Question

Predict the \(\beta\)-elimination product(s) formed when each chloroalkane is treated with sodium ethoxide in ethanol. If two or more products might be formed, predict which is the major product. (a) CC1(Cl)CCCCC1 (b) ClCC1CCCCC1 (c) CC1(C)CCC(Cl)CC1

Short answer

(a) Chloroalkane: CC1(Cl)CCCCC1 Major Product: C=C1CCCCC1 (b) Chloroalkane: ClCC1CCCCC1 Major Product: C=CCC1CCCCC1 (c) Chloroalkane: CC1(C)CCC(Cl)CC1 Major Product: CC1(C)CCC=C(C)C1

Step-by-step solution

Identifying \(\beta\)-Hydrogens

Look for the \(\beta\)-hydrogen atoms in the molecule that are adjacent to the carbon atom attached to the chlorine. In this case, there are two possible locations for the \(\beta\)-hydrogens.

Performing E2 Elimination Reaction

With sodium ethoxide being a strong base, applying the E2 mechanism, we would expect the major product to be the most stabilized alkene. In other words, the alkene with the most highly substituted double bond will be the major product.

Applying Zaitsev's Rule

According to Zaitsev's rule, the major product will be the alkene where the more substituted carbon is doubly bonded to one of the nearest \(\beta\)-hydrogens. In this case, the major product (major alkene) will be: C=C1CCCCC1 (b) ClCC1CCCCC1

Identifying \(\beta\)-Hydrogens

Locate the \(\beta\)-hydrogen atoms in the molecule that are adjacent to the carbon atom attached to the chlorine. In this case, there is only one possible location for the \(\beta\)-hydrogens.

Performing E2 Elimination Reaction

Apply the E2 mechanism with sodium ethoxide as the strong base, which leads us to predict the major product as the most stabilized alkene. In other words, the alkene with the most highly substituted double bond will be the major product.

Applying Zaitsev's Rule

According to Zaitsev's rule, the major product will be the alkene where the more substituted carbon is doubly bonded to one of the nearest \(\beta\)-hydrogens. In this case, the major product (major alkene) will be: C=CCC1CCCCC1 (c) CC1(C)CCC(Cl)CC1

Identifying \(\beta\)-Hydrogens

Find the \(\beta\)-hydrogen atoms in the molecule that are adjacent to the carbon atom attached to the chlorine. In this case, there is only one possible location for the \(\beta\)-hydrogens.

Performing E2 Elimination Reaction

Utilize the E2 mechanism with sodium ethoxide as the strong base, leading us to predict the major product as the most stabilized alkene. In other words, the alkene with the most highly substituted double bond will be the major product.

Applying Zaitsev's Rule

According to Zaitsev's rule, the major product will be the alkene where the more substituted carbon is doubly bonded to one of the nearest \(\beta\)-hydrogens. In this case, the major product (major alkene) will be: CC1(C)CCC=C(C)C1

Key concepts

E2 mechanism

The E2 mechanism, also known as bimolecular elimination, is a reaction pathway used in organic chemistry for alkene formation. This process involves the simultaneous removal of a hydrogen atom (\( \beta \)-hydrogen) and a leaving group, such as chlorine in chloroalkanes.
The reaction is second-order, meaning its rate depends on both the substrate (the molecule undergoing elimination) and the base used in the reaction. In the presence of a strong base like sodium ethoxide, the reaction typically proceeds quickly.

• The base abstracts a hydroxyl hydrogen, resulting in a double bond formation.
• The chlorine group leaves, leading to the formation of an alkene.

This contrast with other mechanisms such as E1, which involves a carbocation intermediate and typically occurs with weaker bases.

Zaitsev's rule

Zaitsev's rule is a guideline used to predict the outcome of elimination reactions like those under the E2 mechanism. It states that the more substituted alkene is usually the major product of an elimination reaction.
In simpler terms, it suggests that the double bond will most likely form in a location that results in the greatest number of alkyl substituents. This occurs because more substituted alkenes are often more thermodynamically stable due to increased hyperconjugation and dispersal of charge.

• Look for the most substituted carbon adjacent to the carbon with the leaving group.
• Apply the rule to predict the major alkene product.

This principle aids students and chemists in predicting the predominant structure formed after a beta elimination reaction.

Chloroalkanes

Chloroalkanes are organic compounds consisting of carbon, hydrogen, and chlorine atoms. They are a type of haloalkane where chlorine is attached to the carbon backbone of the molecule.
In the context of elimination reactions, chloroalkanes serve as substrates because they contain good leaving groups (chlorine atoms) that facilitate the removal process. The presence of a chlorine atom makes the carbon at the \( \alpha \) position an electrophilic center, making it susceptible to various transformation reactions like elimination or substitution.

• They can participate in elimination reactions to form alkenes.
• Chlorine provides stability as a leaving group in these transformations.

Their reactivity and utility in laboratories and industries make them significant, though careful handling is required due to their potential environmental impact.

Sodium ethoxide

Sodium ethoxide is a strong base commonly employed in organic synthesis, particularly in elimination reactions like the E2 mechanism. It is typically represented by the formula \( C_2H_5ONa \).
As a potent base, sodium ethoxide aids in removing hydrogen atoms adjacent to halogens like chlorine in chloroalkanes, promoting the formation of double bonds, or alkenes.

• Prepared by reacting sodium metal with ethanol.
• Beneficial for creating an alkaline environment in ethanol as the solvent.

Its strong baseness and solubility in ethanol make it an ideal choice for inducing E2 eliminations, allowing the prediction and identification of major products based on substrate structure and reaction conditions.

Alkene formation

Alkene formation is a primary result of elimination reactions such as the E2 mechanism in organic chemistry. Alkenes are unsaturated hydrocarbons containing a carbon-carbon double bond, which imparts distinct reactivity and properties compared to alkanes.
The process of forming alkenes involves selectively breaking bonds and rearranging the molecular structure to include double bonds as a result of elimination. These reactions take advantage of the stability and energy considerations influenced by mechanisms and guidelines like Zaitsev's rule.

• Double bonds generated replace single bonds by removing atoms like hydrogen and halogens.
• Zaitsev's rule assists in predicting which alkenes will be most abundant.

Understanding the principles behind alkene formation is key for students and chemists striving to design and synthesize organic products efficiently.