At the end of tape loop n, two copper atoms are moved to permanent storage: copper 2n-2, which came from the previous loop, and copper 2n-1. Copper 2n remains a loose atom and must be passed back and forth between chambers while building the next product.

I believe this output chamber and method of construction is forced, though there are more options for the input chamber. 

The output arm must have conduit, bonder, and exactly one hex of the output arranged contiguously around it. With a partial product, waste and a lone product atom occupying 5 hexes around the arm it is not possible to cycle around the arm without accidental bonds being made. This means that product atoms can only come from the conduit to the bonder in one direction, forcing a construction in which the partial product is flipped over. The space required for this manoeuvre limits the arm and one hex of the bonder to the two central hexes of the chamber.

The bonder is off limits during construction, so waste is restricted to the remaining four hexes. Eventually waste will permanently fill three of these hexes, so one of the conduit and output hexes will be covered at all times. We must be able to switch which one is empty, so the conduit and output must have the empty spaces between them.

It is possible to build the product in 1/2P. Such a solve would get below 200i, and I have submitted a showcase just shy of this which manages to make 3.5 outputs before crashing. I believe it eventually becomes impossible to add waste to storage because of the partial product. However, if I've made a mistake then the winning solve should be at 1/2P.

I also did not manage to find a way to bond the waste together into one large chunk, which would save a lot of shuffling.