According to Hückel's rule, a planar cyclic compound is aromatic if it has:
AAn odd number of double bonds
B$(4n+2) \pi$ electrons in a continuous conjugated system, with $n = 0, 1, 2, \ldots$
C$(4n+1) \pi$ electrons in a continuous conjugated system
D$4n \pi$ electrons
Answer & Solution
Correct answer: B. $(4n+2) \pi$ electrons in a continuous conjugated system, with $n = 0, 1, 2, \ldots$
**Hückel's rule**: a planar, fully conjugated, cyclic molecule is aromatic if and only if it has $(4n + 2)$ $\pi$ electrons, where $n$ is a non-negative integer.
Examples:
- Benzene: 6 $\pi$ electrons ($n=1$). Aromatic.
- Naphthalene: 10 $\pi$ electrons ($n=2$). Aromatic.
- Cyclobutadiene: 4 $\pi$ electrons (not $4n+2$). **Anti-aromatic**, highly unstable.
- Cyclopropenyl cation: 2 $\pi$ electrons ($n=0$). Aromatic.
Aromaticity gives extra stability (resonance stabilisation energy) and shifts properties: NMR ring current, no addition reactions at room temperature, etc.
Related questions
Nitration of toluene (methylbenzene) on the lab gives products mostly at:Benzene satisfies Hückel's rule with $4n + 2$ pi electrons for $n =$:Adding HBr to propene CH$_3$-CH=CH$_2$ in absence of peroxide gives mainly:The general formula for alkanes is:
Benzene is best described by:Wurtz reaction is used to prepare:In the presence of organic peroxides, addition of HBr to propene gives predominantly:
When propene reacts with HBr,