I'm trying to make a circuit that has a load resistance of 2 ohms, load current draw of 2 A and voltage of 5V.
There is something not quite right with your description. I take it you mean that you have a 5 V supply and that you want to provide a constant 2 A through a 2 Ω load.
From Ohm's Law we can calculate the voltage drop across the resistor as \$ V = IR = 2 \times 2 = 4 \ \text V \$. That means that the transistor will need to drop 1 V from the 5 V supply.
Now you have a problem. You have selected a Darlington transistor. You will lose 0.7 V across each of its base-emitter junctions so that to get 4 V at the emitter you will need 5.4 V at the base. You will need an even higher voltage at the left end of the 430 Ω resistor. (Please number your components R1, R2, etc. to aid discussion.) In other words, your circuit will not work.
simulate this circuit – Schematic created using CircuitLab
Figure 1. (a) NPN version. (b) PNP version.
There's more bad news. The TIP120 Darlington transistor will have a voltage drop of 2 V when in saturation (turned fully on). That means that you only have 3 V for your 2 Ω resistor so you will only get 1.5 A through it.
If you replace the Darlington with a regular transistor it will drop < 0.5 V in saturation so you will have 4.5 V or so for the load and get 2.25 A through your load.