JEE Main 2025 Apr 8 Shift 2, Chemistry Q5: Preparation of Alkynes

Step 1: First Dehydrohalogenation

Treatment of 1,2-dibromocyclooctane with alcoholic KOH removes one molecule of HBr via an E2 elimination, forming a bromocyclooctene intermediate.

$\boxed{ \mathrm{1,2\text{-}Dibromocyclooctane} \xrightarrow{\mathrm{alc.\ KOH}} \mathrm{Bromocyclooctene} }$

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Step 2: Formation of Cyclooctyne

The strong base $\mathrm{NaN}{\mathrm{H}}_2$ removes the second molecule of HBr from the vinyl bromide, producing cyclooctyne.

$\boxed{ \mathrm{Bromocyclooctene} \xrightarrow{\mathrm{NaNH_2}} \mathrm{Cyclooctyne} }$

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Step 3: Hydration of Alkyne

Cyclooctyne undergoes acid-catalysed hydration in the presence of $\mathrm{H}{\mathrm{g}}^{2+}/{\mathrm{H}}^{+}$. The enol formed immediately tautomerises to the more stable ketone, cyclooctanone.

$\boxed{ \mathrm{Cyclooctyne} \xrightarrow{\mathrm{Hg^{2+}/H^+}} \mathrm{Cyclooctanone} }$

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Step 4: Clemmensen Reduction

Clemmensen reduction $(\mathrm{Zn}\text{-}Hg/{\mathrm{H}}^{+})$ reduces the carbonyl group to a methylene group, giving cyclooctane.

$\boxed{ \mathrm{Cyclooctanone} \xrightarrow{\mathrm{Zn\text{-}Hg/H^+}} \mathrm{Cyclooctane} }$

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Conclusion

The major product P is cyclooctane.

$\boxed{\text{Correct Answer: Option (2)}}$