c
(c) \(C{{H}_{3}}-\underset{Br\,\,}{\mathop{\underset{|\,\,\,}{\mathop{CH}}\,}}\,-C{{H}_{2}}-C{{H}_{2}}-C{{H}_{3}}+KOH\)

\(\underset{\begin{smallmatrix} 
 \text{Elimination} \\ 
 \,\,\,\,\,\text{reaction} 
\end{smallmatrix}}{\mathop{\xrightarrow{{{C}_{2}}{{H}_{5}}OH}}}\,\underset{\text{2-Pentene}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}{\mathop{C{{H}_{3}}-CH=CH-C{{H}_{2}}-C{{H}_{3}}}}\,\)

When alkyl halide reacts with alc. \( KOH\) then it favours elimination reaction (Dehydrohalogenation). Since, trans pentene \(-2\)  is more symmetrical than cis isomers. Hence, it is main product.