Chemical Forums
Chemistry Forums for Students => Organic Chemistry Forum => Organic Chemistry Forum for Graduate Students and Professionals => Topic started by: ghovinsen on May 30, 2016, 02:46:08 AM
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Scheme here:
(https://www.chemicalforums.com/proxy.php?request=http%3A%2F%2Fi.imgur.com%2FW1OUJ3C.png&hash=6ee314ffcfba8942b3fd647a120eeed9ef10acec)
In the paper describing reaction scheme above, they deprotected the phenol group before amide reduction by a lithium reagent. Question is, this site here: http://www.organic-chemistry.org/protectivegroups/hydroxyl/tbdms-ethers.htm (http://www.organic-chemistry.org/protectivegroups/hydroxyl/tbdms-ethers.htm) describes how TBS is a good protection group against LiAlH4, so I'm a bit confused as to why the authors did not just leave the TBS as is and proceed with amide reduction. So far I can think of 3 possible reasons:
- Phenolic TBS is way less stable (Kocienski, Protecting Groups pp199).
- Steric bulk from TBS might hinder the approach of reducing agent and lower the already mediocre yield.
- With a free phenol, the lithium atom can coordinate with it and the carbonyl carbon, increasing the electrophilicity.
Also, the authors used LiAlH2(OEt)2 instead of the "regular" LiAlH4. Any suggestions would be appreciated :)
Mod edit: img tags added. Dan
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I guess the key is the reason this reaction stops at the aldehyde stage - presumably it relies on the formation of a stable aluminium-hemiaminal chelate that does not collapse until workup. Can you see how the free phenolic OH would be beneficial (and potentially how a TBS ether could be detrimental) to the hemiaminal stability?
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I started to think that the lithium reagent would just reduce the amide into an amine instead of an aldehyde.
I guess I need to think more