Animated Solution for Chemistry - Organic Chemistry: The major product obtained from the following reaction is

Imagine you are looking at a molecular tug-of-war. We have an unsymmetrical alkyne, $p{\text{-NO}}_2{\text{-C}}_6{\text{H}}_4-\text{C}\equiv \text{C}-{\text{C}}_6{\text{H}}_4\text{-}p{\text{-OCH}}_3$, sandwiched between two very different benzene rings. On the left, a ring armed with a nitro group. On the right, a ring equipped with a methoxy group.

The reagents given are ${\text{Hg}}^{2+}/{\text{H}}^{+}$ and ${\text{H}}_2\text{O}$. This is the classic Kucherov reaction, which adds water across the triple bond. But the big question is... which carbon gets the $\text{OH}$, and which gets the $\text{H}$? To answer this, we must dive into the electronic effects.

The nitro group ($-{\text{NO}}_2$) is a powerful electron-withdrawing group due to its $-M$ and $-I$ effects. It pulls electron density away from the alkyne. On the other hand, the methoxy group ($-{\text{OCH}}_3$) is a strong electron-donating group because of its $+M$ effect. It pushes electron density into the system.

Now, imagine the intermediate. The hydration follows Markovnikov's rule, meaning the positive charge will develop on the carbon that can best stabilize it. Because the methoxy group donates electrons, the carbon adjacent to it will happily bear the positive charge. The nitro group, however, would severely destabilize a positive charge next to it.

So, the water molecule attacks this more stable, positively charged carbon. After losing a proton, we get an enol intermediate. Notice how the $\text{OH}$ group is attached to the carbon closer to the methoxy ring.

But we know that enols are generally unstable. It will rapidly undergo keto-enol tautomerization. The double bond shifts, and the hydrogen moves to the adjacent carbon, giving us our final, highly stable ketone product.

And there we have it! The major product has the carbonyl group right next to the methoxy-substituted ring. The final structure is $p{\text{-NO}}_2{\text{-C}}_6{\text{H}}_4-{\text{CH}}_2-\text{C}(=\text{O})-{\text{C}}_6{\text{H}}_4\text{-}p{\text{-OCH}}_3$.

Comparing this with our options, we can clearly see that option (a) is the correct answer. This problem is a beautiful application of electronic effects dictating regioselectivity in organic synthesis.