Question
Why is benzene extra ordinarily stable though it contains three double bonds?
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Answer
Benzene is a hybrid of resonating structures given as:
All six carbon atoms in benzene are $\mathrm{sp}^2$ hybridized. The two $\mathrm{sp}^2$ hybrid orbitals of each carbon atom overlap with the $\mathrm{sp}^2$ hybrid orbitals of adjacent carbon atoms to form six sigma bonds in the hexagonal plane. The remaining $\mathrm{sp}^2$ hybrid orbital on each carbon atom overlaps with the s-orbital of hydrogen to form six sigma C-H bonds. The remaining unhybridized p-orbital of carbon atoms has the possibility of forming three $\pi$ bonds by the lateral overlap of $C_1-C_2, C_3-C_4, C_5-C_6$, or $C_2-C_3, C_4-C_5, C_6-C_1$.
The six $\pi'\text{s}$ are delocalized and can move freely about the six carbon nuclei. Even after the presence of three double bonds, these delocalized $\pi-\text{electrons}$ stabilize benzene.
All six carbon atoms in benzene are $\mathrm{sp}^2$ hybridized. The two $\mathrm{sp}^2$ hybrid orbitals of each carbon atom overlap with the $\mathrm{sp}^2$ hybrid orbitals of adjacent carbon atoms to form six sigma bonds in the hexagonal plane. The remaining $\mathrm{sp}^2$ hybrid orbital on each carbon atom overlaps with the s-orbital of hydrogen to form six sigma C-H bonds. The remaining unhybridized p-orbital of carbon atoms has the possibility of forming three $\pi$ bonds by the lateral overlap of $C_1-C_2, C_3-C_4, C_5-C_6$, or $C_2-C_3, C_4-C_5, C_6-C_1$.
The six $\pi'\text{s}$ are delocalized and can move freely about the six carbon nuclei. Even after the presence of three double bonds, these delocalized $\pi-\text{electrons}$ stabilize benzene.
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