I have been asked to determine the nature of the product for the following reaction:

Oxidation reaction

And as I recognized this reaction to be a case of syn-dihydroxylation, my obvious response was that the product formed will be meso.

But, in the solution to the problem, it says it makes a racemic mixture.

Oxidation reaction with products

Shouldn't both the $\ce{-OH}$ be on the same side and form a meso-compound? And what does the $(sym)$ stand for?

  • 2
    $\begingroup$ The very fact that 'syn' has been written as 'sym' should've heightened your suspicions regarding this answer, which is blatantly wrong. Cold KMnO4 will produce a meso compound in this case. This is a very standard reaction, and it is surprising that a book messed this one up. I strongly suggest you change the book/material you're studying from. $\endgroup$
    – C_Lycoris
    Apr 17, 2021 at 13:43
  • $\begingroup$ May I ask, however which book this came from? $\endgroup$
    – C_Lycoris
    Apr 17, 2021 at 13:45
  • $\begingroup$ @C_Lycoris It's not from any standard organic chemistry book. The book is called "Problems in Organic Chemistry", and is basically a chapter-wise question bank for the JEE exam. $\endgroup$
    – Aniruddha
    Apr 17, 2021 at 13:51
  • $\begingroup$ Referring to the product through its Fischer structure alone was an obvious red flag. Ditch this textbook. $\endgroup$ Apr 18, 2021 at 8:42

1 Answer 1


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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