In a reaction between a chiral substrate and chiral reagent, their asymmetric inductions could cooperate (matched pairs) or compete (mismatched pair). According to Masamune, in a matched pair the degree of asymmetric induction is approximately the product of the selectivities of the substrate and the reagent. In contrast, the asymmetric induction in a mismatched pair is equal to the quotient of the selectivities of the substrate and the reagent.

The following reaction perfectly proves Masamune's rule (syn enolate is produced during enolization):
enter image description here

Why is this rule valid and why is it so accurate? What's the mechanistic explanation behind the multiplicity trend?

  • 1
    $\begingroup$ You might find what you want in this paper (pubs.acs.org/doi/pdf/10.1021/ed067p20). I don't have time to write out a sufficiently detailed answer right now, but let me know if you can't access the paper, and ill answer when I'm free. $\endgroup$
    – NotEvans.
    Jan 10, 2016 at 17:45

1 Answer 1


First, we need to take a look at the possible transition states of the first reaction in your table using $\ce{9{-}BBN}$ as an enolising agent. There are two transition states shown in scheme 1: The upper, favoured and the lower, disfavoured one.

transition states of a *syn*-aldol reaction with 9-BBN
Scheme 1: Aldol reaction with $\ce{9{-}BBN}$ as an enolising agent. Both the favoured (above) and the disfavoured (below) Zimmermann-Traxler transition states are shown along with their respective products.

The difference in both transition states is in the steric clash of the α-carbon and the boron residue: If the aldehyde attacks from the si-side, the stereocentres on said α-carbon can be oriented favourably, namely with hydrogen pointing toward the most crowded area and the largest residue pointing toward the least crowded area. If the aldehyde attacks the enolate’s re-side, methyl must point towards the most crowded area to allow the largest residue to point away from crowded areas. A similar discussion can be made for the orientation of the aldehyde which could attack the enolate’s si-side with its si-side, resulting in an anti-aldol product.

It is important to note here that all four attacks are possible. However, they do not have the same probability of reacting. Based on the observed diastereoselectivity, one can say that $\mathrm{P}_R(F) \approx 92~\%$, while $\mathrm{P}_R(\bar F) \approx 8~\%$ ($R$ is the set reaction occurs and $F$ is the set favoured attack. I have subsumed the two anti-product giving attacks under $\bar F$, since they are nigh irrelevant).

Aldol reaction transition state in the matched case
Scheme 2: Aldol reaction with $\ce{(+){-}(Ipc)2BOTf}$ as the enolising agent. $\ce{Ipc}$’s methyl groups disfavour an aldehyde attack on the enolate’s re-side. This is the same as the substrate’s inherent diastereoselectivity; a matched case.

A similar case can be made for $\ce{DIP}$-mediated aldol additions in the absence of substrate diastereospecificity ($\ce{Me} = \ce{H}$). Here, we need to construct a new set $A$ meaning auxiliary-favoured attack. This auxiliary-favoured attack is shown in scheme 2, although I used the matched case from your examples, i.e. substrate diastereoselectivity is present. The auxiliary-unfavoured attack (same caveat) is shown in scheme 3.

Aldol reaction transition state in the mismatched case
Scheme 3: Aldol reaction with $\ce{(-){-}(Ipc)2BOTf}$ as the enolising agent. $\ce{Ipc}$’s methyl groups disfavour an aldehyde attack on the enolate’s si-side — but substrate diastereoselectivity prefers the si-side attack on the enolate (mismatched case).

Stochastically, we can consider both selectivities separately. Remember that all attack combinations are technically possible but some are just more likely to be productive. It is a good approximation to assume that $\mathrm{P}_R(A)$ and $\mathrm{P}_R(F)$ are independent of one another, which means that $\mathrm{P}_{R,A}(F) = \mathrm{P}_R(A) \times \mathrm{P}_R (F)$. Why? Well, the substrate’s diastereoselectivity is mainly based on the orientation of the residues on the α-carbon. This is pretty much independent of the orientation of $\ce{Ipc}$’s methyl groups, which cause the auxiliary’s enantioselectivity: in both methyl group orientations the same rotamer of the α-carbon has the same probability.

Having taken into account that both sets are independent, we see that we can multiply the probabilities of the different attacks to get the resulting attack’s probability. The substrate will favour (R,S) (counting from ketone to hydroxy group) in each case. To get the probability of (R,S), we multiply a large percentage (substrate selectivity) with a large percentage getting another large percentage (auxiliary selectivity). The probability for (S,R) will be a small percentage (substrate) multiplied by a small percentage (auxiliary) so even smaller. So favouring of the desired (matched) diastereomer is enhanced.

On the other hand, if the auxiliary favours (S,R), to get the probability of (R,S) we must multiply a large percantage with a small one giving a medium result. The same thing for the probability of (R,S). This means that both diastereomers have more equal chances of forming.

(A better discussion of this problem would involve relative reaction rates, as the article NotWoodward mentions does. I feel that the probability argument gets us the same conclusion without all the big kinetic calculations which I am not good at.)


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