= Tetrahedral mirror-image test
Take a central <chemical atom> at the origin and four <substituents> at the vertices of an ideal <regular tetrahedron>:
$$
a=(1,1,1),\quad b=(1,-1,-1),\quad c=(-1,1,-1),\quad d=(-1,-1,1).
$$
These dimensionless <coordinates> describe shape, not measured <bond lengths>. For distinct labels A, B, C and D, define a signed <determinant>
$$
\Delta=\det\begin{pmatrix}|&|&|\\a-d&b-d&c-d\\|&|&|\end{pmatrix}=+16.
$$
It is six times the <oriented volume> of the labelled <tetrahedron>. A <translation (geometry)> cancels from the differences. A <rotation (mathematics)> $Q$ has $\det Q=+1$ and preserves $\Delta$, whereas the mirror <matrix> $S=\operatorname{diag}(-1,1,1)$ has $\det S=-1$ and changes it to $-16$. Consequently no <rotation (mathematics)> 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-turn
$$
Q=\begin{pmatrix}-1&0&0\\0&0&1\\0&1&0\end{pmatrix},\qquad Q^{\mathsf T}Q=I,\qquad\det Q=+1,
$$
about 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 (mathematics)> as well. This is the local reason two identical <hydrogen> <chemical atoms> prevent <glycine>'s <amino acid alpha carbon> from being a tetrahedral <stereocenter>.
\Image[/tetrahedral-chiral-achiral-rotation-test.png]
{title=The same spatial half-turn succeeds only when the exchanged groups are identical}
{description=Original model, mirror model and rotated mirror model. Top: the distinct C and D remain mismatched, so the A/B/C/D model is <chiral>. Bottom: the identical H groups may exchange, so the A/B/H/H model is <achiral>. The operation is an actual three-dimensional $180^\circ$ <rotation (mathematics)> about $(0,1,1)$, with <determinant> $+1$.}
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.
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