6
$\begingroup$

Why do $\ce{LiAlH(O-^{t}Bu)3}$ and similar bulky reducing agents not reduce acidic chlorine further?

$$\ce{CH3-COCl + LiAlH4 + H2O -> CH3-CH=O -> CH3-CH2-OH},$$ but $$\ce{CH3-COCl + LiAlH(O-^{t}Bu)3 + H2O -> CH3-CH=O -> no~reaction}.$$

$\endgroup$
7
$\begingroup$

Without being very specific it looks that $\ce{LiAlH(O-^{t}Bu)_3}$ is a lot less reactive than $\ce{LiAlH4}$, mostly because the $\ce{^{-}O-^{t}Bu}$ is a much weaker base than $\ce{H-}$. Also its much bulkier, further reducing reactivity.

The aldehyde being less reactive than the acyl chloride the reducing agent first eats up all the acyl chloride, and then starts reducing the aldehyde. If you were careful to add stoichiometric amounts, you can stop right when you run out of the acyl chloride.

See also this link.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.