Predicting Products of Organic Reactions: E2 Elimination and Substitution

Q1. Predict the product for the following reaction:

Background

Topic: E2 Elimination Reactions

This question tests your understanding of elimination reactions, specifically the E2 mechanism, which is favored by strong bases (like NaOCH3) and heat. You are asked to predict the organic product formed when a bromoalkane reacts with sodium methoxide in methanol under heated conditions.

Key Terms and Formulas

• E2 Elimination: A concerted mechanism where a base removes a proton (H) from a β-carbon, while the leaving group (Br) departs from the α-carbon, forming a double bond.

• β-Hydrogen: The hydrogen atom on the carbon adjacent to the carbon bearing the leaving group.

• Zaitsev's Rule: The most substituted alkene is usually the major product in E2 eliminations.

Step-by-Step Guidance

1. Identify the α-carbon (the carbon attached to Br) and all β-carbons (adjacent carbons).

2. Determine which β-hydrogens are available for elimination. Draw out the possible alkenes that could form.

3. Apply Zaitsev's Rule: The base (NaOCH3) will preferentially remove a hydrogen from the β-carbon that leads to the most substituted (stable) alkene.

4. Draw the structure of the major alkene product, showing the double bond between the α and β carbons.

Try solving on your own before revealing the answer!

Final Answer:

The major product is the most substituted alkene formed by E2 elimination, following Zaitsev's rule. The double bond will be placed between the α-carbon (where Br was) and the β-carbon with the most substituents.

This occurs because the strong base and heat favor elimination over substitution, and the most stable (substituted) alkene is formed.