Intuition behind Capacitive reactance?
I think you need to consider these basic relationships for a capacitor: -
$$Q = CV \hspace{1cm}\text{(charge = capacitance}\cdot\text{voltage)}$$
$$\dfrac{dQ}{dt} = I\hspace{1cm}\text{(rate of change of charge = current)}$$
$$\text{Therefore,}\hspace{1cm}I = C\cdot \dfrac{dV}{dt}$$
So, in the last formula, current is greater if the rate of change of voltage rises and, current is greater (for the same rate of change of voltage) when capacitance increases.
The last formula above directly leads to the AC reactance of a capacitor being \$\frac{1}{\omega C}\$.
If a sinewave voltage is applied to a capacitor then current will be maximum when the rate of change of voltage is maximum hence, for AC circuits you should be able to visualize that current and voltage waveforms are displaced by a phase angle of 90°.
For an inductor we have: -
$$V = L\cdot \dfrac{di}{dt}$$
So, do you see the similarities and shouldn't you expect a phase angle to be produced in both scenarios?