Why is the following compound not aromatic: $C_4{H}_6$ (cyclobutadiene)?

• A. It does not have a planar structure.
• B. It contains a heteroatom that disrupts aromaticity.
• C. It has $4$ $\pi$ electrons, which does not satisfy Hückel's rule.
• D. It does not have a conjugated system of $\pi$ electrons.

Answer: (C)

Verified step by step guidance

1. Recall that for a compound to be aromatic, it must satisfy Hückel's rule, which requires having a planar, cyclic, conjugated system with (4n + 2) π electrons, where n is a non-negative integer (0, 1, 2, ...).
2. Identify the number of π electrons in cyclobutadiene (C\_4H\_4). Each double bond contributes 2 π electrons, and cyclobutadiene has two double bonds, so it has a total of 4 π electrons.
3. Check if the compound is planar and conjugated. Cyclobutadiene is planar and has a conjugated system because the p orbitals on each carbon overlap continuously around the ring.
4. Apply Hückel's rule to the 4 π electrons: substitute into (4n + 2) = 4 and solve for n. Since 4 does not fit the formula (4n + 2) for any integer n, cyclobutadiene does not satisfy Hückel's rule.
5. Conclude that because cyclobutadiene has 4 π electrons, which corresponds to the 4n rule (where n=1), it is antiaromatic rather than aromatic, explaining why it is not aromatic.