Here is a general solution for n>6.
The rectangles spiral around. To get the solution for a particular n, stop when you have placed the nth rectangle and truncate it so that the whole figure becomes a rectangle. Sadly this solution does not work for n=6, because then rectangles 2 and 6 together form a rectangle.
Here is a proof of why there is no solution for n=6.
1. A single rectangular piece cannot cover two of the corners of the completed figure.
If there were such a piece, then the remaining n-1 pieces would form a sub-rectangle of the figure.
Therefore the four corners of the final figure are from four different pieces.
2. A corner piece must have at least 3 adjacent pieces.
There must be at least one neighbour on each internal side of the corner piece. If a corner piece had exactly two neighbours, there must be exactly one on each side. If the neighbours were both longer than that side, they would overlap. Therefore at least one of those neighbours is the same length, and then the corner piece and that neighbour together form a rectangle. This is not allowed, so a corner piece must have more than two neighbours.
3. Diagonally opposite corner pieces cannot touch.
If diagonally opposite corner pieces touched, then the remaining area of the whole figure would consist of two rectangular areas. If you fill such an area with two or more pieces, those pieces are a sub-rectangle of the figure, which is not allowed. If you fill it with a single piece, then that is a corner piece with exactly two neighbours, which is also not allowed as per #2 above.
Now lets consider n=6 specifically.
Four of those six pieces must be in the four corners of the final figure (#1). Suppose the remaining two pieces are fully internal to the figure. Each of the outside pieces can only expose one side to the internal area, so the internal area is rectangular. Filling it with the remaining two pieces creates a sub-rectangle with 2 pieces.
Suppose on the other hand that all 6 pieces are on the boundary of the final figure, i.e. there are no internal pieces. So we have 4 corner pieces and 2 edge pieces. Suppose a corner piece lies between two edge pieces. It must have a third neighbour (#2), but the only candidate is the diagonally opposite corner, which violates #3. The only other arrangement for the edge pieces is on opposite sides of the final figure, say the left and right sides. The two top corners are adjacent, cannot touch either of the bottom corners, so the only way for them to have 3 neighbours is for both corners to be adjacent to both edge pieces. This is not possible.
The last possibility is that we have 4 corner pieces, 1 edge piece, and 1 internal piece. The two corners next to the edge piece must have the internal piece as their third neighbour. The edge piece has three internal sides and so must have at least three neighbours. The only possibility is that it is also adjacent to the internal piece. In a similar argument to #2, the corners cannot be the same length as the edge piece, and if both were longer then the edge piece and the internal piece have matching lengths and form a rectangle.
All possibilities lead to failure, so n=6 is impossible.