7. Determine if the following will follow an E2 or El pathway. Since base strength plays such a pivotal role in an E2 process, a strong base will favor the E2 pathway. Since the base does not play a crucial role in the El process, a weak base will favor the El pathway. Br NaOH H3C I H H3C (stronger base) H. E2 - anti-periplanar geometry is required Br H3C I H20 (weaker base) H3C © H- H3C. H E1 - Zaitsev control because base is non-bulky intermediate cation does not require anti-periplanar geometry
Reactive Intermediates
In chemistry, reactive intermediates are termed as short-lived, highly reactive atoms with high energy. They rapidly transform into stable particles during a chemical reaction. In specific cases, by means of matrix isolation and at low-temperature reactive intermediates can be isolated.
Hydride Shift
A hydride shift is a rearrangement of a hydrogen atom in a carbocation that occurs to make the molecule more stable. In organic chemistry, rearrangement of the carbocation is very easily seen. This rearrangement can be because of the movement of a carbocation to attain stability in the compound. Such structural reorganization movement is called a shift within molecules. After the shifting of carbocation over the different carbon then they form structural isomers of the previous existing molecule.
Vinylic Carbocation
A carbocation where the positive charge is on the alkene carbon is known as the vinyl carbocation or vinyl cation. The empirical formula for vinyl cation is C2H3+. In the vinyl carbocation, the positive charge is on the carbon atom with the double bond therefore it is sp hybridized. It is known to be a part of various reactions, for example, electrophilic addition of alkynes and solvolysis as well. It plays the role of a reactive intermediate in these reactions.
Cycloheptatrienyl Cation
It is an aromatic carbocation having a general formula, [C7 H7]+. It is also known as the aromatic tropylium ion. Its name is derived from the molecule tropine, which is a seven membered carbon atom ring. Cycloheptatriene or tropylidene was first synthesized from tropine.
Stability of Vinyl Carbocation
Carbocations are positively charged carbon atoms. It is also known as a carbonium ion.
![**Title: Understanding E2 and E1 Reaction Pathways**
**Introduction:**
In organic chemistry, elimination reactions are classified primarily into two types: E2 and E1. The base strength significantly influences which pathway a reaction will follow.
**Reaction Analysis:**
1. **E2 Reaction Pathway:**
- **Starting Material:** The structure provided shows a cyclohexane ring with a bromine (Br) and a methyl group (CH₃).
- **Reagent:** NaOH is used as a stronger base.
- **Product:** The reaction results in an alkene structure with the Br eliminated and a double bond formed.
- **Mechanism Details:** E2 is a bimolecular reaction (second-order kinetics) requiring anti-periplanar geometry, where the leaving group (Br) and the hydrogen (H) to be eliminated are positioned opposite each other. This mechanism occurs in a single concerted step without intermediates.
2. **E1 Reaction Pathway:**
- **Starting Material:** The same brominated cyclohexane ring is used.
- **Reagent:** H₂O is used as a weaker base.
- **Intermediate and Product:** Formation of a carbocation intermediate is shown, followed by a rearrangement to give the final alkene.
- **Mechanism Details:** E1 is a unimolecular reaction (first-order kinetics) that proceeds through a carbocation intermediate. It does not require the anti-periplanar geometry, offering greater flexibility. The Zaitsev rule indicates the most substituted (and stable) alkene is the major product, especially with non-bulky bases like water.
**Conclusion:**
To determine whether an elimination reaction will follow an E2 or E1 pathway, consider the strength of the base involved. Strong bases favor E2 mechanisms due to their ability to facilitate the removal of a proton in a concerted reaction. On the other hand, weaker bases favor E1 mechanisms, allowing the reaction to proceed through a carbocation intermediate without the need for strict geometric alignment. This understanding is crucial for predicting product outcomes and optimizing reaction conditions in synthetic organic chemistry.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa78be049-a055-4ff5-8f71-f33c2a4a40f9%2F508f8e62-bd16-45cc-b6db-344ae0f485c6%2Fhfsgke_processed.jpeg&w=3840&q=75)
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