How to show that if there's a mapping reduction from L to its complement, it doesn't imply that L∈R?
Start with any non-recursive set $A$ of natural numbers. Then $L = \{ 2n \mid n \in A \} \cup \{ 2n+1 \mid n \notin A \}$ is a counterexample, because for all natural numbers $x, x\in L$ iff $x+(-1)^x \notin L.$ ($L$ is clearly not recursive, since $A$ is reducible to it.)
This actually shows that $L$ and $L^c$ are recursively isomorphic, which is stronger than what you were asking for.