Thevenin equivalent circuit with external loop
To find Vth

Currents flow only in close loops. You opened two terminals in the circuit. It means, no current will flow through the lower 2 ohm resistor. So you can discard it. Draw the circuit in an understandable manner then. 



You can any method like KVL, nodal voltage, superposition theorem etc to find Vth. 

Use KVL, assuming I is the current in the circuit:
$$V_1 - IR_1 - IR_2 - V_2 = 0$$

On solving \$I = -66.66 mA\$

Therefore -
$$V_{th} = 2.1 + IR_2 \approx 2 V$$

To find Rth

Go back to the original ckt. Short all voltage sources. Open all current sources. The result is:



Rth = R3 + R1||R2 = 8/3 ohms.

Hence the thevenin eq. ckt -



There is no "open circuit current" source as you drew in the ckt.