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?

  • 4
    $\begingroup$ I have edited your question to include the ce for chemical equations. For example, typing $$\ce{2KClO3 ->[MnO2] 2KCl + 3O2}$$ will be rendered as: $$\ce{2KClO3 ->[MnO2] 2KCl + 3O2}$$ $\endgroup$ – Pritt says Reinstate Monica Jul 2 '17 at 9:45

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.


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