Which of the following compounds is aromatic according to Hückel’s rule (planar, fully conjugated, and containing $4n+2$ $\pi$ electrons)?

• A. Cyclooctatetraene ($C_8{H}^8$)
• B. Cyclobutadiene ($C_4{H}^4$)
• C. Cyclopropenyl anion ($C_3{H}^3$$-$)
• D. Cyclopropenyl cation ($C_3{H}^3$$+$)

Answer: (D)

Verified step by step guidance

1. Recall Hückel's rule for aromaticity: a compound must be cyclic, planar, fully conjugated, and have a total of \$4n + 2\( \)\pi\( electrons, where \)n$ is a non-negative integer (0, 1, 2, ...).
2. Determine the number of \(\pi\) electrons in each compound by counting the electrons in the conjugated \(p\) orbitals. For example, each double bond contributes 2 \(\pi\) electrons, and charges can add or remove electrons accordingly.
3. Check the planarity and conjugation of each compound. Planarity allows for effective overlap of \(p\) orbitals, and full conjugation means every atom in the ring has a \(p\) orbital participating in the \(\pi\) system.
4. Apply the \$4n + 2\( rule to the number of \)\pi\( electrons found in each compound. For instance, if the compound has 2, 6, 10, etc., \)\pi$ electrons, it satisfies the electron count for aromaticity.
5. Compare the results for each compound: cyclooctatetraene, cyclobutadiene, cyclopropenyl anion, and cyclopropenyl cation. Identify which one meets all criteria (planar, fully conjugated, and \$4n + 2\( \)\pi$ electrons) to conclude which is aromatic.