4. As shown in Figure 1, let point $O$ be inside $\triangle A B C$, and $O A + 2 O B + 3 O C = 0$. Then the ratio of the area of $\triangle A B C$ to the area of $\triangle A O C$ is ( ).
(A) 2
(B) $\frac{3}{2}$
(C) 3
(D) $\frac{5}{3}$
4.C.

Let $D$ and $E$ be the midpoints of sides $AC$ and $BC$, respectively, then
$$
\begin{array}{l}
O A+O C=2 O D, \\
2(O B+O C)=4 O E .
\end{array}
$$

From (1) and (2), we get
$$
O A+2 O B+3 O C=2(O D+2 O E)=0,
$$

which means $O D$ and $O E$ are collinear, and $|O D|=2|O E|$.
Therefore, $\frac{S_{\triangle A B C}}{S_{\triangle A D C}}=\frac{3}{2}$. Thus, $\frac{S_{\triangle A B C}}{S_{\triangle A D C}}=\frac{3 \times 2}{2}=3$.