2. If $\alpha$ is an angle in the second quadrant, then $\pi-\frac{\alpha}{2}$ is ( ).
(A) an angle in the first quadrant
(B) an angle in the second quadrant
(C) an angle in the first or third quadrant
(D) an angle in the second or fourth quadrant
2.D.
$$
\begin{array}{l}
\text { Given } 2 k \pi+\frac{\pi}{2}<\alpha<2 k \pi+\pi, \\
\Rightarrow k \pi+\frac{\pi}{4}<\frac{\alpha}{2}<k \pi+\frac{\pi}{2} \\
\Rightarrow -k \pi+\frac{\pi}{2}<\pi-\frac{\alpha}{2}<-k \pi+\frac{3 \pi}{4} .
\end{array}
$$

Thus, $\pi-\frac{\alpha}{2}$ is an angle in the second or fourth quadrant.