Given the following Lagrangian
$$\mathscr{L} = c\frac{g}{m}\bar{\psi}_A\Gamma_5\gamma^\mu\psi_B (i\partial_\mu)\phi$$
where $\Gamma_5 \in \{\gamma_5, 1\}$, for two spin one-half particles $A$ and $B$ and a spin $0$ particle $\phi$.
I'm trying to figure out how to determine when $\Gamma_5 = 1$ and $\Gamma_5 = \gamma_5$
I know the Lagrangian should be a scalar. I also know how the bilinear covariants transform.
Let now $A$ and $B$ both have $J^P = \tfrac{1}{2}^+$ and let $\phi$ have $J^P=0^-$.
If $\Gamma_5=1$, then $\bar\psi\gamma^\mu\psi$ transforms like a vector. $\partial_\mu \rightarrow -\partial_\mu$. To my understanding it follows then that $\bar\psi(\partial\!\!/ ~\phi)\psi$ transforms like a pseudoscalar. Likewise, if $\Gamma_5=\gamma_5$, then $\bar\psi\gamma_5(\partial\!\!/~\phi)\psi$ transforms like a scalar.
How can I factor in the parities above?If I just multiply them, I would find that the pion nucleon nucleon interaction $$\mathscr{L}=\frac{g}{m}\bar\psi\gamma_5(\partial\!\!/~\vec\pi)\cdot \vec\tau ~\psi \tag{2}$$ transforms like a pseudoscalar. But the Lagrangian should be scalar. So what am I missing?
Also, taking the $N(1535)$ into account, which has spin parity $J^P=\tfrac{1}{2}^-$, we do not need a $\gamma_5$. How can I see this rigorously?