How soon do you know the exponent of a FP product? You can’t do the last shift until you know this; its in the critical path. Adding the exponents to the two factors gets you to within 1 of the answer. Looking at n high fraction bits (excluding leading 1 bit) of both operands resolves this ambiguity most of the time. This Java program draws these pretty pictures for n=4, n=5, n=6. The combinations in blue represent when the higher exponent is correct, the red squares are for the smaller exponent, the white squares are when the multiplication must finish to be sure of the answer.

Actually it is not as bad as all that. If the multiply rejects denormal inputs and need not produce them there is only a late-contingent shift of one. If it allows denormals then it must be prepared for a long shift, but it can set up those gates with time to spare. The heavy data is in the critical path but the shift amount is not. In either case there is often a late contingent shift of just one.