Let $X_{1}, \ldots, X_{n}$ be an iid sample from $N(\mu,\sigma^2)$. Then the distribution of $$ Z_{n} = \frac{\bar{X}-\mu}{\sigma/\sqrt{n}} $$ is exactly $N(0,1)$ assuming that $\sigma$ is known.
If $\sigma$ is unknown then the distribution of $$ T_{n} = \frac{\bar{X}-\mu}{S/\sqrt{n}} $$ is a $t$-distribution with $n-1$ degrees of freedom.
If $X_{1}, \ldots, X_{n}$ be an iid sample from a distribution (not necessarily normal) then the distribution of $$ W_{n} = \frac{\bar{X}-\mu}{\sigma/\sqrt{n}} $$ is approximately $N(0,1)$ if $n$ is large by Central Limit Theorem.
But I have not seen any theorems mentioning the distribution of the quantity $$ U_{n} = \frac{\bar{X}-\mu}{S/\sqrt{n}}. $$
If the value of $n$ is large then again by CLT the distribution of $U_n$ is approximately $N(0,1)$ but assume that $n$ is not large enough to apply the CCLT. Then, what is the distribution of $U_n$ (assuming $n$ is not large)? Is it still $t$-distribution with $n-1$ degrees of freedom?
Moreover, what is the distribution of $$ V_{n} = \frac{\bar{X}-\mu}{\sigma/\sqrt{n}} $$ if the sample $X_1, \ldots, X_n$ is not coming from a normal population?
I am totally lost here.