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Ch. 21 - Enolate Chemistry: Reactions at the Alpha-Carbon WorksheetSee all chapters
All Chapters
Ch. 1 - A Review of General Chemistry
Ch. 2 - Molecular Representations
Ch. 3 - Acids and Bases
Ch. 4 - Alkanes and Cycloalkanes
Ch. 5 - Chirality
Ch. 6 - Thermodynamics and Kinetics
Ch. 7 - Substitution Reactions
Ch. 8 - Elimination Reactions
Ch. 9 - Alkenes and Alkynes
Ch. 10 - Addition Reactions
Ch. 11 - Radical Reactions
Ch. 12 - Alcohols, Ethers, Epoxides and Thiols
Ch. 13 - Alcohols and Carbonyl Compounds
Ch. 14 - Synthetic Techniques
Ch. 15 - Analytical Techniques: IR, NMR, Mass Spect
Ch. 16 - Conjugated Systems
Ch. 17 - Aromaticity
Ch. 18 - Reactions of Aromatics: EAS and Beyond
Ch. 19 - Aldehydes and Ketones: Nucleophilic Addition
Ch. 20 - Carboxylic Acid Derivatives: NAS
Ch. 21 - Enolate Chemistry: Reactions at the Alpha-Carbon
Ch. 22 - Condensation Chemistry
Ch. 23 - Amines
Ch. 24 - Carbohydrates
Ch. 25 - Phenols
Ch. 26 - Amino Acids, Peptides, and Proteins
Ch. 26 - Transition Metals
Tautomers of Dicarbonyl Compounds
Acid-Catalyzed Alpha-Halogentation
Base-Catalyzed Alpha-Halogentation
Haloform Reaction
Hell-Volhard-Zelinski Reaction
Overview of Alpha-Alkylations and Acylations
Enolate Alkylation and Acylation
Enamine Alkylation and Acylation
Beta-Dicarbonyl Synthesis Pathway
Acetoacetic Ester Synthesis
Malonic Ester Synthesis

Concept #1: Haloform Reaction


Alright guys. So the haloform reaction is simply a base catalyzed alpha halogenation of methyl ketones, okay? So, it's the same thing as base catalyzed alpha-halogenation except that you specifically have a methyl ketone, let me show you the mechanism. So, guys the reagents are just the same as we were doing this before the only difference being that I'm starting with at least one methyl group on one side, okay? What's going to happen is after the formation of your enolate and attacking the x's you're going to get a polyhalogenated alpha carbon, okay? Make sense so far, but here's the deal, if you have a methyl ketone you've now made an awesome leaving group because you have cx-3 so the next equivalent of base that reacts instead of just sitting around and being like, hey I've got nothing else to halogenate, I'm done, the next equivalent can actually go through an NAS mechanism and it can form a tetrahedral intermediate. So, what you would get is a tetrahedral intermediate that looks like this, O negative, OH, R, C, X, X, X, and guys carbon is usually not a good leaving group but a carbon with three halogens is an extremely good leaving group. So, then I would kick out the cx-3, okay? And, what that's going to give me is we just draw over here that's going to give me carboxylic acid crazy, right? So, that's going to give me carboxylic acid plus it's going to give me cx-3 negative, okay?

Now, I know my head's in the way a little bit but then to end this reaction, the cx-3 negative is going to deprotonate the acid because it's an acid, right? And you're going to wind up getting a carboxylate and the molecules that direction is named for haloform, haloform is just basically a methyl group with three hydrogen's replaced with halogens, okay? So, the entire idea behind haloform is that the alpha carbon of this reaction is transformed into a good leaving group through successive halogenations and eventually the OH negative just kicks it out entirely, which is something that would not happen in a base catalyzed mechanism without it, because if a base were to attack here, it would have nowhere to go because there's nothing to kick out, it doesn't have a good leaving group, but on this reaction it does have a weak good leaving group, which is why you get the final products, okay? Now guys, just extra facts, haloform depends on what you're using, if it was chlorines going to be chloroform, right? If it's iodine going to be iodoform and some of these end up being tests that you have to use in your chemistry lab to test for methyl ketones. So, a lot of times iodoform will precipitate out of the solution and it will be a, it's like a yellow precipitate, and it's a test to see if you had a methyl ketone because if you have that precipitate that means that you formed iodoform, which means that you had a methyl group on their ketone, interesting, right? Cool Guys. So, let's move on to the next reaction.

Practice: Provide the major product for the following reaction.