The product obtained on heating 2-Bromopentane with potassium ethoxide in ethanol is

The correct option is trans-2-pentene.

Reaction Mechanism and Explanation:

When 2-bromopentane is heated with a strong base like potassium ethoxide ($C_2{H}_5{O}^{-}{K}^{+}$) in ethanol ($C_2{H}_{5}OH$), it undergoes an elimination reaction, specifically a dehydrohalogenation via the E2 mechanism.

In this E2 elimination, a hydrogen atom from a beta-carbon (adjacent to the carbon containing the bromine atom) and the bromine atom are removed to form a double bond. Let's analyze the structure of 2-bromopentane:
$\stackrel{\stackrel{H}{|}}{C}{H}_2-\stackrel{\stackrel{C{H}_3}{|}}{C}H-\stackrel{\stackrel{Br}{|}}{C}H-\stackrel{\stackrel{H}{|}}{C}{H}_2-C{H}_3$
Here, carbon-2 ($C_2$) is the alpha-carbon bearing the bromine group. The adjacent carbons are carbon-1 ($C_1$) and carbon-3 ($C_3$), which act as beta-carbons. Dehydrohalogenation can proceed in two ways depending on which beta-hydrogen is abstracted:

1. Abstraction of hydrogen from $C_1$:
Elimination yields 1-pentene (a monosubstituted alkene):
$C{H}_3-C{H}_2-C{H}_2-CH=C{H}_2$

2. Abstraction of hydrogen from $C_3$:
Elimination yields 2-pentene (a disubstituted alkene):
$C{H}_3-C{H}_2-CH=CH-C{H}_3$

According to Saytzeff's Rule (or Zaitsev's Rule), the major product of an elimination reaction is the more highly substituted, and therefore more stable, alkene. A disubstituted alkene (2-pentene) is more thermodynamically stable than a monosubstituted alkene (1-pentene). Hence, 2-pentene is the major product.

Furthermore, 2-pentene exists as diastereomers: cis-2-pentene and trans-2-pentene. The trans-isomer is more stable than the cis-isomer because it has less steric hindrance between the bulkier methyl (-$C{H}_3$) and ethyl (-$C_2{H}_5$) groups on opposite sides of the double bond. Therefore, trans-2-pentene is formed preferentially as the major product.