Consider the smallest possible completely conjugated, monocyclic planar carbocation that obeys Hucke — Hydrocarbons Chemistry Question

💡 Solution & Explanation

The smallest aromatic ring corresponds to Huckel's integer $n=0$, yielding $4(0)+2 = 2$ $\pi$ electrons ($Y=2$). To have a completely conjugated monocyclic ring holding 2 $\pi$ electrons and a formal charge allowing cyclic delocalization, a minimum of 3 carbon atoms ($X=3$) is required (forming the cyclopropenyl cation, $C_3H_3^+$). The sum is $X + Y = 3 + 2 = 5$.