Your winding resistance will go up by 3.44 squared which close to 12 times the original,so will your copper losses,this will be intolerably inefficient,if you used say, twice as many turns instead the increase would be 4 to 1. Lets say your input power is 1 your losses 0.1 and you efficiency is 0,9 in the original case.now the loss is 0.4 and the output 0.6 your back emf is 0.6, which is proportional to rotational speed times number of windings (assuming that the flux is constant)therefore .

0.9/(Sxw)= 0.6/(S'x2w) or 0,9/S= 0.3/S' or S'= S/3 the rotational speed S' will be 1/3 of the original S, if the input voltage and current are the same.The same input power that is,your efficiency will be 60% instead of 90% due to increased copper losses.Your no load speed will be somewhat less than half the original speed, but drop off to 1/3 due to the increased winding voltage drop under load.You end up overloading the motor unless you reduce the current.I made a gues at the original winding voltage drop (0.1Vin),it may have been optimistic,so things could be even worse.The torque will have gone up by a factor 2.twice the torque at 1/3 of the speed means 60% of the original output power. Reducing the operating speed substantially ends up reducing the power output capability of the motor,or thermally overloading it.This is obviously not a good idea.