Question

Which of the following compounds are antiaromatic?

Answer 1

Hint: A compound is said to the anti-aromatic, if it is cyclic, planar, have the hybridization as $s{{p}^{2}}$and obeys the$(4n)\pi e{{s}^{-}}$rule in which n can be any integer such as 1,2,3,---and so on. From the above compounds, which satisfy this condition ,is called the anti -aromatic compound. now identify it.

Complete step by step answer:
First of all, we should know what aromatic compounds are. A compound is said to be aromatic if it satisfies the following conditions:
1. there should be $(4n+2)\pi e{{s}^{-}}$ and here, n can be any integer 1,2,3,3-----and so on.
2. Compounds should be cyclic( i.e. in ring) and must contain unhybridized p-orbitals and the ring should be planar .
3. All the carbon atoms should be $s{{p}^{2}}$ hybridized.
4. Cyclic systems should have conjugation ( i.e. there should be the delocalization of pi-electrons).
On the other hand, anti-aromatic compounds also satisfy all the conditions except the one, that there are $(4n)\pi e{{s}^{-}}$ instead of $(4n+2)\pi e{{s}^{-}}$. And the pi electrons are in the multiple of 4 like
4,8,12,16---and so on.

S.no	Compound	Cyclic	Planar	$s{{p}^{2}}$hybridization	$(4n)\pi e{{s}^{-}}$	Anti -aromatic
1.		cyclic	planar	$s{{p}^{2}}$	Has 2 pi bonds i.e. 4 electrons (n=1) and 2 electrons of the lone pair which is inside the plane i.e. there are a total no of 6 electrons . so, it doesn’t obey this rule.	no
2.		cyclic	planar	One carbon is $s{{p}^{3}}$hybridized	Has 3 pi bonds i.e. 6 electrons=1 and it is not a multiple of 4 and doesn’t obey this rule.	no
3.		cyclic	planar	$s{{p}^{2}}$	Has 4 pi bonds i.e.8 electrons(n=2) and is multiple of 4. So, it obeys this rule.	yes
4.		Cyclic	planar	$s{{p}^{2}}$	Has 2 pi bonds i.e. 4 electrons (n=1) and 1 electron of the negative charge and electrons of lone pair would not be considered because they are outside the plane i.e. there are a total no of 3 electrons . so, it doesn’t obey this rule.	no
5.		Cyclic	planar	$s{{p}^{2}}$	Has 2 pi bonds i.e. 4 electrons (n=1) and 2 electrons of the lone pair i.e. there are a total no of 6 electrons . so, it doesn’t obey this rule.	yes
6.		cyclic	planar	$s{{p}^{2}}$	Has 2 pi bonds i.e. 4 electrons (n=1) and is multiple of 4. So, obey this rule.	yes

Note: in this, formula $(4n)\pi e{{s}^{-}}$, the value of cannot be negative and it always has a positive value and the compounds which doesn’t above any conditions or any one condition of the aromaticity, those compounds are said to be non-aromatic compounds.