The plaintext is:

you all need to alert the government that J L Morrison is the traitor and has been selling state secrets to the Russians go

Sadly, the intelligence was not received in a timely manner and John's employer is now a Russian vassal state. Perhaps the Russians will teach their new officers some proper cryptanalysis.

Take the five strings of hex digits in order and consider them as a single ciphertext:

16e46c8cebc3dc8c6073a8d87159cf1d38aedf7b5fc5f63749e0fb6d17073973eea7a4
9e871680fdb729dd7f391c42a30edd94fd3ddb712cc59578cc418fc4c8903cb5fb0fe3
edf32055eae75b8d66377ce845c85f1f7188c70fbac45c6381a0d78acedb16d5f23b49
ced8744c5164541b732491d1c5cdcff73b6ced458cc48533fed81317ce177051b357c7
dcd7174c4d8f8a8c93b4edcf86c1ede3b078d3b3a416305ea1d71330edf3735c997b3f

Use the Index of Coincidence to detect any periodic phenomena in the hex digits.

The yellow plot (the Kappa test) is similar to Autocorrelation and counts the number of digits that repeat at each distance. This plot tells us that digits tend to recur at distances 7, 14, 21, etc.

The blue plot (the Kullback Test) arranges the text into different widths and for each width counts the average IoC down every column. The plot tells us that on a width of 7, or 14, or 21, or so on, each column contains a lot of repeated hex digits.

Since hex digits can be used to represent binary data, find out if the same period exists at the binary level. Convert the hex strings to binary:

1    6    e    4    6    c    8    c    e    b      [...]
0001 0110 1110 0100 0110 1100 1000 1100 1110 1011   [...]

Repeat the IoC test. Sure enough, the period of 7 shows up at the binary level too, but the period of 14 is an even stronger hit.

Investigate the period of 7 further. Break the binary string into 7-bit "words", and count the number of times a 1 bit appears at each of the 7 positions in that word. (In other words, write the string on a width of 7 and count the 1 bits in each column.)

There is some roughness on bits 2, 4, and 5. Recall that a Hamming(7, 4) code uses bits 2, 4, 5, and 6 as its data bits. Group the binary string into seven-bit groups and attempt to apply the Hamming code to decode each group into a four-bit data group. dCode's online tool can be used for this, though I did write my own to overcome dCode's limit on input length. My own tool also tells me that each group has only a single flipped bit.

Encoded: 0001011 0111001 0001101 1001000 1100111 0101111  [...]
Errors:  0000100 0100000 0000010 0000100 0000001 0100000  [...]
Fixed:   0001111 0011001 0001111 1001100 1100110 0001111  [...]
Data:      0 111   1 001   0 111   0 100   0 110   0 111  [...]

The result is 800 bits long, and the first bit of every group of 8 is always a zero. This is probably ASCII-encoded text, so interpret it as such:

ytgtiraicRooohsasnruuavaoibgesaletntestslerJioetsilrnLsrnatantmMtastoneteohneetsehnredlshgdetrthleeo

There are 100 characters of text here, a square number. Print them on a width of 10. The plaintext appears on the columns.

ytgtiraicR
ooohsasnru
uavaoibges
aletntests
lerJioetsi
lrnLsrnata
ntmMtaston
eteohneets
ehnredlshg
detrthleeo

youallneedtoalertthegovernmentthatJLMorrisonisthetraitorandhasbeensellingstatesecretstotheRussiansgo