The floors
The floor under every comparison sort
No algorithm that sorts by comparing pairs of elements can average fewer than log₂(n!) comparisons. Not one that exists, and not one that ever will. The argument takes three sentences, it is about counting leaves in a tree, and it is one of the few results in this subject that is genuinely about every possible algorithm rather than about a particular one.
How close anything gets to the floor
The interesting question about a sorting algorithm is not its complexity class but its distance from the bound nothing can cross. Merge sort comes within 2.2% of the information-theoretic floor. Heapsort uses 96% more than it needs to. Selection sort uses nineteen times. Those three numbers say more than the classification does.
The floor moves when the question does
Sorting 4,096 elements needs at least 43,250 comparisons. Finding one element among the same 4,096, already sorted, needs at least 13. The difference is a factor of 3,300 and it comes entirely from how many different answers the algorithm has to be able to give. A lower bound is a property of the question, not of any algorithm.