2. For the geometric sequence $\left\{a_{n}\right\}$ with the first term $a_{1}=1536$, and common ratio $q=-\frac{1}{2}$, let $\pi_{n}$ denote the product of its first $n$ terms. Then the largest $\pi_{n}(n \in \mathbf{N})$ is
A. $\pi_{9}$
B. $\pi_{11}$
C. $\pi_{12}$
D. $\pi_{13}$
Notice that $a_{9}=1536\left(-\frac{1}{2}\right)^{8}=6, a_{10}=-3, a_{11}=\frac{3}{2}, a_{12}=-\frac{3}{4}, a_{13}=\frac{3}{8}$. And $\left.\pi_{n}=a_{1}^{n} q^{\frac{n(n-1)}{2}}>0 \Rightarrow 4 \right\rvert\, n(n-1) \Rightarrow n=4 k$ or $n=4 k+1$, thus we only need to consider $\pi_{12}, \pi_{13}$. Also, $a_{10} a_{11} a_{12}=\frac{27}{8}>1$, so the answer is $C$.