[owlpower]

When the molecule undergoes hydration reaction in 2 different manners, as seen in the reaction scheme, I know that there will be stereoisomers formed.

However, I am having trouble determining whether the product(s) would be optically active. At first, I concluded that they are both optically active, but I am starting to doubt that.

For the reaction above, the addition of OH and H is anti, and we get a R,S and R,R product. They are diastereomers, so I don't think there will be any optical activity, because the "S" and "R" part for the OH cancels out each other.

For the reaction below, there is only one chiral center, and both products are the same configuration, thus its not optically active.

I am not too sure about my answer, so please point out any conceptual mistakes I have made, if any!

[sjb]

If you were just asked to look at the products in isolation, would you say they were optically active?

[kriggy]

For the reaction above, the addition of OH and H is anti, and we get a R,S and R,R product. They are diastereomers, so I don't think there will be any optical activity, because the "S" and "R" part for the OH cancels out each other.

For the reaction below, there is only one chiral center, and both products are the same configuration, thus its not optically active.

I am not too sure about my answer, so please point out any conceptual mistakes I have made, if any!

This is not how it works. The "S" and "R" doesnt cancel each other. Its very, very big simplification that is almost useless.

Each compound that has chiral centers, rotates the polarized light by some angle in some direction (+ or -). The thing is (and this is origin of your confusion IMO) that ENANTIOMERS rotate by same angle but in opposite direction ie. R enantiomer shows +X°, then S enantiomer shows -X°. If you have racemic mixture (50:50 R:S ratio) then they effectively cancel each other. HOWEVER, it ONLY applies for ENANTIOMERS and their racemic mixture, if the mixture is not racemic, then you will see some value. As long as your compounds are not enantiomers (they are DIASTEREOMERS) then they have different values and directions of rotation so they dont "cancel each other". (Im sure there are some very specific examples where this happens but its quite corner case and we dont have to bother with it).

So for solving your task: you correctly drew both possible products that can be formed in your reaction, now, you need to compare those two products, to see what is their stereochemical relationship: are they enantiomers or diastereomers? Do you think they are in 1:1 ratio or maybe one of the products predominates in the reaction mixture?

When you figure out if they are enantiomers or diastereomers, then you have it all and can easily tell if you would expect some optical activity or not.

[owlpower]

Thank you very much, after considering all the factors above, I have come to the conclusion that there will be optical activity for both reactions.

[kriggy]

Thank you very much, after considering all the factors above, I have come to the conclusion that there will be optical activity for both reactions.

Exactly