Question

The major product of the following reaction is:

• A. 2-Phenylhepta-2,4-diene
• B. 6-Phenylhepta-3,5-diene
• C. 6-Phenylhepta-2,4-diene
• D. 2-Phenylhepta-2,5-diene

Hint:

In elimination reactions with excess base, the most stable conjugated diene product is favored due to its resonance stability.

Answer 1

The Correct Option is C

To determine the major product of the given reaction, we need to analyze the structure and reactivity of the compound involved. The question provides us with the compound 6-Phenylhepta-2,4-diene. 

Step 1: Analyze the structure: The compound is a conjugated diene with phenyl substitution at position 6. It has the structure:

• Hepta indicates a seven-carbon chain.
• 2,4-diene implies double bonds at carbons 2 and 4.
• Phenyl group at position 6.

Step 2: Electrophilic addition reaction: In electrophilic additions, conjugated dienes like hepta-2,4-diene can undergo 1,2-addition (kinetic control) or 1,4-addition (thermodynamic control).

Step 3: Identify stability: Due to conjugation, electrons can delocalize, making certain products more stable. The reaction typically forms a product where the double bonds remain conjugated, as seen with 6-Phenylhepta-2,4-diene. Conjugated double bonds stabilize the molecule through resonance.

Option	Product Description
2-Phenylhepta-2,4-diene	Phenyl group at position 2, different from input structure
6-Phenylhepta-3,5-diene	Position of double bonds is shifted, may not be favored
6-Phenylhepta-2,4-diene	Original compound, conjugated diene structure, favored by resonance
2-Phenylhepta-2,5-diene	Non-conjugated, phenyl at position 2

Conclusion: The most stable, resonance-stabilized structure is 6-Phenylhepta-2,4-diene, which is the major product.

Answer 2

Step 1: Given reaction.
The compound provided contains two bromine atoms and a phenyl group. The reagent used is alcoholic KOH (excess) under heat (Δ), which indicates that the reaction will proceed through a double dehydrohalogenation (E2 elimination) process to form a conjugated diene.

Step 2: Reaction mechanism.
• The starting compound is a vicinal dibromide (two bromine atoms on adjacent carbon atoms) with a phenyl substituent on one of the carbons.
• Alcoholic KOH in excess promotes β-elimination (dehydrohalogenation), removing two molecules of HBr successively.
• The first elimination forms an alkene.
• The second elimination creates a conjugated diene system.

Step 3: Formation of the major product.
After two successive eliminations, the final product obtained is a conjugated diene attached to a phenyl group.
Thus, the compound formed is:
\[ \text{C}_6\text{H}_5\text{CH}= \text{CHCH}= \text{CHCH}_2\text{CH}_3 \] which corresponds to 6-Phenylhepta-2,4-diene.

Step 4: Final Answer.
\[ \boxed{\text{6-Phenylhepta-2,4-diene}} \]