2. Given a large regular tetrahedron with an edge length of 6, a smaller regular tetrahedron is placed inside it. If the smaller tetrahedron can rotate freely within the larger one, the maximum edge length of the smaller tetrahedron is . $\qquad$
2. 2 .

Given that a smaller regular tetrahedron can rotate freely inside a larger regular tetrahedron, the maximum edge length of the smaller tetrahedron occurs when it is inscribed in the insphere of the larger tetrahedron.

Let the circumradius of the larger tetrahedron be $R$, and the circumradius of the smaller tetrahedron (which is the inradius of the larger tetrahedron) be $r$.
It is easy to see that, $r=\frac{1}{3} R$.
Therefore, the maximum edge length of the smaller tetrahedron is $\frac{1}{3} \times 6=2$.