Finding corner frequency of RC bandpass filter
The second circuit you have proposed as a replacement differs from the original one though it also performs a high-pass function. If you follow the flow I described earlier, you should find the following expression:

\$H(s)=\frac{sR_2C_2}{1+s(C_2(R_1+R_2)+C_1R_1)+s^2C_1C_2R_1R_2}=\frac{\frac{s}{\omega_z}}{1+\frac{s}{\omega_0Q}+(\frac{s}{\omega_0})^2}\$

in which 

\$Q=\frac{\sqrt{C_1R_1C_2R_2}}{C_2(R_1+R_2)+C_1R_1}\$

\$\omega_0=\frac{1}{\sqrt{C_1C_2R_1R_2}}\$ 

\$H_{00}=\frac{1}{\frac{C_1R_1}{C_2R_2}+\frac{R_1}{R_2}+1}\$

The reference transfer function requires to include the Thévenin output resistance made of \$C_1\$ in parallel with \$R_1\$. The curves are given below:



These are the joys of FACTs!