Q.19 The major product obtained in the following reaction is

Introduction (SEO Optimized)

Questions involving alkyl halides reacting with strong bases like sodium ethoxide (NaOEt) are very common in JEE Advanced, NEET, and other competitive organic chemistry exams. Such problems test the student’s understanding of reaction conditions, substrate structure, and the competition between substitution (SN2) and elimination (E2) mechanisms.

In this article, we analyze the given reaction step by step and determine the major product formed when an alkyl bromide reacts with NaOEt in ethanol at 50 °C, explaining why other options are incorrect.

 Key Concepts Used

1️⃣ Nature of Reagent

• NaOEt (sodium ethoxide) is:

  • A strong base

  • A good nucleophile

2️⃣ Reaction Conditions

• Ethanol solvent

• Elevated temperature (50 °C)

 These conditions strongly favor elimination (E2) over substitution.

 Reaction Analysis

Step 1: Identify the Substrate

• The carbon bearing Br is secondary

• Adjacent to:

  • A benzyl group

  • Alkyl substituents

Secondary halides + strong base + heat → E2 dominates

Step 2: Decide Between SN2 and E2

Conclusion: Reaction proceeds via E2 elimination

Step 3: Apply Zaitsev’s Rule

• E2 elimination gives the more substituted alkene as the major product

• Alkene conjugated with a benzene ring is even more stable

 Correct Product

Option (D)

This product:

• Is the most substituted alkene

• Is conjugated with the benzene ring

• Results from β-hydrogen elimination

• Is favored thermodynamically

✔ Major product

 Explanation of All Options

 Option (A)

• Represents an SN2 substitution product

• Formation of ether

• SN2 is disfavored due to:

  • Secondary substrate

  • High temperature

•  Minor product

 Option (B)

• Another substitution (Williamson ether type) product

• Requires backside attack

• Steric hindrance reduces SN2 probability

•  Not major

 Option (C)

• Alkene formed, but:

  • Less substituted

  • Not maximally conjugated

• Violates Zaitsev’s rule

•  Minor elimination product

 Option (D) ✔

• Zaitsev alkene

• Maximum substitution

• Conjugation with aromatic ring

• Formed via E2 mechanism

• ✔ Correct Answer

 Quick Exam Tip

Whenever you see:

• Strong base (NaOEt, KOtBu)

• Secondary alkyl halide

• High temperature

 Immediately think E2 elimination
 Apply Zaitsev’s rule unless bulky base is mentioned

 Conclusion

This reaction is a classic example where reaction conditions override nucleophilicity. Although NaOEt can act as a nucleophile, the presence of heat, a secondary substrate, and a strong base ensures that E2 elimination dominates, producing the most stable, substituted, and conjugated alkene.

Final Answer: Option (D)