How do you form ROMgI from RMgI reacting with ketones?

1 Answer
Truong-Son N.
Jun 24, 2015

Oh, you're referring to a Grignard Reagent variant. Normally you should see −O−MgBr more often. I saw this in second-semester Organic Chemistry.

http://chemwiki.ucdavis.edu/?title=Organic_Chemistry/Aldehydes_and_Ketones/Synthesis_of_Aldehydes_%26_Ketones/Grignard_Reagents

While the formation of R−O−Mg−I from R−Mg−I needs to be dry −MgBr in the presence of water, the hydrolysis process of R−O−Mg−I is actually acid-catalyzed. Also, what's attached to the −Mg−I must be alkyl or aryl (R = alkane or 1-benzene derivative), at least for R−Mg−Br.

Normally, a carbonyl group is a good electrophilic site because basically, it has a high electron density, and it has two p orbitals within its pi bond that have unoccupied space (antibonding atomic orbitals) for electrons in the pi bond to move into (unshaded).

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Magnesium, on the other hand, is a strong Lewis acid, so it's a strong electron acceptor. As a result, −MgI becomes a great electrophilic site, since nucleophiles are electron donors.

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