Question

• A.
• B.
• C.
• D.

Hint:

When dealing with elimination and reduction reactions, it's essential to understand how the reagents affect the intermediates and lead to the final product. Alcohol formation and reduction are key steps in this type of organic transformation.

Answer 1

The Correct Option is D

To determine the major product P formed from 1,2-dibromocyclooctane under the given reaction conditions, let's analyze each step:

1. The first reagent, KOH (alc.), is used for an elimination reaction. In the presence of alcoholic KOH, dehydrohalogenation occurs, where two hydrogen halides (HBr) are removed to form a double bond. Since 1,2-dibromocyclooctane is used, this step will form cyclooctene as an intermediate. The reaction mechanism proceeds through an E2 elimination.
2. The second reagent, NaNH₂, is a strong base used to abstract protons and can generate an alkyne by a double elimination reaction from a dibromo compound. In our case, it will act on cyclooctene to form 1,2-cyclooctadiyne.
3. Steps (iii) and (iv), involving Hg²⁺ / H⁺ and Zn–Hg / H⁺, might be considered steps to reduce or modify unsaturation, but the main product formation stops after step (ii).

Thus, the major product P after these reactions is 1,2-cyclooctadiyne.

Therefore, the correct option is:

This matches with the provided correct answer.

Answer 2

The reactions proceed as follows:

Step 1: In the first step, \(\text{KOH (alc.)}\) induces an elimination reaction, which removes a bromine atom from the carbon chain, resulting in the formation of cyclooctene (a 8-membered ring with a double bond).

Step 2: The second step with \(\text{NaNH}_2\) induces further elimination, removing another hydrogen atom from the carbon-carbon single bonds (a step leading to the formation of cyclooctyne, a 8-membered ring with a triple bond).

Step 3: The reaction with \(\text{Hg}^{2+}/\text{H}^{+}\) induces hydromercuration of the alkyne, resulting in the addition of a mercury ion to the carbon-carbon triple bond.
This creates an alkene intermediate.

Step 4: Finally, the Clemmensen reduction with \(\text{Zn-Hg} / \text{H}^{+}\) reduces the alkene to a fully saturated ring, cyclooctane, which is the final product.
This reaction removes the double bond and saturates the 8-membered ring.