Feedback op-amp voltage regulator help
First of all, a 9V battery will not output 2Amp. This alkaline battery http://data.energizer.com/PDFs/522.pdf can give you 500mA but even then the rated capacity  is halved.

Next, your "Zener circuit" will start conducting above 17V (it starts conducting when Q2 base is 0.7V above ground, so when current through voltage divider reaches 0.7V/4.7kOhm=0.149mA ; this happens when input voltage is 114.7kOhm * 0.149mA = 17V). As-is it is equivalent to the voltage divider alone ; your + input is therefore about 0.2V below 9V. So the op-amp will be fully-on all the time.

Finally, did you use this particular amplifier? That's not an op-amp (gain is 20x) and the output is referenced to Vcc/2 (here to 4.5V). So, when inputs are equal the output voltage for your circuit is 3.8V.

I'd suggest you replace it with a proper op-amp, and choose it so that the output voltage can swing to at least 1V above output (that's 3V below rail) while delivering significant current. Also mind the input voltage range...

Or, if you're not after high precision (as your use of a non-temperature-compensated reference suggests) why not use directly your zener "circuit" and a pass transistor? Something like this :







simulate this circuit &ndash; Schematic created using CircuitLab

This works because the zener steals the current from the transistor base as soon as the voltage is above the zener threshold, which is 0.7V above the output (because of the base-emitter voltage drop of the transistor).  

Obviously for 2A you also need a beefier pass transistor, something like TIP41 instead of poor little BC547... It will have to dissipate max 2A*4V, that's 8W of power!! Then lower R3 to 47 Ohm.