
How many species out of the following are aromatic?

Answer
463.8k+ views
Hint: Aromatic compounds are considered as the compound which contains a conjugated planar ring system with delocalized pi-electrons cloud located at alternate double bond and single bond. For a compound to be aromatic it should follow Huckel’s rule which says that the compound should satisfy 4n+2$\pi$ electrons.
Complete answer:
In the above given compound, 1 $\pi$ bond is present which contains 2 electrons. Due to negative charge 2 electrons are present in the same plane. So, the total number of electrons is 4. This compound does not follow the 4n+2$\pi$rule. Therefore, it is antiaromatic.
In the above given compound, 1 $\pi$ bond is present which contains 2 electrons. It is an aromatic compound as it is cyclic, and planar. Due to positive charge all the carbon atoms become $s{p^2}$ hybridized and it follows Huckel’s rule.
In the above given compound, 3 $\pi$ bond is present which contains 6 electrons. The oxygen atom is not present in the plane, therefore its electrons are not counted. This compound is aromatic as it follows the 4n+2$\pi$rule.
In the above given compound, 3 $\pi$ bond is present which contains 6 electrons. The other cycle is not present in the plane, so its electrons are not counted. This compound is aromatic as it follows the 4n+2$\pi$rule.
In the above compound, total 5 $\pi$ bonds are present which contains 10 electrons. Therefore, this compound is aromatic as it follows the 4n+2$\pi$rule.
In the above compound, total 5 $\pi$ bonds are present which contains 10 electrons but the $\pi$electrons are not present in conjugation. Therefore, it is an antiaromatic compound.
Note:
In the presence of fused ring structure, the cycle with maximum number of conjugated $\pi$ bonds is considered. The lone pairs which are present within the plane only participate in Huckel’s rule, else they are not considered.
Complete answer:

In the above given compound, 1 $\pi$ bond is present which contains 2 electrons. Due to negative charge 2 electrons are present in the same plane. So, the total number of electrons is 4. This compound does not follow the 4n+2$\pi$rule. Therefore, it is antiaromatic.

In the above given compound, 1 $\pi$ bond is present which contains 2 electrons. It is an aromatic compound as it is cyclic, and planar. Due to positive charge all the carbon atoms become $s{p^2}$ hybridized and it follows Huckel’s rule.

In the above given compound, 3 $\pi$ bond is present which contains 6 electrons. The oxygen atom is not present in the plane, therefore its electrons are not counted. This compound is aromatic as it follows the 4n+2$\pi$rule.

In the above given compound, 3 $\pi$ bond is present which contains 6 electrons. The other cycle is not present in the plane, so its electrons are not counted. This compound is aromatic as it follows the 4n+2$\pi$rule.

In the above compound, total 5 $\pi$ bonds are present which contains 10 electrons. Therefore, this compound is aromatic as it follows the 4n+2$\pi$rule.

In the above compound, total 5 $\pi$ bonds are present which contains 10 electrons but the $\pi$electrons are not present in conjugation. Therefore, it is an antiaromatic compound.
Note:
In the presence of fused ring structure, the cycle with maximum number of conjugated $\pi$ bonds is considered. The lone pairs which are present within the plane only participate in Huckel’s rule, else they are not considered.
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