The user wants me to solve a JEE Chemistry question.

1. Analyze the Question:

• Reactant: A vicinal dibromide attached to a six-membered ring. Specifically, it's 1,2-dibromocyclohexane.
• Product: An alkyne within the six-membered ring. Specifically, it's cyclohexyne.
• Goal: Find the reagent(s) that can convert the reactant to the product.

2. Deconstruct the Transformation:

• The reaction is a double dehydrohalogenation. Two molecules of HBr are eliminated from the reactant.
• The conversion involves two successive elimination (E2) reactions.
  • Step 1: Elimination of one HBr to form a vinylic halide.
  • Step 2: Elimination of the second HBr from the vinylic halide to form an alkyne.

3. Evaluate the Options:

• Option A: alcoholic KOH

  • Potassium hydroxide in ethanol (alcoholic KOH) is a strong base commonly used for dehydrohalogenation to form alkenes.
  • It can effectively carry out the first elimination step: $\text{1,2-dibromocyclohexane} \xrightarrow{\text{alc. KOH}} \text{1-bromocyclohexene} + \text{KBr} + \text{H}_2\text{O}$
  • However, the second elimination from the vinylic halide (1-bromocyclohexene) is much more difficult. This is because the C-Br bond is on an sp² hybridized carbon, which is stronger and less reactive towards elimination than a bond on an sp³ carbon. Alcoholic KOH is generally not a strong enough base to perform this second elimination efficiently.

• Option B: alcoholic KOH followed by NaNH₂

  • This is a two-step process.
  • Step 1: alcoholic KOH. As explained above, this reagent converts the vicinal dibromide into a vinylic halide (1-bromocyclohexene).
  • Step 2: NaNH₂ (Sodamide). Sodium amide is an extremely strong base (the pKa of its conjugate acid, NH₃, is about 38). It is strong enough to abstract the vinylic proton and facilitate the second elimination reaction to form the alkyne. $\text{1-bromocyclohexene} \xrightarrow{\text{NaNH}_2} \text{cyclohexyne} + \text{NaBr} + \text{NH}_3$
  • This sequence of reagents is a standard and effective method for preparing alkynes from vicinal dihalides. Thus, this option is the most appropriate choice.

• Option C: aqueous KOH followed by NaNH₂

  • Aqueous KOH favors nucleophilic substitution (Sₙ reactions) over elimination (E reactions). With a secondary halide like 1,2-dibromocyclohexane, aqueous KOH would primarily lead to the formation of substitution products like 1-bromo-2-hydroxycyclohexane, not the desired vinylic halide intermediate. Therefore, this is an incorrect pathway.

• Option D: Zn/CH₃OH

  • Zinc dust in methanol is a reagent for dehalogenation, not dehydrohalogenation. It removes two halogen atoms from adjacent carbons to form a double bond. $\text{1,2-dibromocyclohexane} \xrightarrow{\text{Zn/CH}_3\text{OH}} \text{cyclohexene} + \text{ZnBr}_2$
  • This reaction yields an alkene (cyclohexene), not the desired alkyne (cyclohexyne). So, this option is incorrect.

4. Conclusion: Based on the analysis, the conversion of 1,2-dibromocyclohexane to cyclohexyne is a double dehydrohalogenation. This is best achieved in a two-step process: first, elimination using a moderately strong base like alcoholic KOH to form the vinylic halide, followed by a second elimination using a very strong base like NaNH₂ to form the alkyne. Therefore, option B is the correct answer.