Spot's doghouse has a regular hexagonal base that measures one yard on each side. He is tethered to a vertex with a two-yard rope. What is the area, in square yards, of the region outside of the doghouse that Spot can reach?

$ \text{(A)}\ 2\pi/3 \qquad
\text{(B)}\ 2\pi \qquad
\text{(C)}\ 5\pi/2 \qquad
\text{(D)}\ 8\pi/3 \qquad
\text{(E)}\ 3\pi$
1. **Understanding the problem**: Spot is tethered to a vertex of a regular hexagonal doghouse with a two-yard rope. We need to find the area outside the doghouse that Spot can reach.

2. **Hexagon properties**: A regular hexagon can be divided into 6 equilateral triangles. Each side of the hexagon is 1 yard.

3. **Circle sector calculation**: Spot can move in a circle with a radius of 2 yards around the vertex he is tethered to. However, his movement is restricted by the hexagon. We need to calculate the area of the sectors he can reach.

4. **Main sector**: The main sector is a 240-degree sector (or \( \frac{2}{3} \) of a full circle) because Spot can move around the vertex but is blocked by the hexagon on two sides. The area of this sector is:
   \[
   \text{Area of main sector} = \frac{2}{3} \pi r^2 = \frac{2}{3} \pi (2)^2 = \frac{2}{3} \pi \cdot 4 = \frac{8\pi}{3}
   \]

5. **Additional sectors**: Spot can also reach around the adjacent vertices of the hexagon. Each of these vertices allows Spot to move in a 60-degree sector (or \( \frac{1}{6} \) of a full circle) with a radius of 1 yard (since the side length of the hexagon is 1 yard). There are two such sectors:
   \[
   \text{Area of one small sector} = \frac{1}{6} \pi r^2 = \frac{1}{6} \pi (1)^2 = \frac{\pi}{6}
   \]
   Since there are two such sectors:
   \[
   \text{Total area of small sectors} = 2 \times \frac{\pi}{6} = \frac{2\pi}{6} = \frac{\pi}{3}
   \]

6. **Total area calculation**: Adding the areas of the main sector and the two smaller sectors:
   \[
   \text{Total area} = \frac{8\pi}{3} + \frac{\pi}{3} = \frac{8\pi + \pi}{3} = \frac{9\pi}{3} = 3\pi
   \]

The final answer is \( \boxed{3\pi} \).