Two coupled LC circuits Kirchhoff's laws

Question

I am struggling to apply kirchoff laws to certain quantum circuits. I will illustrate my problem with this toy example:

I want to find the charge $Q_1$ and $Q_2$ of $C_1$ and $C_2$ to quantise the problem. My guess for the Kirchoff voltage loops and current equations  would be

\begin{align}
    \text{Voltage loops: }&\begin{cases} 
      V_{C_1} + V_{L_1} = \frac{1}{C_1} Q_{1} + L_1 \dot{I}_1 = 0 \\
      V_{C_2} + V_{L_2} = \frac{1}{C_2} Q_2 - L_2\dot{I}_4 = 0
   \end{cases}\notag\\
    \text{Current junctions: }&\begin{cases} 
        I_1 = I_2 + I_3 \\
        I_2 = I_4 + I_5. 
   \end{cases}\notag
\end{align}
And by using the current junction conditions we can write $I_1$ and  $I_4$ in terms of the capacitors charge:
\begin{equation}
    \begin{cases} 
      \frac{1}{C_1} Q_{1} + L_1 (\dot{I}_2 + \dot{I}_3) = \frac{1}{C_1} Q_{1} + L_1 (\ddot{Q}_g + \ddot{Q}_1) = 0\\
      \frac{1}{C_2} Q_2 + L_2(\dot{I}_2 - \dot{I}_5) = \frac{1}{C_2} Q_2 - L_2(\ddot{Q}_g - \ddot{Q}_2) = 0. 
   \end{cases}\notag
\end{equation}

Now, I want to eliminate $Q_g$ since I'm only interested in $Q_1$ and $Q_2$. For this I can read off one more loop equation for the center loop

$$L_2 \dot{I}_4 + \frac{Q_g}{C_g} - \frac{Q_1}{C_1} = 0,$$

and substitute in the above to obtain

\begin{equation}
    \begin{cases} 
       \frac{Q_1}{C_1} + L_1(C_g(\frac{\ddot{Q}_1}{C_1} -L_2 \dddot{I}_4) + \ddot{Q}_1) = 0 \\
      \frac{Q_2}{C_2} + L_2(C_g(\frac{\ddot{Q}_2}{C_2} -L_2 \dddot{I}_4) + \ddot{Q}_2) = 0. 
   \end{cases}\notag
\end{equation}

which has the coupling terms wrong since they should be
\begin{equation}
    \begin{cases} L_1 \ddot{Q_1} + 2C_g(\ddot{Q_1} -\ddot{Q_2}) + \frac{1}{C_1}Q_1 = 0\\
L_2 \ddot{Q_2}  +2C_g(\ddot{Q_2} -\ddot{Q_1}) + \frac{1}{C_2}Q_2= 0
   \end{cases}\notag
\end{equation}

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Any ideas??