Unexpected troubles converting char array to float

I have a char array and need to convert it to convert it to float.

I have two 16 bit values:
A = 7C37
B = 428B
and they need to be arranged like 42 8B 7C 37 to represent 69.743 in float but I have troubles to get there, I have tried the following:

#include <stdio.h>
#include <stdint.h>
#include <string.h>

int main() {
    char pData[8] = {0x00, 0x00, 0x00, 0x00, 0x7c, 0x37, 0x42, 0x8b};
    float floatValue;
    uint32_t raw;

    raw = ((uint32_t)pData[5])      | (((uint32_t)pData[4])<< 8) |
         (((uint32_t)pData[7])<<16) | (((uint32_t)pData[6])<<24);

    // Copy the bytes in the correct order into the float variable
    memcpy(&floatValue, &raw, sizeof(floatValue));

    // Print the float value
    printf("The float value is: %f\nThe raw value is 0x%x\n", floatValue, raw);

    return 0;
}

.. but am confused as this unexpectedly produces the following output:

The float value is: -nan
The raw value is 0xff8b7c37

I assume the -nan probably shows because of the 0xff in the left most Byte, it should be set to 0x42, anyone can assist me with this?

UPDATE: I basically want to implement the Float - Big Endian (ABCD) from https://www.scadacore.com/tools/programming-calculators/online-hex-converter/ where 428B7C37 produces 69.74261

Once I got to squeeze the chars into a 32bit uint, it's as easy as:

#include <stdio.h>

union FloatIntUnion {
    float f;
    unsigned int i;
};

int main() {
    union FloatIntUnion fiu;
    fiu.i = 0x428B7C37; // Set the integer value
    printf("%f\n", fiu.f); // Interpret the bit pattern as a float and print it
    return 0;
}