2018 Knoxville Challenger – Doubles
# 2018 Knoxville Challenger – Doubles

Leander Paes and Purav Raja were the defending champions but only Paes chose to defend his title, partnering Miguel Ángel Reyes-Varela. Paes lost in the quarterfinals to Alex Lawson and Jackson Withrow.

Toshihide Matsui and Frederik Nielsen won the title after defeating Hunter Reese and Tennys Sandgren $7–6^{(8–6)},$ 7–5 in the final.

## Seeds

* Leander Paes / Miguel Ángel Reyes-Varela (quarterfinals)
* Marcelo Arévalo / Roberto Maytín (first round)
* Matt Reid / John-Patrick Smith (first round)
* Marcel Granollers / Mark Vervoort (first round)

## Draw

### Key

* Q = Qualifier
* WC = Wild card
* LL = Lucky loser
* Alt = Alternate
* SE = Special exempt
* PR = Protected ranking
* ITF = ITF entry
* JE = Junior exempt
* w/o = Walkover
* r = Retired
* d = Defaulted
* SR = Special ranking

| First round | First round | First round | First round | First round | Quarterfinals | Quarterfinals | Quarterfinals | Quarterfinals | Quarterfinals | Semifinals | Semifinals | Semifinals | Semifinals | Final | Final | Final |
| - | - | - | - | - | - | - | - | - | - | - | - | - | - | - | - | - |
| 1 | L Paes MÁ Reyes-Varela | 6 | 6 | | | | | | | | | | | | | |
| | H Hach Verdugo R Roelofse | 4 | 2 | | 1 | L Paes MÁ Reyes-Varela | 3 | 4 | | | | | | | | |
| | A Lawson J Withrow | 6 | 6 | | | A Lawson J Withrow | 6 | 6 | | | | | | | | |
| WC | JC Aragone B Fratangelo | 4 | 4 | | | | | | | A Lawson J Withrow | 7 | 1 | [3] | | | |
| 4 | M Granollers M Vervoort | 4 | $6^4$ | | | | | | | T Matsui F Nielsen | 5 | 6 | [10] | | | |
| | T Matsui F Nielsen | 6 | $7^7$ | | | T Matsui F Nielsen | 6 | 3 | [10] | | | | | | | |
| WC | C Eubanks M Krueger | $6^0$ | 4 | | WC | T Stodder P Touliatos | 2 | 6 | [5] | | | | | | | |
| WC | Timo Stodder Preston Touliatos | $7^7$ | 6 | | | | | | | | | | | T Matsui F Nielsen | $7^8$ | 7 |
| | P Polansky A Shamasdin | 6 | 3 | [10] | | | | | | | | | | H Reese T Sandgren | $6^6$ | 5 |
| Q | H Heliövaara H Laaksonen | 4 | 6 | [2] | | P Polansky A Shamasdin | $6^4$ | $7^7$ | [10] | | | | | | | |
| | R Gonzales N Lammons | 3 | 6 | [10] | | R Gonzales N Lammons | $7^7$ | $6^4$ | [8] | | | | | | | |
| 3 | M Reid J-P Smith | 6 | 3 | [5] | | | | | | P Polansky A Shamasdin | 4 | 2 | | | | |
| | H Reese T Sandgren | 3 | 6 | [10] | | | | | | H Reese T Sandgren | 6 | 6 | | | | |
| | J Jung S Verbeek | 6 | 3 | [8] | | H Reese T Sandgren | 6 | 6 | | | | | | | | |
| | R Galloway N Pasha | 6 | 3 | [10] | | R Galloway N Pasha | 2 | 4 | | | | | | | | |
| 2 | M Arévalo R Maytín | 4 | 6 | [3] | | | | | | | | | | | | |
