Tetrahedral mirror-image test

Take a central chemical atom at the origin and four substituents at the vertices of an ideal regular tetrahedron:These dimensionless coordinates describe shape, not measured bond lengths. For distinct labels A, B, C and D, define a signed determinantIt is six times the oriented volume of the labelled tetrahedron. A translation cancels from the differences. A rotation $Q$ has $\det Q=+1$ and preserves $\Delta$, whereas the mirror matrix $S=\mathrm{diag}(-1,1,1)$ has $\det S=-1$ and changes it to $-16$. Consequently no rotation can place every distinct label on its mirror partner. This argument uses three-dimensional coordinates, not clockwise order on a page.

Now let C and D both be the same substituent H. Swapping these identical groups is allowed. For this model the half-turnabout the axis $(0,1,1)$ maps the reflected A and B to their original positions and exchanges the two H positions. The models superpose. With distinct C and D, the same half-turn fails precisely at those two labels; the determinant argument rules out every other rotation as well. This is the local reason two identical hydrogen chemical atoms prevent glycine's amino acid alpha carbon from being a tetrahedral stereocenter.

For a flexible molecule, also distinguish conformational isomerism from a persistent enantiomer pair. A particular frozen shape may be chiral while ordinary internal motions interconvert mirror shapes. Turning the entire model is always allowed in a spatial comparison; an internal motion must be considered with its actual accessibility and timescale.