[furaniki18]

i was having trouble with deciding the migratory aptitude of substituents during baeyer-villeger oxidation and the order that i found on most of the sites is:
hydrogen > tertiary alkyl > secondary alkyl > phenyl > primary alkyl > methyl.

i tried to look for a reason for this and this is what i found out on one of the sites http://classes.yale.edu/02-03/chem125a/125webSpring/BaeyerVilliger/migratoryaptitude.htm :

"At first glance this order seems crazy. One might think that if an anion is migrating, the more stable anion should migrate more easily. What we see instead that the more stable cation migrates more easily.

The lesson we learn is that, although we may draw the curved arrow showing an anion migration, the anion never breaks free of the rest of the molecule. What is really important is the availability of the bonding electrons the the C-A (or C-B) bond to mix with the sigma* orbital of the O-O bond and displace the carboxylic acid from the nearer oxygen. The group that would form a more stable cation holds the bonding electrons less tightly and makes them more available (higher HOMO) for attacking the O-O bond."
i can't understand the underlined part..could somebody please help me comprehend it...

[orgopete]

This is how I understand the reaction. Baeyer-Villiger reactions can occur under acidic or basic conditions. What is common to both is the O-O bond must be polarized such that one of the oxygens will leave taking the electrons between them. Therefore, the key to the migration is a donation of electrons to the other oxygen atom. If a RCH(OH)OOR" is created from an aldehyde, then the apparent migrating group is a hydrogen. I think of it as simply a loss of a proton and the product is carboxylic acid rather than an ester. If RR'(OH)OOR" is created from a ketone, then the group best able to donate electrons migrates. Here, it may be helpful to draw the intermediate as an -O(+). Then you may think about which atom can best donate electrons to it, namely a tertiary carbon is better than secondary, primary, or methyl.

For those following the mechanism, I like the analogy of the Baeyer-Villiger oxidation to the oxidation step in a hydroboration reaction. Boron, because it holds its electrons less tightly, can succeed in migrating all carbons. Carbon, because it holds its electrons more tightly, only migrates the most electron rich carbon. So it is the electrons of the original carbonyl group that you must focus upon. It is the most electron rich bond that migrates. Carbons donate electrons to a tertiary carbon increasing its electron density. I interpret CH bond to go by a proton loss. That is the general mechanism of an oxidation reaction.

[james_a]

i was having trouble with deciding the migratory aptitude of substituents during baeyer-villeger oxidation and the order that i found on most of the sites is:
hydrogen > tertiary alkyl > secondary alkyl > phenyl > primary alkyl > methyl.

i tried to look for a reason for this and this is what i found out on one of the sites http://classes.yale.edu/02-03/chem125a/125webSpring/BaeyerVilliger/migratoryaptitude.htm :

"At first glance this order seems crazy. One might think that if an anion is migrating, the more stable anion should migrate more easily. What we see instead that the more stable cation migrates more easily.

The lesson we learn is that, although we may draw the curved arrow showing an anion migration, the anion never breaks free of the rest of the molecule. What is really important is the availability of the bonding electrons the the C-A (or C-B) bond to mix with the sigma* orbital of the O-O bond and displace the carboxylic acid from the nearer oxygen. The group that would form a more stable cation holds the bonding electrons less tightly and makes them more available (higher HOMO) for attacking the O-O bond."
i can't understand the underlined part..could somebody please help me comprehend it...

What this is saying is that the pair of electrons is required to break the weak O-O bond (more complex way of saying this: donate electrons to the sigma* (antibonding) orbital of the O-O bond). Therefore groups that form more stable carbocations will allow for the lone pair of electrons to break the weak O-O bond.

When migration occurs there is a significant buildup of positive charge on the group that is migrating. It's essentially a 3-center, two-electron bond. Therefore groups that stabilize positive charge will migrate more easily (remember carbocation stability: tert-butyl > secondary > primary > methyl).
Granted, understanding the placement of phenyl here is a little hard to justify using this model, but that is the essence of it.