What I know: let's say $I\subseteq\mathbb{R}$ is a compact interval. Then, $C^\infty(I)$ is the set of infinitely smooth functions on $I$ (to talk about distributions, one usually takes compactly supported test functions, but since the domain is already compact here it shouldn't be a problem). We endow this set with the topology corresponding to the convergence $$ \phi_n \to_n \phi \iff \|\partial^\alpha \phi_n - \partial^\alpha \phi \|_\infty \to_n 0 \quad \forall \alpha\in\mathbb{N} $$ and we call it $D(I)$. Then, we take its topological dual $D'(I)$ and we call it the set of distributions, endowed with the weak-* topology.
First question: if I take a bigger space than $D(I)$, say, $C^1(I)$, then intuitively its dual should be smaller (as the linearity and continuity conditions must hold on a bigger set). Is this the case? If so, I don't know how to make this rigorous. If I want to take the dual of $C^1(I)$, I should define a topology on this set, and in order to satisfy this intuition, the topology of $C^1(I)$ restricted to $C^\infty(I)$ should coincide with that of $D(I)$, but it is not intuitive how to do so.
Second question: in the proper setting, can the dual $C^1(I)^*$ be something like the distributions of order 1?