A Row of Card Pairs

Take a single pair each of Aces, 2s, 3s, and 4s from a deck of cards for a total of 8 cards.  Arrange the cards in a row, such that one card is between the aces, two cards are between the 2s, three cards are between the 3s, and four cards are between the 4s.

Extension 1: Find solutions for 3 pairs (six cards: Aces, 2s and 3s) to 8 pairs (16 cards: Aces, 2s, ..., 8s).

Extension 2: Is there an upper bound?  Can you find some N>8, such that N pairs cannot be arranged in this way?

Source: Reader Jimmy Chng Gim Hong, though I've seen it elsewhere.

Solutions were received from Jospeh DeVincentis, Adrian Atanasiu, Toby Gottfried, Alan O'Donnell, Claudio Baiocchi, David Perryman, Sandy Thompson, Al Zimmermann, Erika Brandner, and Denis Borris. Jospeh DeVincentis's solution below is quite thorough.  The solutions for 3 and 4 pairs are unique.  For higher number of pairs there are many solutions.

I don't think there is an upper bound; you get more freedom to arrange 
the different pairs as you increase the number of pairs. However, there 
is a parity restriction on the numbers of pairs you can use.

Each odd pair (like Aces or 3s) must go in two positions which are both 
even or both odd if you number all the positions sequentially. Each even 
pair goes in an odd position and an even position. Thus, if you have an 
odd number of odd pairs (as with 5 or 6) then you need an unbalanced 
number of even and odd positions (at least two more of one type) but in 
fact you actually have the same number of even and odd positions. 
Therefore it can only be done for numbers of pairs equivalent to 3 or 0 
mod 4.

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