Question

Using models, suggest explanations for the following: (a) On E2 elimination with \(\mathrm{t}-\mathrm{BuOK}^{-+} / \mathrm{t}-\mathrm{BuOH}\), both cis-and trans-2-phenylcyclopenty1 tosylates give 1 -phenylcyclo-pentene as the only alkene; the cis isomer reacts 14 times as fast as the trans. (b) On E2 elimination with \(\mathrm{n}-\mathrm{C}_{5} \mathrm{H}_{11} \mathrm{ON}^{-+} \mathrm{a} / \mathrm{n}-\mathrm{C}_{5} \mathrm{H}_{11} \mathrm{OH}\) to give 2-chloronorbornene, II reacts about 100 times as fast as its diastereomer, I.

Short answer

In reaction (a), the cis isomer reacts 14 times faster than the trans isomer because the anti-coplanar orientation required for E2 elimination is more easily accessible for the cis isomer. In reaction (b), diastereomer II reacts about 100 times faster than diastereomer I because the anti-coplanar orientation is readily available in II, while it is less accessible in I due to steric hindrance.

Step-by-step solution

Identify E2 elimination mechanism and requirements

In an E2 elimination reaction, the leaving group (tosylate) and the hydrogen being abstracted must be in an anti-coplanar orientation (180 degrees apart) for the reaction to proceed. Find the transition state that satisfies the requirement.

Compare structures of cis and trans isomers

In the cis isomer, the phenyl group and tosylate group are on the same side of the cyclopentane ring. In the trans isomer, they are on opposite sides. Determine if the required anti-coplanar geometry is easily accessible in each case.

Rationalize observed reactivity

In the cis isomer, the anti-coplanar orientation is readily available as both phenyl and tosylate group are on the same side, and the hydrogen being abstracted can easily adopt the required geometry. In the trans isomer, the phenyl group and tosylate group are on opposite sides, resulting in a less accessible anti-coplanar orientation, making the reaction slower. This explains why the cis isomer reacts 14 times faster than the trans isomer. #Reaction (b)# For the E2 elimination of diastereomers I and II using \(\mathrm{n}-\mathrm{C}_{5} \mathrm{H}_{11} \mathrm{ON}^{-+} \mathrm{a} / \mathrm{n}-\mathrm{C}_{5} \mathrm{H}_{11} \mathrm{OH}\), 2-chloronorbornene is produced, and II reacts about 100 times faster than I.

Identify E2 elimination mechanism and requirements

Like in reaction (a), the E2 elimination reaction requires an anti-coplanar orientation of the leaving group (chlorine) and the abstracted hydrogen. Find the transition state that satisfies the requirement for both diastereomers.

Compare structures of diastereomers I and II

Examine the stereochemical differences between diastereomers I and II, focusing on the positions of the chlorine atom and the hydrogen being abstracted.

Rationalize observed reactivity

In diastereomer II, the anti-coplanar orientation for E2 elimination is readily available, making the reaction considerably faster. In contrast, in diastereomer I, the required anti-coplanar orientation is less accessible due to steric hindrance, which slows down the reaction. This explains why diastereomer II reacts about 100 times faster than diastereomer I.

Key concepts

Anti-Coplanar Orientation

The concept of anti-coplanar orientation is critical in understanding the E2 elimination mechanism. During this reaction, the key players are the hydrogen atom that is being removed and the leaving group, which in the given exercise is a tosylate group.
For the reaction to occur efficiently, these two groups need to be aligned 180 degrees apart in the same plane, which is the anti-coplanar orientation. This positioning allows for the proper overlap of orbitals in the transition state, facilitating the simultaneous removal of the proton and departure of the leaving group.
In cases like cis and trans-2-phenylcyclopentyl tosylates, the three-dimensional structure of the molecule influences how easily this orientation can be achieved. For the cis isomer, this alignment is more naturally attainable because the phenyl and tosylate are on the same side. On the contrary, the trans isomer presents a geometric challenge, making the necessary orientation less accessible and the reaction slower.

Steric Hindrance

Steric hindrance is another important factor that influences the rate of E2 eliminations. It refers to the spatial clash between atoms or groups of atoms within a molecule, which can impede certain reactions.
Returning to the example of the diastereomers I and II used in the elimination to form 2-chloronorbornene, the chemical environment around the leaving group can be crowded or spacious. In diastereomer II, steric hindrance is minimal, allowing the necessary anti-coplanar orientation to be easily achieved. However, diastereomer I presents significant steric hindrance, which obstructs the anti-coplanar alignment and significantly slows down the reaction.
This concept underscores the importance of molecular geometry and how the arrangement of atoms affects chemical reactivity. The more hindered the transition state, the less likely the reaction will proceed swiftly.

Transition State

Finally, the concept of the transition state plays a central role in the E2 elimination mechanism. The transition state is a high-energy, unstable arrangement of atoms that occurs during the transformation from reactants to products.
In an E2 reaction, the transition state involves the simultaneous breaking of the bond between the carbon and the leaving group, and the bond between the carbon and the hydrogen being abstracted. The energy of this state is crucial—it's the peak of the energy hill that the reacting molecules must surmount for the reaction to happen.
The difficulty in achieving this transition state varies with the structure of the molecule involved in the reaction. As we saw in the two scenarios provided, steric hindrance and the anti-coplanar orientation are determining factors for how easily the transition state can be reached and, therefore, influence the reaction rate significantly.