若 $e_{1}, e_{2}$ 是夹角为 $60^{\circ}$ 的两个单位向量, 则 $\boldsymbol{a}=2 e_{1}+e_{2}$ 与 $\boldsymbol{b}=-3 e_{1}+2 e_{2}$ 的
夹角为 ( )
$\because e_{1} \cdot e_{2}=\left|e_{1}\right|\left|e_{2}\right| \cdot \cos 60^{\circ}=\frac{1}{2}$,

$\therefore \boldsymbol{a} \cdot \boldsymbol{b}=\left(2 e_{1}+e_{2}\right) \cdot\left(-3 e_{1}+2 e_{2}\right)=-6 e_{1}^{2}+e_{1} \cdot e_{2}+2 e_{2}^{2}=-6+\frac{1}{2}+2=-\frac{7}{2}$,

$|\boldsymbol{a}|^{2}=\boldsymbol{a}^{2}=\left(2 e_{1}+e_{2}\right)^{2}=4 e_{1}^{2}+4 e_{1} \cdot e_{2}+e_{2}^{2}=4+4 \times \frac{1}{2}+1=7$,

$|\boldsymbol{b}|^{2}=\boldsymbol{b}^{2}=\left(-3 e_{1}+2 e_{2}\right)^{2}=9 e_{1}^{2}-12 e_{1} \cdot e_{2}+4 e_{2}^{2}=9-12 \times \frac{1}{2}+4=7$.

设向量 $\boldsymbol{a}$ 与向量 $\boldsymbol{b}$ 的夹角为 $\theta$, 则 $\cos \theta=\frac{\boldsymbol{a} \cdot \boldsymbol{b}}{|\boldsymbol{a} \| \boldsymbol{b}|}=\frac{-\frac{7}{2}}{\sqrt{7} \times \sqrt{7}}=-\frac{1}{2}$.

又 $0^{\circ} \leq \theta \leq 180^{\circ}$, 所以 $\theta=120^{\circ}$, 故选 C.