D-Threose: reduction and optical activity I've bumped into this exercise: 


  Judging by the mechanism of reaction between D-threose and $\ce{NaBH4}$, do you think the final product will be optically active? 


To me looks like a reduction, since $\ce{NaBH4}$ is a lightly reductive agent, but then why the answer is "No, the product is not optically active"? 

Is it because of the presence of anomers or there are other reasons?
When you reduce the aldehyde (from the open chain threose, since the cycle cannot be reduced as it has only acetal-like functionality) you introduce a symmetry element (both ends of the molecule become $\ce{CH2OH}$). 

The presence of a symmetry element means the reduced molecule is not chiral, but meso.