ISSUE:

Hello good morning!

Attached a couple of sections of some exercises proposed in an exam from years ago at the uni. The doubt is not related to the reaction itself, but to the given solution.

It is assumed that what I attached is already the solution. However, I note that I believe that both reactions are missing the enantiomer

In the first case of bromohydrin, if I'm not mistaken, the enantiomer could be formed, depending on whether the cyclic intermediate is formed "up or down"

and in the case of diels alder, it basically depends on the orientation of the dienophile.

Am I correct when I think that the enantiomers are missing? Or maybe I have overlooked something?

All the best! and thank you

Hello!
That's right, in the halogenation reaction of the alkene two enantiomers are formed as the carbon of the -OH is chiral and the bromonium ion can be formed both above and below.

In the other reaction, Diels-Alder, a racemic mixture would also be obtained when the dienophile approaches the diene from the top or from the bottom.

I wish you good weekend!

Thank you so much!