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Lecture Comments (2)

1 answer

Last reply by: Professor Starkey
Fri Feb 3, 2017 10:03 PM

Post by Selma Cerimagic on January 26, 2017

Hello, I am wondering about practice problem 4 instead of making an alkoxide and an alkyl halide, could you do a oxymercuration directly and convert the alkene to an ether by using the reagent 1)Hg(OAc)2 / HOCH3 and 2) NaBH4. Would there be anything wrong by doing that?  

Transformation Practice Problems

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


Transformation Practice Problems

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

  • Intro 0:00
  • Practice Problems 0:33
    • Practice Problem 1: Transform
    • Practice Problem 2: Transform
  • Practice Problems 7:49
    • Practice Problem 3: Transform
  • Practice Problems 15:32
    • Practice Problem 4: Transform
    • Practice Problem 5: Transform
  • Practice Problems 24:08
    • Practice Problem 6: Transform
    • Practice Problem 7: Transform
  • Practice Problems 33:08
    • Practice Problem 8: Transform
    • Practice Problem 9: Transform

Transcription: Transformation Practice Problems

Hi, welcome back to

We have seen a variety of functional groups up to this point--alkenes and alkynes, alcohols, ethers.0002

Let's take a moment and do some review problems where we can bring them all together and tackle some problem solving strategies.0009

We are going to focus on the transform type problems, the synthesis problems, because these are the ones you typically don't have too much experience with.0016

They are very challenging and are a step above the more simple predict-the-product type problems which you probably do get a lot of examples to work through.0024

Here is our first one; let's say we wanted to start with this alcohol and go to a Grignard reagent.0034

The approach we are going to take for all these transform problems is we are going to look at our product and analyze that product--do a retrosynthesis.0040

Every time we do this special retrosynthesis arrow, it asks what starting materials do I need?--in other words, we need to imagine the reaction that can be used to form the product given.0049

What that requires is a functional group analysis; we take a look, we see that this is a Grignard reagent.0065

Then we have to go back and ask we made Grignard reagents before; how did we do that?--where did that come from?--we added magnesium metal.0070

What was there in the first place?--the halogen was there in the first place; in other words, you need to start with a halide to form a Grignard reagent.0081

By working backwards a little bit, we can come up with a structure that hopefully we can get to in the forward direction from the starting material that we were actually given.0095

In this case, we were given an alcohol; what we would need to do is convert this alcohol to an alkyl halide; then we would convert the halide to the Grignard.0104

A lot of these transform type problems, multiple step transform type problems, all come down to finding this key intermediate structure.0115

One or two structures that you can both get to from the starting material and go from to the product.0123

It helps by working backwards a little bit, working forwards a little bit, and hopefully meeting somewhere in between.0128

Both of these are transformations we have seen before; we have seen that we can go from an alcohol to an alkyl halide; actually there are quite a few possibilities for that.0134

One possibility would be something like HBr; but the problem with a strongly acidic HBr reagent is it very much favors carbocation formation.0144

Because this is on a primary carbon, and right next to it we have this benzylic carbon, it is far too easy for rearrangement to occur and for the bromine to end up in the wrong position.0154

That is probably not the best strategy; instead we want a strategy that is going to not have a chance of rearrangement.0166

Maybe a reagent like PBr3 would be better because that would make the bromide exactly where the OH used to be; there is no chance of rearrangement there.0174

Maybe also to count on an Sn2 reaction, you could do the tosylate; you could make tosyl chloride pyridine to make the tosylate, make the OH a good leaving group.0187

Then step two, you can add in sodium bromide to do an Sn2 on that good leaving group.0197

That would be another strategy where there is no chance of rearranging and the bromine exactly replaces the OH wherever it used to be; either of those reagents are fine.0202

What you will find is many of these transform type reactions, there are multiple possibilities for a correct solution either sometimes in strategy but most definitely in the reagents that you choose.0211

Then how do we go from the alkyl halide to the Grignard?--we simply add in magnesium metal.0223

Sometimes you will see this with the solvents as well; but just the magnesium, that is definitely the most important part.0227

How about our next one?--we need to transform; we are starting with an alcohol; we are going to an ether.0238

We also notice that there is some stereochemistry that will be relevant; our oxygen was on a wedged bond; this oxygen is also on a wedged bond.0247

We are going to need to take that into consideration when we come up with our strategy; but let's consider our retrosynthesis and ask what starting materials do we need?0256

We consider our functional group that we have in our target molecule is an ether; how do we make an ether?--we need two ingredients; we need an alkoxide and an alkyl halide.0265

We need to do a disconnection; we can either do a disconnection on this side of the oxygen or this side of the oxygen.0278

If we were to form this carbon-oxygen bond as our new bond, that is definitely going to have an effect on the stereochemistry of that bond.0286

Since the stereochemistry is set and we want to keep it fixed, let's avoid that bond; let's look at this disconnection instead.0297

If fact this is the even better disconnection because this leads to a carbon that is less sterically hindered.0305

We are going to break this bond; the oxygen was our nucleophile; the carbon was our electrophile; that is how we come back to our alkoxide nucleophile and an alkyl halide.0312

The way we make this carbon electrophilic is we add on a halogen, fluoride, chloride, bromide, iodide, your choice... I'm sorry not fluoride; chloride, bromide, iodide is your choice.0332

Alkyl halide would be our electrophile; when we combine an alkyl halide and an alkoxide, we call that the Williamson ether synthesis.0343

That is going to be a very good way to synthesize ethers; not the only way, but one of the ways that you should probably consider first.0354

Because this now is a strong nucleophile, we expect it to do an Sn2; that will definitely give us our target molecule.0364

This is what we need; we need this alkoxide and this alkyl halide; we come back to what we have; we have this alcohol; how do I convert an alcohol to an alkoxide?0371

It looks like I just have to deprotonate that hydrogen; we have to deprotonate the alcohol; what kind of reagent will do a deprotonation?--a base, a strong base.0383

How about sodium hydroxide as a strong base?--would that completely deprotonate my alcohol to give an alkoxide?--no, that is not a strong enough base.0393

That is too similar in reactivity to the product you are trying to form; we learned about two other reagents that would be better for this.0403

We could use either sodium hydride which would be an irreversible reaction that gives off the hydrogen gas.0410

Or we could use sodium metal, Na0; that would do a redox reaction to do that same reaction; either of those reagents is fine.0416

Just pick one and go with it; once we have our alkoxide, we would then throw in our ethyl iodide or iodoethane and do our Sn2.0424

At this point, you make sure you ask yourself is this a good Sn2?--would that work?--we have a primary alkyl halide; that is the perfect kind of Sn2.0436

Remember this is a very strong base; so we can't do an Sn2 if it is secondary or tertiary; we want to make sure of that; that is part of the decision we make here.0447

If both of these disconnections were possible, then we would use the steric hindrance of the potential Sn2 as the way to guide which one is the better disconnection.0459

How about this one?--we are starting with an alkyne; both of our double bonds are gone; we now have a diol; we have two OHs in our structure.0472

We also have a particular stereochemistry shown where this is a wedge and this is a wedge; when we do a retrosynthesis on this, we ask what starting materials do I need to give this product?0483

Have we ever seen a reaction that gives as a product a structure with an OH on each carbon?--what starting material did you have for that?0496

You started with an alkene; if you have an alkene, we could do a reaction that made a diol; do you remember this one?--what kind of reagent did we do for that?0509

That was an oxidation; that was KMnO4; just a little reminder here of the problem solving we need to think about.0520

We need to remember the reaction that we saw, what reagent was it, and more importantly what functional group did it start with?--so we could do our retrosynthesis.0528

Tell me about the stereochemistry of that dihydroxylation; remember the KMnO4 or the osmium tetroxide delivered both oxygens from the same face; we call that syn addition.0535

Before we plan what starting material we need, how about rotating this around so that we see it with the OH group cis to each other, coming from the same face.0548

This right-hand carbon, we haven't rotated; but this left-hand carbon, we rotated the OH up 180 which takes the ethyl group that was pointing towards you and moves it to be pointing away from you.0565

This is a better way to view that molecule because now you see that if you wanted to do syn dihydroxylation, you would get this product; what starting material would it have come from?0580

What is the stereochemistry of the alkene that you would need initially?--you see how the methyl and ethyl group are trans to each other?--that is the stereochemistry we need in our starting material.0591

You can verify this as doing like a predict-the-product; if I had this alkene, would it give this product in dihydroxylation?--sure, the OHs come in from the same face.0606

The trans relationship of the ethyl and methyl are seen in the product; that helps us get the stereochemistry right; how do we go from here to here?0617

How do we go from an alkyne to a trans alkene?--have we seen that transformation before?--it is a partial reduction of the alkyne; we call this dissolving metal reduction.0626

If we use sodium metal and NH3, liquid ammonia, we do a dissolving metal reduction; we go from the alkene to the trans alkene.0637

Once we have the trans alkene, how do we go to the diol target molecule?--we add in our KMnO4.0646

This not only changes the functional groups that we want, but it also gives us the proper stereochemistry.0658

One other thing to consider, another possibility is have we ever seen a reaction that adds two OH groups to the opposite face?--we have.0667

What if we did it via an epoxide intermediate structure?--if we had an epoxide, one oxygen would come from the top face; the new oxygen would come from the bottom face.0678

Let me redraw this so we can see it again; what if we wanted to do an anti dihydroxylation, what alkene would we need to get the stereochemistry that is shown here?0693

Now these are already drawn in the anti relationship, the trans relationship; this has the ethyl and the methyl on the same side.0714

If we had them cis to each other and then we did an anti dihydroxylation, that would successfully give us our target molecule.0727

Another possibility to get this same target molecule is to reduce this, not to the trans alkene, but to the cis alkene.0737

How do we do that?--dissolving metal reduction gives the trans; to get the cis, we need to do catalytic hydrogenation.0750

But we can't just do plain catalytic hydrogenation because that would get rid of all our carbon-carbon π bonds.0758

Instead we use Lindlar's catalyst; that is our poison catalyst that stops at the cis alkene; we know how to get the cis alkene.0764

How do we then go to the target molecule with the proper stereochemistry?--how do we do our anti dihydroxylation?0774

We want to make an epoxide, then open up the epoxide; this can actually all be done in one pot if we use peracetic acid as our epoxidation reagent and we do this in water.0788

This makes the epoxide; and this opens the epoxide; so this is a one-step process for doing anti dihydroxylation where you add two OHs and they come from the opposite face.0806

Syn hydroxylation, because it is a concerted mechanism with KMnO4; anti dihydroxylation because we are forming the epoxide and then we are opening it up.0825

If you don't remember that we can do this all in one step, you can also do it stepwise we could say step one, let's make the epoxide; what oxidizing agent does an epoxidation?0833

We need some kind of peroxy group like mCPBA; then we can follow that up; after we make the epoxide, then we can... here I can draw this out so you can follow along.0845

We could react--this is mCPBA which should make the epoxide with the two groups syn to each other, cis to each other.0861

Then we can open that up with water; we could either do that with basic conditions or acidic conditions.0874

Let's say NaOH, water for example; that would open up the epoxide ring with hydroxide; we would get our two OHs coming in opposite directions.0882

Remember this is racemic; I have drawn just one enantiomer here; but both enantiomers would be formed; when we open this up, we would get both enantiomers as a possibility.0893

So we could either do one step with the peracetic acid or two steps with mCPBA followed by NaOH, H2O.0902

A few different ways to get this particular stereochemistry; these both options are possible because we know how to go from the alkyne to either the cis or trans alkene.0911

We also know how to add two OHs either with a syn mechanism or with an anti mechanism; that is why we get two possible roots here.0921

Here we have another ether as our target molecule; when we think about how to make an ether, we need to do a disconnection from one side or the other of our oxygen bond.0935

When we are doing a Williamson ether synthesis, our oxygen is going to be our nucleophile always as an alkoxide.0955

But when we consider our possible electrophiles, we want to have an electrophile that has the least steric hindrance possible.0965

This electrophile is a better one because it is just a methyl group; we will do that disconnection instead; which means we need this alkoxide plus this alkyl halide.0974

That is our two ingredients for the Williamson ether synthesis; RO- is our nucleophile; RX is our electrophile.0994

Just very briefly, why didn't I make that other choice?--the other possibility of having chloride, bromide, iodide, whatever you have here and methoxide?1002

If you wanted to do this Williamson ether synthesis, because this is a secondary alkyl halide and because this is a strong base, we are not going to get the Sn2 as the major product here.1014

These are going to combine to do the E2 elimination; we are going to get the alkene product out; guess what?--all that hard work, we would go back to our starting material.1025

We don't want to make the alkene; we want to make the ether; this is the best Sn2 on the methyl iodide; we need this alkoxide.1033

Where does an alkoxide come from?--once again I am going to ask what starting materials do I need to make this alkoxide?1042

Where have we seen alkoxides come from?--instead of where we had an O-, we needed an OH; if we had an OH there, then we could convert it to the alkoxide.1049

I think our plan is coming here where we have done our planning; we thought things through; we have an alkene; now we need to get to this alcohol.1065

That looks like we added an OH; we also added an H here, just so you can see the difference; we need to add an H and an OH to our alkene; that looks like a hydration reaction.1077

How about using H3O+?--would that give the proper regiochemistry?--how would you describe the regiochemistry?1091

We want to add water, just a little note here; our regiochemistry, our hydrogen goes to the carbon with more hydrogens; we get Markovnikov regiochemistry.1101

H3O+ would do that; but what mechanism do you have with H3O+?--we protonate the alkene; we get a carbocation; that carbocation is free to move.1114

It is not going to have a big driving force to move because we would go from a secondary to another secondary; but it certainly can move and will move; we would probably get a mixture of alcohols.1123

A better reagent for adding water with Markovnikov regiochemistry, the better strategy in the lab if there is a possibility of getting more than one alcohol product, is to do oxymercuration-reduction.1133

Those reaction conditions are as follows--mercuric acetate, HgOAc, taken twice in water; we do our oxymercuration; we add mercury and oxygen in that first step.1146

Then step two, we do some kind of reduction, demercuration; sodium borohydride does that, replaces the mercury with a hydrogen.1157

Oxymercuration-demercuration or oxymercuration-reduction does exactly this; it adds water; it gives it with Markovnikov regiochemistry; very good.1164

How do we convert this alcohol to an ether now?--this is similar to one we just saw; we need to deprotonate completely; we can use something like sodium hydride to deprotonate completely.1176

Sodium metal would work fine here too; again your choice to pick any reagent you want if there is more than one choice that will work.1188

We make the alkoxide; then we add in our alkyl halide; we wanted the methyl group; remember methyl iodide is the one methyl halide that is a liquid rather than a gas.1196

That is the one that is better in this case rather than using a bromide or a chloride; and we make our ether; very good.1207

Our next one is interesting; we have a four carbon chain and another four carbon chain; we haven't done anything to that; but we now have a carbonyl.1217

There have not been many reactions we have seen that give carbonyls as a product; again let's think back; we haven't really talked about synthesis of carbonyls yet.1226

We will be doing that a little later when we are talking about aldehydes and ketones; let's ask what starting materials do I need?--what reaction have we ever seen that gives a carbonyl as a product?1236

We have seen ozonolysis; that would break our carbon chain apart; we want to keep all our carbons here so that is no good; what other oxidation reactions?1251

A carbonyl is pretty highly oxidized; it has two C-O bonds; what other oxidation reactions have we seen?--how about if we had an alcohol?1259

Can an alcohol be oxidized to a carbonyl?--that was one of the oxidation reactions we studied; that would be a way to make the carbonyl.1269

That would be something we could get to from the chloride; let's try our transformation; how would we go from a chlorine to an OH?--it looks like some kind of substitution reaction.1284

We have a choice; do you want to do Sn1 mechanism, Sn2 mechanism?--no stereochemistry is shown; that is not an issue; but how about our carbon bearing our leaving group?1299

This is a primary alkyl halide; we have a primary leaving group; Sn1 is carbocation; Sn2 is backside attack; which one is better for a primary leaving group?1310

Absolutely Sn2; it is great because there is no steric hindrance; the carbocation would be awful if it was primary; it would be guaranteed to rearrange.1324

We definitely want an Sn2; that means we need some kind of strong nucleophile; we need an OH; that is a strong nucleophile; all we need is hydroxide.1332

NaOH will do an Sn2 here because it is a primary alkyl halide; we have a good yield of that; once we have the alcohol, we are going to oxidize this to the aldehyde.1341

We have seen a variety of different oxidation reactions; Jones, PCC, Swern; which of those if any would be suitable here?--could we use any one of them or do we have to be careful in this case?1356

Remember this is a primary alcohol; if we were to react it with a very strong oxidizing agent like Jones conditions, sodium dichromate, an acid, chromic acid oxidation.1369

That would oxidize this carbon not just up to the carbonyl but all the way to the carboxylic acid; that would not give us the product we want.1381

What oxidizing agent gives us just the aldehyde?--those are the other two, PCC or Swern; Swern has a whole laundry list of reagents.1391

When I am doing a synthesis problem and it is my job to pick a reagent, guess which one I pick?--PCC, because it is so much easier and so much shorter.1399

PCC stands for the reagent pyridine chlorochromate; this is something you could actually go to the stockroom and say can I have some PCC?--this two-step transformation looks pretty good.1408

The nice thing about these transforms is when you are done, every step should be a reaction you have seen before.1420

It should be something you could predict the product and verify that the transformation is a valid one; that is the frustrating thing about transforms as well.1425

Because once you see the solution, it seems so simple; you think how could I not have seen that from the very beginning?1435

But of course when we are looking at a blank piece of paper, it is very challenging; that is why it is really worthwhile to go through these examples.1442

Speaking of that, here is our blank piece of paper; what is our simple solution?--again here is our product; I want you think about that product.1450

Think about the functional group in that product; ask yourself in this retrosynthesis, what starting material do I need?1458

What reaction have I ever seen that gives this functional group as a product?--how would you describe the functional group?--it is an epoxide.1465

What reaction have we seen that gives an epoxide as a product and what functional group did we start with in that case?1474

We started with an alkene; if I had this alkene, I could oxidize it to the epoxide; this looks like I just need to make this alkene.1483

We are starting with an alcohol; we have seen that transformation before; we have seen conversion of an alcohol to an alkene.1495

For example, we have something like H2SO4 and heat; we did dehydration--is a way to lose water from an alcohol.1504

But before we think we have solved this problem and move on to the next step, let's treat it as a predict-the-product and ask will this work?1518

If you were to dehydrate this alcohol by something like H2SO4 and heat, what product would it give you?-1525

Remember Zaitsev's rule?--Zaitsev's rule states that you get the most stable alkene product possible, especially in a dehydration where you have a carbocation intermediate.1533

The carbocation can move anywhere it wants; for sure we would get the tri-substituted alkene as our major product; that is not going to work in this synthesis.1544

How can we force the elimination over in this direction, when eliminating over here would clearly give the more stable alkene?1557

If we are trying to do an elimination, remember we have two mechanisms to pick from; it can be either an E1 elimination or an E2 elimination.1567

Before you select your reagents, you need to think about which mechanism is going to work in this case or which one is better; sometimes it doesn't matter; but this is a case where it really does.1575

If we did an E1 elimination, introducing a carbocation guaranteed we are going to eliminate to the right and get the more stable alkene product; how about if we did an E2?1584

Why would an E2 elimination not give the same product, not give this tri-substituted alkene?--do you remember about the stereochemistry of an E2?1595

Remember an E2 is the one-step mechanism where you attack the β hydrogen, form the π bond, and kick out the leaving group.1605

Do you remember anything about the stereochemistry of the leaving group and the β hydrogen, what stereochemical relationship do they have to have?1611

They had to be anti to one another; this is anti elimination; if we were to do an anti elimination here, if the OH or whatever leaving group we have is a wedge, the only anti hydrogen is over here.1619

That is the only place that you can do an E2 elimination; I think we figured out how to handle our regiochemistry; how about if we just treated this with t-butoxide?1639

E2 elimination, we do t-butoxide, strong bulky base; throw in some heat; how does that work?--that doesn't work very well; what is the problem?1653

We have a strong base--grab the proton, form the π bond, kick off the... leaving group; there is no leaving group.1662

An alcohol is not a good leaving group; we can't do an E2 elimination with an alcohol; we can only do the dehydration E1 elimination.1671

If we need to do an E2, we need to have a leaving group; what we are going to do instead is we are going to convert this into a leaving group.1679

We need to keep the stereochemistry so that this is still a wedged bond; what leaving group can we turn OH into, where we would retain the stereochemistry?1690

How about making it a tosylate?--this is a perfect situation for a tosylate because it doesn't affect this carbon at all; it just adds the tosyl group onto the oxygen.1701

If I did tosyl chloride and base like pyridine, tosyl chloride pyridine makes the tosylate; now t-butoxide and heat; if you want really any base.1714

Sodium hydroxide would work here; anything would work because it is secondary; any base, even sodium hydroxide and heat.1726

This would favor the E2 elimination; it has to go in this direction to do anti elimination; how do we convert this to the epoxide?1737

What is our reagent for the epoxidation?-some kind of peroxide reagent; mCPBA again is typically a favorite because it is easy to remember and works very well; it is very commonly used.1751

That was a difficult one; lots going into that one; how about our next one?-here we have as a target molecule an alkyl halide; we have an alkyl halide.1765

We need to think when we do our retrosynthesis and ask what starting materials we could have started with?--we need to think about what reactions have we seen that give alkyl halides as products?1779

We have seen examples where we have added HCl across a π bond; that would be a way to introduce a halogen to our molecule; we could have an alkene plus HCl for example.1790

Another way to make an alkyl halide is maybe we can have an alkane plus Cl2 and hν; do free radical halogenation; that is a way we have seen to make a halide.1808

What is another way?--we saw that if you have an alcohol, we can convert that to an alkyl halide either with HCl or with SOCl2.1824

A few different choices here; we can use a double bond or even just an alkane or an alcohol; which of these syntheses would be best in the given situation?1839

Now we look to see what starting material is given; we are actually given an alkene; we want to add HCl; can we just add HCl?--would that work?--is it that simple?1849

Let's predict that product; if you predict the product, you break the π bond; you add an H and a Cl; you add one thing to each carbon; where do the hydrogen go?1861

Hydrogen goes to the carbon with more hydrogens; remember Markovnikov's rule; that won't work because we want anti-Markonikov; we need anti-Markovnikov.1869

Isaw this trick before with HBr; how about if I did HCl but in the presence of peroxides?--would that then initiate a radical mechanism and cause anti-Markovnikov addition of HCl?1881

Unfortunately no, this is for HBr only; this is for HBr only; we can't do that for the chlorination reaction.1894

Instead I think we are going to want to go through this route and say let's put an OH group here first; if I had an OH here, then I could convert that OH to a halide, to a chloride.1906

Let's fill in our reagents here; now that we have done our retrosynthesis and our forward synthesis a little bit, let's fill in our reagents.1925

We want to add water across a π bond with anti-Markovnikov regiochemistry; this is a reaction we do know how to do.1932

That is hydroboration-oxidation; BH3-THF is hydroboration; oxidation is H2O2, NaOH.1939

It adds an H and an OH; it adds it with anti-Markovnikov regiochemistry; no stereochemistry is shown here; so that is not relevant.1950

How do we from an OH to a Cl?--once again if we try to use HCl right next to this tertiary carbon center, it is very likely to get rearrangement with the carbocation intermediate.1959

Strong acids are difficult because they really favor carbocation formation; a better reagent would be SOCl2 because that directly replaces the OH with a chloride, no chance of rearrangement.1973

Just a few more; let's think of this one; we have an ether that we want to make; we have an ether; how do we make an ether?--we could do a Williamson ether synthesis.1991

We could disconnect here or we could disconnect here; I am wondering we could maybe... we could do this problem... we could disconnect in either direction.2009

Because these are both primary, they will both work; this is a reasonable possibility in either direction.2019

Then we can maybe look at both syntheses; either we have this as our alkyl halide and this as our alkoxide; or this is our alkoxide and this is our alkyl halide.2029

Both of those would be totally reasonable syntheses because they are both primary and they would both give good yields of the Williamson ether synthesis.2048

We either want to put an oxygen at this end carbon or a bromine at this end carbon; but both of these are going to be anti-Markovnikov once again.2056

We could either do that with the bromine by doing HBr and peroxides; then we can treat this with sodium ethoxide to do our Sn2; that would be a simple way to do it.2065

Or we could do anti-Markovnikov addition just like we did before of water with hydroboration-oxidation, BH3-THF, H2O2 base.2085

That is anti-Markovnikov; then to go the ether, we have to take our alcohol, we have to convert it to an alkoxide.2098

That means step one we are going to use NaH to make the alkoxide; then step two, we can put in our alkyl halide, ethyl bromide; very good.2106

Let's look at one more example here; we have another transformation; it looks like an alkyl halide.2124

We just talked about the some of the options that we have for putting a halogen in this structure; we could start with an alkane; we could start with an alkene; we could start with an alcohol.2131

But here when we look at our starting material, we see that we are starting with an epoxide; we already have an oxygen in this structure.2140

I think this is going to be one of the examples where our simplest synthesis is going to be where that alkyl halide came from having an OH in our structure initially; then we convert it to a Br.2148

Now we have this alcohol we need to make; we are starting with an epoxide; there were two carbons in the epoxide; these are the two carbons that are still here.2165

This phenyl group clearly wasn't there in the starting material; we have to come back and add in that phenyl; one of these groups was my nucleophile; one of these groups is my electrophile.2176

In order for them to come together, let's think about the carbon as it was in an epoxide; what kind of reactivity do we have for an epoxide carbon?2187

This is partially positive so it is an electrophile; this carbon was my electrophile as the epoxide; what we need here is for the phenyl to be a nucleophile.2197

How do we make a phenyl group nucleophilic?--if I had a phenyl minus for example, that would be good; that would be a good way to open up the epoxide.2214

How do we get a phenyl minus?--we use an organometallic reagent like phenyl magnesium bromide or phenyl lithium; those both work great, a Grignard or an organolithium.2224

That is how we are going to form this new carbon-carbon bond; et's do our synthesis then; we have already started the epoxide; now we will add in phenyl magnesium bromide.2235

That can't be our complete reaction though because our phenyl minus would open up the epoxide ring; we have phenyl minus opening up the epoxide ring.2247

That would form this bond, but it would give an O- here; every time we use a Grignard, it is always a two-step process.2257

Step one is the Grignard or the organolithium; then step two is H3O+; you always need this aqueous workup to protonate the O- that you form in that first step.2262

That is going to give us an OH and a phenyl; my phenyl is added to one carbon; the OH ends up on the next carbon over.2273

Then how do I go from the OH to the bromide?--I use something like PBr3 to do a substitution.2287

In each case for our transformations, sometimes you see it right away and you can move forward and do your transformation, your multistep synthesis with no problem.2294

But by in large, the best approach you are going to have is going to be looking carefully at your product, identifying the functional groups in that product.2304

Then recalling what reactions you have seen that will give that functional group as a final result.2313

In doing that retrosynthesis and working backwards a little, you can bridge that gap and find out how you can work forwards to this common structure and go forward from there.2322

But a good synthesis is really about planning; that is why this retrosynthetic technique is going to be something that really pays off down the line.2331