Which of the following compounds is aromatic?

• A. Benzene ($C_6{H}_6$)
• B. Cyclobutadiene ($C_4{H}_4$)
• C. Cyclooctatetraene ($C_8{H}_8$)
• D. Cyclohexane ($C_6{H}_{12}$)

Answer: (A)

Verified step by step guidance

1. Recall the criteria for aromaticity: a compound must be cyclic, planar, fully conjugated (p orbitals at every atom in the ring), and follow Hückel's rule, which states it must have (4n + 2) π electrons, where n is a non-negative integer (0, 1, 2, ...).
2. Analyze each compound to determine if it is cyclic and planar. Benzene, cyclobutadiene, cyclooctatetraene, and cyclohexane are all cyclic, but planarity must be confirmed or assumed based on typical structures.
3. Check for conjugation: benzene has alternating double bonds with continuous overlap of p orbitals, cyclobutadiene and cyclooctatetraene have conjugated double bonds but may have different geometries affecting planarity, and cyclohexane has only single bonds, so no conjugation.
4. Count the number of π electrons in each compound: benzene has 6 π electrons, cyclobutadiene has 4 π electrons, cyclooctatetraene has 8 π electrons, and cyclohexane has 0 π electrons.
5. Apply Hückel's rule to each: benzene with 6 π electrons fits (4n + 2) where n=1, making it aromatic; cyclobutadiene with 4 π electrons fits 4n (n=1), which is antiaromatic if planar; cyclooctatetraene with 8 π electrons fits 4n (n=2), typically non-aromatic due to non-planarity; cyclohexane is non-aromatic as it lacks π electrons.