Calculate expected voltage around a resistor
From the Ohm's law, we know that the current in the circuit will be of (considering only resistor has any resistance)


Na-ah, the LED has a resistance too. Common LEDs use 2.5V at 20mA. We can calculate the LED's resistance (on 2.5V) with \$R=\frac{V}{I}=\frac{2.5}{0.02}=125\Omega\$. 

Now, when it comes to your circuit, you shouldn't just use a 2K resistor, you should think about that and calculate the needed value. Now, the voltage over the LED probably won't be 2.5V, which means the LED has a different resistance, different current is flowing, etc. You cannot calculate the current flowing in this circuit and therefore you cannot calculate the voltage over the resistor (we'll come to that later). Here's how you should design your circuit:

You want a 0.02A current through the circuit (the current is the same everywhere as it is a series circuit). That means the total resistance has to be \$R_{tot}=\frac{V}{I}=\frac{6}{0.02}=300\Omega\$. We already saw the resistance of the LED was \$125\Omega\$, that means your resistor has to be \$300-125=175\Omega\$. 

Now, how do you calculate the voltage over the resistor? Take Ohm's law, the current through the resistor (0.02A) and the resistance (\$175\Omega\$):

$$V=I\cdot{}R = 0.02\cdot175 = 3.5V$$

Note that this also equals \$6V-2.5V\$.