The document discusses the application of Hückel molecular orbital (HMO) theory to monocyclic conjugated systems. Key points include calculating the energy levels of π electrons using HMO theory, determining the Hückel delocalization energy, using Frost-Musulin diagrams to visualize molecular orbital energies, and understanding Hückel's (4n+2) rule for determining aromaticity in cyclic conjugated hydrocarbons and heterocycles.

1. Application of HMO theory to
monocyclic conjugated systems,
Frost-Musulin diagrams and
Hückel's (4n+2) and 4n rules.
1
 Imran Laiq
 Roll No. 37
 MSc Part 1
 Semester I (2019-2020 )
 Date: 0/11/2019

2. Index
 Application of Hückel Molecular orbital theory to
monocyclic conjugated systems
 Energy levels for  electrons
 The Huckel delocalization energy (HDE)
 Frost-Musulin diagrams
 Aromaticity
 Hückel's (4n+2) and 4n rules
 Conclusion
 references

3. • Hückel Molecular orbital theory is applied to
conjugated systems. This theory is based on the
assumption that the pi-system can be treated
independently of the sigma frame work in conjugated
planer molecules. And the pi-system determining the
chemical and spectroscopic properties of conjugated
polyenes and aromatic compounds. This theory
provides a good qualitative description of pi-molecular
orbitals in both cyclic and acyclic conjugated systems.
• By HMO theory we can calculate:
 Energy levels for  electrons
 Delocalization energy in conjugated systems
Application of Hückel Molecular orbital
theory (HMO theory) to monocyclic
conjugated systems:

4. Energy level for a  electron in monocyclic conjugated
system can be calculated by the following equation.
𝐸𝑘 = 𝛼 + 2𝛽 𝐶𝑜𝑠
2𝐾𝜋
𝑛
Where,
K=0,±1, ±2, ±3….
n= numbers of carbon atoms in cycle
𝛼 and 𝛽 are constant and have different values and value
of v is always negative.
𝜋=180 degree
But energy level for a  electron in linear conjugated
polyenes is calculated by:
𝐸𝑘 = 𝛼 + 2𝛽 𝐶𝑜𝑠
𝐾𝜋
𝑛 + 1
1-Energy levels for  electrons:

5. 5
Energy levels for  electrons in cyclobutadiene:
Here are four energy levels in cyclobutadiene (𝐸0, 𝐸1, 𝐸−1and 𝐸2)
For 𝐸0
𝐸0 = 𝛼 + 2𝛽 𝐶𝑜𝑠
2×0 𝜋
4
𝐸0 = 𝛼 + 2𝛽 𝐶𝑜𝑠 0 as we know 𝐶𝑜𝑠 0 = 1
so 𝐸0 = 𝛼 + 2𝛽
For 𝐸1
𝐸1 = 𝛼 + 2𝛽 𝐶𝑜𝑠
2×1×180
4
𝐸1 = 𝛼 + 2𝛽 𝐶𝑜𝑠 90 as we know
𝐶𝑜𝑠 90 = 0 so 𝐸1 = 𝛼 + 2𝛽 0 𝐸1 = 𝛼
For 𝐸−1
𝐸−1 = 𝛼 + 2𝛽 𝐶𝑜𝑠
2×(−1) ×180
4
𝐸−1 = 𝛼 + 2𝛽 𝐶𝑜𝑠 −90 as we know
𝐶𝑜𝑠 −90 = 0 so 𝐸−1 = 𝛼 + 2𝛽 0 𝐸−1 = 𝛼
For 𝐸2
𝐸2 = 𝛼 + 2𝛽 𝐶𝑜𝑠
2×2×180
4
𝐸2 = 𝛼 + 2𝛽 𝐶𝑜𝑠 180 as we know
𝐶𝑜𝑠 180 = -1 so 𝐸2 = 𝛼 + 2𝛽 −1 𝐸2 = 𝛼 − 2𝛽

6. Cyclo
butadiene
Antibonding
Bonding
We got 𝛼 + 2𝛽, 𝛼, 𝛼, 𝛼 − 2𝛽 values of energy
the lowest energy value is 𝛼 + 2𝛽
and the highest energy value is 𝛼 − 2𝛽
𝛼 + 2𝛽
𝛼
𝛼
𝛼 − 2𝛽
Degenerate
energy levels

7. 7
Energy levels for  electrons in Benzene:
Here are six energy levels in cyclobutadiene (𝐸0, 𝐸1, 𝐸−1 𝐸2, 𝐸−2 and 𝐸3)
For K=0
𝐸0 = 𝛼 + 2𝛽 𝐶𝑜𝑠
2×0 𝜋
6
𝐸0 = 𝛼 + 2𝛽 𝐶𝑜𝑠 0 as we know 𝐶𝑜𝑠 0 = 1
so 𝐸0 = 𝛼 + 2𝛽
For K=1
𝐸1 = 𝛼 + 2𝛽 𝐶𝑜𝑠
2×1×180
6
𝐸1 = 𝛼 + 2𝛽 𝐶𝑜𝑠 60 as we know
𝐶𝑜𝑠 60 =
1
2
so 𝐸1 = 𝛼 + 2𝛽
1
2
𝐸1 = 𝛼 + 𝛽
For K=-1
𝐸−1 = 𝛼 + 2𝛽 𝐶𝑜𝑠
2×(−1) ×180
6
𝐸−1 = 𝛼 + 2𝛽 𝐶𝑜𝑠 −60 as we know
𝐶𝑜𝑠 −60 =
1
2
so 𝐸−1 = 𝛼 + 2𝛽
1
2
𝐸−1 = 𝛼 +𝛽
For K=2
𝐸2 = 𝛼 + 2𝛽 𝐶𝑜𝑠
2×2×180
6
𝐸2 = 𝛼 + 2𝛽 𝐶𝑜𝑠 120 as we know
𝐶𝑜𝑠 120 = -
1
2
so 𝐸2 = 𝛼 + 2𝛽 −
1
2
𝐸2 = 𝛼 − 𝛽

8. Benzene
Antibonding
Bonding
For K=-2
𝐸−2 = 𝛼 + 2𝛽 𝐶𝑜𝑠
2×(−2)×180
6
𝐸−2 = 𝛼 + 2𝛽 𝐶𝑜𝑠 −120 as we know
𝐶𝑜𝑠 −120 = -
1
2
so 𝐸−2 = 𝛼 + 2𝛽 −
1
2
𝐸 −2= 𝛼 − 𝛽
For K=3
𝐸3 = 𝛼 + 2𝛽 𝐶𝑜𝑠
2×3×180
6
𝐸3 = 𝛼 + 2𝛽 𝐶𝑜𝑠 180 as we know 𝐶𝑜𝑠 180 = -1
so 𝐸3 = 𝛼 + 2𝛽 −1 𝐸3 = 𝛼 − 2𝛽
We got 𝛼 + 2𝛽, 𝛼 + 𝛽, 𝛼 + 𝛽, 𝛼 − 𝛽, 𝛼 − 𝛽, 𝛼 − 2𝛽 values of energy the
lowest energy value is 𝛼 + 2𝛽 and the highest energy value is 𝛼 − 2𝛽
𝛼 + 2𝛽
𝛼 + 𝛽
𝛼 + 𝛽
𝛼 − 𝛽
𝛼 − 𝛽
𝛼 − 2𝛽

9. 2-The Huckel delocalization energy (HDE)
for cyclo butadiene:
HDE= Total energy in conjugated system – (Energy of ethene unit)×n
Where,
n= number of ethene unit
Energy of ethene unit= 𝟐 𝜶 + 𝜷
𝟐 𝜶 + 𝟐𝜷 + 𝟐𝜶 − 𝟐 𝜶 + 𝜷 × 𝟐
𝟐𝜶 + 𝟒𝜷 + 𝟐𝜶 − 𝟒𝜶 − 𝟒𝜷
HDE= 0
Because of this cyclo butadiene
Is not aromatic
9
𝛼 + 2𝛽
𝛼
𝛼
𝛼 − 2𝛽

10. 10
In Frost-Musulin diagram ,Frost’s circle is a useful mnenmonic device
For quickly setting down the HMO’s for cyclic system with out the use of
Mathematics.
The resulting diagram gives energy levels of the molecular orbitals of
cyclic, conjugated systems. If any energy levels presented on medal horizontal
dashed line, they must be non-bonding MOs. All those MOs below non-bonding Mos
Will be bonding Mos and above non-bonding Mos will be anti-bonding Mos.
Frost-Musulin diagram for Benzene:
Frost-Musulin diagrams:
Benzene
6 -electrons
non-bonding level
bonding MO's
anti-bonding MO's
𝛼 + 2𝛽
𝛼 + 𝛽
𝛼 − 𝛽
𝛼 − 2𝛽
Energy

11. 11
Frost-Musulin diagram for Cyclobutadiene:
Main point: For anti- aromatic compounds, such as cyclobutadiene and
cyclooctatetraene, there will be unpaired electrons in bonding,
non-bonding or antibonding MO's.
Cyclobutadiene
4 -electrons
non-bonding MO's
bonding MO
anti-bonding MO
non-bonding MO's
bonding MO
anti-bonding MO
Cyclooctatetraene
8 -electrons
Frost-Musulin diagram Cyclooctatetraene:
Energy
𝛼 + 2𝛽
𝛼
𝛼 − 2𝛽
Energy
𝛼 + 2𝛽
𝛼 + 1.4𝛽
𝛼
𝛼 − 1.4𝛽
𝛼 − 1.4𝛽

12. Aromaticity:
 Aromaticity is a property of the sp2 hybridized planar
rings in which the p orbitals (one on each atom) allow
cyclic delocalization of  electrons.
 The Criteria for Aromaticity : four structural criteria
must be satisfied for a compound to be aromatic.
[1] A molecule must be cyclic.
[2] A molecule must be planar.
[3] A molecule must be completely conjugated
(Each atom in the ring must be sp2-hybridized).
[4] A molecule must satisfy Hückel’s 4n + 2 rule
12

13. It was German scientist Hückel who in1931restricted his analysis to planar,
completely conjugated, monocyclic polyenes.
he formulated a simple rule which, states that the conjugated, monocyclic
polyenes will be aromatic, if it contains 4n + 2  electrons and if it contains
4n  electrons will be anti-aromatic where n is an integer
(n=0,1,2,3,4…etc.).
Anti-aromatic
aromatic Anti-aromatic
4 p electrons 8 p electrons
4n)= 4
n = 1
(4n + 2)= 6
4n = 6-2
4n =4
n=1
6 p electrons
4n= 8
n = 2
Hückel's (4n+2) and 4n rules:

14. 14
From Hückel's Rule monocyclic polyenes are divided by three types:
 Aromatic:
Cyclic
Conjugated: “alternating single and double bonds”
Planar: maximum overlap between conjugated  -bonds
Must contain 4n+2 -electrons (where n is an integer).
 Anti-aromatic:
cyclic, conjugated, planar molecules that contain
4n -electrons (where n is an integer).
 non-aromatic:
are every other molecule that fails one of these conditions.
Continuous…
Anti-aromatic
Anti-aromatic
4n= 4
n = 1
4n= 8
n = 2
(4n + 2)= 6
4n = 6-2
4n =4
n=1
Aromatic

15. larger Rings and Hückel's Rule
• Completely conjugated monocyclic rings larger than
benzene are also aromatic if they are planar and have 4n +
2  electrons.
• Hydrocarbons containing a single ring with alternating
double and single bonds are called annulenes.
• To name an annulene, indicate the number of atoms in the
ring in brackets and add the word annulene.

16. • [10]-Annulene has 10  electrons, which satisfies Hückel's
rule, but a planar molecule would place the two H atoms
inside the ring too close to each other. Thus, the ring
puckers to relieve this strain.
• Since [10]-annulene is not planar, the 10  electrons can’t
delocalize over the entire ring and it is not aromatic.
Continuous…

17. Heterocyclic Aromatic Compounds and
Hückel's Rule
• Heterocyclics containing oxygen, nitrogen or sulfur, can also be aromatic.
• With heteroatoms, we must determine whether the lone pair is localized
on the heteroatom or part of the delocalized  system.
• An example of an aromatic heterocyclic is pyridine.

18. Pyrrole:
 Pyrrole has a p orbital on every adjacent atom, so it is completely
conjugated.
 Pyrrole has six  electrons—four from the  bonds and two from the
lone pair.
Pyrrole is cyclic, planar, completely conjugated, and has 4n + 2 
electrons, so it is aromatic.
Continuous…

19. Furan:
 Furan has six  electrons—four from the  bonds and two from
the lone pair.
 two lone pairs on oxygen one pair is in a p orbital and is part of
ring p system; other is in an sp2 hybridized orbital and is not
part of ring p system.
O
••
••
(4n + 2)= 6
4n = 6-2
4n =4
n=1
Continuous…

20. • Histamine is a biologically active amine formed in many tissues. It is
an aromatic heterocycle with two N atoms—one which is similar to
the N atom of pyridine, and the other which is similar to the N atom of
pyrrole.
Continuous…

21. monocyclic ions and Hückel's Rule
Both negatively and positively charged ions can be aromatic if they
possess all the necessary elements.

22. although five resonance structures can also be drawn for the
cyclopentadienyl cation and radical, only the cyclopentadienyl
anion has 6  electrons, a number that satisfies Hückel’s rule.
Continuous…

23. 23
Conclusion:
 Hückel Molecular orbital theory is applied to
conjugated systems to get energy levels for  electrons
and delocalization energy of them.
 Frost-Musulin diagram is a useful mnemonic device for
quickly setting down the HMO’s for cyclic system with
out the use of mathematics. By Frost-Musulin diagram
we can easily found that the cycle is aromatic or not.
 Hückel's (4n+2) and 4n rules states that the
conjugated, monocyclic polyenes will be aromatic or
anti aromatic.

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1. Vollhardt, Peter, and Neil E. Schore. Organic Chemistry: Structure and
Function. 5th ed. New York: W. H. Freeman & Company (2007).
2. Berson, Jerome. Chemical Creativity: Ideas from the Work of Woodward,
Hückel, Meerwein, and Others. New York: Wiley-VCH (1999).
3. Badger, G.M. Aromatic Character and Aromaticity. London, England:
Cambridge University Press (1969).
4. Lewis, David and David Peters. Facts and Theories of Aromaticity. London,
England: Macmillan Press (1975).
5. Jonathan Clyden, Nick Greevens and Stuart Warren , Organic Chemistry,
2nd edition, Oxford university press (2012).
References:

25. Acknowledgment:
I am thankful to and fortunate enough to get
constant encouragement, support and guidance
from all Teaching staffs of Department of
Chemistry especially Prof. who
helped me in successfully completing my
presentation work. Also, I would like to extend
my sincere esteems to all non teaching staff
and classmates for their timely support.

26. 26