Playing Settlers of Catan with a D12 instead of 2D6
Significantly.  The thing about 2d6 is that you've got a range of numbers that have a certain probability of them showing up.

When you're rolling 2d6, your available numbers are 2-12.  You're dropping off the number 1.  Also, you have an escelating scale of percentage probability.  2's and 12's will happen 2.77% of the time (each).  3's and 11's will happen 5.55% of the time.  4's and 10's happen 8.33% of the time.  5's and 9's happen 11.11% of the time.  6's and 8's will happen 13.88% of the time.  And 7's will happen 16.66% of the time.

Result    Rolls with a 2d6        ≠  Rolls with a 1d12
------------------------------------------------------
 1        (impossible)                1
 2        1+1                         2
 3        1+2 2+1                     3 
 4        1+3 2+2 3+1                 4
 5        1+4 2+3 3+2 4+1             5 
 6        1+5 2+4 3+3 4+2 5+1         6
 7        1+6 2+5 3+4 4+3 5+2 6+1     7
 8        2+6 3+5 4+4 5+3 6+2         8
 9        3+6 4+5 5+4 6+3             9
10        4+6 5+5 6+4                10
11        5+6 6+5                    11
12        6+6                        12


As you can see, the closer you get to 7, the greater chances you have of something on those hex's from happening.  Because of this, the game has been weighted so that certain places will not show up as often as others.  That's part of the resource management and placement.  It's what makes the game interesting.

By using a 1d12, you have two things working against you.  One, you have a dead number that you won't be using.  And two, each other number will only show up 8.33% of the time (as much as 4's and 10's used to show up).  And they all have the same chance, so strategy would be very different.  I would think it would make it a whole new game.