a) Draw both chair conformations of the starting alkyl bromide. Indicate which conformation, if either is more stable. Briefly justify your answer. b) The starting alkyl halide has two chiral carbons. What are the configurations (R/S) at each of these carbons? c) Identify the mechanism (E1/E2) which reactions 1, 2, and 3 proceed by. d) Provide mechanisms that explain the products formed in reactions 1, 2, and 3. All electron movement must be shown in order to receive full credit. e) Indicate which product is the major and which product is the minor in reactions 2 and 3. Briefly justify your answer.
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.
a) Draw both chair conformations of the starting alkyl bromide. Indicate which conformation, if either is more stable. Briefly justify your answer.
b) The starting
c) Identify the mechanism (E1/E2) which reactions 1, 2, and 3 proceed by.
d) Provide mechanisms that explain the products formed in reactions 1, 2, and 3. All electron movement must be shown in order to receive full credit.
e) Indicate which product is the major and which product is the minor in reactions 2 and 3. Briefly justify your answer.
![This image depicts an organic chemistry reaction, specifically an elimination reaction involving the starting compound, a halogenoalkane, reacting with sodium ethoxide (Na+ -OCH2CH3) under heat.
**Reactants:**
- **Starting Compound:** The structure on the left is a bromocycloalkane with a bromine atom (Br) and a methyl group (CH3) attached.
- **Reagent:** Sodium ethoxide (Na+ -OCH2CH3)
**Reaction Conditions:**
- Heat is applied to the reaction mixture.
**Products:**
1. **Product 1:** A cycloalkene with a double bond forming on one of the carbons, with a methyl group attached.
2. **Product 2:** Another cycloalkene isomer with a double bond, but with the methyl group oriented differently.
3. **By-products:** Sodium bromide (NaBr) and ethanol (CH3CH2OH).
**Explanation:**
The reaction involves the elimination of hydrogen bromide (HBr) from the starting compound, facilitated by sodium ethoxide, which acts as a base. This leads to the formation of two alkenes (double-bonded hydrocarbons) as major products, with the specific isomer formed depending on the position of the double bond. Sodium bromide and ethanol are produced as by-products.
This type of reaction is a classic example of an E2 (bimolecular elimination) mechanism, where the elimination of HBr occurs in a single step, leading to the formation of the alkene.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa71b61ff-4034-408f-951d-aa29cd0015fa%2Faacd027b-0ecb-4d36-b281-6d05caca5073%2F2mjet8_processed.png&w=3840&q=75)
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