Structure of Compound Y in Bromination and Elimination Reaction Sequence

Understanding the structure of compound Y reaction sequence bromination elimination is a common organic chemistry concept.
When an alkene reacts with bromine followed by a strong base such as KO^tBu, halogen addition occurs first and then elimination produces multiple double bonds.
Let us solve this step-by-step.

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Given Reaction

Cyclohexene + Br₂ / AcOH → X

X + KO^tBu → Y

Find the structure of compound Y.

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Step 1: Bromination of Alkene

Reagent:

Br₂ in acetic acid (AcOH)

Concept:

• Halogen adds across C=C double bond
• Anti-addition takes place
• Forms vicinal dibromide

Product formed (X):

1,2-dibromocyclohexane

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Step 2: Reaction with KO^tBu

Reagent Nature:

• Strong base
• Bulky base
• Favors E2 elimination

What happens?

• First elimination removes one HBr → forms bromo-alkene
• Second elimination removes another HBr → forms second double bond

Final product:

1,3-cyclohexadiene

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Final Answer

Y = 1,3-Cyclohexadiene → Option (C)

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Explanation of All Options

Option	Structure	Reason
(A)	Bromocyclohexene	Only one elimination. Reaction actually undergoes double elimination.
(B)	Cyclohexene	Starting compound, not final product.
(C)	1,3-Cyclohexadiene	Two eliminations form two double bonds. Correct answer.
(D)	Benzene	Requires aromatization/oxidation, not possible by simple elimination.

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Important Exam Concepts

• Vicinal dihalide + strong base → double elimination
• E2 elimination removes HX
• Two HX removals → two double bonds
• Strong bulky bases favor elimination over substitution

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Quick Memory Trick

Two halogens + strong base = two eliminations → diene formation

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Conclusion

Cyclohexene first forms a vicinal dibromide with bromine. Treatment with KO^tBu causes two successive E2 eliminations, producing
1,3-cyclohexadiene. Therefore, the correct structure of compound Y is Option (C).