This seems like the answer should be trivial but when decomposing the direct product of 4 spin-$\frac{1}{2}$ states into a direct sum, one gets two singlets, namely $$\frac{1}{\sqrt{2}} \left(\mid{\uparrow\downarrow\uparrow\downarrow}\rangle \pm \mid{\downarrow\uparrow\downarrow\uparrow}\rangle\right).\tag{1}$$ This is assuming one follows the normal textbook procedure using the Clebsch Gordan coefficients to combine the spins one at a time. However, if one hastily combines the spins into pairs of singlets, i.e. $$\frac{1}{\sqrt{2}}\left( \mid\uparrow\downarrow\rangle - \mid\downarrow\uparrow\rangle\right) \otimes \frac{1}{\sqrt{2}}\left( \mid\uparrow\downarrow\rangle - \mid\downarrow\uparrow\rangle\right) \tag{2}$$ then one obtains a state that superficially is annihilated by $\vec{S}_1 + \vec{S}_2$ and $\vec{S}_3 + \vec{S}_4$ and so one expects it to be annihilated by $\vec{S}_1 + \vec{S}_2 + \vec{S}_3 + \vec{S}_4$. Yet this state is clearly not part of the above representations.
My suspicion is that while the singlets are invariant under global $SU(2)$ rotations that is not the case for local ones, and that the quick and dirty method ends up in a different choice of basis.
So my question is why does combining spins iteratively produce a more symmetric state than combining them as pairs, and where does that break down?
