Let $\displaystyle s(x)=exp(-(\frac{x}{2})^c)$
$\displaystyle x\ge 0$, $\displaystyle c>0$
How does the distribution look like? Can I conclude that $\displaystyle s(x)$ takes the form of an exponential distribution?
Since there is no normalising constant (which will be $\displaystyle \frac{1}{\Gamma \left(\frac{c+1}{c}\right)}$, by the way) it does not define a pdf. Overlooking that, I don't think it has any special name. It certainly is NOT an exponential distribution. The closest I can think of is the exponential power distribution (except it's not).
Try different values of c eg. c = 1/2, 1, 2, 3. Plot the graphs. What do you find?
c = 1/2: plot y = Exp[-(x/2)^1/2] - Wolfram|Alpha
c = 1: plot y = Exp[-(x/2)^1] - Wolfram|Alpha
c = 2: plot y = Exp[-(x/2)^2] - Wolfram|Alpha
c = 3: plot y = Exp[-(x/2)^3] - Wolfram|Alpha