Proof by Reduction: From Empty Language to Halting Problem on Empty Input
See (Blank tape halting problem vs. Emptiness problem ($H_0$ vs. $E_{TM}$)).

$H$ is Turing-recognizable but not co-Turing-recognizable.

$E$ is co-Turing-recognizable but not Turing-recognizable.

Their undecidabilities are of different kind. So we need to reduce $M$ to $N$ in such a way that:

$\langle M \rangle \in E$ if and only if $ \langle N \rangle \not\in H$.

Consider the following reduction from $M$ to $N$. Turing Machine $N$ on empty input will simultaneously generate all input and run $M$  as a universal Turing Machine on that input (again simultaneously) and halts (either accept of reject, doesn't matter) if some simultaneous thread of $M$ accepts on some input. On non-empty input it does not matter what $N$ will do.

We can easily see that if $M$ does not accept anything then it implies $N$ will never halt on empty input.
If $M$ accepts something then it implies that $N$ will halt on empty input. 

Therefore, $\langle M \rangle \in E$ if and only if $ \langle N \rangle \not\in H$.

So, we have got a Turing  reduction (in fact, a polynomial time Karp reduction).

$E \leq_M \overline{H}$