**Step 1: Recall the dot product formula** The dot product of two vectors \(\vec{p}\) and \(\vec{q}\) is given by: \[\vec{p} \cdot \vec{q} = |\vec{p}| |\vec{q}| \cos{\theta}\] where \(\theta\) is the angle between the two vectors.

**Step 2: Apply the dot product formula to the given vectors** We want to find the angle between \(2\vec{a}\) and \(-\vec{b}\). Let this angle be \(\theta\). Then: \[(2\vec{a}) \cdot (-\vec{b}) = |2\vec{a}| |-\vec{b}| \cos{\theta}\] \[-2(\vec{a} \cdot \vec{b}) = 2|\vec{a}| |\vec{b}| \cos{\theta}\]

**Step 3: Substitute the given values** We are given that \(|\vec{a}| = 1\), \(|\vec{b}| = 2\), and \(\vec{a} \cdot \vec{b} = \sqrt{3}\). Substituting these values into the equation: \[-2(\sqrt{3}) = 2(1)(2) \cos{\theta}\] \[-2\sqrt{3} = 4 \cos{\theta}\]

**Step 4: Solve for \(\cos{\theta}\)** \[\cos{\theta} = \frac{-2\sqrt{3}}{4} = -\frac{\sqrt{3}}{2}\]

**Step 5: Find the angle \(\theta\)** We need to find the angle \(\theta\) such that \(\cos{\theta} = -\frac{\sqrt{3}}{2}\). Since the range of the angle between two vectors is \([0, \pi]\), the angle is: \[\theta = \frac{5\pi}{6}\]