Which is the stronger base among $NaOEt$ $"NaOEt"$ (sodium acetate) and $LDA$ $"LDA"$ (lithium diisopropylamide)?

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Answer 1 · 2016-03-15T02:06:00

In decreasing basicity

$LDA > NaOEt > NaOAc$ $"LDA" > "NaOEt" > "NaOAc"$

Sodium acetate is written as $NaOAc$ $"NaOAc"$ , with chemical formula

https://upload.wikimedia.org/wikipedia/commons/4/45/Sodium-acetate-2D-skeletalpng

Not $NaOEt$ $"NaOEt"$ , which is called sodium ethanoxide and has a chemical formula

https://upload.wikimedia.org/wikipedia/commons/4/4a/Sodium_ethoxidepng

To compare whether $LDA$ $"LDA"$ or $NaOEt$ $"NaOEt"$ is more basic, we can consider two reactions

The first reaction proceeds much further than the second one.

LDA can only be formed from treatment of diisopropylamine with n-butyllithium, which is much more basic than sodium ethanoxide.

From this, we conclude that LDA is more basic than sodium ethanoxide.

So the final result, in decreasing basicity

$LDA > NaOEt > NaOAc$ $"LDA" > "NaOEt" > "NaOAc"$

For stronger evidence, please refer to their $p K_{a}$ $"p"K_"a"$ values.

Which is the stronger base among NaOEt"NaOEt" (sodium acetate) and LDA"LDA" (lithium diisopropylamide)?