Given that the function $f(x) = \sin(2x + \phi)$ is an even function,

By the definition of even functions, we have $f(-x) = f(x)$, which implies

$$\sin(-2x + \phi) = \sin(2x + \phi).$$

Using the property of sine function that $\sin(A) = \sin(B)$ if and only if $A = B + 2k\pi$ or $A = \pi - B + 2k\pi$ for some integer $k$, we obtain

$$-2x + \phi = 2x + \phi + 2k\pi \quad \text{or} \quad -2x + \phi = \pi - (2x + \phi) + 2k\pi.$$

Solving the first equation gives us $x = -k\pi$, which is not possible since $x$ is a variable. Solving the second equation gives us

$$\phi = \dfrac{\pi}{2} + k\pi.$$

Since we need $\phi$ to be a single value, we take $k = 0$, and thus $\phi = \dfrac{\pi}{2}$.

Therefore, the answer is $\boxed{\text{C}: \dfrac{\pi}{2}}$.