Chemistry - GOC Question with Solution | TestHub

Reaction (A): Non-aromatic cyclooctatetraene accepts two electrons from sodium ($2Na$) to form a stable, planar $10\pi$-electron aromatic dianion.

Reaction (B): Terminal propyne is a stronger acid than ammonia ($N{H}_3$), allowing the strong base $NaN{H}_2$ to easily deprotonate it in a favorable acid-base equilibrium.

Reaction (D): Cyclopentadiene is highly acidic due to resonance, and losing a proton to $NaH$ yields a highly stable $6\pi$-electron aromatic anion.

Conclusion: Reactions A, B, and D favor the forward direction because they drive toward significantly more stable aromatic species or a weaker conjugate acid-base pair.