these three: we have an alkyl amine with just an ethyl group (ethylamine or ethanamine), and compare that to aniline or a substituted aniline.0604

Now, if we wanted to try and determine the difference between their basicity, we want to think about the lone pairs on these nitrogens, and think about how willing those lone pairs are to be protonated--to be a base--that means you are accepting a proton.0615

OK, well, this amine--the alkyl amine--there is nothing special going on there; but as soon as you put a nitrogen onto a benzene ring, that lone pair is now benzylic, and it has resonance.0631

The lone pair is no longer just residing locally on that nitrogen: it is now being delocalized through the π system of the aromatic ring.0644

In fact, there are more resonance forms, right?--let's just say "etc." here.0655

We can move that negative charge down to this para position; we can move it up to this ortho position; there are several resonance forms we can draw.0659

That is going to have an effect on the basicity; and furthermore, here we have an electron withdrawing group; we have a carbonyl placed in the para position.0667

So, if we think about the resonance that this has, we can have a resonance form where the lone pair comes in, the π bond moves over, the π bond moves over, and actually, we could put that excess electron density all the way up on the oxygen of the electron withdrawing group.0675

This not only has the same resonance forms that the aniline one would, but it has this extra resonance form that puts the negative charge on an oxygen.0690

We know that is a great place for a negative charge to be.0701

There are varying degrees of resonance here: what is the effect of that resonance on the basicity?0704

Well, we could say that this last one--we could say it has extra resonance with the electron withdrawing group, right?--it has extra resonance involving the electron withdrawing group, compared to just the aniline derivative.0711

This is the most stable amine.0727

If it is most stable, that means it is the least reactive; it is the least reactive, and therefore, weakest base.0736

We could describe the lone pair as being tied up; the lone pair is tied up, delocalized; it is not available to be a base.0749

The fact that we can draw these additional resonance forms where there is no lone pair on nitrogen means that that lone pair is delocalized over the π system, and therefore it is not readily available to deprotonate.0764

OK, this one has some resonance, but we come to the alkyl amine, and this has no resonance at all.0778

That makes this the least stable amine; remember, resonance always adds stability, so this is the least stable amine, and therefore the strongest base.0789

The lone pair in this case is very available; it's localized right on this nitrogen; it's not spread out; it's not stabilized; it would love to be protonated.0803

We could just kind of put that in here as a parenthetical: we could say that this lone pair is very available; it is quite available to act as a base.0813

OK, now the answers actually lie right here in the bases, but we should be able to confirm these base strengths by looking at the conjugate acids and seeing their pKas.0828

OK, we would predict that this amine--because this is the strongest base, we would predict that this has the weakest parent acid (or conjugate acid, because we haven't drawn the conjugate yet--we are looking directly at the bases).0838

This has the weakest conjugate acid, and we would predict that this one has the strongest conjugate acid.0860

OK, in other words, this conjugate acid--if we protonated this, and we looked at that acid, that would be a relatively strong acid; it would want to give up its proton to get back to this resonance-stabilized conjugate base.0871

Remember, the key to acid strength is: you always want to have a stabilized conjugate base.0882

So, because this is the most stable conjugate base, we are expecting the conjugate acid to be the strongest conjugate acid.0888

Let's take a look at those three structures.0894

And, in fact, when we look at those NH3 groups, the pKa of an alkyl amine is somewhere around 10; but an aryl amine--the pKa of a protonated aniline is 4.6--so it's significantly more acidic.0897

Why is it so much more acidic--why is it so much more willing to lose its proton?--because, once it loses the proton, this lone pair can be delocalized throughout the π system; so that is going to be very stable.0912

And it's even more acidic here--because, when this deprotonates, this lone pair can now be delocalized throughout this electron withdrawing group.0924

You can also discuss how, as an electron withdrawing group, its effect on this acid is to withdraw electron density; what does that do to this positive charge?0931

Putting some positive character here--this destabilizes (this electron withdrawing group destabilizes) the acid.0956

So, this is the least stable; that makes it the most reactive; that makes it the strongest acid; and that, of course, is exactly what we see with the pKa.0965

We can see that an electron withdrawing group would be decreasing the stability of this acid and making it even more likely to lose a proton so that it can go back to the very resonance-stabilized conjugate.0981

How about if we put on a methoxy group--where do you think that would fit in pKas here--how would that compare?0997

A methoxy group, of course, is an electron donating group, right?1004

So, that would stabilize this acid, and it would also destabilize the resonance forms of the conjugate base where the lone pair was trying to add into the same benzene ring; it would be competing for that.1011

This, now, is going to be more stable; what does that do to your pKa?1029

That is going to increase your pKa; it's going to be a weaker acid, compared to just having aniline, where there is no electron donating group (this is an electron donating group).1034

And sure enough, the pKa is somewhere in between here; it is 5.3.1044

So, putting on an electron donating group would decrease the acidity of this ammonium group, and putting on an electron withdrawing group would increase the acidity of the ammonium group.1052

Let's review some reactions of amines that we have seen before: they can, as a good nucleophile--we have seen SN2 about if they reacted with a ketone or an aldehyde?1064

Well, we can imagine that acting as a nucleophile and doing an addition to the carbonyl--that would give this kind of a structure, where we have a nitrogen and an OH; but that is not very stable.1073

This would continue to collapse, at some point, and lose water to give an imine product.1083

A reaction of a primary amine (meaning we have NH2 group), plus an aldehyde or a ketone, gives an imine.1095

And so, we replace the C-O double bond with a C-N double bond, and the one alkyl group or aryl group--the one group we had initially is still there.1103

OK, and what is interesting about these imines is that every one of the steps in this mechanism is reversible; so we H3O+; it can undergo hydrolysis, in which we replace the C-N bonds with C-O bonds.1112

Now, if we had a secondary amine instead--here we have a nitrogen with two alkyl groups; we call that a secondary amine.1129

We can imagine doing this same reaction: we replace the C-O bond with a C-N double bond.1137

However, this nitrogen now has four bonds; this is unstable, and it is not going to go there; we need to lose a proton--we need to lose an H+ to neutralize this.1142

And the only place we can go to find a proton is one of these neighboring carbons; so we can make this a CH2; so we can see one of the hydrogens, and sure enough, we can deprotonate here.1155

That would be a way of getting rid of that positive charge.1169

This product is called an enamine, because it has an alkene (carbon-carbon double bond) and an amine on the same carbon, kind of like an enol.1178

We have seen enols, where you have an OH here; it would be the same sort of behavior, same sort of reactivity, but this one is stable--you can isolate it; you can work with it.1195

And, in fact, you can even use it in a synthesis.1205

Where it comes in handy there is: it is going to be reacting like an enolate would, and so we can use it as the synthetic equivalent of an enolate.1210

So, instead of using the enolate, we could use an enamine instead.1218

Let's see an example of that: if we took this ketone and reacted it with a secondary amine, we would get out this aminium.1221

We can go kind of halfway, and imagine replacing the C-O double bond with a C-N double bond.1231

And then, to make the enamine, we could remove this α proton or this α proton; and we are going to want to remove this α proton, because the resulting double bond is going to be conjugated with the benzene ring.1238

OK, so what we find when we form enamines is that we get thermodynamic control; we are going to get the more stable enamine when we form it.1263

So, if we can have a more highly substituted double bond (or in this case, we can have a conjugated double bond), that is going to be the major product.1274

OK, well, this kind of looks like an enol and kind of looks like an enolate; it turns out that this carbon is nucleophilic, and so, just like an enolate would do, if we reacted this with an electrophile (like methyl iodide), we can have a reaction.1281

We are going to form this bond here between the nucleophilic carbon, α carbon, and the electrophilic methyl group (that is a methyl group).1296

And the mechanism we have for that is just like an enolate: we start with this lone pair; it forms a π bond, kicks this π bond out, and does an SN2 and kicks off the leaving group.1305

And so, we can do an alkylation; we can introduce an alkyl group in this position.1321

That brings us back to our unstable aminium ion: OK, but when we work this up with H3O+, we could do a hydrolysis to get rid of the aminium group altogether--get rid of the nitrogen group altogether, and get back our carbonyl.1327

So, this strategy--this enamine strategy--took our carbonyl, activated it as an enamine, so that we can react with the α carbon, and then we got rid of the amine group to go back to the carbonyl.1347

Now, if you look at the transformation we did, we just did an α alkylation, so you might say, "Well, can't I do it just using LDA?--we used lithium diisopropylamide as a strong base, and it would deprotonate, and then we could methyl iodide to alkylate."1361

OK, well, yes, you could do that, although remember: there is a choice of where your LDA can deprotonate, and LDA is something that is going to be under kinetic control; that is not a reversible reaction, so it's governed by sterics.1375

The deprotonation by LDA is governed by sterics, so in this case, it might go to the same position; but in other cases, it wouldn't, so there is a complementary nature between enamines and LDA enolates.1391

But also, it predates LDA, so this is something that chemists used before lithium diisopropylamide was developed.1403

And also, something to keep in mind is: if you are looking at reactions in the body, if you want to functionalize the α carbon of the carbonyl (for example, in the body, your body can't just take out a bottle of LDA and use it), it needs some other way of functionalizing the α carbon, and this would be a way that you could do it.1410

You could use a secondary amine nucleophile in the body to react with a ketone, and the enamine intermediate is what can react with electrophile and form new carbon-carbon bonds.1429

So, this is nice, because it is closer to maybe a biological model of doing an α alkylation.1441

Another interesting reaction that is unique to amines is a reaction called a Hofmann Elimination; you can take an amine, and you can eliminate it and form a carbon-carbon double bond.1452

Now, what is interesting about this is: usually, in elimination, we are losing a leaving group and a β hydrogen.1461

We would not normally consider NH2 to be a good leaving group, and so that is part of the strategy of a Hofmann Elimination: to first convert it into a good leaving group, and then do an elimination.1469

It is a 2-step process: first we react this with methyl iodide; we would use an excess of methyl iodide, so we are going to do an SN2, and we are going to continue doing SN2s until the nitrogen has added as many groups as possible (which, in fact, would be 3, in this case) to get a quaternary ammonium salt.1479

Remember, nitrogen is capable of forming 4 bonds, but then, it would have a positive charge: 1, 2, 3, 4--nitrogen wants 5, so it's missing an electron there.1498

OK, now we just made a good leaving group.1509

It's going to be the iodide salt, because we used methyl iodide; when we treat it with silver oxide, this replaces the I- with hydroxide.1513

The silver grabs onto the iodide, takes that out, and puts an OH- in its place; so now, we have this quaternary ammonium salt.1530

I'm going to draw in this β hydrogen, because we're going to use it--actually, let's match our product; we use the other side; in this case, it's symmetrical, so it wouldn't matter.1543

OK, but we now have a hydroxide salt instead of the iodide salt; well, hydroxide, of course, is a strong base, so when we heat this mixture, it will force an E2 elimination to take place, where, in a single-step concerted mechanism, we lose our leaving group and form a carbon-carbon double bond.1551

This is called the Hofmann Elimination of an amine.1572

Let's predict a product: in a case where regiochemistry is concerned, we will find that the kinetic product is major.1577

That means the product that is formed the fastest.1585

And so, in this case, we do a 2-step Hofmann Elimination; we do excess alkylation and then silver oxide.1590

Let's just draw this intermediate that we are going to get, just as a reminder.1600

We are going to attach three methyl groups on here, so we have a good leaving group; and now, when that hydroxide comes in, where can it deprotonate?1607

It can is a β carbon with a β hydrogen; here is a β carbon with a β hydrogen.1613

And what happens is: we get the kinetic control, which is like a steric control (now, there might be more going on with the Hofmann than just sterics, but just paying attention to sterics is enough to get the right answer here).1620

It goes for the less sterically hindered proton, and so the product we get out is this one; it's our major product.1635

We get something like 91% of the terminal alkene, and just the balance is the internal alkene.1648

Now, what is interesting about this product mixture is: this is not the typical mixture we have observed for elimination reactions, right?1659

Typically, our elimination reactions (the E1s and E2s we have seen with a traditional leaving group, like a halide) have been following Zaitsev's Rule; and that is where we looked at thermodynamic control, where we get the most stable product.1665

We would describe this as the Zaitsev product.1679

OK, but Hofmann control gives the opposite product, where instead of getting the most stable product, you get the least stable product--again, partially because of the sterics of the situation.1682

So sometimes we can get a Hofmann Elimination by starting with an amine as our potential leaving group.1693

Another interesting thing you can do with a Hofmann Elimination is: you could use it to do some structural analysis to learn something about the structure of your molecule.1703

OK, so for example, if I had this starting material, and I did a Hofmann Elimination--I react it with an excess of methyl iodide to make a good leaving group; I treat it with silver oxide to get a good base--now, what does my elimination product look like?1711

Here is my leaving group: these are the carbons that the leaving group is attached to; so where are the carbons next to it, the β hydrogens that I can do the elimination?1726

Well, it's one of these; and it's symmetrical, so we could pick either side.1734

Here is the reaction that is going to happen: when we treat it with silver oxide, I'm going to grab the proton, form the π bond, and kick off the leaving group.1739

I can break open the ring with this Hofmann Elimination and form a carbon-carbon double bond.1752

Now again, because it was a ring, the leaving group, even though it left, is still attached to the rest of the structure.1758

And now, if I were to repeat the Hofmann Elimination--if I did another treatment with methyl iodide to make this a good leaving group, and silver oxide again, I would be able to eliminate this second one.1765

And I would get a second elimination product, along with trimethylamine as my leaving group.1778

OK, and so what is interesting is: you could use the Hofmann Elimination and say, "Well, how many times did I have to do a Hofmann Elimination before my carbon chain was free of nitrogen?"1792

And that way, you can see, "Well, because it took me two rounds of Hofmann Elimination, there must have been two positions of attachment to that nitrogen; it must have been part of a ring, so that it took not just one cleavage, but a second cleavage, to remove that nitrogen completely."1802

That is kind of an interesting way that the Hofmann Elimination can be used, especially because many natural products contain nitrogen atoms, and this is a way to get into doing some of that structural determination.1818

OK, and let's take a look at some of these amines: I'll wrap up this lesson by looking at some interesting amines that are biologically active.1831

Because amines are such good bases, because they are such good nucleophiles, they have all sorts of interesting biological activities.1840

Many of the drugs we take (legal, illegal, all sorts of things that interact with the central nervous system)--when you look at their structures, you find that they have nitrogens in their compounds, and that nitrogen is key to the biological activity.1849

For example--just a very brief look at some of these--this structure is the structure of adrenaline: that is a hormone that gets released in your body when you are in a panic or you are all excited.1863

That is what gives you the energy to have that "fight or flight" response; that is adrenaline.1874

You can see, it is a secondary amine here.1878

This is an alkaloid that is taken from the peyote cactus; an alkaloid is a nitrogen-containing natural product that is alkaline--it is basic, so it is removed from the natural product by acidic extraction; and so those classes of compounds are called alkaloids.1882

And so again, why is it basic?--well, you see it has a nitrogen here--a pretty similar structure, isn't it?1900

It is not all that different from adrenaline; but mescaline is something that causes hallucinations and has that kind of effect on your central nervous system.1907

Of course, amino acids are the building blocks to build proteins, which have all sorts of uses in our bodies and in nature, from structure to messaging to catalysts and so on.1917

And so, we already saw the structure, today, of amino acids: it has an amino group; it has a carboxylic acid; when they combine, you can form amides.1934

This is an amide linkage, and that is what puts together a protein; so nitrogen is a critical component of proteins.1943

I should also mention: it is a critical component of DNA; we talked about how nitrogen can hydrogen bond as something that effects its boiling point--well, when you look at the two strands of DNA and how they are held together, it is nitrogens that are interacting that are one of the components of that hydrogen bond.1950

So again, it's critical to have nitrogens in biological molecules.1966

This is morphine--an excellent painkiller: again, another alkaloid; it is isolated from opium, and this is a very good painkiller.1971

Unfortunately, it is also extremely addictive; but for terminal cancer patients, or patients like that, who need extreme pain relief and don't have to worry about the addictive nature of it, morphine is something that is really critical for end-of-life care.1984

Again, look at our structure here: we have that nitrogen; it is definitely essential to its activity.2001

This is epibetadine: it is the component in the poison dart frog that is toxic (so not all amines are good for us--this one is something that is a toxin).2008

And again, a couple of nitrogens in there--it's interesting.2025

This is the structure of nicotine: again, it has a pyridine ring and another--a tertiary amine; nicotine is the component in cigarettes that makes them so addictive--so again, it clearly has a big impact on our central nervous system.2029

Here is a molecule called choline; this is a quaternary ammonium salt.2044

We talked about how stable these structures are and how happy nitrogen is having a positive charge and four alkyl groups.2049

And so, this is a cation that exists in nature, and it is used as something that could help conduct electricity; it is something that is used for nerve impulses.2056

You can use choline to travel from one to the other and help send messages.2069

So, nitrogen, in all shapes and forms, have really interesting structures.2073

And so, the next time you come across a little pamphlet for a drug, or you look in a Merck index, you can look up some interesting molecules there, and you will be really surprised on how many of these molecules contain nitrogen compounds.2078

That wraps it up for our discussion of amines today.2091

I hope to see you again soon at; thanks very much.2093