Why Does Calculating Torque Away from the Center of Mass Cause Confusion?

[transparent]

Tell me where I've gone wrong (here r(1,2) means position of point 1 with respect to point 2 etc):

ω x r(1,2)=v(1,2)
Differentiating both sides with respect to time:
ω x v(1,2) + $\alpha$ x r(1,2) = a(1,2)
=>ω x (ω x r(1,2)) + $\alpha$ x r(1,2) = a(1,2)

Now let us imagine a uniform rod in free space of length l and mass m, lying with the origin at its centre. A force F(+$\hat{j}$) is applied at one end (+l/2$\hat{i}$). We observe the motion about centre of mass first at time 0:
I_cm=ml²/12
$\tau$=+l/2$\hat{i}$ x F$\hat{j}$=F*l/2$\hat{k}$.
$\alpha$=6F/(m*l) $\hat{k}$
Now since ω=0,
a(cm, positive end)=6F/(m*l) $\hat{k}$ x l/2$\hat{i}$= 3F/m $\hat{j}$

Now we observe the motion about the positive end:
$\tau$=0 (since the force is applied at that very point)
=> $\alpha$=0
ω=0
Hence a(positive end, cm)=0
=>a(cm, positive end)=0

What have I done wrong? :(
Edit: By positive end I mean the end which is on the positive side of the x axis.

 

[Doc Al]

Now we observe the motion about the positive end:
$\tau$=0 (since the force is applied at that very point)
=> $\alpha$=0
ω=0
Hence a(positive end, cm)=0
=>a(cm, positive end)=0

What have I done wrong? :(

Things aren't so simple when you take torques with respect to an accelerating point other than the center of mass. You cannot set net torque equal to Iα in such a case.

 

[transparent]

Things aren't so simple when you take torques with respect to an accelerating point other than the center of mass. You cannot set net torque equal to Iα in such a case.

Why? And isn't the centre of mass accelerated?

Edit: Didn't read "other than the centre of mass" part. But still, why is it so?