Capacitor - debugging basic LED?
What you are missing is that the capacitor can only charge up to the LED forward voltage. If LED1 is a red LED then it will have a VF (forward voltage) of 1.8 to 2.0 V and that is the voltage you will see on C1. (You should be able to measure this.) Now when you switch off Q1 you have only just enough voltage to light the LED for a very short while before the voltage drops below the LED visible range.

Wire C1 between 3 and 1 and leave out R3. You can get a very rough idea of the discharge time from the time constant \$ \tau = R_1 C_1 = 1k \times 400\mu = 0.4 \ \text s \$.