See image — GOC and Organic Chemistry Basics Chemistry Question

💡 Solution & Explanation

# Analysis of Aromaticity To determine which species is not aromatic, apply **Hückel's Rule**: A cyclic, planar, conjugated compound is aromatic if it has $4n+2$ π electrons (where $n = 0, 1, 2, ...$). ## Checking Each Species: **(A) Pyrylium cation** ($C_5H_5O^+$) - 5-membered ring with oxygen - π electrons: benzene-like structure with 6 π electrons - $6 = 4(1) + 2$ ✓ **AROMATIC** **(B) Indole** ($C_8H_7N$) - Benzene ring fused to pyrrole - Combined aromatic system with 10 π electrons - $10 = 4(2) + 2$ ✓ **AROMATIC** **(C) Boron-containing compound** - Three boron atoms bonded through oxygen bridges with methyl groups - This is **not a cyclic conjugated π system** - Boron atoms are sp² hybridized but lack continuous π-orbital overlap - Does not satisfy aromaticity criteria ✗ **NOT AROMATIC** ## Conclusion: **(D) None of these** is correct because **option (C) is the non-aromatic species**. The compound lacks the continuous cyclic conjugated π-electron system required for aromaticity, making it non-aromatic while (A) and (B) are both aromatic.