Question
By using the rules of aromaticity, explain whether the following compounds are aromatic or non$-$aromatic.
$i.$ Benzene $ii.$ Naphthalene $iii.$ Cycloheptatriene
$i.$ Benzene $ii.$ Naphthalene $iii.$ Cycloheptatriene
✓
Answer
$i.$ Benzene:
$a.$ It is cyclic and planar.
$b.$ It has three double bonds and six $\pi$ electrons.
$c.$ It has a $p$ orbital on each carbon of the hexagonal ring. Hence, a continuous overlap above and below the ring is possible.
$d.$ According to Huckcl rule, this compound is aromatic if, $4n + 2 =$ Number of $\pi$ electrons.
$4n + 2 = 6,$
$\therefore 4n = 6 – 2 = 4$
$n = 4/4 = 1,$ here, $‘n \ ’$ comes out to be an integer.
Hence, benzene is aromatic.
$ii.$ Naphthalene:
$a.$ It is cyclic and planar.
$b.$ It has $5$ double bonds and $10 n$ electrons.
$c.$ It has a $p$ orbital on each carbon atom of the ring. Hence, a continuous overlap around the ring is possible.
$d.$ According to Huckel rule, this compound is aromatic if, $4n + 2 =$ Number of $\pi$ electrons.
$4n + 2 = 10,$
$\therefore 4n = 10 – 2 = 8$
$n = 8/4 = 2,$ Here $‘n’$ comes out to be an integer.
Hence, naphthalene is aromatic.
$iii.$ Cycloheptatriene:
$a.$ It is cyclic and planar.
$b.$ It has three double bonds and $6 \pi$ electrons.
$c.$ But one of the carbon atoms is saturated $(sp^3$ hybridized$)$ and it does not have a $p$ orbital.
$d.$ Hence, a continuous overlap around the ring is not possible in cycloheptatriene. Hence, it is non$-$aromatic.
$a.$ It is cyclic and planar.
$b.$ It has three double bonds and six $\pi$ electrons.
$c.$ It has a $p$ orbital on each carbon of the hexagonal ring. Hence, a continuous overlap above and below the ring is possible.
$d.$ According to Huckcl rule, this compound is aromatic if, $4n + 2 =$ Number of $\pi$ electrons.
$4n + 2 = 6,$
$\therefore 4n = 6 – 2 = 4$
$n = 4/4 = 1,$ here, $‘n \ ’$ comes out to be an integer.
Hence, benzene is aromatic.
$ii.$ Naphthalene:
$a.$ It is cyclic and planar.
$b.$ It has $5$ double bonds and $10 n$ electrons.
$c.$ It has a $p$ orbital on each carbon atom of the ring. Hence, a continuous overlap around the ring is possible.
$d.$ According to Huckel rule, this compound is aromatic if, $4n + 2 =$ Number of $\pi$ electrons.
$4n + 2 = 10,$
$\therefore 4n = 10 – 2 = 8$
$n = 8/4 = 2,$ Here $‘n’$ comes out to be an integer.
Hence, naphthalene is aromatic.
$iii.$ Cycloheptatriene:
$a.$ It is cyclic and planar.
$b.$ It has three double bonds and $6 \pi$ electrons.
$c.$ But one of the carbon atoms is saturated $(sp^3$ hybridized$)$ and it does not have a $p$ orbital.
$d.$ Hence, a continuous overlap around the ring is not possible in cycloheptatriene. Hence, it is non$-$aromatic.
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