JEE Chemistry 2025 Question with Solution

Given: 1,2-dibromocyclooctane is treated with $\text{KOH (alc.)}$, $\text{NaNH}_2$, $\text{Hg}^{2+}/\text{H}^+$, and $\text{Zn-Hg}/\text{H}^+$.

Find: The major product $P$.

The solution indicates that the first step is dehydrohalogenation with alcoholic KOH. This gives an unsaturated ring intermediate.

In the next step, $\text{NaNH}_2$ acts as a strong base and further elimination gives a cyclic alkyne intermediate.

The later steps, $\text{Hg}^{2+}/\text{H}^+$ followed by $\text{Zn-Hg}/\text{H}^+$, convert the unsaturated intermediate into the fully saturated ring. Thus the final product corresponds to cyclooctane.

Therefore, the correct option is D.

Given: The reaction sequence is read from the figure and the solution.

Find: Which of the four drawn structures is $P$.

From the second approach on the solution:

1. $\text{KOH (alc.)}$ causes elimination.
2. $\text{NaNH}_2$ gives further elimination to form cyclooctyne.
3. $\text{Hg}^{2+}/\text{H}^+$ reacts with the alkyne.
4. $\text{Zn-Hg}/\text{H}^+$ reduces the intermediate to cyclooctane.

Among the four structures shown in the image, the fully saturated eight-membered ring is option 4.

The first approach text on the page contains inconsistent intermediate discussion, but the page explicitly states "The Correct Option is D". Therefore the answer is taken as D.