The major product of the following reaction is :
`C_(6)H_(5)CH_(2)-underset(Br)underset(|)overset(CH_(3))overset(|)C-CH_(2)-CH_(3)underset(C_(2)H_(5)OH)overset(C_(2)H_(5)O"N"a)to`

Options

A. `C_(6)H_(5)CH=underset(overset(|)(CH_(3)))(C)-CH_(2)-CH_(3)`
B. `C_(6)H_(5)CH_(2)-underset(overset(|)(CH_(3)))(C)=CHCH_(3)`
C. `C_(6)H_(5)CH_(2)-underset(overset(|)(OC_(2)H_(5)))overset(underset(|)(CH_(3)))(C)-CH_(2)CH_(3)`
D. `C_(6)H_(5)CH_(2)-underset(overset(|)(CH_(2)CH_(3)))(C)=CH_(2)`

Answer

To determine the major product of the given reaction, we will follow these steps:

### Step 1: Identify the Reactants
The reactant is a compound with a benzyl group (C₆H₅CH₂-) attached to a carbon that has a bromine (Br), a methyl (CH₃), and an ethyl (C₂H₅) group.

### Step 2: Understand the Reaction Conditions
The reaction is carried out in the presence of sodium ethoxide (C₂H₅O⁻Na⁺) in an alcoholic medium. This suggests that the reaction will proceed via an elimination mechanism (E2), where a hydrogen atom and a leaving group (Br) are removed to form a double bond.

### Step 3: Identify the Beta Hydrogens
The carbon attached to the bromine is the alpha carbon, and the adjacent carbons are the beta carbons. In this case, there are two beta hydrogens available for abstraction:
1. One hydrogen on the beta carbon adjacent to the methyl group.
2. One hydrogen on the beta carbon adjacent to the ethyl group.

### Step 4: Elimination Mechanism
The base (C₂H₅O⁻) will abstract one of the beta hydrogens while the bromine will leave, forming a double bond. We can visualize the two possible elimination pathways:
- **Pathway A**: Removal of the hydrogen from the beta carbon next to the methyl group.
- **Pathway B**: Removal of the hydrogen from the beta carbon next to the ethyl group.

### Step 5: Draw the Possible Products
1. **Product A**: If the hydrogen from the beta carbon adjacent to the methyl group is removed, we get a double bond that is conjugated with the benzene ring.
2. **Product B**: If the hydrogen from the beta carbon adjacent to the ethyl group is removed, we get a double bond that is not conjugated with the benzene ring.

### Step 6: Analyze Stability of Products
- **Product A** is more stable due to resonance stabilization from the conjugated double bond with the benzene ring.
- **Product B** is less stable since it lacks this conjugation.

### Step 7: Conclusion
The major product of the reaction is the one that is more stable, which is **Product A**.

### Final Answer
The major product of the reaction is the alkene formed by the elimination of the hydrogen adjacent to the methyl group, resulting in a conjugated system with the benzene ring.

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