# Migratory aptitude in Baeyer-Villiger reaction

In short, I do not understand how or why the migratory aptitude is as listed in every textbook I've read.

Firstly, they talk about a positive charge being built up that is stabilised by the movement of the most substituted group.

I cannot fathom such a build up in my mechanisms.

The migratory aptitude list for the Baeyer-Villager oxidation is as it is because that is how the molecules undergoing the reaction behave. When we are learning about new reactions from a textbook, we often read about the mechanism and theoretical explanations of chemo-, regio-, and stereoselectivity. It is easy to forget that the experimental observations came first.

The migratory aptitude list $(\ce{H>R3C>R2HC}\approx\ce{Ph(>H)>RCH2>Me})$ is the result of many experiments. Nor is it a linear list as is usually presented. $\ce{H}$ migrates more readily than tertiary groups, which migrate more readily than secondary groups, which migrate more readily than phenyl groups (which migrate more readily than $\ce{H}$).

The mechanism and theoretical explanations came later. The partially complete mechanism from the Wikipedia page is shown below. The key step (center) is a concerted collapse of the tetrahedral intermediate with a [1,2]-sigmatropic shift of $\ce{R}^1$ and ejection of the carboxylate leaving group. Note that this mechanism shows none of the proton transfer steps.

Since the who thing is supposedly concerted, I can understand your objection to the build-up of $\delta^+$ on $\ce{R}^1$. However, with the exceptions of the anomalous behavior of $\ce{H}$ and $\ce{Ph}$, the migratory aptitude list is the same as the carbocation stability list. Thus, researchers looked for a link between the two phenomena and a way to relate one to the other. The link may not be real, but the trends are there. To a student learning organic chemistry, it is much friendlier at first to apply an existing trend in reactivity to a new reaction (with an exception), than to learn a whole new trend that is unrelated but cosmetically similar.

• How is $\ce{H}$ present twice in the list? – Jan Jun 6 '16 at 13:32
• That is mysterious, isn't it. – Ben Norris Jun 8 '16 at 0:53
• I guess, if the intermediate shown above is protonated to any degree (conceivable under acidic conditions), that makes the $\delta^+$ more obvious. Without the protonations, the way I like to think of it is that the leaving group ($\ce{RCO2-}$) bears a negative charge, so the remainder of the molecule must have a positive charge spread out over it to compensate. – orthocresol Dec 9 '16 at 14:12
• @BenNorris Can you please give the reason behind this mystery? I don't seem to understand this circular train of thought here. "The migratory aptitude list (H>R3C>R2HC≈Ph(>H)>RCH2>Me)' Why is the H present twice here? And how's that even mathematically possible? – Gaurang Tandon Mar 19 '18 at 0:54
• @GaurangTandon - I may need to amend the answer. The source of this circular order is my undergraduate textbook, but like many textbooks no references are given. I have not been able to find references that MCPBA or other peroxyacids will oxidize benzaldehyde to phenyl formate and not benzoic acid. – Ben Norris Jun 19 '18 at 12:32