analog shut down whilst turning on an LED
simulate this circuit &ndash; Schematic created using CircuitLab

Figure 1. Voltage follower. Figure 2. "Open collector" switch.

The way you have connected results in the situation in Figure 1. The base current flows through R1 which decreases the voltage at the base of Q1 to maybe about 3.6 V. The emitter will be 0.6 V less so should read about 3 V. You can check these with your multimeter. So, your circuit is losing voltage all the way.

You could make Figure 1 a little better by removing R1 so that it becomes an emitter follower. The emitter voltage would then be one diode voltage drop less than the output of OA1.

The normal way we do this is to ground the emitter and switch a load in the collector circuit as shown in Figure 2. The advantage is that with the 5 V drive from OA2 we can turn Q2 "hard on" giving minimum resistance between collector and emitter. This also has the advantage that we could use the transistor to switch a relay or lamp powered by a higher voltage - 12 V, for example.