The Wittig Reaction

Concept: Concept: Box-Out Method and Full-Mechanism

16m
Video Transcript

In this section, I’m going to teach you possibly the most important reaction of this entire chapter. Are you listening yet? Good, because you need to. This is called the Wittig reaction and it’s definitely going to be in your test. Let's get going.
The Wittig reaction is a special way to make new carbon-carbon bonds between aldehydes and ketones. What it’s going to do is it’s going to make regiospecific alkenes. What do I mean by regiospecific? Regio just talks about locations. It just means that I can make a custom alkene with whatever R groups on it I want. That's actually pretty hopeful because most reactions that make alkenes have to do with elimination. Sometimes the R groups are kind of in different places. In this case I can pick exactly where I want those R groups to be. In general, I’m going to show you an easy way to do things. I’m also going to show you the whole mechanism of course. The easy way to think of it is that you have a reaction of a carbonyl and a molecule called an ylide. I'll define that more in a second. When a carbonyl and an ylide come together, obviously it's kind of a weird mechanism but the R groups wind up attaching to each other through an alkene and I get my regiospecific alkene as a product.
If you were to get a question in your exam about a Wittig reaction and it's not asking for a mechanism, the answer to this question would be extremely easy because all you have to do is use the box out method. This is something that is like Johnny-special, Clutch-special. What's the box out method? The box out method says if a professor gives you a ketone or an aldehyde plus an ylide, all you do is you take them and you face them towards each other so that the phosphorus from the ylide and the oxygen from the carbonyl are almost touching. Then what you do is you take your little combination there. You draw a box around the phosphorus and the oxygen. You start scratching out. You squint a little bit. Pretend like that’s actually a big alkene in the back and that's your answer. Without a mechanism, without any crazy memorization, look at that. We get our answer which would be an H and a R in this case on one side. That came from the carbonyl and then whatever R group you had on your ylide. Really straight forward. It’s one of the funniest reactions to use because it’s an easy reaction to learn.
However, we know that Organic Chemistry 2 professors are so mechanistic. They want you to know all the arrows. In case your professor is one of those, let's go through the entire mechanism. The first part is the formation of the ylide. It turns out that the ylide actually comes from an alkyl halide. That alkyl halide is going to react with a molecule called a triphenylphosphine. A triphenylphosphine has a very nucleophilic lone pair. That nucleophilic lone pair is actually going to be able to do an SN2 backside attack reaction. You would take your alkyl halide and you would use that as your leaving group and you would kick out the halogen. This is just an SN2 mechanism from your Orgo 1 glory days. What you wind up getting is now the X is gone, the phosphorus is there but now the phosphorus has a plus charge because it's not too happy with four bonds. It wants three bonds and a lone pair like nitrogen. Everyone cool so far? We don’t have an ylide yet, by the way. This is just the triphenylphosphine doing the SN2 attack.
Keep in mind everything you learned about SN2 applies here. It’s not going to work on what type of alkyl halide? You guys happen to remember this? If you’re going to learn anything in organic chemistry, know your backside attack. You can never forget SN2. It even comes up in med school, pharmacy school, what have you. You're going to need to know SN2. You’re shocked that I said that but it's true. It’s not going to work on tertiary alkyl halides because it’s too blocked up. I can’t access the backside, etc. Also, think that it's going to work fastest. It’s going to work the worst on tertiary then a little better on secondary, much better on primary and then obviously the best ono methyl. If you were asked a question of which alkyl halide would yield the fastest Wittig reaction, you would go with the one that does an SN2 reaction the fastest. That’s another type of question that you could see about which Wittig is the best. It will be the one with the least substituted alkyl halide. But I digress. Let's move on to the next part.
Now I've got my phosphorus with the positive charge and I’m going to deprotonate it. Theoretically, I could use a lot of different strong bases to do this, but for some reason there's a very popular base that’s used specifically with this reaction that you don't see a lot in most of organic chemistry. That is BuLi. What is BuLi? BuLi is just an organometallic but it's a really strong. Remember that organometallics are strong bases. It’s butyllithium. It literally looks like this. It's a four-carbon chain with a negative charge and a lithium with a positive. Remember that’s a super ionic bond, so you can just draw them as anion and cation if you want. Butyllithium, extremely strong base, is going to pull off your hydrogen and it's going to deposit a lone pair onto the carbon. What this is going to do is it’s going to give me two charges that are next to each other on the same compound. I’m going to get a positive charge on my P, a negative charge on my carbon. This is what we call an ylide. Just so you know, an ylide means that you have charges adjacent to each other. Got that so far? Charges adjacent to each other is an ylide. I’ve got my positive and my negative right next to each other.
Depending on what type of question this is, there's two resonance structures that you could use for this problem. First of all, let me just draw it. What would be the resonance structure for this? Remember that phosphorus actually has an expanded octet. It can expand to hold five bonds which is unusual. We don't usually deal with these that often. But you could use this negative charge to make a double bond. This phosphorus is actually just fine like that because now it has a double bond with three phenyl groups. It has valence of 5 because that’s five bonds, one electron for each bond. Phosphorus is happy in that arrangement.
These two different resonance structures are good for two different things. This resonance structure is the first one, the ylide, is the one that I want to use for my mechanisms. If you’re asked to draw the full mechanism this is the one you use. If you just have to determine the product, so if you’re given an alkyl halide with butyllithium, etc. and it just to predict the product, then by far definitely use this one because this one's going to work for your box out. This one you can just put next to your carbonyl, draw your box and you're done. The ylide however, that's good for the whole mechanism.
Now let's move on to the last step. The last step is I want to react this ylide with a carbonyl. How does that work? Let's just take 2-butanone. I’ve got 2-butanone and I’m going to draw it like this, 2-butanone. I've got this ylide next to it. Let me just draw that down here. What's going to happen? What we're actually going to form in this situation is a nucleophilic addition. This is a really interesting nucleophilic addition because I’ve got a negative charge and I’ve got an electrophilic carbon. This follows the same rules for nucleophilic addition that we’ve been using this whole time. The negative would attack the positive, kick up the O. I'm going to get this crazy looking tetrahedral intermediate but it is a tetrahedral intermediate. Look at it. I'm going to have O negative. For the sake of drawing the next step, I’m going to draw these R groups a little strange. I’m going to draw that now the ethyl is there and the methyl is here. is that cool with you guys? Because I need to draw everything connected. I’m just kind of shifting it over. Then I’m going to draw the new bond. This is the new bond that was created between the carbons. This is the other carbon that’s attached to a methyl and a methyl. The I’ve got my PPh3 positive. This is the ugliest tetrahedral intermediate in the world but it still is a tetrahedral intermediate because this is nucleophilic addition, my friends.
Typically, with nucleophilic addition, what would happen is that I protonate and I’d be done with it. But this is the weird part. Because I’ve got the oxygen and the phosphorus with those charges so close to each other, a new type of reaction can happen that actually basically decomposes the entire structure. By the way, this tetrahedral intermediate that I showed actually does have a specific name. You know what the name is? This might come up. This is called a betaine. What is a betaine? I’ve showed you so many weird words. Ylide means that you have charges adjacent. Betaine means that you have charges that are not adjacent. This would be a betaine because my charges are not adjacent to each other. Now you’re waiting for the next step. What’s the next step?
The next step is going to be that my O can actually attach to the phosphorus. I can do something like this. Now it's going to take care of the positive charge again. Look what I'm going to get. I’m going to get an intermediate that looks like this. OPPh3, carbon, methyl methyl, methyl, ethyl. This intermediate has another name. This intermediate, notice there's no charges so it’s not a betaine anymore. This is an oxaphosphetane. I'm pretty sure I’m spelling that right. I might be one vowel off. Spelling is not my forte, unfortunately. Orgo is. But I'm pretty sure that’s right, oxaphosphetane. It’s this four-membered intermediate.
Now we’re finally ready to be done this mechanism. What we're going to do is we’re going to take the bond between the phosphorus and the carbon and make that into a double bond here. That's going to make a double bond at the bottom but we have to break a bond. We're going to break these electrons up to the O. What this is going to give us at the very end, check out what just happened. I just made a new double bond at the bottom. What does double bond have on it? It has two methyls on one side and it has a methyl and an ethyl on the other. Why does it have a methyl and ethyl? Because I chose that ketone. I could've chosen any ketone I wanted but that's the one I wanted so that's the alkene that I get. This will make alkene product and then you also get a byproduct with the phosphorus which is really unimportant and I'm not going to even bother to name it because I’ll probably get it wrong but it's going to be the PPh3 bonded to O negative. That's going to be a positive. That's just going to kind of go away in the solution. I don't care about it. What I care about is that now I got my alkene. Isn’t that so cool? This isn’t as cool as the box out method, I’ll admit. Box out is the best because you can just put your ylide, remember which version? This version of your ylide. You put that ylide next to your carbonyl, draw a box, pretend like you can’t see what’s in the back. That’s your alkene. But if you do have to draw the full mechanism, this is everything you need to know. This is literally you’d make the ylide. You make the betaine. You do the nucleophilic addition, make the betaine, make the oxaphosphetane and you’re down to the alkane.
I’m just going to throw in one more little tidbit here which is that this reaction is not stereospecific. There’s research and there's literature on some stereospecific versions of Wittig but it's not confirmed and especially in your textbooks. Your textbooks do not talk about it being stereospecific. If there's a mixture of products, if you could make ENZ isomers, notice that in this case I would just get one product. But what if I could get multiple products? You’d have to draw both of them. You’d have to draw that you're going to get, let's say this is one of them. Let's just say that it had ethyl and methyl on both sides. Then I would get that version and I would also get this one. If you recall, if you had your large groups on the same side that’s Z, and if your large groups are on different sides, that’s E. You would have to draw both. If your professor specifically says that he wants you to draw stereospecific version, by all means go for it. But the standard Wittig reaction is not supposed to be very stereospecific. Regiospecific, yes. You can choose what's on both sides but not sterospecific. I really hope you found that video helpful. If you did, let your friends know about us, about me specifically. Let's go on and move on to the next type of topic. 

Problem: Determine the carbonyl and ylide that formed the following product. 

3m

The Wittig Reaction Additional Practice Problems

Complete the following reaction supplying the missing product and showing correct regio- and stereochemistry where applicable. If a racemic or diastereomeric mixture forms show all stereoisomers; if no reaction takes place, write N.R.

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Predict the product for the following reaction. 

 

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Provide both the reagents (in square boxes) and the product (rounded box) for this synthesis. (Note, that a second reagent may be required in the square boxes. For example, the acid step of a Grignard addition.) In the rounded rectangle, draw the structure of the compound from the first reaction.

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What is the product of the reaction sequence below?

a. 2-methyl-1-hexene

b. 2,3-dimethyl-2-pentene

c. 2-methyl-2-hexene

d. 3-methyl-1-hexene

e. none of these

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Reactions in context: Following is a Wittig reaction used in the published synthesis of a pharmaceutical called Ifetroban used to treat hypertension. Draw the product of the reaction.

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Consider the following reaction, circle the statement below that is true.

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Texas two-step: Provide both the reagents (in square boxes) and the product (rounded box) for these synthesis. In some of the steps, there is more than one correct method. (Note, that a second reagent may be required in the square boxes. For example, the acid step of a Grignard addition-please include it.)

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In the Wittig reaction of compound 1, several possible intermediates are shown. Indicate the order in which these intermediates would appear during the conversion of 1 into 2.

1) 1 → I → IV → III → 2

2) 1 → IV → II → 2

3) 1 → IV → III → I → 2

4) 1 → I → III → 2

5) 1 → I → IV → 2

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Draw the structure of the products from this reaction and circle the major. (Hint: Triphenylphosphine oxide is a by-product.) 

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In the Wittig reaction of compound  1, several possible intermediates are shown. Indicate the order in which these intermediates would appear during the conversion of 1 into 2.

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Predict the product of the following reaction. Assume an acid quench if necessary.

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Provide a complete mechanism for the following transformation.

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