Trouble analyzing BJT circuits when we change β from very high to 100
You forget about \$R_B = 47\textrm{k}\Omega\$ resistor and the base current. 

The base current is \$I_B = \frac{2\textrm{mA}}{101} = 19.8\mu\textrm{A}\$ therefore  the voltage drop across \$R_B\$ resistor is \$VR_B = 0V - 19.8\mu\textrm{A}\cdot47\textrm{k}\Omega = -0.930\textrm{V}\$ 

and the voltage at the emitter is \$-0.930\textrm{V} - 0.7\textrm{V} = -1.63\textrm{V}\$