What happens to the capacitance of a system if capacitors are charged in parallel and then put into a series circiut
Each capacitor stores energy which is conserved.  The energy stored in one of the capacitors is

$$\frac{1}{2}CV^2 = \frac{1}{2}(2\:\mathrm F) (2\:\mathrm V)^2 = 4\:\mathrm J$$

for a total of \$8\:\mathrm J\$ of stored energy.  Whether the capacitors are placed in parallel or series, the amount of energy stored is the same.

If the charged capacitors are placed in parallel appropriately, the voltage across the combination is \$2\:\mathrm V\$ and the energy stored is \$8\:\mathrm J\$.

Thus the equivalent capacitance is

$$C_\text{EQ} = 2\frac{8\:\mathrm J}{(2\:\mathrm V)^2} = 4\:\mathrm F$$

If the charged capacitors are placed in series appropriately, the voltage across the combination is 4V and the energy store is \$8\:\mathrm J\$.

Thus, the equivalent capacitance is

$$C_\text{EQ} = 2\frac{8\:\mathrm J}{(4\:\mathrm V)^2} = 1\:\mathrm F$$

Yes, there is a \$3\:\mathrm F\$ difference but asking "what happened to the other 3 farads?" is like asking "what happened to the other 3 ohms?" when comparing series and parallel connected 2 ohm resistors.

No capacitance has 'vanished'.  Both capacitors still have \$2\:\mathrm F\$ each of capacitance.  What has changed is the configuration of the capacitors and, thus, the equivalent capacitance as seen by an external circuit.