Question

Draw structural formulas for the alkene(s) formed by treatment of each haloalkane or halocycloalkane with sodium ethoxide in ethanol. Assume that elimination occurs by an E2 mechanism.

Short answer

Question: Draw the structural formula(s) for the alkene(s) formed when treating haloalkanes or halocycloalkanes with sodium ethoxide in ethanol via E2 elimination mechanism.

Step-by-step solution

Understanding E2 Elimination Mechanism

E2 elimination (bimolecular elimination) is a one-step process in which the leaving group (halogen atom) and a hydrogen atom are removed from the substrate, forming an alkene product. A strong base like sodium ethoxide (NaOEt) promotes the E2 elimination mechanism.

Recognition of Leaving Group

Haloalkanes or halocycloalkanes contain a halogen atom, which acts as a leaving group when subjected to E2 elimination conditions. Identify the halogen atom, which will be removed from the structure during the reaction.

Locate the Beta-Hydrogen

Identify the beta-hydrogen in the substrate. Beta-hydrogen is a hydrogen atom that is bonded to the carbon adjacent to the carbon containing the leaving group (halogen atom).

Base Abstraction of the Beta-Hydrogen

Sodium ethoxide in ethanol acts as a strong base (ethoxide ion, EtO-) and abstracts the beta-hydrogen from the substrate, forming a new double bond (alkene) between leaving group's carbon and the beta-carbon. The halogen atom will be removed simultaneously.

Draw The Structural Formula(s) for The Alkene(s)

After the base abstracted the beta-hydrogen and the halogen left, draw the structural formula for the new alkene(s). If there is more than one beta-hydrogen in different orientations, then it's possible that there are more than one alkene products (cis/trans or E/Z isomers). Note: For specific substrate examples and step-by-step solutions, please provide the specific haloalkane or halocycloalkane structure.