Solution:

The three operations can be ordered in $3!=6$ ways, but performing the addition first or last leaves the two multiplications interchangeable, so at most four values arise.

Indeed $$(2\times3)+(4\times5)=26,\quad (2\times3+4)\times5=50,$$$$2\times(3+4\times5)=46,\quad 2\times(3+4)\times5=70$$ are four distinct values.

Thus, the correct answer is C.