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Hint: (1) A compound is said to be aromatic if it is cyclic and has planar ring, and also it must have cyclic delocalized pi-electron clouds lying above and below the plane of the ring.
(2) It must also follow Huckel’s $(4n+2)\pi$ electron rule where n is any integer.
(3) For antiaromaticity, the rule is that the molecule must have ${\text{4n\pi }}$ electrons where n is any integer.
Complete step-by-step answer: Both the cyclopentadienyl cation and cyclopentadienyl anion are formed from cyclopentadiene by the loss of a hydride ion and the abstraction of a proton respectively. Cyclopentadiene is not an aromatic compound because of the presence of a ${\text{s}}{{\text{p}}^{\text{3}}}$ hybridized ring carbon on its ring due to which it does not contain an uninterrupted cyclic pi-electron cloud. When a hydride anion $\left( {{{\text{H}}^{\text{ - }}}} \right)$ is removed from the ${\text{s}}{{\text{p}}^{\text{3}}}$ hybridized ring carbon of cyclopentadiene, the cyclopentadienyl cation will be formed and the hybridization of the ${\text{s}}{{\text{p}}^{\text{3}}}$ hybridized ring carbon will be changed from ${\text{s}}{{\text{p}}^{\text{3}}}$to${\text{s}}{{\text{p}}^2}$. As a result of this conversion of hybridization, the cyclopentadienyl cation formed will be planar and will contain a cyclic pi-electron cloud above and below the ring. However, it fails to meet the Huckel’s rule of aromaticity as it does not have $(4n+2)\pi$ electrons and so it is not aromatic. But, it does have ${\text{4n\pi }}$ electrons (n is equal to 1 as there are 4 pi electrons). Hence, it is antiaromatic.
When a proton is abstracted from the ${\text{s}}{{\text{p}}^{\text{3}}}$ hybridized ring carbon of cyclopentadiene, the cyclopentadienyl anion is formed and the hybridization of the ${\text{s}}{{\text{p}}^{\text{3}}}$ hybridized ring carbon will be changed from ${\text{s}}{{\text{p}}^{\text{3}}}$ to ${\text{s}}{{\text{p}}^2}$. As a result of this conversion of hybridization, the cyclopentadienyl anion formed will be planar and will contain a cyclic pi-electron cloud above and below the ring. Moreover, it also satisfies the Huckel’s rule for aromaticity as it has $(4n+2)\pi$ electrons (n is equal to 1 as there are 6 pi electrons) and so it is aromatic. Thus, the cyclopentadienyl anion is an aromatic compound.
Thus, the given statement is true and so 1.
Note: In the $(4n+2)\pi$ rule, n is not a property of the molecule. This rule is applied just to generate a series: n is equal to 2, 6, 10, 14 etc. and if our pi electron value matches any number in this series, then it possesses aromaticity.
(2) It must also follow Huckel’s $(4n+2)\pi$ electron rule where n is any integer.
(3) For antiaromaticity, the rule is that the molecule must have ${\text{4n\pi }}$ electrons where n is any integer.
Complete step-by-step answer: Both the cyclopentadienyl cation and cyclopentadienyl anion are formed from cyclopentadiene by the loss of a hydride ion and the abstraction of a proton respectively. Cyclopentadiene is not an aromatic compound because of the presence of a ${\text{s}}{{\text{p}}^{\text{3}}}$ hybridized ring carbon on its ring due to which it does not contain an uninterrupted cyclic pi-electron cloud. When a hydride anion $\left( {{{\text{H}}^{\text{ - }}}} \right)$ is removed from the ${\text{s}}{{\text{p}}^{\text{3}}}$ hybridized ring carbon of cyclopentadiene, the cyclopentadienyl cation will be formed and the hybridization of the ${\text{s}}{{\text{p}}^{\text{3}}}$ hybridized ring carbon will be changed from ${\text{s}}{{\text{p}}^{\text{3}}}$to${\text{s}}{{\text{p}}^2}$. As a result of this conversion of hybridization, the cyclopentadienyl cation formed will be planar and will contain a cyclic pi-electron cloud above and below the ring. However, it fails to meet the Huckel’s rule of aromaticity as it does not have $(4n+2)\pi$ electrons and so it is not aromatic. But, it does have ${\text{4n\pi }}$ electrons (n is equal to 1 as there are 4 pi electrons). Hence, it is antiaromatic.
When a proton is abstracted from the ${\text{s}}{{\text{p}}^{\text{3}}}$ hybridized ring carbon of cyclopentadiene, the cyclopentadienyl anion is formed and the hybridization of the ${\text{s}}{{\text{p}}^{\text{3}}}$ hybridized ring carbon will be changed from ${\text{s}}{{\text{p}}^{\text{3}}}$ to ${\text{s}}{{\text{p}}^2}$. As a result of this conversion of hybridization, the cyclopentadienyl anion formed will be planar and will contain a cyclic pi-electron cloud above and below the ring. Moreover, it also satisfies the Huckel’s rule for aromaticity as it has $(4n+2)\pi$ electrons (n is equal to 1 as there are 6 pi electrons) and so it is aromatic. Thus, the cyclopentadienyl anion is an aromatic compound.
Thus, the given statement is true and so 1.
Note: In the $(4n+2)\pi$ rule, n is not a property of the molecule. This rule is applied just to generate a series: n is equal to 2, 6, 10, 14 etc. and if our pi electron value matches any number in this series, then it possesses aromaticity.
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