1, 2-dibromocyclooctane $\xrightarrow[\substack{\text { (ii) } \mathrm{NaNH}_{2} \\ \text { (iii) } \mathrm{Hg}^{2+} / \mathrm{H}^{+} \\ \text {(iv) } \mathrm{Zn}-\mathrm{H}^{2} / \mathrm{H}^{+}}]{\text {(i) } \mathrm{KOH} \text { (alc.) }} \underset{\text { (Major product) }}{\mathrm{P}}$ (iv) $\mathrm{Zn}-\mathrm{Hg} / \mathrm{H}^{+}$ ' P ' is

JEE Main · 2025 · 08 Apr 2025 · Shift 2 · Hydrocarbons

1, 2-dibromocyclooctane

(i) KOH (alc.) (ii) NaNH_{2} (iii) Hg^{2 +} / H^{+} (iv) Zn - H^{2} / H^{+} (Major product) P

(iv)

Zn - Hg / H^{+}

' P ' is

(A)(see structure)
(B)(see structure)
✓ Correct
(C)(see structure)
(D)(see structure)

Answer: (B)

Solution

Alcoholic potassium hydroxide removes one molecule of HBr from 1,2-dibromocyclooctane to give a bromocyclooctene, and sodium amide then eliminates the second HBr to form cyclooctyne. Acid-catalysed hydration with mercury(II) adds water across the triple bond, and the resulting enol tautomerises to cyclooctanone. Clemmensen reduction with zinc amalgam and acid finally converts the carbonyl group to a methylene unit. The major product is cyclooctane.

JEE Main · 2025 · 08 Apr 2025 · Shift 2 · Single correct option

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