When we deprotonate we generate enolates or enols. Mechanism for deprotonation: Resonance form of the anion:

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1 Lecture 5 Carbonyl Chemistry III September 26, 2013 Ketone substrates form tertiary alcohol products, and aldehyde substrates form secondary alcohol products. The second step (treatment with aqueous acid) is necessary to convert the alkoxy lithium intermediate into the desired alcohol. The Grignard reagents can be formed from treating an alkyl (or aryl) bromide with magnesium i.e. We can also generate nucleophiles from carbonyl compounds themselves the alpha carbon (carbon next to the carbonyl) is susceptible to deprotonation because of the electrophilic nature of the carbonyl itself: When we deprotonate we generate enolates or enols. Mechanism for deprotonation: Resonance form of the anion: It is most stable to put the negative charge on the oxygen (because oxygen is more electronegative), but since the enolate is almost always going to react as if the carbon is negatively charged, then for this class you can just draw the anion at that position. Another interesting carbonyl related nucleophile is called malonic acid (or malonate ester). This compound is even more susceptible to deprotonation at the alpha carbon because it has two adjacent carbonyl groups. Deprotonated malonate:

2 H H H Any of these carbonyl related nucleophiles can react with another carbonyl group that is electrophilic. We will talk about some of those reactions during the next class. First, though, I want to make it clear that enolates can act as nucleophiles and react with a whole variety of standard electrophiles. Example 1: Allyl bromide: Example 2: Epoxides: Example 3: Primary alkyl halides: Today we are going to continue our discussion from last time by talking about reactions of enolates, and in particular, the Aldol and Michael reactions. (a) Aldol reaction occurs between the enolate of one carbonyl and the electrophilic carbon of the second carbonyl.

3 The aldol reaction can be catalyzed either by acid or base. In the first step, you form either the enolate (under base conditions), or the enol (under acidic conditions). This carbon based nucleophile then attacks the carbonyl carbon on another molecule, to generate a hydroxyl group attached to the beta carbon. Under most aldol reaction conditions, the initial alcohol product will dehydrate to form the alkene. In the particular example shown below, there are three possible alkene products that can be formed: Compounds B and C are identical. Which of these alkenes are most likely to form? Compound A it turns out is almost always the dominant product. Why? 1. Loss of a second alpha hydrogen is more favorable than loss of the hydrogen from either of the terminal methyl groups. 2. The resulting double bond is more highly substituted and therefore more stable. 3. The resulting double bond is also in conjugation with the carbonyl and therefore gets stability that way. So in general, you should know that the final aldol product always has the double bond in conjugation with the carbonyl group. Let s do a real example of an aldol reaction: The overall reaction (in this case intra molecular aldol) is shown below: And we can draw either a base catalyzed or an acid catalyzed mechanism for this reaction. Base catalyzed mechanism: Acid catalyzed mechanism:

4 (b) Michael reaction Carbon nucleophiles can also react in a Michael reaction involving carbonyl compounds. Background for the Michael reaction: Let s say I have a generic nucleophile that wants to react with this alpha, beta unsaturated carbonyl compound. It has two possibilities. It can either react directly with the carbon of the carbonyl, which is called 1,2 addition (path a), or since the double bond is also in conjugation with the carbonyl, it can react here as well, which is termed 1,4 addition (path b), or conjugate addition. A Michael reaction is really just a specific type of conjugate addition that involves a carbonyl enolate adding to an alpha beta unsaturated carbonyl. The general reaction is shown below: In general, if you look at a nucleophile, how do you know if it is going to add 1,2 or 1,4 to an α,β unsaturated ketone? This is related to hard soft acid base theory which was first developed by Pearson for acids and bases, and later extended to nucleophiles and electrophiles. Hard nucleophiles/electrophiles are things that have more concentrated charge. Soft nucleophiles/ electrophiles have less, or more diffuse, negative charge. This classification is useful for comparing two different things (and deciding which is harder or softer) rather than trying to define species in absolute terms. For example: 1. R is harder than RS, because the oxygen is smaller and therefore the single negative charge is more concentrated. 2. Cu2+ is harder than Cu+ because it has more charge 3. CN is an interesting nucleophile because it can react from either end. The nitrogen end is typically harder than the carbon end. 4. In general, carbon nucleophiles are considered soft. So back to 1,2 vs. 1,4 addition consider this example electrophile:

5 Where will it react? There are two options, depending on the hard/soft nature of the nucleophile: softer nucleophile harder nucleophile soft Nu - or - Nu hard So when we think of Michael reactions, the enolate nucleophile is considered soft and so will attack at the 4 position. Here s an example of a Michael reaction: verall reaction: Mechanism: First, you deprotonate at the acidic alpha carbon. This carbon based nucleophile than attacks the a,bunsaturated ketone at the 4 position to generate an enolate that is then reprotonated to give the final product. I want to also introduce the idea of retrosynthesis, which is basically a way of looking at a compound and identifying strategic disconnections. You should have been introduced to this idea in your previous organic chemistry classes, but let s do an example. Consider the compound below: and try to retrosynthetically disconnect it. The first step is to realize that the α,β unsaturated ketone could have come from an aldol reaction where the initial alcohol product was dehydrated, so we can disconnect it back one step. Then we can actually break the bond that was formed during the aldol reaction, and say that this came from an enolate that attacked a ketone:

6 This brings us back to a diketone starting material. I will post a number of retrosynthetic practice problems on the website. You should take these problems seriously. The only way to get good at retrosynthesis is to practice. Next time we are going to continue our carbonyl discussion by talking about esters, amides, and carboxylic acids.

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