5. (1999 National High School Mathematics League Question) Given that the base of the tetrahedron $S-ABC$ is an equilateral triangle, the projection of point $A$ on the side face $SBC$ is $H$, which is the orthocenter of $\triangle SBC$, the plane angle of the dihedral angle $H-ABC$ is $30^{\circ}$, and $SA=2\sqrt{3}$. Then the volume of the tetrahedron $S-ABC$ is . $\qquad$
5. Given that the dihedral angle $H-AB-C$ has a plane angle of $30^{\circ}$, we can first construct the plane angle of the dihedral angle, and then find the height and the area of the base.

From the problem, $AH \perp$ plane $SBC$. Draw $BH \perp SC$ at $E$. By the theorem of three perpendiculars, we know that $SC \perp AE, SC \perp AB$, hence $SC \perp$ plane $ABE$. Let the projection of $S$ on plane $ABC$ be $O$, then $SO \perp$ plane $ABC$. By the inverse of the theorem of three perpendiculars, we know that $CO \perp AB$ at $F$. Similarly, $BO \perp AC$. Therefore, $O$ is the orthocenter of $\triangle ABC$. Since $\triangle ABC$ is an equilateral triangle, $O$ is also the centroid of $\triangle ABC$, thus $SA=SB=SC=2\sqrt{3}$.

Since $CF \perp AB, CF$ is the projection of $EF$ on plane $ABC$, by the theorem of three perpendiculars, $EF \perp AB$, so $\angle EFC$ is the plane angle of the dihedral angle $H-AB-C$, hence $\angle EFC=30^{\circ}, OC=SC \cdot \cos 60^{\circ}=2\sqrt{3} \cdot \frac{1}{2}=\sqrt{3}, SO=\sqrt{3} \cdot \tan 60^{\circ}=\sqrt{3} \cdot \sqrt{3}=3$. Also, $OC=\frac{\sqrt{3}}{3} AB$, so $AB=\sqrt{3} OC=\sqrt{3} \cdot \sqrt{3}=3$.
Therefore, $V_{S-ABC}=\frac{1}{3} \cdot \frac{\sqrt{3}}{4} \cdot 3^{2} \cdot 3=\frac{9}{4} \sqrt{3}$.