The temperature of a chemical compound was 2 C at 4​:00 A.M. During a​ reaction, it dropped 3 C per minute until 4​:12 A.M. What was the temperature at 4​:12 ​A.M.?

From 4:00 A.M. to 4:12 A.M., there are 12 minutes in total.

Since the temperature dropped 3 C per minute, the total temperature drop is 3 C * 12 minutes = <<3*12=36>>36 C.
Therefore, the temperature at 4:12 A.M. is 2 C - 36 C = -34 C. Answer: \boxed{-34}.

To find the temperature at 4:12 A.M., we need to calculate the number of minutes that have passed from 4:00 A.M. to 4:12 A.M.

To do this, we subtract the two times:

4:12 A.M. - 4:00 A.M. = 12 minutes

Now we know that 12 minutes have passed.

Next, we need to calculate the temperature drop over this time period. The temperature dropped 3°C per minute, so we multiply the number of minutes by the rate of temperature drop:

12 minutes x 3°C/minute = 36°C

Therefore, the temperature at 4:12 A.M. was 2°C - 36°C = -34°C.

Note that the negative sign (-) indicates a drop in temperature.