You have an array of $n$ elements. Suppose you implement quicksort by always choosing the central element of the array as the pivot. Then the tightest upper bound for the worst case performance is

In Quicksort, choosing any fixed position (first element, last element, or the middle/central element) as the pivot does not prevent the worst-case scenario from occurring.

There exist input permutations (such as certain adversarial or already partitioned sequences) where the central element chosen as the pivot is consistently the smallest or largest element in the current subarray, resulting in an unbalanced partition of sizes $0$ and $n-1$.

The recurrence relation in the worst case is:

Solving this recurrence yields:

Hence, the tightest upper bound for the worst-case performance is $O(n^2)$.