Does Halts reduce to all other undecidable languages?
Recall that $A \leq_m B$ iff $\bar A \leq_m \bar B$. Indeed, the same reduction function works for both cases.

So, let $H$ be the halting problem. Both $H,\bar H$ are undecidable. We prove $H \not\leq_m \bar H$.

By contradiction, assume $H \leq_m \bar H$. Hence, by the property above, $\bar H \leq_m H$. Hence $\bar H$ reduces to some RE problem, making $\bar H$ RE as well. Since $H,\bar H$ are both RE, we conclude that both are decidable. Contradiction.