jmc

If $y=\frac{1-x}{2x+3}$, then $1-x=(2x+3)y=2xy+3y$. We can rearrange to $1-3y=x(2y+1)$. When $2y+1=0$ or $y=-\frac{1}{2}$, the left hand side is nonzero while the right hand side is zero, so the value $y=\boxed{-\frac{1}{2}}$ is unattainable.