5. Let the ellipse $C: \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$ have its left and right foci as $F_{1}, F_{2}$, with a focal distance of $2 c$. The point $N\left(\frac{3 c}{2}, \frac{\sqrt{2} c}{2}\right)$ is inside the ellipse, and point $M$ is a moving point on the ellipse $C$, such that $\left|M F_{1}\right|+|M N|<2 \sqrt{3}\left|F_{1} F_{2}\right|$ always holds. Then the range of the eccentricity of the ellipse $C$ is
A. $\left(0, \frac{\sqrt{3}}{3}\right)$
B. $\left(\frac{\sqrt{3}}{3}, 1\right)$
C. $\left(\frac{4 \sqrt{3}}{21}, 1\right)$
D. $\left(\frac{4 \sqrt{3}}{21}, \frac{\sqrt{3}}{3}\right)$
Solve:
$$
\left|M F_{1}\right|+|M N|=2 a-\left|M F_{2}\right|+|M N| \leqslant 2 a+\left|N F_{2}\right|=2 a+\frac{\sqrt{3} c}{2} \text {, }
$$

Thus $4 \sqrt{3} c>2 a+\frac{\sqrt{3} c}{2} \Rightarrow \frac{7 \sqrt{3} c}{2}>2 a \Rightarrow e>\frac{4}{7 \sqrt{3}}=\frac{4 \sqrt{3}}{21}$;
Also $\frac{9 c^{2}}{4 a^{2}}+\frac{c^{2}}{2 b^{2}}0$
$\Rightarrow e^{2}<\frac{1}{3} \Rightarrow e<\frac{\sqrt{3}}{3}$. Therefore, the answer is $D$.