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: Predict the alkene product(s) formed by treating a haloalkane or halocycloalkane with sodium ethoxide in ethanol through an E2 mechanism, and provide the structural formula.

Step-by-step solution

Understanding an E2 Mechanism

E2 stands for bimolecular elimination, and it is a type of elimination reaction that forms alkenes from haloalkanes or halocycloalkanes. In an E2 reaction, the leaving group (halide) and a hydrogen atom are removed from the reactant molecule. This leads to the formation of a double bond (alkene) and the formation of the halide anion. Sodium ethoxide (NaOEt), in this case, acts as a strong base and promotes the E2 elimination process in ethanol as a solvent.

Identify the Reactants

Before starting, make sure you know the structure of the haloalkane (or halocycloalkane) given in the exercise. In this exercise, we are given a general problem statement to draw the structural formulas for alkene(s) formed from the treatment of haloalkanes and halocycloalkanes with sodium ethoxide. Therefore, specific haloalkane/halocycloalkane structures are not provided. However, we can demonstrate the process with generic R-X, where R represents an alkyl or cycloalkyl group, and X represents a halide (like Cl, Br, or I).

Determine the Location of the Double Bond in the Alkene Product

To form the alkene product, the E2 reaction requires the elimination of the halide and a hydrogen atom on the beta carbon, which is one carbon away from the one attached to the halide. After identifying the beta carbon(s), anticipate where the double bond will form.

Apply Zaitsev's Rule (if applicable)

In some cases, there may be more than one possible alkene product, resulting from the reaction. To predict the major product, apply Zaitsev's rule, which states that the major product is the alkene with the more substituted double bond. This means that the double bond will form in a way that leads to more carbons attached to the alkene carbons.

Draw the Structural Formula of the Alkene

Once the location of the double bond is determined, draw the structural formula of the alkene product, keeping the rest of the molecule unchanged. This will show the final product of the E2 reaction between the given haloalkane/halocycloalkane and sodium ethoxide in ethanol.