There are plenty of them, for example take two languages A and B such that A is non-recursive (that is, undecidable), $A \leq_{T} B$ (where $\leq_{T}$ denotes Turing reducibility) but not $A \leq_{m} B$.
For a rather trivial example (since it can be solved considerably easier), take $A=HALT$ and $B=\overline{HALT}$. Then, clearly $A \leq_{T} B$ (to compute if $x \in B$, just run the oracle to ask if $x \in A$ and say yes if the answer of the oracle is no, and say no if the answer of the oracle is yes). But if $A \leq_{m} B$, we have a recursive function f such that $x \in A \iff f(x) \in B$, and then we can show that B is recursively enumerable: to check if $x \in B$ just compute f(x) and generate HALT, when (if it happens) f(x) appears in the generation of HALT, stop. This procedure is a semidecision procedure for $B=\overline{HALT}$, hence B is recursively enumerable. Since we know HALT itself is recursively enumerable, we would have HALT recursive, a contradiction. Thus, A does not m-reduces to B, but nonetheless, since A T-reduces to B, and we know that A is undecidable, we can immediately conclude that B is undecidable.