When the transistor is turned on in that manner it can be switched into saturation if the base current is sufficient. The collector-emitter emitter voltage will drop to about 0.2 V. This voltage will be applied to the LED.
Figure 1. Current through various colours of LED as a function of forward voltage. Image source IV curves.
Figure 1 shows that none of the LEDs from infrared to ultra-violet will pass any significant current at 0.2 V. There just isn't enough voltage to get the charge carriers to jump the P-N junction.
Figure 2. A water check-valve analogy. Image source: What is an LED?.
If you look at the check-valve in the figure above, it should be clear that the spring normally keeps the ball in position and prevents back-flow. When “forward-biased” the ball shut-off can be moved against the spring but it will take some initial pressure to move the ball. This results in a pressure drop across the valve: the pressure downstream will be less than the inlet pressure.
In a similar manner the PN junction in an LED causes a voltage drop. For a red LED it is about 1.5 V to 2.0 V. You need to exceed the Vf to get enough current to flow and light the LED.
The links are to articles by me and may help you further.