Let $G$ be a finite group and let $M$ be a perfect $\mathbb{Q}[G]$-complex.
Suppose that $M\otimes\_{\mathbb{Q}[G]}\mathbb{Q}$ is quasi-isomrphic to $0$ can we conclude that $M$ is quasi-isomorphic to $0$ ? It will be nice if one provide a concrete counterexample in the negative case. 



Perfect $Q[G]$-complex
Since all $\mathbb{Q}[G]$-modules are projective, every bounded complex of finitely generated modules is perfect. For $M\otimes\_{\mathbb{Q}[G]}\mathbb{Q}$ to be acyclic you just need that the homology of $M$ has no nonzero trivial summand. So, for example, take $G=C\_2$, and $M$ the one-dimensional $\mathbb{Q}[G]$ module on which a generator of $G$ acts by multiplication by $-1$, regarded as a complex concentrated in degree zero.


