Why is the following compound not aromatic? $C_4{H}_4$ (cyclobutadiene)

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

Answer: (D)

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. Each double bond contributes 2 π electrons, so with two double bonds, cyclobutadiene has 4 π electrons in total.
3. Check if the compound is planar and conjugated. Cyclobutadiene is planar and has a conjugated system because the p orbitals overlap around the ring.
4. Apply Hückel's rule to the 4 π electrons: 4 does not fit the formula (4n + 2), since for n = 0, 1, 2, ... the allowed numbers are 2, 6, 10, etc., but not 4.
5. Conclude that because cyclobutadiene has 4 π electrons, it does not satisfy Hückel's rule and therefore is not aromatic; instead, it is antiaromatic and thus unstable.