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+ Stereochemistry
+ Stereoselectivity
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+ Hydrocarbons
Posted by
Alan Jia

Does syn-dihydroxylation form racemic mixture?

Armand Wong Sr.  Follow

Oxidation of alkene to 1,2-syn-diol by cold basic $\ce{KMnO4}$ is a common classification test for alkene in undergraduate organic chemistry laboratory courses. OP had correctly recognized that this reaction is a case of syn-dihydroxylation, and OP's conclusion of the product to be meso is also correct since the alkene in hand is a cis-but-2-ene. The schematic representation of the proposed mechanism is given below:

Mechanism for oxidation by KMnO41
Source of the scheme

Since but-2-ene structure is planer and a prochiral molecule, it has two faces (Si- or Re-face; above or below the plane) to get attacked by a reactant, here it is $\ce{KMnO4}$. The above mechanism displays the attack by only one face. The oxidation gives butan-2,3-diol by syn-attack on that face. The product has two stereocenters and the stereochemistry of that product is $(2R,3S)$. As indicated in the diagram, the product also has a plane of symmetry, and hence, the $\ce{C}$2 stereocenter is the mirror image of the $\ce{C}$3 stereocenter. Therefore, the product is a meso-compound.

If the $\ce{KMnO4}$ attack is done on the opposite plane, the product would be $(2S,3R)$-butan-2,3-diol, if numbering has maintained the same. This is essentially the same meso-product, as we discussed above.

If the starting substrate is trans-but-2-ene, the $\ce{KMnO4}$ attack from one face gives you $(2S,3S)$-butan-2,3-diol, while the $\ce{KMnO4}$ attack from opposite face gives you $(2R,3R)$-butan-2,3-diol. Since possibility of attacking from the Si-fase and/or Re-face is $0.5$ (or $50\%$), you get essentially a racemic mixture.

Based on above facts, OP's textbook description of the reaction is erroneous, not only on stereochemistry, but also on the notation of the reaction (it is syn- not sym-addition).

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