|  | Chapter 15: Alcohols, Diols and Thiols |  | 
Reductions of Carboxylic Acids and Esters
 
 
    
 
 
    Summary
| REACTION OF 
        
        LiAlH4 WITH AN ESTER  | |
| 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. |   | 
| 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. | |
| 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. | |
| 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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|  | © Dr. Ian Hunt, Department of Chemistry |  |