Question

Which one of the following pairs is correctly matched? (a) Saytzeff rule, least substituted alkene (b) Hoffmann rule, most substituted alkene (c) \(\mathrm{E}_{1}\) cb reaction, Hoffmann elimination (d) \(\mathrm{E}_{1}\) reaction, Hoffmann elimination

Short answer

Option c is correctly matched.

Step-by-step solution

Understand the Saytzeff Rule

The Saytzeff (or Zaitsev) rule predicts the outcome of eliminations reactions, stating that the more substituted alkene, where more alkyl groups are attached to the double bond, is usually favored. Therefore, the Saytzeff rule correlates with the most substituted alkene, not the least.

Recall the Hoffmann Rule

The Hoffmann rule is generally applicable to elimination reactions that form the least substituted alkene. This happens when bulky bases are used or in specific elimination conditions that favor Hofmann products.

Assess E1cb and Hoffmann Elimination

An E1cb (unimolecular elimination conjugate base) reaction involves the formation of a carbanion intermediate. Hoffmann elimination, however, is a different elimination reaction that results in a less substituted alkene and primarily involves the use of quaternary ammonium compounds.

Evaluate E1 Reaction and Hoffmann Elimination

The E1 reaction is a two-step process where a carbocation intermediate is formed. This reaction does not correlate with Hoffmann elimination, which is a separate type of elimination process.

Match Each Option to Its Correct Rule

Review each pair: 1. Saytzeff rule leads to the most substituted alkene. 2. Hoffmann rule leads to the least substituted alkene. 3. E1cb reaction is not commonly related to Hoffmann elimination. 4. E1 reaction does not produce Hoffmann products. Option (b) is incorrectly described, and option (c) is correctly matched with Hoffmann elimination.

Key concepts

Saytzeff Rule

The Saytzeff Rule, also known as the Zaitsev Rule, guides us in predicting the major product of an elimination reaction. In organic chemistry, elimination reactions often involve the removal of adjacent atoms or groups from a molecule, resulting in the formation of a double bond. According to the Saytzeff Rule, the more substituted alkene will be the major product. This means the alkene with more alkyl groups (such as methyl or ethyl groups) connected to the double-bonded carbons is preferred.

Why does this happen? More substituted alkenes are more stable due to the electron-releasing effects of the additional alkyl groups. These groups donate electron density towards the double bond, stabilizing it through a concept known as hyperconjugation. It’s like having more helpers to share the workload!

When students come across elimination reactions, it's crucial to look for the carbon that, when bonded to different groups, results in the most stable double-bonded structure. The Saytzeff Rule thus helps predict that this will be the favored product.

Hoffmann Rule

The Hoffmann Rule gives us guidance on another type of elimination reaction, predicting that the least substituted alkene will form as the major product. This is somewhat opposite to the Saytzeff prediction and occurs under specific conditions.

The Hoffmann elimination becomes significant when bulky bases are involved or certain reaction conditions favor less substituted products. Here, steric hindrance plays a major role: bulky bases cannot easily access more substituted (and thus more crowded) carbons, so they prefer removing a hydrogen atom from a less hindered carbon, leading to a less substituted alkene.

This rule often comes into play in reactions involving quaternary ammonium compounds, where due to the steric factors, the removal of hydrogen results in the least substituted alkene being formed.

E1 Reaction

The E1 reaction, or unimolecular elimination reaction, is a classic elimination process. It involves a two-step mechanism, distinctly characterized by the formation of a carbocation intermediate. Here's how it works:

• In the first step, the leaving group departs, creating a carbocation. This step is slow and forms the rate-determining step, as the reaction's speed depends solely on this step (hence, unimolecular).
• In the second step, a base abstracts a proton from the carbocation, resulting in the formation of a double bond, completing the creation of the alkene product.

E1 reactions are favored in situations where the substrate can easily form a stable carbocation, typically seen in tertiary alkyl halides. As opposed to the Hoffmann elimination, E1 reactions typically result in the more stable, more substituted alkene, aligning with the Saytzeff prediction.

E1cb Reaction

The E1cb reaction, or unimolecular elimination conjugate base reaction, involves the formation of a carbanion intermediate. It's a specialized elimination reaction mechanism you might encounter under specific conditions.

Here’s the step-by-step process:

• First, a base removes a proton, typically adjacent to a carbonyl group, to form a carbanion. This carbanion is stabilized by resonance or electron-withdrawing groups, which makes this step feasible.
• Following the carbanion formation, the leaving group is expelled, forming a double bond.

An important factor in E1cb reactions is the presence of a strong base and a good leaving group, often found in conjugated systems where the intermediate carbanion's stability is crucial.

The Hofmann rule is not typically related to this mechanism, as E1cb reactions depend more on the stabilization of intermediates rather than steric hindrance or base bulkiness.