6.17
E2 Reaction: Stereochemistry and Regiochemistry
Elimination reactions of alkyl halides can yield one or more alkenes depending on the specific regiochemical and stereochemical considerations. While…
Elimination reactions of alkyl halides can give one or more alkenes based on specific regiochemical and stereochemical requirements.
While regiochemistry governs the location of the double bond in the product, stereochemical requirements often influence the geometry.
Consider the E2 elimination of 2-bromo-2-methyl-butane, which has two different beta hydrogens. In the presence of a strong base like sodium ethoxide, it gives two regioisomeric alkenes.
The major product is the more substituted alkene called the Zaitsev product, whereas the less substituted alkene is the Hofmann product, and each product is formed through a different pathway
Each reaction has a transition state showing a partial double bond character. Since the more substituted carbon-carbon double bond has a higher stability, the transition state leading to the Zaitsev product requires less energy and proceeds faster.
Thus, the Zaitsev product is more stable and readily formed.
However, the regiochemical outcome of an E2 elimination reaction can be controlled using a sterically hindered base, like potassium tert-butoxide, to give the less substituted alkene in higher yields.
Stereochemical requirements strongly influence the outcome of E2 reactions. Here, the filled carbon-hydrogen σ bonding orbital and the empty carbon-halogen σ* antibonding orbital, lie in the same plane and overlap to give a π bond.
Both conformations — one where the hydrogen and halide are anti-coplanar and staggered, and the other where they are syn-coplanar and eclipsed — fulfill the coplanarity requirement to form an alkene.
E2 eliminations preferentially occur via a lower energy anti-coplanar transition state, where the base and the leaving group are farther apart and the two orbitals are fully parallel, allowing for best overlap.
The stereochemistry of E2 reactions depends on the number of β hydrogens. Alkyl halides with two β hydrogens undergo stereoselective elimination, forming the more stable E-alkene as the major product.
However, an alkyl halide with only one beta hydrogen gives a single stereospecific isomer. Depending upon the stereochemistry of the substrate, either E-alkene or Z-alkene is formed as the product, irrespective of their relative stabilities.
View the full transcript and gain access to JoVE Core videos
Q1: What are the Zaitsev and Hofmann products in E2 elimination reactions?
The Zaitsev product is the more substituted alkene formed as the major product in E2 eliminations, while the Hofmann product is the less substituted alkene. When 2-bromo-2-methyl-butane reacts with a strong base like sodium ethoxide, both products form through different pathways. The Zaitsev product predominates because its transition state has greater double bond character, requiring less energy and proceeding faster.
Q2: How does base strength control regiochemistry in E2 reactions?
Strong bases like sodium ethoxide favor the Zaitsev product due to lower transition state energy. However, sterically hindered bases like potassium tert-butoxide create crowding around the transition state leading to the Zaitsev product, making the reaction regioselective for the less substituted Hofmann product instead.
Q3: Why is the anti-coplanar conformation preferred in E2 eliminations?
E2 eliminations preferentially occur via the anti-coplanar transition state because the base and leaving group are farther apart, and the filled carbon-hydrogen σ orbital and empty carbon-halogen σ* antibonding orbital are fully parallel. This geometry allows maximum orbital overlap to form the π bond with lower energy than the syn-coplanar alternative.
Q4: What orbital interactions are required for E2 elimination to occur?
The filled carbon-hydrogen σ bonding orbital and the empty carbon-halogen σ* antibonding orbital must lie in the same plane and overlap to form a π bond. Both anti-coplanar and syn-coplanar conformations fulfill this coplanarity requirement, though anti-coplanar is energetically favored because it positions the orbitals parallel for optimal overlap.
Q5: How does the number of beta hydrogens affect E2 stereochemistry?
Alkyl halides with two β hydrogens undergo stereoselective elimination, forming the more stable E-alkene as the major product. However, substrates with only one β hydrogen give a single stereospecific isomer—either E-alkene or Z-alkene depending on substrate stereochemistry—regardless of relative alkene stability.
Q6: Why is the Zaitsev product more stable than the Hofmann product?
The Zaitsev product is more substituted, making its transition state have greater double bond character and higher stability. Since the transition state resembles the product, the pathway to the more substituted alkene requires less activation energy and proceeds faster, making the Zaitsev product the thermodynamically favored major product.
Q7: Can E2 reactions proceed through syn-coplanar conformations?
While E2 eliminations preferentially occur via the lower energy anti-coplanar transition state, some rigid molecules can proceed through syn-coplanar conformations where the hydrogen and halide are eclipsed. Both conformations satisfy the coplanarity requirement for orbital overlap, though anti-coplanar remains energetically favorable in most cases.