Is there any direct approach to generate discrete 2 partitions of a set of having 2^n elements for a given n ?

Assume that you have a set S of having 2^2 elements first, let S={0,1,2,3}
Then the desired 2 partitions would be
1-{{0,1},{2,3}}
2-{{0,2},{1,3}}
3-{{0,3},{1,2}}

If S={0,1,2,3,4,5,6,7} having 2^3 elements then similarly the 2 partitions would be

1- {{0,1},{2,3},{4,5},{6,7}}
2- {{0,2},{1,3},{4,6},{5,7}}
3- {{0,3},{1,5},{2,6},{4,7}}
4- {{0,4},{1,6},{2,5},{3,7}}
5- {{0,5},{1,4},{2,7},{3,6}}
6- {{0,6},{1,7},{2,4},{3,5}}
7- {{0,7},{1,2},{3,4},{5,6}}

So it seems that for S having 2^n elements the initial table would like

1: {{0,1},{2,3},{4,5},{6,7},...,{(2^n)-2,(2^n)-1}}
2: {{0,2},... }
.
.
.
(2^n)-1: {{0,(2^n)-1},... }

I am just wondering whether there is an direct approach to generate the above table for a given n.

Thanks