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Is there anyone who can give me a convincing argument about the ketone, or the carbonyl oxygen in the ketone being nucleophile or not? . I really don't know what to write anymore. Everyone says something different. My instructor said that the Ketone oxygen can be protonated, thus reacting as a nucleophile. Others says that Protonating the ketone in fact makes it more electrophilic.

Here is the molecule I'm working with (attached) and I have to mention all the nucleophilic sites - I've done that. This is the only one that really is confusing me. Hoping for some clarification.

http://imgur.com/a/20NAw

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Everyone says something different

Well, actually they don't. A ketone has a nucleophilic and an electrophilic center. The carbonyl carbon is electrophilic and can be attacked by nucleophiles, the oxygen itself can act as a nucleophile and can, for example, be protonated. And yes, if we protonate the oxygen the "ketone", or better the carbonyl carbon, get's more electrophilic.

An example would be the Luche reduction. In this case the carbonyl oxygen coordinates to cerium chloride, thus acting as nucleophile, and makes the carbonyl carbon even more reactive towards nucleophiles.

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  • $\begingroup$ Thank you the explaining this. It makes more sense now. However, I don't have this reaction the my textbook (The Luche reduction). Do you know any other more common kind of reactions to be used in this situation? $\endgroup$ – sunshine257 May 21 '17 at 17:28
  • $\begingroup$ As I have to indicate a reaction regarding this nucleophilic site. $\endgroup$ – sunshine257 May 21 '17 at 18:04
  • $\begingroup$ Well, I've linked the wikipedia article so you can look it up there. Another example would be the acid catalyzed synthesis of ketals. Or the acid calayzed formation of Cyanhydrine ( de.wikipedia.org/wiki/Cyanhydrine ) , which is actally quite interesting since you can use lithium salts instead of acids to catalyze that. Li+ acts quite similar to H+. $\endgroup$ – DSVA May 21 '17 at 18:10
  • $\begingroup$ Thank you so much for your great answers, DSVA! Now it makes more sense and I can carry on with my assignment. $\endgroup$ – sunshine257 May 21 '17 at 18:41

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