The Question is from neet-chemistry haloalkanes-haloarenes , and it was created by the author THIRUPATHI REDDY

Question

The reaction of 2-bromopentane with sodium ethoxide in ethanol predominantly yields:

THIRUPATHI REDDY · Accepted Answer

## CORE INFORMATION
✓ **Answer**: C
## CONCEPT FRAMEWORK
	**Primary Concept**: E2 Elimination Reaction
	**Related Topics**: Zaitsev's Rule, Alkene Stability, SN2 Reaction
	**Prerequisites**: Organic Chemistry - Elimination and Substitution Reactions
## SOLUTION PATHWAY
**Given Information**:
	Reactant: 2-bromopentane
	Reagent: Sodium ethoxide (NaOEt)
	Solvent: Ethanol (EtOH)
**Method & Steps**:
	1. Identify the reaction type:
		Reasoning: Sodium ethoxide is a strong base and ethanol is a polar protic solvent, favoring E2 elimination over SN2 substitution.
	2. Apply Zaitsev's rule:
		Working: Zaitsev's rule states that the major product of an elimination reaction is the more substituted alkene.
		Key Point:  In this case, removal of a proton from either the C1 or C3 position of 2-bromopentane leads to alkene formation.  Removal from C3 is favored as it leads to a more substituted alkene (2-pentene).
	3. Consider the possibility of multiple alkene isomers:
		Result: While 2-pentene is the major product, 1-pentene can also be formed as a minor product.
		Verification: The reaction proceeds via the formation of a transition state where the base abstracts a proton and the bromide leaves simultaneously, leading to the formation of a double bond.
## OPTION ANALYSIS
[A] :x: 1-pentene is a minor product, not the predominant one.
[B] :x: 2-pentanol would be the product of an SN1 or SN2 reaction, which are less favored under these conditions.
[C] ✓: A mixture of alkenes is formed, with 2-pentene as the major product due to Zaitsev's rule.
[D] :x: While a mixture of alkenes is formed, 1-pentene is the minor product, not the major one.
## LEARNING AIDS
	**Common Mistakes**:
		1.Ignoring Zaitsev's Rule:  Remembering that the more substituted alkene is usually the major product in elimination reactions is crucial.
		2.Confusing SN2 and E2: Strong bases and sterically hindered alkyl halides favor E2.  Strong nucleophiles and primary alkyl halides favor SN2.
	**Quick Tips**:
		1.Zaitsev's Rule: 'The rich get richer' – the more substituted carbon gets the double bond.
		2.Strong base/protic solvent: Think E2 elimination.

THIRUPATHI REDDY · Answer

1-pentene

THIRUPATHI REDDY · Answer

2-pentanol

THIRUPATHI REDDY · Answer

The correct Option is "A mixture of alkenes with 2-pentene as the major product"

THIRUPATHI REDDY · Answer

A mixture of alkenes with 1-pentene as the major product

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