What would happen if you reduced the coupling of $SU(2)$ in the standard model to zero? Ultimately, my goal is to find a free parameter that you could change in order to significantly reduce the strength of, or eliminate, the weak interaction. Would such a modification leave other parts of the Standard Model unchanged?
 A: It is straightforward to see, even though your ultimate vision should be in trouble. I assume you mean decrease the coupling g of just SU(2), and leave the EM coupling e and the Higgs v.e.v. v alone, which cannot be done.
You then just look at the formulas:
$$\cos \theta_\text{W} = \frac{g}{\,\sqrt{g^2+g'^2\,}\,}, \qquad     \sin\theta_\text{W} = \frac{g'}{\,\sqrt{g^2+g'^2\,}\,}  \\
 e=g\sin\theta_W= g'\cos \theta_W\\
 m_\text{Z} = \frac{m_\text{W}}{\,\cos\theta_\text{W}\,}, \qquad  m_W= {ev\over 2\sin\theta_W}, \qquad G_F= 1/(v^2\sqrt{2}). $$

As $g\to 0$, the Weinberg angle increases to π/2, its cosine vanishes, and its sine goes to 1.

*

*But note, the EM charge e  cannot stay invariant, since $e\to g$.

The Fermi constant stays put; the mass of the W  goes to zero (it stops coupling!), and the mass of the Z goes  to $g'v/2$.

NB aside: If you are seeking decoupling, the opposite limit, $g\to \infty$, paradoxically is better behaved, and often taught in class: in that case, e can stay unchanged, $e=g'$, since $\theta_W=0$, the cos is 1, and $m_Z=m_W=\infty$, while the Fermi constant is what it always was: the old Fermi theory! So you may think of EW unification as a descent from infinite to a finite g...
