Why is 0x1 interpreted as less than 0xC0000000?

I'm learning about binary representation of integers and tried to write a function that returns an int multiplied by 2 using saturation. The thought process is if the value overflows positively the function returns INT_MAX, and conversely if it overflows negatively it returns INT_MIN. In all other cases the binary value is left shifted by 1.

What I'm wondering is why I have to cast the value 0xC0000000 as an int in order to get my function to work correctly when I pass the argument x = 1.

Here is my function:

int timestwo (int x){
    if(x >= 0x40000000) // INT_MAX/2 + 1
        return 0x7fffffff; // INT_MAX
    else if(x < (int) 0xC0000000) // INT_MIN/2
        return 0x80000000; // INT_MIN
    else
        return x << 1;
    return 0;
}

Hexadecimal (and octal) literals in C are typed using the smallest promoted (=int or a higher ranking type) type, signed or unsigned, that can accommodate the value. This differs from decimal literals, which stay within signed types if they don't have the u/U suffix, or within unsigned types otherwise (6.4.4.1p5):

This makes 0xC0000000 on a system with 32-bit integers unsigned and comparing (or otherwise pairing by means of an operator) an unsigned with a signed of the same rank forces the signed to become unsigned (6.3.1.8), so without the (int) cast you get an implicit (unsigned int)x < (unsigned int) 0xC0000000.