Transistor biasing problem
The voltage from V1 is divided between R1 and the B-E junction of the transistor. Since that junction is a forward-biased diode, the voltage drop across it is only about 0.65V; the rest of the voltage appears across R1, setting the magnitude of the base current.

The voltage from V2 is divided among R2, the C-B junction of the transistor, and the B-E junction. Since we've already established that the last one of those is 0.65V, we can say that V2 - 0.65V is divided across R2 qand the C-B junction. As long as this quantity is positive (i.e., V2 is greater than 0.65V), the C-B junction is reversed-biased.