Which of the following heterocycles is aromatic?

• A. Pyridine
• B. Azete
• C. Furan
• D. Cyclobutadiene

Answer: (C)

Verified step by step guidance

1. Step 1: Recall the criteria for aromaticity, which are: (a) the molecule must be cyclic, (b) the molecule must be planar or nearly planar, (c) the molecule must have a continuous ring of p-orbitals (conjugation), and (d) the molecule must follow Hückel's rule, having \$4n + 2\( π electrons, where \)n$ is a non-negative integer.
2. Step 2: Analyze Pyridine: It is a six-membered ring with five carbons and one nitrogen atom. The ring is planar, fully conjugated, and contains 6 π electrons (one from each of the five carbons and one from the nitrogen's lone pair in the p orbital). This satisfies Hückel's rule with \(n=1\) (since \$4(1) + 2 = 6$).
3. Step 3: Analyze Azete (azetidine or azete depending on context): Azete is a four-membered ring with one nitrogen atom. Typically, four-membered rings are strained and may not be planar. Also, the number of π electrons and conjugation must be checked. Azete usually does not have a continuous conjugated system with \$4n + 2$ π electrons, so it is generally not aromatic.
4. Step 4: Analyze Furan: Furan is a five-membered ring with four carbons and one oxygen atom. It is planar and conjugated. The oxygen contributes two π electrons from its lone pair, and the ring has 4 π electrons from the double bonds, totaling 6 π electrons, satisfying Hückel's rule with \(n=1\).
5. Step 5: Analyze Cyclobutadiene: It is a four-membered ring with alternating double bonds. Although it is cyclic and conjugated, it has 4 π electrons, which fits \$4n\( (with \)n=1\() rather than \)4n + 2$. This makes it antiaromatic and unstable, so it is not aromatic.