2. For the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ with eccentricity $e \in$ $\left[\frac{2 \sqrt{3}}{3}, 2\right]$. The range of the angle $\alpha$ between the two asymptotes of the hyperbola is ( ).
(A) $\left[\frac{\pi}{6}, \frac{\pi}{3}\right]$
(B) $\left[\frac{\pi}{6}, \frac{\pi}{2}\right]$
(C) $\left[\frac{\pi}{3}, \frac{\pi}{2}\right]$
(D) $\left[\frac{\pi}{3}, \frac{2 \pi}{3}\right]$
2.C.

Let the inclination angle of the asymptote $y=\frac{b}{a} x$ be $\beta$. Note that
$$
\begin{array}{l}
1+\frac{b^{2}}{a^{2}}=e^{2} \in\left[\frac{4}{3}, 4\right], \\
\tan \beta=\frac{b}{a} \in\left[\frac{1}{\sqrt{3}}, \sqrt{3}\right], \beta \in\left[\frac{\pi}{6}, \frac{\pi}{3}\right] .
\end{array}
$$

Thus, $a=\min \{2 \beta, \pi-2 \beta\} \in\left[\frac{\pi}{3}, \frac{\pi}{2}\right]$.