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Organic Chemistry 4e Carey | |||||
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Reactions of Esters |
Chapter 20: Carboxylic Acid Derivatives. Nucleophilic Acyl Substitution |
Interconversion Reactions of Esters
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Reaction type: Nucleophilic Acyl Substitution
Summary
Summary
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| Step 1: The hydroxide nucleophiles attacks at the electrophilic C of the ester C=O, breaking the p bond and creating the tetrahedral intermediate. |
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| Step 2: The intermediate collapses, reforming the C=O results in the loss of the leaving group the alkoxide, leading to the carboxylic acid. |
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| Step 3: An acid / base reaction. A very rapid equilibrium where the alkoxide functions as a base deprotonating the carboxylic acid (an acidic work up would allow the carboxylic acid to be obtained from the reaction). |
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Reaction under ACIDIC conditions:
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| Step 1: An acid/base reaction. Since we only have a weak nucleophile and a poor electrophile we need to activate the ester. Protonation of the ester carbonyl makes it more electrophilic. |
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| Step 2: The water O functions as the nucleophile attacking the electrophilic C in the C=O, with the electrons moving towards the oxonium ion, creating the tetrahedral intermediate. |
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| Step 3: An acid/base reaction. Deprotonate the oxygen that came from the water molecule. |
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| Step 4: An acid/base reaction. Need to make the -OCH3 leave, but need to convert it into a good leaving group first by protonation. |
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| Step 5: Use the electrons of an adjacent oxygen to help "push out" the leaving group, a neutral methanol molecule. |
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| Step 6: An acid/base reaction. Deprotonation of the oxonium ion reveals the carbonyl in the carboxylic acid product and regenerates the acid catalyst. |
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Reduction
of Esters
(review of Chapter 15)
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| Step 1: The nucleophilic H from the hydride reagent adds to the electrophilic C in the polar carbonyl group of the ester. Electrons from the C=O move to the electronegative O creating an intermediate metal alkoxide complex. |
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| Step 2: The tetrahedral intermediate collapses and displaces the alcohol portion of the ester as a leaving group, this produces a ketone as an intermediate. |
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| Step 3:
Now we are reducing an aldehyde. The nucleophilic H from the hydride reagent adds to the electrophilic C in the polar carbonyl group of the aldehyde. Electrons from the C=O move to the electronegative O creating an intermediate metal alkoxide complex. |
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| Step 4: This is the work-up step, a simple acid/base reaction. Protonation of the alkoxide oxygen creates the primary alcohol product from the intermediate complex.
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Reactions
of RLi and RMgX with Esters
(review of Chapter 14)
Summary:
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| Step 1: The nucleophilic C in the organometallic reagent adds to the electrophilic C in the polar carbonyl group of the ester. Electrons from the C=O move to the electronegative O creating an intermediate metal alkoxide complex. |
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| Step 2: The tetrahedral intermediate collapses and displaces the alcohol portion of the ester as a leaving group, this produces a ketone as an intermediate. |
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| Step 3: The nucleophilic C in the organometallic reagent adds to the electrophilic C in the polar carbonyl group of the ketone. Electrons from the C=O move to the electronegative O creating an intermediate metal alkoxide complex. |
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| Step 4: This is the work-up step, a simple acid/base reaction. Protonation of the alkoxide oxygen creates the alcohol product from the intermediate complex.
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