4. The system of inequalities about $x$
$$
\left\{\begin{array}{l}
x^{2}-x-2>0, \\
2 x^{2}+(2 k+5) x+5 k<0
\end{array}\right.
$$

has a unique integer solution -2. Then the range of the real number $k$ is ( ).
(A) $[-3,2)$
(B) $\left(2, \frac{5}{2}\right)$
(C) $[-2,2]$
(D) $\left[-3, \frac{5}{2}\right]$
4. A.

From $x^{2}-x-2>0 \Rightarrow x>2$ or $x-2>-\frac{2}{5}$.
Thus, $-\frac{5}{2}<x<-k$.
Since -2 is the only integer solution that satisfies inequality (2), then $-2 \leqslant -k \leqslant 3$.
Therefore, $-3 \leqslant k<2$.