Is it true that $(\Delta\varphi)\phi=\nabla \cdot(\nabla\phi)\varphi-\nabla\phi\cdot\nabla\varphi$ for complex-valued functions $\phi$ and $\varphi$?

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I know from vector calculus that for a complex-valued functions $\phi$ and $\varphi$ say, the following identity hold: $$\phi\Delta\varphi=\nabla \cdot(\varphi\nabla \phi)-\nabla \phi \cdot \nabla \varphi.$$ This is an obvious identity from $$\nabla \cdot(\varphi\nabla \phi)=\phi\Delta\varphi+\nabla \phi \cdot \nabla \varphi$$

Now, what about if I want simplify $(\Delta \varphi)\phi$, would it be something like $$(\Delta\varphi) \phi=\nabla \cdot(\nabla \phi)\varphi-\nabla \phi \cdot \nabla \varphi \quad ?$$ If yes, why?

It might be easy to figure out, but it is not apparent to me.