Question

The major product (A) formed in the reaction given below is :
\includegraphics[width=0.6\linewidth]{image43.png}

• A. \includegraphics[width=0.3\linewidth]{image43.a.png}
• B. \includegraphics[width=0.3\linewidth]{image43.b.png}
• C. \includegraphics[width=0.3\linewidth]{image43.c.png}
• D. \includegraphics[width=0.3\linewidth]{image43.d.png}

Hint:

In substitution vs. elimination reactions, always look for factors that can stabilize the elimination product, such as conjugation. The formation of a conjugated \(\pi\)-system (alkene conjugated with a phenyl ring, another alkene, or a carbonyl group) can make the E2 reaction the major pathway even in cases where S\(_N\)2 might otherwise be expected to dominate (like with primary halides).

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
We are asked to predict the major product of the reaction between 1-bromo-2-phenylbutane and sodium methoxide in methanol. This involves a competition between substitution (S\(_N\)2) and elimination (E2) reactions.
Step 2: Detailed Explanation:
1. Analysis of Reactants:
- Substrate: 1-bromo-2-phenylbutane is a primary (1\(^\circ\)) alkyl halide. Primary halides generally favor S\(_N\)2 reactions.
- Reagent: Sodium methoxide (CH\(_3\)O\(^-\)Na\(^+\)) provides the methoxide ion (CH\(_3\)O\(^-\)), which is a strong nucleophile and also a strong, unhindered base.
- Solvent: Methanol (CH\(_3\)OH) is a polar protic solvent, which can favor both S\(_N\)1 and E1 for tertiary halides, but for primary halides, it doesn't prevent S\(_N\)2 or E2.
2. S\(_N\)2 vs. E2 Competition:
For a primary halide with a strong base/nucleophile, both S\(_N\)2 and E2 are possible.
- S\(_N\)2 Path: The methoxide ion attacks the primary carbon bearing the bromine, displacing Br\(^-\) to form an ether. Product: 1-methoxy-2-phenylbutane.
- E2 Path: The methoxide ion acts as a base and removes a proton from the \(\beta\)-carbon (the carbon adjacent to the one with the bromine). In this substrate, the \(\beta\)-carbon is the one bonded to the phenyl group.
3. Favoring the E2 Pathway:
Although the substrate is primary, several factors strongly favor the E2 elimination in this specific case:
- Acidity of \(\beta\)-Proton: The proton on the \(\beta\)-carbon is benzylic, meaning it is adjacent to the phenyl ring. Benzylic protons are more acidic than typical alkyl protons, making them easier to remove by a base.
- Stability of the Product: The alkene formed by the E2 elimination is 2-phenylbut-1-ene. In this molecule, the newly formed double bond is conjugated with the aromatic \(\pi\)-system of the phenyl ring. This conjugation provides significant stabilization to the product.
The formation of a highly stable, conjugated product is a powerful thermodynamic driving force that makes the E2 pathway the major one, even for a primary halide.
4. Identifying the Product:
The E2 product is 2-phenylbut-1-ene (structure shown in option B). The S\(_N\)2 product is 1-methoxy-2-phenylbutane (structure shown in option A). Due to the formation of a conjugated system, the E2 product is the major product.
Step 3: Final Answer:
The major product (A) is 2-phenylbut-1-ene.