The major product formed in dehydrohalogenation reaction of 2-Bromo pentane is Pent-2-ene. This product formation is based on ?

The correct option is Saytzeff’s Rule.

Step-by-Step Explanation:

1. Understanding the Reaction:
The reaction described is the dehydrohalogenation of 2-bromopentane. Dehydrohalogenation is an elimination reaction (specifically an $E2$ or $E1$ mechanism) where a hydrogen halide ($H\mathrm{Br}$) is removed from an alkyl halide to form an alkene.

2. Analyzing the Reactant (2-Bromopentane):
The structure of 2-bromopentane is:
${\mathrm{CH}}_3-\mathrm{CH}\left(\mathrm{Br}\right)-{\mathrm{CH}}_2-{\mathrm{CH}}_2-{\mathrm{CH}}_3$
The carbon attached to the bromine atom (C-2) is the $\alpha$-carbon. The adjacent carbons, C-1 and C-3, are $\beta$-carbons and have hydrogen atoms (called $\beta$-hydrogens) that can be eliminated:
- C-1 is a primary carbon with three $\beta$-hydrogens: $-{\mathrm{CH}}_3$
- C-3 is a secondary carbon with two $\beta$-hydrogens: $-{\mathrm{CH}}_2-$

3. Possible Elimination Products:
Depending on which $\beta$-hydrogen is removed along with the bromine atom, two different alkene products can be formed:
- Route A (elimination from C-1): Removal of a hydrogen from C-1 and bromine from C-2 yields Pent-1-ene:
${\mathrm{CH}}_2=\mathrm{CH}-{\mathrm{CH}}_2-{\mathrm{CH}}_2-{\mathrm{CH}}_3$ (a monosubstituted alkene).
- Route B (elimination from C-3): Removal of a hydrogen from C-3 and bromine from C-2 yields Pent-2-ene:
${\mathrm{CH}}_3-\mathrm{CH}=\mathrm{CH}-{\mathrm{CH}}_2-{\mathrm{CH}}_3$ (a disubstituted alkene).

4. Applying Saytzeff’s (Zaitsev's) Rule:
Saytzeff's Rule states that in an elimination (dehydrohalogenation) reaction of an alkyl halide, the preferred major product is the more highly substituted alkene, which is thermodynamically more stable.
- Pent-2-ene is disubstituted (two alkyl groups attached to the double-bonded carbons).
- Pent-1-ene is monosubstituted (only one alkyl group attached to the double-bonded carbons).
Therefore, according to Saytzeff's Rule, Pent-2-ene is formed as the major product (about 81%) because it has more alkyl substituents attached to the double bond, making it more stable than Pent-1-ene.