The \( \beta \)-elimination of 2-bromopentane results in the formation of pent-2-ene as the major product, not pent-1-ene. This is because the most favorable elimination (E2) reaction occurs at the beta position with the most accessible hydrogen.
Thus, the correct answer is (C).
| LIST I | LIST II | ||
|---|---|---|---|
| A | Lyman | I | Near IR |
| B | Balmer | II | Far IR |
| C | Paschen | III | Visible |
| D | p-fund | IV | UV |

