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Timeline for Carbonyl oxygen as nucleophile

Current License: CC BY-SA 3.0

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Nov 18, 2021 at 13:50 comment added orthocresol @SasQ Yes. But there's nothing wrong with that, is there? [Heads up: You need to tag me if you want me to get a notification.]
Nov 18, 2021 at 2:06 answer added Oscar Lanzi timeline score: 4
Nov 18, 2021 at 1:17 comment added SasQ Well, in that case we would have to treat the entire conjugate base as a leaving group that leaves the proton, wouldn't we?
Nov 18, 2021 at 0:50 comment added orthocresol @SasQ Those labels aren't mutually exclusive. H+ is an electrophile, and anything that gets protonated is acting as a nucleophile towards it.
Nov 18, 2021 at 0:49 comment added SasQ @orthocresol Isn't protonation of the carbonyl an acid-base reaction though? (i.e. not a nucleophile-electrophile reaction)
Apr 2, 2018 at 3:03 history tweeted twitter.com/StackChemistry/status/980641827018756096
S Oct 16, 2015 at 18:44 history suggested Mithoron
added tag
Oct 16, 2015 at 18:08 review Suggested edits
S Oct 16, 2015 at 18:44
Sep 27, 2015 at 20:57 comment added permeakra >Is there any reason for the low reactivity of the oxygen? || Yes. It is in $sp^2$ hybridization state, meaning it binds its lone pairs extremely strong. It seems, however, that it can react with strong electrophyles like proton. It also should act as one in synthesis of pyrilium salts, but I was unable to find a proposed mechanism. Also, technically the nucleophilic oxygen in Wittig synthesis is derived from carbonyl oxygen, but is not one at the moment of attack
Sep 27, 2015 at 19:38 history edited bon CC BY-SA 3.0
edited body; edited tags; edited title
Sep 27, 2015 at 19:37 answer added bon timeline score: 6
Sep 27, 2015 at 14:23 comment added orthocresol It does react, for example in the formation of enol ethers, and technically the protonation of a carbonyl group is also an example. It's a much harder nucleophile than the alpha carbon and therefore reacts with different things.
Sep 27, 2015 at 14:14 history asked Charles CC BY-SA 3.0