Why wouldn't the diode in this transistor scheme turn on?
The transistor is off since it's base-emitter junction is reverse-biased.  That allows no current to flow thru the collector, which keeps the LED off.

The B-E junction looks like a diode to the external circuit.  The special thing that a BJT (bipolar junction transistor) does is allow C-E current some multiple of the B-E current.  That multiple is the gain.

Since you have the B-E "diode" reverse biased, no current flow thru it.  That, in turn, does not allow for any collector current.

Being a silicon diode, the B-E junction drops about 600-750 mV when conducting useful current.  Depending on the value of the base resistor, you could start to see the LED come on with the input being about 1 V or more.

Let's work thru some numbers to pick good values for the resistors.  Let's say this is a common green LED rated for 20 mA, and that it drops 2.1 when fully on.  The C-E drop of the transistor in saturation is about 200 mV.  After the LED and the C-E drop, that leaves 6.7 V across the LED resistor.

We want 20 mA to flow, so use Ohm's law:  (6.7 V)/(20 mA) = 335 &Omega;, which is the minimum resistance to not over-drive the LED.  I'd round that up to at least the next common value of 360 &Omega; to give some margin.  Working backwards, the LED current would be 18.6 mA.  You wouldn't notice the difference in brightness between that and 20 mA, even in a side by side comparison.

Let's require the transistor to have at least a gain of 30, which is very easy to find in a small signal NPN.  You want 20 mA collector current, so you need at least (20 mA)/30 = 670 &micro;A base current.

Let's assume you want to drive this LED from a typical digital output that is either 0 V for off and 3.3 V for on.  Figure the B-E junction will drop 700 mV, so that leaves 2.6 V across the base resistor.  Again using Ohm's law, the base resistor needs to be no more than 3.9 k&Omega;.  Round down to the next common value of 3.6 k&Omega;.