135.Aromaticity(28) – Practice problems(2).

5)

1

Is it cyclic ?

Conjugated double bonds ?

Is it planar?

×

# of π electrons

8

# of lone pair of electrons involved in conjugation

0

Total π electrons

8

4n+2 ?

×

This a 8-membered ring and due to ring strain it is not planar. Thus, this molecule is non-aromatic.


6)

1351

The above compound has 10 π electrons. There is delocalisation of these electrons due to conjugation.This molecule has a bridge head carbon atom , which prevents the ring from bending. So, the structure is planar. While considering aromaticity , we just look at the peripheral carbon atoms i.e we consider the larger ring(10 Carbon atom ring) and we ignore the bridge head.

1352.jpg

 

Is it cyclic ?

Conjugated double bonds ?

Is it planar?

(due to the bridgehead carbon)

# of π electrons

10

# of lone pair of electrons involved in conjugation

0

Total π electrons

10

4n+2 ?

∴ The structure is aromatic.

 


7)1.jpg

The above compound looks complicated .However, we shall just count the no.of π electrons it has and check if it is aromatic. The molecule has both double and triple bonds. A triple bond has sp hybridization .So, the carbon atom has two hybridized sp orbitals and two unhybridized p-orbitals. The p-orbitals in a triple bond ,which lie in the plane of the p-orbitals of the other double bonds, are considered for conjugation. 

The above molecule thus has 9 multiple bonds in conjugation. Thus, there are 18 π electrons.

Is it cyclic ?

Conjugated double bonds ?

Is it planar?

# of π electrons

18

# of lone pair of electrons involved in conjugation

0

Total π electrons

18

4n+2 ?

Thus, this molecule is aromatic.


8)

1.jpg

All derivatives of benzene are aromatic as they have the parent benzene ring in their structure. The above molecule, benzoic acid, is aromatic too.


9)

1.jpg

This molecule is a component of nucleic acid. It has two rings and there is conjugation between the single and double bonds ,throughout the molecule. According to the exception rule , all N’s are sp2 hybridized. Though the lone pairs on nitrogen atoms are not shown, we should be able to discern them. As there is no formal charge on nitrogen atoms, they all have one lone pair respectively(refer post 122).However, only the lone pair on nitrogen attached to -H (shown in red color)takes part in aromatic sextet.

1.jpg

 

Is it cyclic ?

Conjugated double bonds ?

Is it planar?

# of π electrons

8

# of lone pair of electrons involved in conjugation

2

Total π electrons

10

4n+2 ?

Thus, this molecule is aromatic.


Q:Arrange the following compounds in descending order of their resonance energies –

1.jpg

More the conjugation in a molecule, higher will be its resonance energy. Napthalene (1) has more conjugation than benzene (3).Thus, naphthalene has the highest resonance energy. Benzene is more aromatic than furan(2). So , its resonance energy is higher than furan. Thus, the order of resonance energy for the above three compounds is –

Naphthalene > Benzene > Furan.


From the next post onwards we will start discussing a new topic. Till then ,

Be a perpetual student of life and keep learning…..

Good day!

 

References –

1)https://www.youtube.com/watch?v=KU6YSsss4UI

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