No, they are not. You can only construct the angles with $k=2^{\alpha}p_1...p_s$, where the $p_i$'s are (distinct) Fermat primes.
The proof is not hard but a bit long, make comment if you want more details. I denote by $\mathcal{P}~$the set of $k$ such that $\frac{2\pi}{k}$ is constructible :
1) Show that if $k\in \mathcal{P}$, then $2k\in\mathcal{P}$.
2) Show that if $n\in \mathcal{P}$, then any divisor $d$ of $n$ (different from $1$) lies also in $\mathcal{P}$.
3) Show that if $n$ and $m$ are coprime and both belong to $\mathcal{P}$, then $mn$ also belnogs to $\mathcal{P}$.
All these elementary and easy questions show that it remains to answer the following question :
Let $p$ be an odd prime. When does $p^\alpha$ belong to $\mathcal{P}$ ?
The answer is that $p^{\alpha}$ lies in $\mathcal{P}~$ if and only if $p$ is a Fermat prime and $\alpha=1$.
Answering this question is less elementary and (as far as i know) needs to use some Galois theory.
Best Answer
The only possibility is indeed numbers of the form $2^k$.
We use the famous characterization of constructible regular polygons. The $\frac{360^\circ}{N}$ angle is straight edge and compass constructible if and only if $N$ is of the shape $$N=2^k p_1\cdots p_s,\tag{1}$$ where the $p_i$ are distinct Fermat primes (possibly none).
This theorem rules out immediately all numbers $N$ not of the shape (1). But it also rules out the numbers of shape (i) where the number $s$ of Fermat primes in the factorization is non-zero.
For the theorem says that if $N$ involves one or more Fermat primes, then the $\frac{360^\circ}{N}$ angle cannot be straight-edge and compass divided into $N$ equal parts.