For the coefficients $2^n$ the equidistribution theorem fails. In fact it is easy to exhibit an irrational $a$ such that the sequence $(2^na)_{n=0}^\infty$ is not even dense in $(0,1)$ modulo $1$. For example, take an $a\in (0,1)$ whose binary expansion consists of increasing blocks of $1$'s seperated by $0$'s: $$ a=0.101101110111101111101111110\dots $$
Added. For algebraic irrationals such as $a=\sqrt{2}$, it is not even known if they are normal in a given base, let alone $(2^n a)_{n=0}^\infty$ being dense or even equidistributed in $(0,1)$ modulo $1$.