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

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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?