$D_8$ as a derived subgroup

Question

Every undergraduate student knows that there are (exactly) two non-abelian groups of order 8: Dihedral ($D_8$) and Quaternion ($Q_8$). The group $Q_8$ has many interesting properties; simple of them are: it has unique element of order $2$, all subgroups are normal, etc. A property, which is strange and not obvious, the group $Q_8$ has, is that there is no group $G$ such that $G/Z(G)$ is isomorphic to $Q_8$. On the other hand, $D_8$ has no such property; there are (infinitely many) groups $G$ such that $G/Z(G)$ is isomorphic to $D_8$; example - $D_{16}$.

The question, I would like to ask, is related to a property of $D_8$, which is not shared by $Q_8$. There are (infinitely many) groups whose derived subgroup is $Q_8$; example- $SL_2(3)$. For groups $G$ of order <100, one can check on GAP that $D_8$ is not a derived subgroup of any group. The natural question is then:

Question 1: Does there exist a finite/infinite group whose derived subgroup is equal to $D_8$?

Question 2: Does there exist a finite/infinite group whose derived subgroup is equal to (general) dihedral group $D_{2n}$ of order $2n$ ($n>2$)?

Answer 1

Suppose that $G$ is a group with a dihedral commutator subgroup $G^\prime=D_{2n}$ for $n\geq 3$. The subgroup $R$ of rotations is characteristic in $G^\prime$, and therefore normal in $G$, so $G$ acts on $R$ by automorphisms. Since $R$ is cyclic, its automorphism group is abelian, and thus the kernel of this action must contain $G^\prime$. In particular, $G^\prime$ centralizes $R$, so $R\leqslant Z(G^\prime)$. But we know that rotations are not centralized by reflections in dihedral groups, so this can't be true.