Consider a capacitor (capacitance = C), which is wired to a battery, is charged to potential V volts, and has Q coulombs charge. Now the battery is removed, and a 2nd identical capacitor (uncharged) is connected to the 1st capacitor.
We find that (due to charge conservation in parallel config), the final potential across the system is (V/2) volts, and system capacitance is (2C).
This means that the system's energy is (1/2) * (2C) * (V/2)² = (1/4) CV².
However, the energy stored by the initial capacitor is (1/2) CV².
Where did the remaining energy go ???