Organic Chemistry Substitution Reactions

Welcome back to Educator.0000

Next we are going to talk about a class of reactions called substitution reactions.0002

Let's take a look at an example of such a reaction; here we have hydroxide reacting with chloromethane.0007

Our product has, where the chlorine used to be on the carbon chain, we've replaced it now with the OH group; and the chlorine is now on its own as chloride.0016

This is described as a substitution since we have replaced one group with another group; we've substituted; and let's define some of the players in this reaction.0027

This first species I've labeled Nu:--that stands for nucleophile; so I am going to be using the abbreviation; Nu with a lone pair.0039

A nucleophile is defined as something that is electron rich; so typically we will see a lone pair on a nucleophile; we will also see examples where a π bond can behave as a nucleophile.0047

It can have a negative charge; like our nucleophile in this case is hydroxide; we see a negative charge.0058

But a very nice abbreviation for nucleophile is Nu: with a lone pair; that lone pair right away gives you some clue as to the behavior of this nucleophilic species.0063

Because it is electron rich, a nucleophile is going to be seeking an electron deficient species.0073

The next part in this reaction is described as the electrophile; a very handy abbreviation for electrophile is E⁺.0081

Again that tells us something about the species; because electrophiles are things that are electron poor.0090

It might be possible that you will see a positive charge to show the electron deficiency; or more likely what we will usually have is a partial positive (δ+).0095

That is true for this electrophile here; it is a neutral species; there are no charges.0106

But because this chlorine is pulling electron density away from the carbon, the carbon is in fact partially positive; and that is what makes him a good electrophile.0111

An electrophile is something that seeks an electron rich species; oh, I should mention, where does the name electrophile come from?0123

It is called an electrophile because it is electron loving; right?--and that makes sense; if it electron deficient, it is going to be seeking electrons, loving electrons.0130

A nucleophile is called such because it is nucleus loving; and what sort of things do we have in the nucleus of an atom?--we have protons; we have positively charged species.0143

A nucleophile is something that is seeking out electron deficient species; and electrophiles are something that are electron deficient and seeking out electron rich species.0154

What we have here with the nucleophile-electrophile pair really is just a perfect match.0163

We are going to see that nucleophiles and electrophiles coming together are going to explain the vast majority of the organic reactions we are going to be seeing down the road.0168

The other part in this equation is in this case played by the chloride; it is called the leaving group; which I am going to abbreviate LG; sometimes it is just abbreviated with an L.0177

This as the name implies is a group that leaves; and as it leaves, it takes its two electrons with it.0189

You can see that this bond ended up leaving with the chloride because now it has four lone pairs of electrons.0195

That is going to be true of all of our leaving groups; they will take electrons with them.0202

We are also going to be learning about what makes a good leaving group; we will find that stable groups, things that are stable on their own, make very good leaving groups.0206

Things that are weak bases are typically pretty stable; so we will look for that, those sorts of features.0214

A general reaction... and I should also mention that this leaving group we had here is a chloride.0219

For example, X^- makes a very good leaving group; we will see chloride, bromide, iodide as a very common leaving group that we can have; these are all weak bases because it is conjugate.0227

Look at HCl; how do we describe HCl?--we know that is a strong acid; so that tells us that the conjugate base chloride is a very weak base; very stable; and so they make very good leaving groups.0242

Overall the general reaction for the substitution reactions we are going to be studying in organic chemistry look like this.0254

We have some kind of nucleophile reacting with some kind of carbon chain; this R represents a carbon chain; that has a leaving group on it.0259

When the reaction is done, that R group is going to have the nucleophile bonded to it in place of the leaving group.0267

The leaving group is going to now be on its own; and typically with an extra lone pair and a negative charge.0275

If this our general reaction, what are some things we can study about it?0287

What if we wanted to understand the mechanism by which these starting materials get converted to these products?--what are the possible mechanisms for a substitution reaction?0290

There is really two reasonable possibilities; one possibility would be a simultaneous mechanism.0299

Or in other words, a concerted mechanism in which the nucleophile attacks the carbon, starts to form a bond with the carbon group, at the same time as the leaving group leaves.0307

That would be just a single step mechanism that would accomplish the transformation; and this mechanism actually does exist.0320

We will be studying this; this mechanism is called the Sn2 mechanism--a simultaneous concerted single step mechanism.0326

But another possibility is that we have a stepwise mechanism in which the leaving group leaves first.0334

If the leaving group leaves, what does that leave behind on this carbon?--this carbon is now missing a bond so this would end up being a carbocation.0343

Then that carbocation could combine with the nucleophile to give our substitution product; and in fact that mechanism happens as well; this mechanism is known as the Sn1 mechanism.0357

We are going to be learning both about this concerted mechanism and this stepwise mechanism; and they are called the Sn2 and the Sn1.0373

We will start with the Sn2 mechanism; and here is an example of such a mechanism; we said this is the one-step mechanism; so what does it look like?0383

We could show arrows to follow the electron path, the electron movement, the bonding; our nucleophile... I'm sorry, we should label these guys.0391

This group that is coming in and doing the substituting, we describe a nucleophile; here again we have something, we have the lone pair and a negative charge.0404

The alkyl halide here is going to be our electrophile; again why is it electrophilic?--I don't see any electron deficient species.0413

Well sure, that carbon bearing the leaving group, bearing the halogen, is in fact partially positive.0421

There is going to be an attraction between this electron rich nucleophile and this electron deficient carbon; and that is what is going to help form this bond.0427

As that bond is formed, this carbon-iodine bond is going to break; and those electrons are going to go to the iodide.0435

There is our mechanism; just two arrows; single step; and we get our two products--this is our substitution product and this is our leaving group; remember we labeled him a leaving group.0445

The direction of attack of the nucleophile onto the carbon bearing the leaving group is described as backside attack.0457

The nucleophile has to come in from the opposite face of where the leaving group is; that is the only possible attack to get that leaving group to leave and have the orbitals interact properly.0463

We are going to see the consequences of that shortly; let's talk about the kinetics of the reaction--things that affect the rate of the reaction.0475

The rate of the Sn2 is given by the following rate expression: the rate is proportional to both the concentration of the methoxide, the CH₃O^-... sorry.0484

The CH₃O^-, our nucleophile, and the concentration of the ethyl iodide; let's just abbreviate this CH₃CH₂--we could abbreviate that as ethyl; ET for ethyl.0497

In other words, in the rate expression, we find both the nucleophile and the electrophile; that is why this reaction is described as a bimolecular reaction.0515

You add up the exponents; each of these are to the first power; so it is a bimolecular reaction; and that is where the Sn2 comes from.0525

That is what the 2 represents in the name--is that it is the bimolecular reaction; it has both the nucleophile and the electrophile coming together in the one step of the reaction.0533

We will find that sterics play a big role in the rate of the Sn2; so let's see some examples of that.0543

How about if we study this Sn2 reaction where we have an alkyl bromide; so we have our leaving group here; and let's use cyanide as our nucleophile.0552

Take a look at this substitution reaction where the cyanide has replaced the leaving group to do an Sn2.0564

If we vary the type of alkyl group here, the type of R group we have, we can measure the rate and see what effect that has on the rate of the reaction.0572

Here is some different R groups that we can have; this first one where it is a CH₃, we just describe that as a methyl group; we would just describe that as a methyl group.0582

It is handy to take a look at some abbreviations that organic chemists often use; MeBr represents methyl bromide; so CH₃Br could be a short hand notation, can be MeBr.0593

How about this one?--if we have two carbons, that is now an ethyl group; so a common name for this would be ethyl bromide and abbreviation could be EtBr.0609

How would you describe the type of carbon bearing the leaving group?--well, we could use terms like primary, secondary, tertiary to describe the number of carbons attached to that carbon.0620

Because this carbon has just one carbon group attached, it is a primary carbon; the leaving group is on a primary carbon for ethyl bromide.0630

How about this next one?--what do we call the group when you have a three-carbon chain and you have an attachment at the middle carbon?0640

That is one of the common names for the propyl groups; this is the isopropyl group; so a common name for this would be isopropyl bromide.0647

Isopropyl bromide is a nice example of a secondary carbon bearing the leaving group because this carbon has two carbon groups attached.0655

Our next one is a great example of a tertiary carbon; this carbon bearing the leaving group--so right here is where we are imagining our leaving group is.0664

It has one, two, three carbons attached; and what do we call this arrangement of four carbon atoms?--this is called the t-butyl group; so we can also run this reaction using t-butyl bromide.0676

What is happening as we are moving down this column, down this list, is we are steadily replacing the three hydrogens on this carbon with more and more and more carbon groups.0690

What we are doing is we are increasing the steric hindrance.0703

This last one is kind of an interesting molecule; how would you describe the carbon bearing the leaving group here?0711

It looks like it is just a primary carbon, right?--because there is just one carbon group attached; but what is interesting about that carbon group that is attached?--it is a tert-butyl group.0720

Even though it has just one group attached to it, it is a very bulky group; and so this is also a really sterically crowded center.0729

This is called a... it is not really just a simple primary; this is known as the neopentyl group; that is another common name.0737

The neopentyl group is uniquely known because of its large amount of steric hindrance; let's take a look at the rates of these Sn2 reactions.0745

If we describe the primary carbon as having a relative rate of 1, it turns out that the methyl bromide reacts 30 times faster; so really fast Sn2 when you are attacking just a methyl group.0753

By making it a secondary, it slows it down... I'm sorry, when we make it primary, it slows it down0769

When you make it secondary, it now goes to .02 so it is 50 times slower by having isopropyl instead of ethyl.0776

When you go to tert-butyl, our number becomes about approximately 0.0004; so now we are looking at a reaction that is so slow, it is barely measurable.0785

How about neopentyl?--and neopentyl is even slower actually; there is one, two, three, four 0s and then a 1.0801

What we have done is we have increased the branching on the carbon bearing the leaving group.0810

We describe this first one as having alpha (α) branching because the branchings occur at the same carbon as the bromine.0814

For neopentyl, we don't have any branching at the α carbon; but right away on the beta (β) carbon, the very next carbon over if we had lettered these, we have branching.0825

This effect is called... is due to what is described as beta branching.0834

But clearly both of these are very sterically crowded; and we see that we definitely have a decreasing rate as we move down this list and increase our steric crowding.0840

Let's take a quick look at just some extremes here.0854

These stick models don't really do justice to the system because when we just represent the electron cloud with this skinny little line, it seems like there is plenty of room.0860

But remember that these are all clouds of electrons; and when you look at a space filling model, you could see a much better picture on why there is such a big difference in steric hindrance.0871

So make sure you take a look at some of those illustrations.0882

This would be an example of methyl bromide; there is a bromine, one carbon, and just three hydrogens attached.0884

The nucleophile coming in... the nucleophile coming in has to come in from opposite the bromine; this is what we called backside attack.0891

When you view the molecule from the backside, you see that all we have our hydrogens here; very accessible; very easy for the nucleophile to approach.0898

If we compare that to a tert-butyl group, a tert-butyl group has replaced those three hydrogens with three other methyl groups each of which is tetrahedral.0906

Now... and once again, in this model, you don't see it quite as dramatically as if the space filling model.0917

Clearly when you try coming here, you are now going to experience some steric hindrance, some steric crowding, something blocking the nucleophile trying to get in here.0925

Even with the neopentyl, if the crowding is right here, you get that same effect.0933

In a space filling model, when you show the actual regions of electron density for these bonds, the carbon that you are attacking isn't even visible; so huge steric hindrance for the tert-butyl.0937

We could maybe draw that out here; I've shown the tert-butyl group here, trying to show how each of these carbons is tetrahedral; so they are projecting out in all directions.0950

My nucleophile is trying to get in from this direction--from the backside; but it is experiencing some steric hindrance from the methyl groups that are attached to the carbon.0961

That is a way we can show it; we can describe it as having a crowded backside approach; and that is going to be true for both the tertiary and the neopentyl.0971

Overall the rate of the Sn2 by considering the type of the alkyl halide, the fastest that we have, the best Sn2 that is the quickest, is going to be when you have a leaving group on a methyl carbon.0985

That is faster than primary; that is faster than secondary; and actually we have this huge rate increase; let's make this a double arrow--double greater than here.1002

There is a huge jump in tertiary and neopentyl; and the fact that these are so slow that we actually consider them usually to be a no reaction.1014

It takes so long for any amount of it to react that it is just... we describe it as having no reaction at all.1023

We will keep that in mind when it comes to backside attack, this rate based on sterics; and tertiary and neopentyl are going to be so slow that we say no reaction.1029

How about an energy versus progress of reaction diagram for an Sn2 reaction?--what would that look like?--let's come back to our example here of this substitution reaction.1047

Let's just say one-step mechanism; so a one-step mechanism starts at our starting materials; we can put them at some energy level.1058

Then when we go to draw the energy level of our products, typically our products are going to be lower in energy.1067

That is what makes a substitution favorable; and we will actually talk about that today on what makes a substitution a favorable reaction.1076

Then when it is a single step reaction, we just have a single transition state to go through before our starting materials go to products.1085

It looks kind of like a typical single step mechanism that is going to be endothermic.1095

Let's take a look at this transition state a little more closely, that I just labeled up here as TS.1107

Remember a transition state is something that is a high energy species; sometimes it is called an activated complex where you are starting to form bonds, you are starting to break bonds.1113

What is characteristic of a transition state is partial bonds, partial charges.1126

We know that our mechanism is this: our oxygen is attacking the carbon in the process of breaking the iodide; so what does it look like in the meantime?1131

How we can represent that bond that is being formed between the carbon and the oxygen is as a partial bond... partial bond, one that is being formed; in this case, this is a bond forming.1140

What do we have on the other side?--we have this carbon-iodine bond that is on its way out; so this is also a partial bond because this is a bond that is breaking.1161

We represent both forming bonds and breaking bonds with a dashed line, a partial bond.1171

What happens with these three groups?--we have a hydrogen and a hydrogen and a methyl; you could see tetrahedrally, they are pointing all off in one direction now.1178

But in the product, notice they are pointing all off in the other direction; we are going to look at that stereochemistry in just a second.1186

What happens in the transition state is they are going to be all planar to each other; they are all going to be flattened out.1192

As they transition from one direction, they are going to go planar, and then they are going to go in the other direction.1200

This is a good looking transition state; but let's take a look at the charges that are going to be developing or dissipating in the transition state.1208

We see that in our starting material, our oxygen here is negatively charge; but in the product when it is bonded to the carbon, it is going to be neutral.1215

What does it look like in the meantime in the transition state?--it is going to be somewhere between -1 and 0; so we describe that as a partial minus (δ-); a δ- means we have a partial charge.1222

Are there any other partial charges?--sure, the iodine starts out neutral and it ends up with a negative charge in the product.1233

This is going to be again somewhere between 0 and -1; so we will put a δ- on this iodine as well.1239

This is a pretty good looking transition state--partial bonds, partial charges; and that is what we will have to fill in our energy versus progress of reaction diagram.1246

Let's think about this transition state and consider its implications in the kinetics that we just learned about.1260

The fact that sterics, right?--if the carbon bearing the leaving group is too sterically hindered, what effect is that having on the reaction?1268

That means that these, instead of a hydrogen and a hydrogen and a carbon, let's say all three of these were carbons; it was a tertiary carbon.1280

That is going to have a lot of steric hindrance; that is going to have a lot of crowding; and that is going to take that transition state energy and raise it up.1287

What does that do to a reaction?--it slows down the reaction because it has a higher energy of activation.1295

The transition state energy is higher in energy; and the energy of activation is higher therefore; and so the reaction is too slow.1301

This transition state is going to explain the rate effects that we just saw in the previous slide.1311

Let's also consider what the hybridization of this carbon looks like; this is kind of a strange carbon, isn't it?--it looks like it has five bonds.1318

What kind of... we saw sp2, sp3... what do you expect if you have five bonds to carbon?--is this sp4 hybridized?--there is no such thing; that doesn't exist.1325

In fact, carbon can't have five full bonds; we know it cannot accommodate that because that would violate the octet.1339

Instead what arrangement have you seen for a carbon that has three groups that are planar and then something else that is perpendicular to that plane?--it looks to me like it is sp2 hybdridized.1346

In fact, if we draw that picture, if we replace those simple partial bonds and charges with a p orbital.1361

We can see how it is that this carbon can accommodate both the incoming nucleophile and the outgoing leaving group.1371

An sp2 hybridized transition state accounts for that very nicely; and that is what is happening as we are starting to form one bond and starting to break the other.1388

Knowing that there is a p orbital that develops on the carbon that is undergoing the Sn2 also explains another observation we see for the Sn2 reaction.1404

We find that the presence of a neighboring p orbital, conjugated p orbitals, would be good for the Sn2.1415

For example, if you had an allylic leaving group... the word allylic means that you are next to a π bond.1422

Here we have a carbon that is not on the double bond; it is next to the double bond; this allylic leaving group would be a great Sn2.1429

This allylic center is a very fast very favorable Sn2; and that is because right next door to this carbon that is going to have a p orbital.1443

In the transition state is another p orbital and another p orbital; so you can get delocalization; you can get some resonance energy; some stabilization because of that.1453

We will take a look at those Sn2 reactions; on occasion, we will see that if the leaving group happens to be an allylic one, that is very favorable for the Sn2.1462

Let me take a quick look back to our energy versus progress of reaction diagram.1471

We said this is what our transition state looks like; and let's imagine now if we were to have a more sterically crowded carbon bearing the leaving group.1476

Let's say this is a primary alkyl halide; what if you had a tertiary alkyl halide?--well, that transition state is now going to be higher in energy for a tertiary alkyl halide.1486

The path leading up to that would be much harder; and in fact for tertiary, it is typically going to be so high that it prohibits the reaction; it is slow, we say it is no reaction.1499

How about if our leaving group was allylic?--then instead of being here, that is going to be something that stabilizes the transition state.1511

That might be what we find if we have an allylic leaving group; and so what effect would we see?--that would speed up the reaction; so the Sn2 would be a faster reaction if it was allylic.1519

I just wanted to point out that what we are looking at are kinetic effects here; these are things that affect the rate of the reaction.1532

Something that either destabilizes or stabilizes the transition state is going to affect the rate.1538

One more thing to discuss about the Sn2 is the stereochemistry of the Sn2; we said that it is backside attack--the nucleophile has to come in 180 degrees from where the leaving group is located.1549

The result of that is we get what is known as an inversion of stereochemistry; we say inversion takes place.1566

Let's consider this chiral carbon; it has four unique groups on it; if our nucleophile comes in to do an Sn2, we know it has to come in from this direction.1573

It is going to be coming in from the left; and when it comes in and forms a bond now, it is going to be staying on the left; it is still going to be bonded on the left.1584

What is going to happen to these three groups that were on the right?--I think I'm mixing up my directions here... on the left?1594

They are going to go up and become flat or planar in the transition state; and then they are going to come down to be in the other direction.1600

It is like you have an umbrella; if you think of these three groups as forming an umbrella, you are going to straighten them out and then it is going to flip in the other direction in the product.1610

This is only observable... this happens every time you do an Sn2; but it is only observable when you have a chiral carbon.1625

If there were two hydrogens on here, if this was a symmetrical carbon, you would not be able to see that an inversion has taken place; you would have no experimental evidence for that.1631

This Sn2 usually changes the configuration as a result; so in other words, if you started out with the R configuration, you would now have S; and vice versa, an S would end up R.1641

I say usually; that is not always true; because this assumes that the four groups you prioritized in the beginning have essentially the same priorities in the end; and that is most often the case.1652

For example, if your leaving group is priority group #1, which it very often is if it is a halide.1665

And the nucleophile coming in is also a priority #1 compared to the others, then absolutely your R is going to get converted to an S; or your S is going to get converted to an R.1671

But because this prioritization has to be redone with the new four groups on the product, there is sometimes where you will find an exception to that.1681

Your configuration happens to come up with the same name when you do it.1690

Let's see an example here--we can see this when our leaving group is on the right and the nucleophile has to come in from the left.1696

How about in this line drawing?--in this case, our leaving group is a wedge; and so that means it is projecting out towards you.1703

What does backside attack mean?--backside attack means it has to come in from behind the board and attack from behind; which means the nucleophile is going to end up being bonded in the back position.1711

When my nucleophile comes in, it has to approach from behind the board; and it is going to end up as a dashed bond as a result.1726

We might see the inversion as a left to right; we might see it as a wedge going to a dash or a dash going to a wedge; those are the sorts of different ways that we can observe this inversion of stereochemistry.1742

Let me just throw in the hydrogen here; with our line drawing, a lot of times we leave off the hydrogen.1757

I know the hydrogen is still here; it must be a dashed bond in this position right behind the chlorine; so what happens to that hydrogen now in the product?--it can't be a dash anymore.1764

Again it is like the umbrella is flipping; the hydrogen was behind; but when the nucleophile attacked, the hydrogen straightened out and then it flipped forward; and there was your inversion.1775

The hydrogen as a result will now be in a wedged position; so there you can see the inversion of stereochemistry again.1786

Let's summarize what we know about the Sn2 reaction; this is a one-step mechanism; we describe it as backside attack.1799

My one step is my nucleophile attacks the carbon and kicks off the leaving group; that is it; one step--one concerted process.1807

That backside attacks results in inversion of stereochemistry; and what are the things that are required for the Sn2 where maybe we can predict that an Sn2 is going to happen?1816

We need an unhindered leaving group or electrophile; the carbon bearing the leaving group has to be unhindered in order for that backside of the carbon to be available.1824

The leaving group has to be on a tetrahedral carbon; so when we are picturing this inversion of stereochemistry in the backside approach, it is implying that we have a tetrahedral carbon.1837

This is for alkyl groups to undergo Sn2s; and it has to have a good nucleophile.1847

If the nucleophile is going to be coming out and attacking the carbon and displacing the leaving group, you have to have a pretty aggressive nucleophile to do that.1856

We are going to talk about what it means to be a strong nucleophile shortly; and you most definitely need a strong nucleophile in order to do the Sn2 reaction.1865

The rate... remember once again that this is based on sterics; and our fastest then is going to be the methyl; that is the best Sn2; that is great.1875

Our slowest is going to be the tertiary; tertiary is so slow, you should just go ahead and say no reaction.1887

If you ever find yourself doing an Sn2 mechanism on a tertiary center, you are probably making a mistake; you need to back up and think about something else.1894

Let's go through some examples of Sn2 mechanisms and see if we can predict what the product is going to look like.1904

We will start by looking for a nucleophile and an electrophile; I see my alkyl halide here; there is my leaving group.1910

Once I identify my leaving group, that tells me I have my electrophile, my E⁺; remember it is the carbon bearing the leaving group that is partially positive.1919

Where do I have a potential nucleophile?--right here; ammonia is actually a great nucleophile; this nitrogen has a lone pair.1928

Because it is just a single step mechanism with two arrows, I always recommend drawing that mechanism as part of your predict-the-product exercise.1937

Because that is really going to help you draw the product more accurately.1947

It looks now like we have formed a new bond between the nitrogen and the carbon; you can put the hydrogens on either side here; I just wanted to clearly show the new nitrogen-carbon bond.1950

I have kicked off my leaving group; I have Br^- here; some instructors really emphasize showing that leaving group after it leaves.1962

But for others, they are more interested only in the organic product--what happens to the carbon chain as a result of the mechanism.1970

Looking at this product, it looks a little strange to me; because when I look at this Lewis structure, something doesn't look like a typical configuration.1979

I think it is this nitrogen, right?--usually nitrogen likes to have three bonds and a lone pair to be neutral; so I better check this formal charge.1988

What do we have for this nitrogen?--it has one, two, three, four bonds; four electrons; and nitrogen wants five; so it is missing an electron; so actually this is an N⁺.1994

It is possible in certain cases to get charged products after an Sn2; especially with things like amines--they love to do the Sn2; so they would give this ammonium salt as a product.2007

How about the next one?--do we have a leaving group?--can we find a leaving group?--there it is; our iodide is the leaving group; so here is our electrophile.2019

Sodium hydroxide, NaOH; as soon as I see that sodium in there, that means I have an Na⁺ and O^-.2029

The Na⁺ is just a spectator ion; we could just scribble him out and stick with just the hydroxide.2037

I know the hydroxide is the species with the electrons; that is going to be doing the reaction.2043

I have hydroxide as a nucleophile; it has a lone pair; that is what we are looking for as a nucleophile.2048

We have our leaving group; what kind of carbon is the leaving group on?--how would you describe this carbon?--this has two carbons attached so it is secondary.2054

Anything else about this carbon?--it is also next to a π bond--secondary allylic; it looks like this would be a great Sn2 to take place; secondary is okay; allylic makes it even better.2065

What is going to happen is this is going to attack the carbon and kick off the leaving group.2077

Then finally tell me about the stereochemistry of this attack; how is that hydroxide going to be approaching the alkyl halide?2084

In this case, because the iodide is pointing away from us, the hydroxide, the nucleophile, has to be coming in from the top of the page; and that is where it is going to end up being bonded.2092

The hydroxide will be a wedge; there is where we see our inversion of stereochemistry; and that looks like a good Sn2.2103

How about the next one?--how would you describe this first structure, the bromide?--this looks like an electrophile because we have our leaving group here.2113

KCN--again any formula with potassium in it must be ionic so what we have is CN^-; so this is also a very good nucleophile that we should get familiar with and be used to seeing.2124

He loves to do the Sn2; but tell me about this electrophile; does this electrophile love to do an Sn2?--how would you describe the carbon bearing the leaving group in this case?2139

This carbon has one, two, three carbons attached; how would you describe that?--we describe that as tertiary.2149

What happens on an Sn2 on a tertiary center?--nothing happens; no reaction because we have a tertiary RX; so we really have to know all those details about the Sn2 in order to accurately predict the product.2156

How about our next one?--this is NaOCH₃; so once again that means I have OCH₃^-.2173

That looks like he would be a good nucleophile; it looks pretty similar to some that we have seen before; like a hydroxide, it has an O^-.2179

We have a leaving group here as the chlorine; and how would you describe the carbon bearing that leaving group?--we could describe that as a secondary carbon; it has two carbons attached.2187

Or is it something different than that?--yes, it is actually on a benzene ring; so we call this an aromatic, AR for aromatic, leaving group.2200

Or vinyl is just a general way to describe a leaving group or something that is directly attached to a carbon-carbon double bond.2211

How could I do backside attack on this molecule?--I would have to be coming in through the plane of the benzene ring in order to do that backside attack; it is impossible.2221

This is no reaction because we have an sp2 hybridized, sp2 carbon that the leaving group is on; and so we need to remember that Sn2 reactions take place on ordinary alkyl groups, sp3 hybridized carbons.2230

Finally this last one is quite interesting; we have hydroxide again--very good nucleophile; so you might think that he can come over here and attack and kick off the leaving group; that would be an Sn2.2253

That reaction certainly is reasonable; but there is a second reaction that can take place here.2270

Because hydroxide can not only act as a nucleophile; if this is acting as a nucleophile, it would be attacking a carbon and kicking out a leaving group.2276

But it can also act as a base; do we have any protons on this structure that might be reasonably acidic?--look here we have a carboxylic acid functional group; we know those are pretty good acids.2285

What happens is, instead of doing an Sn2, we are going to do the very very fast acid-base reaction instead.2299

If an acid-base reaction is ever possible, is ever something that can be considered, that is what is going to happen first; so let's consider that first.2308

In other words, the hydroxide is going to go after this proton; that is what it means to be a base; it is going to remove that proton.2316

What does that leave behind?--let me redraw this... as an O^-; and then we have a CH₂, CH₂, CH₂, and then a chlorine.2323

I am redrawing this because I want to illustrate something; I am redrawing it as a line drawing because I want to illustrate that what we have now is an O^-; that is a nucleophile.2338

In the same molecule, as we have a δ+ down here, this an electrophile.2350

Do you think it might be possible for an intramolecular reaction to take place?--where a nucleophile and an electrophile that are tethered can come around and react??it is possible.2356

We would describe this as an intramolecular Sn2 or a backside attack; an intramolecular backside attack where my O^- attacks the carbon, kicks off the leaving group.2367

What size ring did we just form?--we are going to form a ring every time we do an intramolecular reaction.2384

Oxygen is one, two, three, four, five; we just formed a five membered ring; so we could draw a very nice five-membered ring; oxygen was one, two, three, four, five.2389

Intramolecular reactions, Sn2s in this case, are very favorable reactions; very favorable especially entropically because we don't have to worry about bringing the nucleophile and electrophile together.2407

They are already together; they are already tethered; so the likelihood that they are going to collide and react is very high; so these are very favored reactions.2418

They are favored when we are forming either a three or a five or a six-membered ring; for the most part, only those size rings are going to be easily prepared this way.2430

Why are five and six-membered rings good?--because they have so little ring strain; we took a look at those structures.2450

We saw that they are very stable; they are a little flexible; so it is very easy for the ends of that carbon chain to come around and have those two meet.2456

A three-membered ring we know has a lot of ring strain; but because the nucleophile and the electrophile are so close.2466

They are so closely aligned that it is impossible for them not to react; it is a very rapid reaction; very easy to happen.2474

But a four-membered ring, now you are further apart; and so now trying to bring them together is not going to be able to overcome that ring strain.2480

If it is a seven-membered ring or larger, now these nucleophile and electrophile are so far away, that it is going to be hard.2488

They are flopping around; it is going to be difficult now for them to come together; and those reactions are usually disfavored.2495

So three or five or six-membered rings are things that we will keep an eye out for; and that is when we have within the same structure, we have a nucleophile and we have an electrophile.2501

Let's take a look at another mechanism; we saw the Sn2 mechanism--single step substitution mechanism; let's consider the following reaction.2514

Here we have another nucleophile with an O^-; we have a carbon bearing a leaving group so this would be an electrophile--an alkyl halide; it could act as an electrophile.2524

We had a leaving group leave; and in the place of that leaving group, we have that nucleophile; so is this a substitution reaction?2538

It looks exactly like a substitution; we have a nucleophile replacing a leaving group.2549

Could this be an Sn2 mechanism?--could this follow the mechanism that we just studied?--the one-step mechanism?2555

Backside attack where the nucleophile attacks the carbon, kicks out the leaving group?--does that seem reasonable in this case?2563

It doesn't; no, it cannot be; because what kind of carbon is bearing the leaving group?--it is a tertiary carbon; we have a tertiary alkyl halide.2571

It is still a substitution; overall we still get a product where our leaving group has been replaced by a nucleophile.2582

But it must have a different mechanism; there must be a different path taken to go from the starting materials to the products.2594

That second mechanism is known as the Sn1 reaction; that mechanism is called the Sn1; I just remembered I forgot to define the Sn2.2602

The S is substitution; n is nucleophilic; just like the S and the n in the Sn2; we have a nucleophilic substitution reaction; but in this case it is unimolecular.2614

The Sn2 is bimolecular; the Sn1 is going to be unimolecular; let's see what that looks like.2625

How can we go from these starting materials to these products without doing it in a single backside attack?2634

There is going to be always at least two steps; we will see a little later that there is very often a third step; but here is what happens in this step-wise mechanism.2641

The first thing is loss of our leaving group; so our leaving group is just going to pick up on its own and be lost; so I can show my Cl^- here.2650

What happens to this carbon?--this carbon just lost a bond; so there is going to be some kind of charge associated with that.2672

It has got one, two, three for its electron count; carbon wants four; so this is going to be a C⁺; it is going to be a C⁺--we call that a carbocation.2681

We are going to get a carbocation; that is not our final product; so this is known as an intermediate because it is formed in the reaction but then it is consumed right away.2698

Because that carbocation is going to continue reaction in our second step--step two is addition of our nucleophile; so our nucleophile is going to come in here, our O^-.2709

Do you think this carbocation would be a good electrophile?--I think it would be a great electrophile because it is super reactive; it is missing its octet; it has got a full positive charge.2723

This is most definitely electron deficient and looking forward to reacting to a nucleophile; so this reaction would be a great one where the nucleophile attacks the carbon.2734

Now that carbon gets its fourth bond back so it is neutral; the oxygen has two bonds so it is neutral; and there is our product.2747

We have achieved our products; we've made the chloride in the substitution product; what we did in this two-step process.2759

When you have a multiple step reaction, we try and identify what the rate determining step would be--which of these two steps is the rate going to depend on?2767

When we compare the two, I would describe this first step as a very slow step, very difficult; and that would be what determines our rate then; while the second step is very fast and easy.2778

What is it about this about this first step that makes it endothermic?--that makes it something that you have to climb uphill?2789

We are going to a very unstable carbocation intermediate so that is one clue; and the other thing is that you are breaking a bond.2797

The process of breaking a bond is always an endothermic one; you always have to put energy in to break a bond.2806

We go uphill for this step; and then the second step is a piece of cake; the nucleophile is going to love bonding with the carbocation; it satisfies both parts and you get to form a bond.2813

Bond formation is an exothermic process so you release energy here; so once we form the carbocation, it is very easy for the species to continue onto product.2826

How would we define the kinetics and what could we tell about this reaction?--well, like we said, it is going to be dependent on just this one step, this rate determining step.2839

What is involved in that step?--all we have is the alkyl halide, t-butyl chloride in this case, losing the leaving group.2849

It turns out that the rate for the Sn1 is dependent only on the concentration of the t-butyl chloride; t-butyl chloride in this case or the electrophile in general.2857

The rate determining step involves the electrophile only; that is why this is called a unimolecular reaction; that is for the Sn1, where the 1 comes from.2871

It is observed that the rate is independent of the nucleophile; so it doesn't matter how much of that nucleophile you had around; you could double the amount, triple the amount.2882

It has no effect on the rate because we are just waiting for that leaving group to be lost; that is the only thing that defines the rate.2889

What will really have a big impact on the rate though is that a more stable carbocation will be formed faster; so we are going to look for systems where we can get good carbocations being formed.2897

If the carbocation is formed faster, that was our rate determining step, so what does that tell us?--we have a faster Sn1 as a result.2914

What do we know is a good carbocation, is a stable carbocation?--well, things like benzylic, allylic, tertiary; those are all very stable carbocations.2926

Because they are stable carbocations, they are going to be fast Sn1 reactions.2936

When I put a leaving group in one of those positions, it is going to very readily undergo this substitution mechanism by this reaction.2943

Secondary is okay; that is still a good carbocation; but when we got to primary or methyl, those were very very high energy carbocations.2951

Because those are unstable carbocations, we describe that essentially as having no reaction for Sn1; again it is really a kinetics effect; because it is so slow, we describe it as no reaction.2961

If you find yourself forming a carbocation in a primary carbon or a methyl carbon and having no other place for it to go, you are probably moving the wrong direction.2975

There is probably a better solution that you should consider.2986

How about the energy versus progress of reaction diagram for this reaction?--we are starting again at some combined energy for our starting materials, our nucleophile plus our alkyl halide.2991

When we are all done, our products are going to have some combined energy that will typically be lower in energy if it is a favorable substitution reaction.3004

But it is no longer a single step reaction; it is at least two steps; because what we are going to be doing is we are going to be going through this intermediate which is a carbocation.3014

Step one is we are going to go from our starting material to our intermediate; and step two, we are going to go from our intermediate down to our product.3032

As usual, for every step, you need to go through a transition state; so this is transition state 1; and then going from the carbocation to the product, again you go through some transition state.3041

Our energy diagram will look something like this; now when we take a look at this energy diagram, can we identify what the rate determining step would be?3059

Remember we consider how fast the reaction is; we need to consider its energy of activation; so for the starting material to go up to the transition state here, energy of activation 1.3068

For the carbocation to do its reaction, it starts at a high energy; so it is a very short path to go the next transition state; we would call that energy of activation 2.3080

So this is our slow step; that is our rate determining step; rate determining step is that endothermic reaction where we are just breaking a bond and forming that high energy carbocation intermediate.3094

Let's see if we can figure out what the transition states will look like; we have two transition states; every step has its own transition state.3108

What is happening in our first step of the reaction?--remember a transition state has partial bonds, partial charges.3114

That first step is simply we are breaking that carbon-halogen bond; our leaving group is leaving.3122

Because we are starting with tert-butyl bromide in this case, we are going to be breaking that carbon-bromine bond; it is going to be shown as a partial bond because it is breaking.3131

Do we have any partial charges?--what does it look like after this step?--after this step, the leaving group is going to be bromide, Br^-; so we have a δ- here.3143

After the leaving group leaves, our carbon is going to be a carbocation; so we have a δ+ here; so as that bromine is leaving, it is pulling its electrons toward itself.3154

There is a negative charge building up on that leaving group and a positive charge building up on the carbon; that is going to ultimately lose those electrons and become a carbocation.3166

That is our first transition state; what does our second transition state look like?3175

Our second step was our carbocation combining with the nucleophile; so in that step, we are going to be forming a new bond.3180

Our nucleophile in this case was this guy; so he is now going to be forming a bond to the t-butyl group.3192

He came in as an O^-, but he will be neutral in the product; so he is losing his charge as he is sharing those electrons to form the bond with the carbon.3203

This is now going to be losing its positive charge as it is gaining those electrons; it is gaining back its fourth bond; and so that is our second transition state for the second step of the Sn1.3212

What can we know about the stereochemistry of the Sn1?--or what happens if we have a chiral carbon, a leaving group on an asymmetric carbon that undergoes Sn1 substitution?3227

What happens to the stereochemistry there?--remember Sn2, we saw the stereochemistry; that was the backside attack mechanism; so we saw the stereochemistry got inverted.3241

What happens in an Sn1?--well, the way we describe an Sn1 reaction is we describe it as undergoing racemization; that is what we refer to when we describe losing stereochemistry.3250

That is because, with this stepwise mechanism, we go through a carbocation; and the carbocation is achiral.3262

Here we have a chiral starting material; but once this leaving group leaves, this carbon now has just four groups around it.3270

What hybridization do we expect for a carbocation?--this is sp2 hybridized; it is planar; and any planar species has that plane of symmetry; so this is achiral.3279

Now when my nucleophile comes in, the question is: is it going to attack the top face of that carbocation or the back face of that carbocation?3297

The answer is both; it is going to attack from both directions to give an equal amount of the two possible products.3306

Let's take a look at a different perspective here; let's twist this around on its side; we have a CH₃ and a CH₂CH₃.3313

Our carbocation is sp2 hybridized so it has this p orbital here; it has an empty p orbital; and it is that empty p orbital that our nucleophile is going to attack.3326

Here we have sodium iodide as our nucleophile so let's bring iodide in here; and what we will find is that can attack from this face; let's call this path A.3338

If it attacked from this direction, that means that the iodide would be bonded off to the left.3350

We still have our ethyl group pointing out and our methyl group pointing back; but they are going to be off to the right a little bit.3358

Or it can attack the opposite face, let's call that path B; and how does that product differ?3367

It is now going to have the iodide off to the right; and again, the ethyl is still out and the methyl is still back.3372

There are two possible pathways here leading to two different structures; what is the relationship of those two structures?--are they identical?3384

They kind of look like mirror images, don't they?--are they superimposable?--nope, these are chiral centers; these are asymmetric carbons with four unique groups on it.3393

Guess what?--we just formed one with the R stereochemistry; the other with the S stereochemistry; the relationship of these two are enantiomers.3403

We are going to get equal amounts of enantiomers because the likelihood of the iodide attacking one face or the other face is statistically equivalent.3415

We are going to get a 1:1 ratio of the two possible approaches; so we will get a 1:1 mixture of our two products.3431

We have a special name for a 1:1 mixture of enantiomers; what is that called?--it is called a racemate or a racemic mixture.3440

That is why we describe this process as one that undergoes racemization; because even though we started out with a chiral starting material, we ended up with a racemic mixture product; a racemic mixture.3451

We could also describe it as having 50% of the reaction went with inversion; I know we are looking at it from different angles; but if you compare this structure.3465

If you tip it over and compare it to your starting structure, you see that this is the one where the nucleophile came in from the opposite face of the leaving group.3476

But this other one, we describe as retention; this product has the same configuration as the starting material.3487

We get equal amounts of retention and inversion based on attack of the two faces of this achiral carbocation intermediate.3495

What have we seen so far for the Sn1 mechanism?--this was a stepwise mechanism via the carbocation.3507

Knowing something about that carbocation then is going to tell us something about the rate of the reaction; because the more stable the carbocation, the faster the reaction.3517

The best looking Sn1 reactions we can have are for the best carbocations; things like benzylic--next to a benzene ring, allylic, or tertiary; let's just review what benzylic means.3525

Benzylic means that you are next to a benzene ring; allylic means that you are next to a double bond; and then tertiary of course means that you have three alkyl groups attached to the carbocation.3542

Those, you should recognize, is all very very good carbocations; so these substrates would love to do Sn1s if they had a leaving group in those positions.3556

These would be the fastest; and these are the best carbocations, fastest reaction; and primary and methyl are poor carbocations and very very slow or even no reaction Sn1.3565

Because we have lost that leaving group, we no longer know whether it used to be a wedge or a dash or where it used to be.3585

Racemization occurs in that situation and we end up getting a racemic mixture of products.3592

Another thing we will find is that because the rate determining step simply was based on how easily the leaving group was lost, how easily that carbocation was formed.3598

We don't need a strong nucleophile at all; in fact, most Sn1 reactions have no strong nucleophile; and we will talk about nucleophilic strength in just a moment.3607

But very often, the Sn1 is simply a reaction with a very weak nucleophile like a solvent when we use water or some kind of alcohol; those are very common Sn1 reactions.3617

In those cases, we call the Sn1 the solvolysis reaction because it is reaction with solvent; and we will take a look at that mechanism too.3632

But before we do that, let's compare the Sn1 and the Sn2 and think about their requirements and see if for the following reactions, can we decide whether an Sn1 or an Sn2 might be favored.3642

We just saw the Sn1 was the stepwise reaction; so let's put down here... let's make a little note up here... Sn1.3655

If you can remember one thing when you think about the Sn1, if you remember the word carbocation, that is going to answer a ton of questions for you about the Sn1; that is all about a carbocation.3665

How about Sn2?--Sn2 is the single-step mechanism that was backside attack; so if you remember that the Sn2 is about backside attack, that is going to explain a lot of things such as sterics.3676

Remember sterics are really important; that is another very good word to remember when thinking about the Sn2 mechanism.3693

Let's look at these various substrates; we have a leaving group here; he is on a methyl carbocation.3700

Let's think about Sn2; is Sn2 possible with a methyl leaving group?--is backside attack good when you have a leaving group on a methyl group?--it is excellent because there is no steric hindrance whatsoever.3708

How about an Sn1?--Sn1 means carbocation; would this be a good carbocation if the carbon just left on its own to give a methyl positive?--absolutely not; because we can't have H₃C⁺.3723

This substrate looks like it loves to do Sn2 reactions; do we have a strong nucleophile?--if you want to do backside attack, you need to have a strong nucleophile.3741

Sure, we have here KOH which is hydroxide; so yes, I think Sn2 is going to happen; what does that product look like?--we are simply going to replace our leaving group with our nucleophile.3750

How about the stereochemistry?--can we see that inversion that is taking place here?--well no, not in this case because it is just a methyl group.3763

It is not a chiral center so there is really no stereochemistry to show here; there is only one stereoisomer of this product; so we just draw it as shown.3771

How about our next one?--how would you describe the carbon bearing the leaving group?--we have a tertiary alkyl halide in this case because we have one, two, three carbon groups attached.3780

Tell me, how does that look for the Sn2?--how does backside attack look on a tertiary center?--remember we need to think about sterics; and tertiary is way too sterically hindered.3791

We are not going to be able to do the Sn2; how about the Sn1 though?--carbocation?--is tertiary good for a carbocation?--it is excellent for a carbocation.3804

That looks good here; so what does our product look like?--well, our nucleophile now is iodide; our leaving group is the bromide; so our product will replace the bromine with an iodine.3814

Tell me about the stereochemistry here; do I need to be concerned with whether I put the iodine up here or down here or a wedge or a dash?--no, because this is a symmetrical molecule; this is not a chiral center.3828

Again whether you draw it as a wedge or a dash, it would be the same molecule; so there is no stereochemistry to worry about in this particular case.3839

How about the next one?--this time our leaving group is on a secondary alkyl halide.3847

Tell me about backside attack; is backside attack okay for a secondary?--it is actually; it is not great but it is possible; so Sn2 is okay.3855

How about being a carbocation?--would a carbocation on a secondary carbon be acceptable?--absolutely; again not the best carbocation possible but certainly stable enough to be formed.3864

Secondary halides, secondary electrophiles are where we are going to have some of our biggest challenges because both mechanisms are going to be possible in that case.3877

We are going to have to look beyond, think about something else to decide which one is going to be favored.3886

Let's take a look at our nucleophile in that case; tell me about this nucleophile; I have O^-; I know we haven't quite talked about nucleophilicity yet.3893

But I see a negative charge; this tells me I have a good nucleophile; so something that would love to come out and attack.3901

Guess what?--if an Sn2 can happen, it will happen; it is a very good... it will be the major product; it is a very favorable reaction.3911

It doesn't mean Sn1 is impossible; you might get a mixture of some racemization; some carbocation formation here.3927

But typically we are looking for a major product when we are doing predict-the-product; so in this case, we would go with Sn1... I'm sorry, excuse me; Sn2 the backside attack.3936

I am going to be replacing my leaving group with the nucleophile; and tell me about the stereochemistry here.3946

Here we do finally have a chiral center that we are dealing with; our leaving group is a wedge; that means it is coming out towards you.3951

Where does the nucleophile have to approach?--backside attack; it has to come from behind the molecule; so we need to draw that as a dashed bond here.3958

It was the oxygen that had the negative charge; so it is the oxygen that is now attached to the carbon chain; very nice.3970

How about the last one?--how would you describe the carbon bearing the leaving group in this case?--it is a primary carbon; but it is not just primary; it is also next to a double bond--primary allylic.3978

What do you think about primary allylic?--Sn2; Sn2 means backside attack; primary allylic good for that?3995

It is excellent; not only does it have very little sterics; remember this neighboring p orbital helps stabilize that transition state; so this is a great Sn2.4003

How about Sn1?--how about Sn1?--primary; I know primary sounds bad for a carbocation; but remember, because it is allylic, it is now going to be resonance stabilized.4013

This is also an excellent carbocation; so here is another case where we can have both Sn1 and Sn2.4028

Let's look at our nucleophile in this case now though; our nucleophile is methanol; it is a neutral molecule; it is no longer negatively charged like we've seen most of our nucleophiles will have.4034

We will learn very shortly that we would describe this as a weak nucleophile; and a weak nucleophile, what do you think about that backside attack?4047

Do you think a weak nucleophile is going to want to come out and attack?--I don't think so; so the weak nucleophile means that the Sn2 is not possible.4055

Instead what happens?--we have an Sn1; what does that mean?--it means that our leaving group just leaves on its own to give a carbocation.4064

What product do we get from that carbocation?--if methanol is going to add, if this is the group that is going to add to it... we will take a look at this mechanism in just a moment actually.4076

The OCH₃ group is going to be what replaces the leaving group; we will see how that mechanism happens in just a second.4090

But there is something else that is interesting about this carbocation intermediate; because it is allylic, it is resonance stabilized.4100

When we draw the other resonance form that is possible, we find that we have a different carbon--a unique carbon that is electrophilic.4108

My methanol, my nucleophile, can add there as well; and that will give me a different product; and that product will also be formed.4119

So in the case of a resonance stabilized carbocation, we need to take a look to see if those different resonance forms might possibly lead to a different substitution product.4132

In fact sometimes that can happen like in this case; so let's see what that mechanism looks like.4147

I just showed that if methanol was your nucleophile, then you would end up forming this methoxy group--this OCH₃ group.4156

This mechanism is going to be a little more complicated; it is going to be three steps.4165

This happens very often in the Sn1; especially if your nucleophile is like we showed before--either water or alcohol; we call that solvolysis because it is reaction with solvent.4170

What happens in this case is a final deprotonation is required; so let's see if we can do the mechanism here; it looks like a substitution, right?4183

Which substitution mechanism do you think we will want to do here?--let's assume that we don't already anticipate that it is an Sn1; how can you verify that it should be an Sn1 mechanism?4200

Sn1 means carbocation; because our leaving group is on a tertiary carbon, then that for sure tells me right off the bat, I cannot do a backside attack; I can only do a carbocation.4213

Furthermore I have a weak nucleophile; so that also tells me that I can't do an Sn2; I would have to do an Sn1.4226

So Sn1 means carbocation; how do we get there?--our mechanism?--our leaving group leaves, right?4237

If there is no nucleophile that can come and attack, the leaving group just leaves on its own to give a carbocation.4245

That is never going to be our final step; our carbocation is a great electrophile so it looks around for a nucleophile that it can react with.4254

What nucleophiles do I have in the reaction?--well, actually I just made bromide; so what we can note here, this is a good thing to note, is this mechanism is going to be reversible.4261

This is an equilibrium that we have; the bromide can leave; it can come back in; so that would be possible.4271

But if methanol is our solvent, that means that methanol is surrounding every species in the reaction mixture including the carbocation.4277

So actually statistically, it is more likely that methanol is going to be the nucleophile that is going to react.4286

Why not use methoxide?--wouldn't it be great to use methoxide, CH₃O^-?--because I want to add just an OCH₃ group.4300

Wouldn't it be great if I could just use the CH₃O^- and then I would be done?--well, that would be great but what is the problem?4307

Take a look at your reaction conditions; where do you see that CH₃O^- has been added?4314

This is not an ionic compound; we can't just lose that proton; this is not a salt; this is a covalent molecule and the molecule must be used as a whole.4319

So methanol is the nucleophile; I have a new bond between oxygen and carbon; and what else is on this oxygen?--it still has a hydrogen and a methyl.4328

Does it have any lone pairs?--it had two; but one is now used--that is these two electrons that are being shared as a bond; so it still has one lone pair left.4345

How about a charge on that oxygen?--let's count; we have one, two, three, four, five; oxygen wants six; so it is missing an electron; this is an O⁺.4357

Here it is; we did our two-step mechanism, right?--we lose our leaving group; and then step two is we add our nucleophile.4372

We did our two steps; and we are not done; we are not done because this does not look like a stable molecule does it?--does oxygen like having a positive charge?--no way.4385

What we have to do is we have to do something to get rid of that positive charge; we know that oxygen wants to have two bonds and two lone pairs to be neutral.4394

Which of these bonds can it lose?--what group can it lose?--well, it is this proton that can easily come and go; we want to lose that proton; we call that step--deprotonate, when you lose a proton.4402

Something is going to come in and deprotonate; you could just say B: as some kind of generic base.4419

But again, who is most likely to do this deprotonation?--it is going to be the solvent because it is the solvent that is surrounding everything.4427

I typically use methanol as my base; a lot of textbooks will use this bromide as the base to tidy things up because your other product that you are forming here when you balance your reaction is HBr.4434

But whatever species you use, something is going to come and take that proton and then leave those electrons behind.4454

Proton transfer reaction is always two arrows; grab the proton, leave the electrons behind, and now we have our neutral product.4464

This is going to be a little more common for an Sn1 because usually our nucleophile is not something negatively charged; usually our nucleophile is neutral.4474

So after it adds to the carbocation, it now will have a charge; and we have to do a third step to get rid of that charge.4482

Let's talk about one additional complication we might have with a carbocation; and that is the fact that it can undergo something known as rearrangement.4492

Let's take a look the following substitution reaction; I have a leaving group and it has been replaced by a nucleophile; but take a look at where that nucleophile has come in.4502

This was the carbon that was bearing the leaving group; that carbon is not the carbon that received the nucleophile.4514

What has happened is the carbocation intermediate that is involved in this reaction has undergone a rearrangement; let's go see what that looks like.4521

First of all, once again let's confirm our mechanism; does this look like it should be an Sn1 mechanism?--could it be an Sn2?4531

Our leaving group is on a secondary carbon; that is okay for an Sn2; but tell me about your nucleophile; this looks like a pretty weak nucleophile because it is neutral; it is stable.4541

So we are not going to have an Sn2; instead the only thing that can happen is our leaving group can just leave on its own; so let's do that--the bromide can just leave.4551

You can show that leaving if you want; I often don't show leaving groups after they have left so forgive me if your instructors like to see that; you need to get into that habit then.4562

I lose my leaving group and I form a carbocation; now what happens here is this molecule undergoes a rearrangement where this carbon here with a positive charge is missing electrons; it is unstable.4580

It looks next door and sees this carbon-hydrogen bond with two electrons and it steals it; it takes it; this hydrogen picks up and moves with its two electrons and shifts over to the next carbon.4597

We call this a 1,2-hydride shift... we call this a 1,2-hydride shift; and let's see what the result is of this movement.4612

The hydrogen is now going to be attached to the next carbon over; so this carbon is happy; he now has four bonds again.4628

But what about this carbon?--this carbon just lost that hydrogen; so it now only has three bonds and our carbocation is now in a new position.4636

As you can see, the way I've drawn this equilibrium and the major product that is observed, is this is a pretty favorable rearrangement.4646

This is a pretty favorable equilibrium moving in the forward direction; so why is that?--why do carbocations undergo rearrangements such as these?4655

Well how would you describe this first carbocation?--what kind of carbon is it on?--it is on a carbon that is attached to two other carbons; we would call that a secondary carbocation.4664

How about the new carbocation after the rearrangement?--the positive charge is now on a tertiary carbon; so we call this a tertiary carbocation.4674

Which one is more stable?--the tertiary; the more alkyl groups you have, the more stable; so this is why rearrangements occur; this is why we need to be on the lookout for rearrangements.4684

If there is any way for a carbocation to rearrange itself to relocate the positive charge to a more stable center, it will do so; so we need to be on the lookout for that.4699

This process is called a rearrangement; and I want to be very careful to point out that it is not resonance.4711

Because we just saw an example where a positive charge moved, right?--when it was an allylic carbocation.4721

Remember resonance is just moving lone pairs and π bonds; no atoms move; you are just shifting around electrons and p orbitals.4726

Here we change the arrangement of the atoms; we've broken bonds and formed new bonds; so this is definitely not resonance.4735

Be careful with your arrows here; this is an equilibrium; a rearrangement is how we would describe it.4742

So this is the carbocation; this tertiary carbocation is going to be the one that we have in highest concentration; so where do we go from here?4750

This is an electrophile like all carbocations; so let's look around for a nucleophile; water is our solvent; water is what we have all around; so water is the nucleophile we are going to use.4758

What does that product look like?--let me just combine these two hydrogens now together; so we just formed a new bond between carbon and oxygen.4776

What is on this oxygen?--it still has the two hydrogens; and lone pairs?--it had two; but once again this lone pair is now being shared as the bond.4784

So we get an oxygen with three bonds and a lone pair; this is an O⁺; one, two, three, four, five; oxygen wants six.4797

Very good; we lost our leaving group; we did our rearrangement; we added our nucleophile; we are still not done because we need to get to a neutral product; so how do we get to this final product?4808

We just need to lose one of these bonds; so we could bring water back in as our base; and deprotonate as our final step to get back to the neutral substitution product.4819

So rearrangement is something to be on the lookout for anytime we have a mechanism involving a carbocation; so far the only example of that is an Sn1 substitution reaction.4833

Let's see one more example of a carbocation rearrangement; we saw an example where a hydride, a hydrogen, picked up and moved with its two electrons.4849

It turns out that also alkyl groups can shift; carbon chains can shift; and so let's see an example of that.4859

Here we have a leaving group; now if ethanol, EtOH, was our nucleophile, what group ends up adding and replacing the leaving group?--let's see if we could predict the product here.4867

If I asked you to predict the product of this substitution reaction, you would replace that leaving group with what?--would it be an OH?--no, this ethyl group isn't going to disappear in some mechanism.4879

An OH is what you get when you have water as your nucleophile; when you have ethanol as your nucleophile, then it is the ethoxy group that is going to be in the final substitution product.4895

That looks like a good idea; but unfortunately, that is not the product that is observed; the actual product that is observed is this one.4907

Where now, not only is the nucleophile on a different carbon from the leaving group, but my carbon backbone, my carbon structure, is actually different.4917

So this clearly must be some kind of rearrangement involving a carbocation; and let's see how that looks.4926

Once again our leaving group just leaves because we have no strong nucleophile to kick it out; so we have a five-membered ring with a positive charge here.4933

Let's draw in these methyl groups here; instead of a line drawing, let me draw them in as CH₃s.4947

Once again our carbocation is electron deficient, looking for some way to stabilize itself.4952

It looks around and it sees this carbon with its electrons; and this entire carbon group, this methyl group, picks up and shifts over.4958

We call this a 1,2-methyl shift; 1,2 just because it is starting on one carbon, it is ending up on the very next carbon; so a 1,2-methyl shift.4966

Because it is a methyl group in general, you could just call it an alkyl shift; and now let's see what we have as a result.4976

This carbon now has the methyl group attached; the top carbon still has one methyl group attached; and what happened to my carbocation?4984

It is now going to be on this top carbon; it used to have four bonds; now it only has three bonds; so this is another example of a carbocation rearrangement.4993

Now my ethanol... I will draw out ethanol so that we know what that abbreviation is standing for.5006

Our ethanol is our nucleophile; it will attack at the new location of the carbocation, followed by deprotonation.5015

I won't fill out that entire mechanism; that would be good practice on your own; but eventually, we would get to this rearranged substitution product.5023

We also get some very interesting reactions where this rearrangement involves a ring expansion to alleviate some ring strain; you will see all sorts of very interesting examples of carbocation rearrangements.5032

Not only can you have one of these rearrangements, but then that new carbocation intermediate can continue rearranging; and your carbocation can end up in all sorts of interesting places.5048

Anytime we are doing a mechanism with a carbocation, we want to be on the lookout for these rearrangement products.5059

Let's talk a little bit more about the leaving group that is involved; in both the Sn1 and the Sn2, it requires that it is a nucleophilic substitution of the a leaving group.5068

What I alluded in the beginning is that weak bases are things that make good leaving groups.5078

Way back when we talked about acids and base strengths, what we looked for to be a weak base is we looked for something that was going to be stable.5084

Really what we are just looking for is a species that is stable on its own; things like halides; not so much fluoride for an Sn2--that is a pretty lousy leaving group.5092

But chloride and bromide and iodide are all good leaving groups; typically iodide is the best leaving group; meaning it is the most stable leaving group.5104

Why would a negative charge be more stable on an iodine than a chlorine?--what is the difference between chlorine and bromine and iodine?5125

You get bigger and bigger and bigger; so the iodide remember is larger; so that negative charge is more delocalized on the larger surface area; it is a large delocalized negative charge here.5135

Alkyl iodides are very hot electrophiles; they really love to undergo substitution reactions; but bromides are fine and chlorides are as well.5151

We can use leaving group ability as a way to predict whether or not a substitution reaction is going to be favored; so let's take a look at the following substitution reaction.5161

Here we have methanol reacting with sodium chloride; let's get rid of that sodium right away and just focus in on the reactive species--we have Cl^- going to chloromethane and hydroxide.5172

This forward reaction can be described as a substitution, right?--you used to have OH and it has been replaced by the chlorine.5187

The reverse reaction can be described as a substitution; you used to have a chlorine and now that OH has come into it.5194

Let's compare the two potential leaving groups; because it turns out that the best substitution reaction has the best leaving group leaving.5200

When we compare our two leaving groups, chloride and hydroxide, which do you think is the better leaving group?--which of these would you describe as a very stable weak base?5212

The chloride of course; the halides are what we have been seeing as great leaving groups; this is a great leaving group; in other words, a weak base.5223

Hydroxide is a very poor leaving group; this is a strong base, right?--most students remember from their general chemistry days that hydroxide is an example of a very strong base.5233

So this is not stable; this is very reactive; that makes it a very poor leaving group.5248

Which reaction is the favored reaction?--which is the reaction that has chloride, the good leaving group, leaving and getting replaced by a nucleophile?5252

It is the reverse reaction that has the leaving group getting displaced by the nucleophile.5263

This is a great leaving group on the carbon; where this is not a leaving group on the forward reaction; it is no good.5273

In this case, the reverse reaction has the better leaving group; and therefore it is favored.5283

Is that something that would make sense?--would we rather have this combination of products?--yes, if this is more stable, if this is the weaker base.5300

We are always going to be favoring more stable products of a favorable reaction; so that is what we are going to be looking for.5307

We could say that the best substitution reaction is the reverse reaction that has the ΔH that is less than 0, the ΔG that is less than 0, when you have the stable products.5316

For the reverse reaction... of course the products for the reverse reaction are drawn here on the left... that is going to be the one that is going to be more favorable.5336

Because this is an equilibrium, when we describe this reaction, when we say that the reverse reaction is favored, remember our K_eq expression?5346

K_eq is always the concentration of products over the concentration of the reactants; so how would describe the K_eq for the given equilibrium?5359

If the reverse reaction is favored, the starting materials, these reactants, are the ones we are going to have in the highest concentration; these are the most stable.5373

That makes this the higher number; so K_eq is going to be less than 1 for this substitution reaction.5382

So anytime we have an alkyl halide, that is typically going to be very good substitution and very favorable because it gives a very stable halide leaving group.5390

We just showed how hydroxide is a poor leaving group because it is so unstable; it is such a strong base; but there are some strategies that will make it better.5401

It is possible to do a substitution reaction starting with an alcohol, starting with an OH on a compound; and there is two methods for this.5410

The first method is you can add a strong acid; so HCl, HBr, HI--those are all very strong acids; these work well in this situation.5420

What happens is when you add an acid, it protonates the alcohol and then a reaction takes place.5432

For example, let's a look at the reaction we just saw was this one; we said what happens if methanol reacts with sodium chloride?5441

This is the one where we said the reverse substitution would be the favorable one.5448

So here we would say that there is no reaction here; hydroxide is too poor of a leaving group; chloride could never displace it.5452

Either Sn1 or Sn2, it doesn't matter which mechanism we are looking at here because now we are just looking at the energy of the starting materials to products.5460

It doesn't matter the mechanism we take to get there; however instead of using sodium chloride which means you just have Cl^-, if instead we use HCl.5467

The difference here is HCl has both the acid part and it has Cl, then you do get a substitution reaction take place; you can replace the OH with the Cl.5478

How does that work?--let's see how HCl could affect the substitution on an alcohol.5495

What happens is as usual every time we see HCl, HBr, HI, H₂SO₄--all those guys, those strong acids; we know they are strong acids.5503

That is the very first they are going to do is they are going to look around for a base; they are going to look around for someone to donate a proton to; that in fact is the first step that happens.5511

Our mechanism for that proton transfer is the same every time; we start at the base; we grab that proton; we leave the conjugate; and so we are going to protonate the alcohol.5522

What does that do for us... that also gave us a Cl^-; let's show that because we are going to be using it.5545

What does that do for us?--why does protonating the alcohol make a difference?5552

Because now... let's take a look at what is attached to that carbon now; instead of just an OH, what do you have?--you have a positively charged H₂O.5556

When this leaves, what leaving group would we be kicking out?--we'd be kicking out water; can you get more stable than water?--that is an excellent leaving group.5566

By protonating, we make a great leaving group; and once you have a great leaving group, a substitution can take place.5576

Is it going to be... let's think about that mechanism for a minute; it could maybe be an Sn2; it could be an Sn1; what do you think is going to be favored in this case?5584

Sn2 means backside attack; is this a good carbon for doing backside attack?--great carbon; very little steric hindrance.5594

How about Sn1?--do you think the water will just leave on its own to give a carbocation?--does that look favorable in this case?5604

No way, not for a methyl group; so it looks like this is going to be an Sn2 mechanism to do the substitution.5611

That means backside attack, single-step; attack the carbon, kick off the leaving group; and we get our substitution reaction.5619

The other product we are forming is water; so why was this substitution allowed to occur?--why is this overall a favorable reaction?5629

Because water is incredibly stable; it is a very weak base; and that is what makes it a good leaving group; so that would be a favorable reaction.5639

That is one way to make an OH a good leaving group; let's take a look at another strategy; another strategy is making what is known as a tosylate.5653

What we are going to do is we are going to react the alcohol with TsCl; that is called tosyl chloride--it is an abbreviation; we will look at these structures in just a second.5664

If you take tosyl chloride and base, we replace the proton on the oxygen with a tosyl group--we just abbreviate Ts; and this is called a tosylate... this is called a tosylate.5677

Again just like a halide, this OTs group is a great leaving group; so once you have converted an alcohol to the tosylate, you can now react this with any nucleophile you want.5696

It can either do an Sn1 or an Sn2 depending on your carbon chain that you are dealing with; and you can get a substitution to take place.5712

The product you are kicking out is TsO^-; and this is a great leaving group.5724

Let's take a look at what this tosylate looks like; the Ts abbreviation stands for this part of the molecule; so tosyl stand for para-toluene.5735

Toluene is when you have a methyl substituted benzene ring like this; so it is para-toluenesulfonyl--is what the tosyl is short for.5747

If you just have a chlorine here, that chlorine can get displaced by the alcohol group.5757

That mechanism looks like it might be Sn2; but it is not as a simple mechanism like that so I am not going to talk about the mechanism for that substitution.5763

What happens is you attach this tosyl group onto your oxygen; then that now with the oxygen is what makes TsO the leaving group.5775

This is the same as this structure; we have a CH₃OTs; we have CH₃OTs.5790

What would happen if this saw a nucleophile?--well, we think of this big whole other thing as our leaving group.5797

Let's say we want to do an Sn2; it attacks the carbon, kicks off the leaving group; there it is; so that gives me my substitution product.5805

What does the tosylate, TsO^- look like?--well, we have our tosyl group, SO, double bond, O^-.5815

This is our leaving group; why is this more stable?--this is a leaving group so it must be pretty stable; why is this more stable than hydroxide?5830

We said hydroxide was a lousy leaving group because it was such a strong base; here is another O^-; why is this one more stable?--can you see what the difference is?5840

Is there any chance for resonance stabilization here?--absolutely; this O^-, this lone pair, is allylic.5853

It can be distributed up to this top oxygen, this bottom oxygen; in fact all three oxygens share it equally; so it is very delocalized; it is highly stabilized by resonance.5861

Tosylates are really a great strategy for taking an alcohol... alcohols make very good starting materials; we have a lot of those commercially available.5872

These make very good starting materials; so it would be very useful to be able to reliably turn it into a good leaving group; and tosyl chloride is one example of an application of that.5883

Let's try a synthesis problem; let's do some organic synthesis; let's say we wanted to synthesize this molecule starting with this alcohol.5896

Our challenge is provide the reagents necessary to transform the given starting material into desired product; these problems are described as transform reactions.5909

Really we are just asking for the reagents needed; very often more than one step is going to be needed to do this.5921

I want to get rid of the OH; I want to put in an iodide; so how about I just use sodium iodide.5928

Why can't I just use I^- and do an Sn2 reaction?--have the I^- just kick out the leaving group?--what is the problem with that?5935

Who would your leaving group be in that case?--the leaving group would be hydroxide; is hydroxide a good leaving group?--no way.5945

Our challenge here is that this is not a leaving group; so if we want to do a substitution, we need to remedy that.5953

One strategy we can have for that is instead of using sodium iodide, we could use hydrogen iodide, HI.5962

What is the difference in that case?--now this is a strong acid in addition to being a source of the iodide nucleophile.5973

We know we are going to protonate first and make a good leaving group that could then be displaced by the iodide.5980

Just a choice of different reagents in this case will enable the substitution reaction to take place.5991

What was the other strategy we saw for making the OH a good leaving group?--the other strategy was reaction that converts the OH to an OTs.5999

If I had an OTs there instead, then that would also be a good leaving group and allow for a substitution; so what were the reaction conditions here?6011

We know we need this Ts group; and we need something on the Ts group that can be replaced with the oxygen; it is called tosyl chloride, TsCl.6020

You will notice in this reaction we are going to lose that chloride; we also need something to deprotonate that proton.6031

We are going to lose HCl so we need some kind of base in this reaction--usually pyridine or very often pyridine.6036

It is very common to see the reaction conditions as tosyl chloride pyridine; you might get used to working with that.6044

Either way, once I have my tosylate, now it would be simple to convert that tosylate to the iodide.6051

We have our leaving group; all we need is the appropriate nucleophile; so now we can come in with sodium iodide and expect the Sn2 to take place to displace our good leaving group.6058

Finally let's talk about what it takes to be a good nucleophile; we've been using nucleophiles in both the Sn1 and the Sn2 mechanisms to replace our leaving groups.6072

Let's just talk about a few factors that we've really already come across.6082

One thing, remember a definition of a nucleophile back at the very beginning?--the nucleophile is something that is electron rich seeking an electron deficient species.6087

The more electron rich you are, the better the nucleophile; so let's compare here we have this is called methanol; this alcohol with the methyl group here--this is methanol.6097

Here we have a protonated form of methanol so there is a positive charge; here is a deprotonated form of methanol so there is a negative charge.6111

We call this methoxide; when you have a methyl and an O^-, it is called methoxide.6117

The more electron rich you are, the better the nucleophile you are; so methoxide is described as a strong nucleophile.6126

Methanol, because it is neutral, it is stable, still can be a nucleophile; it still has lone pairs; but this is described as a weak nucleophile.6138

How about this species?--it still has a lone pair.6148

But because it already has a positive charge, there is no way it is going to want to share those electrons with something else and have a double positive charge.6152

We would describe this as not a nucleophile; so it could be neutral or it could be negatively charged; we could even do a double arrow there because there is such a huge difference there.6158

What kind of mechanism, of the substitution reactions, what kind of mechanism do you think would be suitable for something like methoxide?6172

If we have a strong nucleophile, what reaction?--maybe backside attack?--I think backside attack; so this is good for the Sn2; it would love to do the Sn2.6179

But methanol now is neutral; it is stable; it is not a good nucleophile; so it wouldn't do the Sn2; so this something that would be good for an Sn1 reaction.6190

If I had a carbocation, then methanol would be a good enough nucleophile to attack that.6199

We already saw a few examples where we made this distinction between them; and that will be something that we will do in the future as well.6204

So really most good nucleophiles, if you take a look at nucleophiles, like we've seen RO^- and HO^-; we saw cyanide is a good nucleophile; N₃^-.6212

We will see a lot of the nucleophiles; we will see X^- so we could have a halide being a nucleophile; most of them have negative charges.6226

The exception to this are the amines, R₃N, and the phosphines, R₃P; if you take a look at the periodic table, phosphorus is right there underneath nitrogen.6236

These guys, even though they are neutral, they are strong nucleophiles; and they are great for doing the Sn2.6251

We saw an example of that when we had ammonia; it did an Sn2 even though it ended up having a positive charge in the product; that was just fine.6262

That is because these are pretty electropositive; they are very happy sharing their electrons.6269

When you see an amine or a phosphine, that is something that you should know--that an Sn2 is quite possible.6278

Finally how about some periodic trends for nucleophilicity?--if we move across a row, we will see that we decrease in our nucleophilicity.6287

Ammonia... we just saw actually; we said ammonia is a very good nucleophile; it loves to do an Sn2; but water is a pretty weak nucleophile and wouldn't do an Sn2.6298

We've seen an example of that trend; and that is going to be a general trend--ammonia is described as a strong nucleophile and water is described as a weak nucleophile.6309

What is the difference there?--what is changing as you move across the row on the periodic table?--it is the electronegativity as we move closer to fluorine.6322

Because oxygen is more electronegative... oxygen is more electronegative than nitrogen; what does it mean to be electronegative?--it means you pull electron density toward yourself.6331

That means oxygen is holding onto its lone pair of electrons very tightly... oxygen holds onto lone pair tightly; not willing to share them.6349

That means the nitrogen is a better nucleophile; it is more willing to share those electrons and form a bond with an electrophile.6364

Once again just a little reminder; this is one of those facts of organic chemistry that is going to pay off knowing down the road.6375

It is that nitrogen is a great nucleophile even without a negative charge.6382

In fact when you put a negative charge on nitrogen, it is too unstable to handle a negative charge; and it doesn't want to be a nucleophile then; it will end up being a strong base instead.6390

For it to be a great nucleophile, we just have a neutral nitrogen and that would be excellent.6402

The last trend in nucleophilicity is looking down a family; and this one actually is not so clear cut; it can depend on the solvent; there are things you can do to change this trend.6408

But one thing that is very common is that as you move down a family, you will see an increase in nucleophilicity; remember we just talked about what happens as you move down a family--you increase in size.6418

What we say here is that I^- is larger; and as a result, it is more polarizable; it is more polarizable.6431

To be polarizable means that it has this huge cloud of electron density; it can stretch that electron density, reach it out toward an electrophile, start to form a bond.6446

That is going to make it a faster reaction, a better reaction; and so therefore iodide is the better nucleophile; large nucleophiles are good nucleophiles.6456

Again that explains too that nitrogen and the phosphorus; the phosphorus is bigger so phosphorus is an excellent nucleophile; it loves to do the Sn2 reaction.6469

That wraps it up for substitution reactions; hope to see you back at Educator soon; thank you.6479