The code that originally made me discover this is:
if ((a & 0x1) == 0x1) { .. }
In this case we don't actually have any branch probability information and should not assume to have any. LLVM transforms this into:
%and = and i32 %a, 1 %tobool = icmp eq i32 %and, 0
So, in this case, the result of a bitwise and is compared against 0. Again, we don't really have information. We could make this check more narrow, e.g. by testing whether the other and-operand is 1 or whether it has a single bit set.
If you need braces, just put them around the entire case.