Can a stereospecific / stereoselective reaction produce a racemic mixture of enantiomers?

I think I have my terms mixed up but I'm confused about the concept of stereoselectivity / stereospecificity as they relate to reactions with alkenes.


For example, hydroboration is a stereospecific reaction. It leads to *syn* addition across an alkene. However, you can form a racemic mixture from it because you can add from either the top or bottom faces. 


[![Hydroboration of isoprene](https://i.stack.imgur.com/zkf01.png)](https://i.stack.imgur.com/zkf01.png)


I think this is a good depiction of what I mean. It seems to me that hydroboration is called stereospecific because it does *syn* (rather than *anti* and *syn*). But it still forms two different stereoisomers, so how can we still call it stereospecific?




You haven't mixed up your terms: a stereoselective reaction is [defined](https://doi.org/10.1351/goldbook.S05991) as one which preferentially forms one stereoisomer over another. Likewise, all stereospecific reactions are [also stereoselective](https://doi.org/10.1351/goldbook.S05994), also stereospecificity has stricter criteria then plain old stereoselectivity.


However, you have to remember that the term stereoisomer encapsulates both enantiomers and diastereomers. The hydroboration reaction is stereospecific because reacting one particular *diastereomer* of the alkene (either *cis* or *trans*) leads to one particular *diastereomer* of the product. It doesn't necessarily lead to a particular *enantiomer*. Basically, hydroboration leads to the formation of a racemic mixture of a particular diastereomer.


The alkene you have chosen, 2-methylbut-2-ene, is not a good example substrate that can actually show how the reaction is stereospecific. Since one of the alkene carbons is symmetrically substituted, there's no diastereomerism present in the products. If we use (2­*E*)-3-methylpent-2-ene as an example substrate, then it becomes clearer:


[![Hydroboration-oxidation of (2E)-3-methylpent-2-ene](https://i.stack.imgur.com/Jv36H.png)](https://i.stack.imgur.com/Jv36H.png)


Note how neither of the (2​*R*,3​*R*) or the (2​*S*,3​*S*) stereoisomers are formed. However, the two products that are formed are enantiomers of each other.


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