If $f(n)=O(g(n)),$ Then It means that
$f(n)\leq C_{0}. g(n)$; Where $C_{0}$ is constant, $C_{0} > 0,$$n \geqslant n_{0} ,$ and $n_{0}\geq 1$
$2^{f(n)}\leq 2^{(C_{0}.g(n))}$ ;Raising power to both side ----------- 1
If $2^{f(n)}=O(2^{g(n)}),$ Then it means that
$2^{f(n)}\leq C_1.2^{g(n)}$ ;Where $C_{1}$ is constant, $C_{1} > 0,$$n \geqslant n_{0} ,$ and $n_{0}\geq 1$
$2^{f(n)}\leq C_1.2^{g(n)}$ -------------2
Since 1 & 2 are not the same, Then we can say that it's not true.