Why can't I solve the 4x4x4 centers as a 2x2x2 ??

Schmidt

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If you do F R U R' U' F' on a solved 2x2x2 cube it gets to a position where you need to do y2 and the same alg. again.

If you do f r u r' u' f' on a 4x4x4 cube it is the same.

So why can't I solve the 4x4x4 centers on a scrambled cube like I would solve a scrambled 2x2x2 ( using the algs. used in the Ortega method) ?
 
Centers on a 2x2x2 are connected to each other. But 4x4x4 is not- just do something like r U on your 4x4x4. r makes it reachable by 2x2x2, but U shows the centers are independent to each other.
 
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