Let $a,b$ be sets, we write $a\leq^\ast b$ if either $a=\varnothing$ or there exists a surjection $f\colon b\to a$. With the axiom of choice this is a linear ordering equivalent to the usual ordering of cardinals $a\leq b$ if there is an injection from $a$ into $b$).

Without the axiom of choice this order need not be linear, or even a partial order. Indeed there are models where $a < b$ (so trivially $a\leq^\ast b$) but also $b\leq^\ast a$.

Let $\Theta(a)$ denote the least ordinal $\alpha$ such that $\alpha\nleq^\ast a$. This is a dual notion to $\aleph(a)$ which is the least ordinal $\alpha$ such that $\alpha\nleq a$. It is not very hard to see that both these ordinals are in fact cardinals, and that they exist for every set in $\mathrm{ZF}$. In the case where $a=\mathbb R$ we simply write $\Theta$ for $\Theta(\mathbb R)$.

It is consistent that for some set $a$ it holds $\aleph(a)<\Theta(a)$. For example if $V$ is a Solovay model then $\aleph(\mathbb R)=\aleph_1\leq^\ast\mathbb R$.

We also know that if $a\leq^\ast b$ then $a\leq 2^b$, where the injection is simply the sending a point in $a$ to its fiber under a fixed surjection from $b$. It follows, if so, that $\Theta(a)\leq\aleph(2^a)$. Even in $\mathrm{ZFC}$ one can see that it is consistent that there is an equality, and it is consistent that there is no equality (e.g. $a=\omega$ and take a model in which $\frak c=\aleph_1$ and another where it is is $\aleph_2$).

Question:Suppose that $V$ is a model of $\mathrm{ZF+AD}$, we know that if $\alpha\leq^\ast\mathbb R$ then $2^\alpha\leq^\ast\mathbb R$, and therefore $\alpha^+\leq^\ast\mathbb R$. Is it consistent that $\Theta<\aleph(2^\mathbb R)$?Equivalently, suppose $\alpha<2^\mathbb R$, can we find a surjection from $\mathbb R$ onto $\alpha$?