Consider the smallest possible completely conjugated, monocyclic planar carbocation that obeys Hucke — Hydrocarbons Chemistry Question
Question
Consider the smallest possible completely conjugated, monocyclic planar carbocation that obeys Huckel's rule and exhibits true aromaticity. Let $X$ be the number of carbon atoms in this ring, and $Y$ be the number of $\pi$ electrons it contains. What is the value of $X + Y$?
Answer: 5
💡 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 $X + Y = 3 + 2 = 5$.
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