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Ch. 20 - Carboxylic Acid Derivatives: NAS 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
Carboxylic Acid Derivatives
Naming Carboxylic Acids
Diacid Nomenclature
Naming Esters
Naming Nitriles
Acid Chloride Nomenclature
Naming Anhydrides
Naming Amides
Nucleophilic Acyl Substitution
Carboxylic Acid to Acid Chloride
Fischer Esterification
Acid-Catalyzed Ester Hydrolysis
Lactones, Lactams and Cyclization Reactions
Decarboxylation Mechanism
Additional Guides
Carboxylic Acid

Base-catalyzed Transesterification occurs when an ester is exposed to an alkoxide base with a dissimilar alkyl group. Now, do we think this a good thing? Stay tuned for the answer!  

Concept #1: General Reaction


On this page, we’re going to discuss a reaction called transesterification.
Base-catalyzed transesterification occurs when an ester is exposed to an alkoxide base with a dissimilar alkyl group. If I’m using an alkoxide base, let’s say that I was in a basic environment because it’s base-catalyzed. Imagine that my alcohol instead of having an R group, the R group that I have on my alkyl group, I have R prime. R prime just means different. It's something different, maybe one is a methyl and one is an ethyl. What’s going to wind up happening is after it reacts, we're going to wind up getting another ester. It seems like nothing changed but wait, the R group is going to be different. I’m going to get one R group transferring with another or substituting with another. This happens because in equilibrium, all these OR groups are going to be constantly substituting back and forth and back and forth.
If you have different alkyl groups in different positions, they’re going to wind up blending together and you’re going to wind up getting OR groups of both types on your ester. That’s very problematic because when you have an ester, you want to make sure that it has all the same alkyl group. You don't want a bunch of different random alkyl groups. There is one way this can be avoided. The way this can be avoided is simply to only expose esters to alkoxides with the same R group. Imagine that now I have let's say R1 and I’m exposing it to an alkoxide that is OR1 negative. What's going to happen? In equilibrium, it's going to switch and it’s going to do this whole mechanism that I’m going to show you. But are we going to be able to actually tell that it's happening? The answer is no reaction in terms of ‘Does it have any effect on my product?’ No, because I am going to get substitutions taking place. But since my R groups are identical, I won't be able to notice this difference. I won’t be able to notice that a reaction is actually taking place because it's not really mattering for my reaction. The only time it would matter is if I have a different R group. Let’s say it was R1 and R2, now I’m going to get a mixture of R groups transesterifying. In this next video, I’m going to show you guys the mechanism and heads up, it’s easy. 

Concept #2: General Mechanism


Alright, so let's say that I do mess up and I have my methyl ester, okay? And I forget that it's a methyl ester and I expose it to methoxide. So, O, CH2, CH3 negative, okay? So, I wasn't supposed to do that I should have only used methoxide here, because I know methoxide would avoid this situation but let's say I use methoxide by mistake? Well, this is what happens, you wind up getting a nucleophilic acyl substitution, you wind up forming a tetrahedral intermediate and that's going to have now OCH3, OCH2CH3 and now what happens? Well, now which group gets kicked out, in my Na s reaction it could be either one, the point here is that it's going to be a mixture of both. So, I'm going to wind up getting two different esters, I'm going to wind up getting the ester that forms, when it kicks out the methyl, right? So, that would be, it's going to wind up being an ethyl ester, right? But I'm also going to get the one that forms, if it kicks out the original molecule, okay? So, it would be a methyl ester, okay? So anyway, you should never draw multiple arrows like that on the same reagent, okay? So, that's kind of me just kind of showing you conceptually what's going on, let's just say that you are going forwards you are transesterification. So, you're kicking out the OCH3? Well, then you're going to get this as a product, you're going to wind up getting an ethyl ester, okay? As a product. So again, this is problematic guys and it's going to be important, later on there's going to be more reactions that we're going to use esters for and so it's going to be important to always use an alkoxide of the same base of the same R group, because if I can keep those R groups consistent then transesterification doesn't matters to me anymore, because even if the reaction is taking place I can't appreciate it because nothing is actually happening, okay? So, I hope that made sense guys, really easy mechanism, let's move on to the next video.