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Question 55

1,2-dibromocyclooctane is treated with 

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'P' is :

To determine the major product P, we analyze each step of the reaction sequence systematically.

The first two steps involve the conversion of 1,2-dibromocyclooctane into an alkyne. Treatment with aqueous $$KOH$$ initially causes dehydrohalogenation to form a vinyl bromide, and the subsequent treatment with the strong base $$NaNH_2$$ removes another molecule of $$HBr$$ to produce cyclooctyne through double dehydrohalogenation.

In the third step, cyclooctyne is treated with $$Hg^{2+}/H^+$$. Under these conditions, the alkyne undergoes acid-catalyzed hydration to form an enol, which rapidly tautomerizes to the corresponding ketone. Thus, cyclooctyne is converted into cyclooctanone.

In the final step, cyclooctanone is subjected to Clemmensen reduction using $$Zn\text{-}Hg/H^+$$. The carbonyl group is completely reduced to a methylene group $$(-CH_2-)$$, converting the ketone into the corresponding saturated hydrocarbon.

Therefore, the overall transformation is

$$\text{1,2-dibromocyclooctane} ;\longrightarrow; \text{cyclooctyne} ;\longrightarrow; \text{cyclooctanone} ;\longrightarrow; \text{cyclooctane}.$$

Hence, the final major product P is cyclooctane, corresponding to the eight-membered saturated ring shown in option B

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