May 20, 2022, 02:05:48 PM
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Topic: Mechanism: Reduction of carboxylic acids via LiAlH4 (choice of steps)  (Read 286 times)

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Offline oilpaintedclouds

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Hey, first post on this forum so let me know if I'm doing anything wrong.

I'm learning about reducing carboxylic acids via LiAlH4 but find the steps of the mechanisms less intuitive than another way I think *should* work.

I've attached a screenshot of the common mechanism within the attachments.

Instead, I find the following steps more intuitive for me:

1. Nucleophilic attack by the Hydride anion to the carbonyl carbon, double bond thus shifts electrons up to the oxygen giving it a negative charge.

2. Introduce our H+ (water or H3O) which will act as an acid to OH making it H20 (which will act as a leaving group)

3. To stabilize our negative charge on our oxygen, we move our lone pair of electrons back down to form a double bond, expelling the leaving group we just made (H2O)

4. Since waters gone we re-introduce our Hydride, which attacks the carbonyl carbon protonating it, whilst shifting the double bonds electrons onto the oxygen (again) forming a negative charge.

5. Stabilize the negative charge on the oxygen by protonating it thus forming the alcohol. 

Would the above mechanism steps be a correct way to achieve the same goal? Thanks!  :)

Offline mjc123

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Re: Mechanism: Reduction of carboxylic acids via LiAlH4 (choice of steps)
« Reply #1 on: January 09, 2022, 04:34:30 PM »
Proton transfer is fast, so if you have an acidic hydrogen, the first thing that happens is it will react with the hydride.
Likewise, if you introduce water when hydride is still around, they will react to give H2.

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