---
title: "Hückel Theory and Conjugated Systems"
book: "University Chemistry — Year 2"
subject: chemistry
language: en
chapter: 16
exercises: 12
source: https://one-course.com/books/chemistry/3/en/chapter/16-huckel-theory-and-conjugated-systems
license: CC-BY-NC-SA-4.0
credit: "One Chemistry Book, One Course (one-course.com)"
---

# Chapter 16 — Hückel Theory and Conjugated Systems

Add hydrogen to cyclohexene and $119\,\mathrm{kJ}$ are released per mole. Benzene, with three double bonds in a ring of six carbons, should release three times as much on its way to cyclohexane; it releases $150\,\mathrm{kJ}$ less. That missing energy is the stability of an [aromatic](#def-b2-huckel-aromatic) ring, the reason why benzene resists the additions that alkenes undergo. In 1931 Erich Hückel computed it with a determinant, after reducing the molecule to its $\pi$ electrons and their neighbours. His method is crude, but it explains why rings of six $\pi$ electrons are special, why long conjugated chains absorb visible light, and where a conjugated molecule carries its charges.

**You already know.**

[Chapter 14](https://one-course.com/books/chemistry/3/en/chapter/14-molecular-orbitals-of-diatomic-molecules#ch-b2-diatomic-mos): [Coulomb integral](https://one-course.com/books/chemistry/3/en/chapter/14-molecular-orbitals-of-diatomic-molecules#def-b2-diatomic-mos-integrals) $\alpha$, [resonance integral](https://one-course.com/books/chemistry/3/en/chapter/14-molecular-orbitals-of-diatomic-molecules#def-b2-diatomic-mos-integrals) $\beta$, [secular determinant](https://one-course.com/books/chemistry/3/en/chapter/14-molecular-orbitals-of-diatomic-molecules#def-b2-diatomic-mos-integrals). [Chapter 15](https://one-course.com/books/chemistry/3/en/chapter/15-fragment-orbitals-and-polyatomic-molecules#ch-b2-fragment-orbitals): $\pi$ orbitals of ethene and allyl. The Year 1 volume: conjugated systems, resonance, chromophores and absorption maxima.

## 16.1 The Hückel method

In a planar conjugated molecule, the $\sigma$ orbitals are symmetric and the p orbitals perpendicular to the plane antisymmetric with respect to that plane: they do not mix ([Proposition 14.7](https://one-course.com/books/chemistry/3/en/chapter/14-molecular-orbitals-of-diatomic-molecules#prop-b2-diatomic-mos-symmetry)), and the $\pi$ electrons can be treated on their own.

**Definition 16.1 (Hückel method).**

The *Hückel method* computes the $\pi$ orbitals of a planar conjugated system as combinations $\psi = \sum_r c_r p_r$ of one p orbital per atom, with the approximations: overlaps $S_{rs} = 0$ for $r \neq s$ (and 1 for $r =
s$); [Coulomb integrals](https://one-course.com/books/chemistry/3/en/chapter/14-molecular-orbitals-of-diatomic-molecules#def-b2-diatomic-mos-integrals) all equal to $\alpha$; [resonance integrals](https://one-course.com/books/chemistry/3/en/chapter/14-molecular-orbitals-of-diatomic-molecules#def-b2-diatomic-mos-integrals) equal to $\beta < 0$ between bonded neighbours and zero otherwise.

With $x = (\alpha - E)/\beta$, the secular equations become, for each atom $r$, $xc_r + \sum_{s \text{ bonded to } r} c_s = 0$, and the energies are the roots of a determinant with $x$ on the diagonal and 1 between bonded atoms. Equivalently, $E =
\alpha + m\beta$ where the $m$ are the eigenvalues of the adjacency matrix of the molecule’s carbon skeleton; since $\beta < 0$, a positive $m$ is a bonding level.

**Method 16.2 (A Hückel calculation).**

1. Number the conjugated atoms; write the matrix with 0 on the diagonal and 1 for each bonded pair.
2. Find its eigenvalues $m_k$ (by the closed forms below, or numerically): $E_k =  \alpha + m_k\beta$ .
3. Fill the levels from the most bonding, two electrons each: $E_\pi = \sum_k n_k E_k$ .
4. From the normalised coefficients, compute charges and bond indices.

**Proposition 16.3 (Trace).**

The energies of the $n$ $\pi$ orbitals of a Hückel system add up to $n\alpha$.

**Proof.** The sum of the eigenvalues of a matrix is its trace; the Hückel matrix has $\alpha$ on each of its $n$ diagonal entries. ∎

## 16.2 Linear polyenes

**Theorem 16.4 (Levels of a linear chain).**

A linear conjugated chain of $n$ atoms has the levels and coefficients

$$
E_k = \alpha + 2\beta\cos\frac{k\pi}{n + 1}, \qquad
  c_{jk} = \sqrt{\frac{2}{n + 1}}\,\sin\frac{jk\pi}{n + 1}, \qquad k = 1, \dots, n .
$$

**Proof.** The secular equations are $c_{j-1} + xc_j + c_{j+1} = 0$ for $j = 1, \dots, n$, with $c_0 = c_{n+1} = 0$. Try $c_j = \sin j\theta$: $\sin(j - 1)\theta + \sin(j + 1)\theta = 2\sin
j\theta\cos\theta$, so the equations hold with $x = -2\cos\theta$, and $c_0 = 0$ automatically. The end condition $c_{n+1} = \sin(n + 1)\theta = 0$ gives $\theta =
k\pi/(n + 1)$. Then $E = \alpha - x\beta = \alpha + 2\beta\cos\theta$. The normalisation uses $\sum_{j=1}^n\sin^2 j\theta = (n + 1)/2$. ∎

**Example 16.5 (Butadiene).**

For $n = 4$: $E = \alpha \pm 1.618\beta$, $\alpha \pm 0.618\beta$. Four $\pi$ electrons fill the two bonding levels: $E_\pi = 4\alpha + 2(1.618 + 0.618)\beta = 4\alpha + 4.472\beta$. The lowest orbital has coefficients 0.372, 0.602, 0.602, 0.372, the next 0.602, 0.372, $-0.372$, $-0.602$.

![The four π orbitals of butadiene, lobes drawn in proportion to the Hückel coefficients (computed) and coloured by their sign; energies increase upwards, with 0, 1, 2, 3 nodes. The two lower orbitals are filled.](https://one-course.com/images/onecourse/chapters/chemistry-3/b2-huckel/fig-469ad203d12e.svg)

*The four $\pi$ orbitals of butadiene, lobes drawn in proportion to the Hückel coefficients (computed) and coloured by their sign; energies increase upwards, with $0, 1, 2, 3$ nodes. The two lower orbitals are filled.*

**Definition 16.6 (π\piπ charges and bond indices).**

With $n_k$ electrons in orbital $k$ (0, 1 or 2) and real coefficients $c_{rk}$, the *$\pi$ charge* on atom $r$ is $q_r = \sum_k n_kc_{rk}^2$ (the number of $\pi$ electrons it carries), and the *$\pi$ bond index* of the bond $rs$ is $p_{rs} = \sum_k n_kc_{rk}c_{sk}$.

For butadiene, $p_{12} = 2(0.372 \times 0.602 + 0.602 \times 0.372) = 0.894$ and $p_{23} =
2(0.602^2 - 0.372^2) = 0.447$: the end bonds carry most of the $\pi$ bonding and are the shorter, the central bond has partial double-bond character. All the charges are 1.

**Proposition 16.7 (Alternant hydrocarbons).**

In a neutral hydrocarbon whose conjugated atoms can be coloured in two sets with no two neighbours of the same set (no odd ring), every $\pi$ charge equals 1.

**Proof.** *Admitted at this level.* ∎

The result (Coulson and Rushbrooke) is checked on butadiene above and on benzene by symmetry; it is the reason why such hydrocarbons have no permanent dipole from their $\pi$ electrons.

**Definition 16.8 (Delocalisation energy).**

The *delocalisation energy* of a conjugated system is the difference between its $\pi$ energy and that of the same number of isolated double bonds, each with $E_\pi = 2\alpha + 2\beta$.

For butadiene it is $4.472\beta - 4\beta = 0.472\beta$; for hexatriene, $0.988\beta$.

## 16.3 Rings and aromaticity

**Theorem 16.9 (Levels of a ring).**

A planar ring of $n$ conjugated atoms has the levels $E_k = \alpha + 2\beta\cos(2\pi k/n)$, $k = 0, 1, \dots, n - 1$: one lowest level $\alpha + 2\beta$, then pairs of degenerate levels, and for even $n$ one highest level $\alpha - 2\beta$.

**Proof.** In a ring every atom has two neighbours, with indices taken modulo $n$: $c_{j-1} + xc_j + c_{j+1} = 0$ for all $j$ and $c_{j+n} = c_j$. Try $c_j = \mathrm e^{\mathrm
i jq}$: the equation gives $x = -(\mathrm e^{-\mathrm iq} + \mathrm e^{\mathrm iq}) = -2\cos q$, and periodicity requires $nq = 2\pi k$. The levels $k$ and $n - k$ have the same energy; their sum and difference are real orbitals. Only $k = 0$ (and $k = n/2$ for even $n$) is single. ∎

**Corollary 16.10 (Frost’s circle).**

The levels of a ring are read on a circle of radius $2|\beta|$ centred at $\alpha$, in which a regular $n$-gon is inscribed with one vertex at the bottom: each vertex is at the height of a level.

**Proof.** The vertex $k$ of the polygon is at angle $-\pi/2 + 2\pi k/n$, at height $-2|\beta|\cos(2\pi k/n) = 2\beta\cos(2\pi k/n)$ relative to the centre. ∎

![Frost’s circles. The vertices of each polygon, inscribed with a vertex at the bottom in a circle of radius 2| |, give the levels of the ring. With six π electrons, C5H5-, C6H6 and C7H7+ fill the lowest level and one degenerate pair: closed shells. Cyclobutadiene’s four electrons leave two electrons in a degenerate nonbonding pair (drawn).](https://one-course.com/images/onecourse/chapters/chemistry-3/b2-huckel/fig-bf946aa7ee69.svg)

*Frost’s circles. The vertices of each polygon, inscribed with a vertex at the bottom in a circle of radius $2|\beta|$, give the levels of the ring. With six $\pi$ electrons, $\ce{C5H5-}$, $\ce{C6H6}$ and $\ce{C7H7+}$ fill the lowest level and one degenerate pair: closed shells. Cyclobutadiene’s four electrons leave two electrons in a degenerate nonbonding pair (drawn).*

**Definition 16.11 (Aromaticity).**

A cyclic, planar, fully conjugated system is *aromatic* when its $\pi$ electrons form a closed shell with a large [delocalisation energy](#def-b2-huckel-delocalisation), and *antiaromatic* when its $\pi$ electrons half-fill a degenerate pair, which makes it less stable than the corresponding open chain.

**Theorem 16.12 (Hückel’s rule).**

A planar monocyclic conjugated system has a closed shell of $\pi$ electrons if and only if it holds $4N + 2$ of them ($N = 0, 1, 2, \dots$); with $4N$ it has two electrons in a half-filled degenerate pair. This is *Hückel’s rule*.

**Proof.** By [Theorem 16.9](#thm-b2-huckel-ring), the levels from the bottom are one single level and then degenerate pairs (for the occupied part). A closed shell fills the single level (2 electrons) and a whole number $N$ of pairs (4 electrons each): $4N + 2$. With $4N$ electrons, the last pair holds only two, one in each of its orbitals (Hund’s rule). ∎

**Method 16.13 (Is it aromatic?).**

1. Is the system a ring with a p orbital on every atom (counting carbocation and carbanion centres and lone pairs of heteroatoms)?
2. Can it be planar?
3. Count the $\pi$ electrons: $4N + 2$ [aromatic](#def-b2-huckel-aromatic) , $4N$ [antiaromatic](#def-b2-huckel-aromatic) if planar.
4. An [antiaromatic](#def-b2-huckel-aromatic) ring avoids planarity or conjugation when it can (cyclo-octatetraene is tub-shaped and behaves as a polyene).

**Example 16.14 (Benzene).**

Six $\pi$ electrons fill $\alpha + 2\beta$ and the pair $\alpha + \beta$: $E_\pi = 6\alpha + 8\beta$. Three isolated double bonds give $6\alpha + 6\beta$: the [delocalisation energy](#def-b2-huckel-delocalisation) is $2\beta$, about twice that of the open chain of six carbons, hexatriene ($0.988\beta$): closing the ring is worth as much again. By symmetry all bond indices are equal ($p = 2/3$), and so are the six bond lengths, $139.7\,\mathrm{pm}$, between those of ethene ($133.9\,\mathrm{pm}$) and ethane ($153.6\,\mathrm{pm}$).

![Enthalpies of hydrogenation to cyclohexane in the gas phase, from tabulated enthalpies of formation (bars, height = heat released). Two conjugated double bonds release 10\, kJ/ mol less than twice one; benzene releases 150\, kJ/ mol less than three times one: its aromatic stabilisation.](https://one-course.com/images/onecourse/chapters/chemistry-3/b2-huckel/fig-07baee509514.svg)

*Enthalpies of hydrogenation to cyclohexane in the gas phase, from tabulated enthalpies of formation (bars, height = heat released). Two conjugated double bonds release $10\,\mathrm{kJ}/\mathrm{mol}$ less than twice one; benzene releases $150\,\mathrm{kJ}/\mathrm{mol}$ less than three times one: its [aromatic](#def-b2-huckel-aromatic) stabilisation.*

**History — Kekulé’s ring.**

![](https://one-course.com/images/onecourse/chapters/chemistry-3/b2-huckel/img-97ea6f34d5a4.jpg)

In 1865 August Kekulé proposed that the six carbon atoms of benzene form a ring with alternating single and double bonds, and soon after that the two alternations exchange so fast that every bond is the same. The equal bonds found by X-ray diffraction, and the $\pi$ orbitals spread over the whole ring, gave his intuition its modern form; [Hückel’s rule](#thm-b2-huckel-huckel-rule), in 1931, explained why six electrons, and not four or eight, make a ring special. (Portrait, 1873, unknown author, public domain, Wikimedia Commons.)

## 16.4 Conjugation and light

**Proposition 16.15 (The gap of a polyene).**

In a linear polyene of $n$ carbons ($n$ even), the gap between the highest occupied and the lowest empty $\pi$ level is $4|\beta|\sin(\pi/(2(n + 1)))$: it decreases as the chain lengthens.

**Proof.** The highest occupied level is $k = n/2$, the lowest empty $k = n/2 + 1$. With $\theta_k =
k\pi/(n + 1)$, $E_{n/2+1} - E_{n/2} = 2|\beta|(\cos\theta_{n/2} - \cos\theta_{n/2+1}) = 4|\beta|
\sin\frac{\theta_{n/2} + \theta_{n/2+1}}{2}\sin\frac{\theta_{n/2+1} - \theta_{n/2}}{2}$, and the first sine is $\sin(\pi/2) = 1$. It decreases with $n$ since $\sin$ increases on $[0, \pi/2]$. ∎

![Left: the Hückel gap of linear polyenes falls as the chain grows, the reason why long conjugated chains absorb at longer wavelengths, up to the visible. Right: computed levels of ethene, allyl, butadiene, hexatriene (chains) and of cyclobutadiene and benzene (rings); bonding levels below .](https://one-course.com/images/onecourse/chapters/chemistry-3/b2-huckel/fig-a7973ed41704.svg)

![Left: the Hückel gap of linear polyenes falls as the chain grows, the reason why long conjugated chains absorb at longer wavelengths, up to the visible. Right: computed levels of ethene, allyl, butadiene, hexatriene (chains) and of cyclobutadiene and benzene (rings); bonding levels below .](https://one-course.com/images/onecourse/chapters/chemistry-3/b2-huckel/fig-9b9128c2591c.svg)

*Left: the Hückel gap of linear polyenes falls as the chain grows, the reason why long conjugated chains absorb at longer wavelengths, up to the visible. Right: computed levels of ethene, allyl, butadiene, hexatriene (chains) and of cyclobutadiene and benzene (rings); bonding levels below $\alpha$.*

A heteroatom is brought in by changing its [Coulomb integral](https://one-course.com/books/chemistry/3/en/chapter/14-molecular-orbitals-of-diatomic-molecules#def-b2-diatomic-mos-integrals), $\alpha_X = \alpha + h\beta$, and the [resonance integrals](https://one-course.com/books/chemistry/3/en/chapter/14-molecular-orbitals-of-diatomic-molecules#def-b2-diatomic-mos-integrals) of its bonds: a model with adjustable parameters, which keeps the qualitative results (an electronegative atom, $h > 0$, draws the $\pi$ density towards itself) but whose numbers are fitted, not derived.

## 16.5 Exercises

**Exercise 16.1 ★.**

Give $E_\pi$ for the allyl cation, radical and anion.

**Solution of Exercise 16.1.**

Allyl levels ($n = 3$): $\alpha + 1.414\beta$, $\alpha$, $\alpha - 1.414\beta$. Cation (2 electrons): $2\alpha + 2.828\beta$; radical (3): $3\alpha + 2.828\beta$; anion (4): $4\alpha + 2.828\beta$. The extra electrons go into the nonbonding level and change only the $\alpha$ part.

**Exercise 16.2 ★.**

Compute $E_\pi$ of butadiene from the levels $\alpha \pm 1.618\beta$, $\alpha \pm 0.618\beta$.

**Solution of Exercise 16.2.**

$2(\alpha + 1.618\beta) + 2(\alpha + 0.618\beta) = 4\alpha + 4.472\beta$.

**Exercise 16.3 ★.**

[Aromatic](#def-b2-huckel-aromatic), [antiaromatic](#def-b2-huckel-aromatic) or neither (if planar): benzene, cyclopentadienyl anion, cyclopentadienyl cation, tropylium cation $\ce{C7H7+}$, cyclobutadiene, planar cyclo-octatetraene.

**Solution of Exercise 16.3.**

Benzene (6), cyclopentadienyl anion (6), tropylium (6): [aromatic](#def-b2-huckel-aromatic). Cyclopentadienyl cation (4), cyclobutadiene (4), planar cyclo-octatetraene (8): [antiaromatic](#def-b2-huckel-aromatic) if planar (real cyclo-octatetraene is tub-shaped and [non-aromatic](#def-b2-huckel-aromatic)).

**Exercise 16.4 ★.**

Draw Frost’s circle for planar cyclo-octatetraene and place its eight $\pi$ electrons.

**Solution of Exercise 16.4.**

An octagon in the circle: levels $\alpha + 2\beta$, $\alpha + 1.414\beta$ (twice), $\alpha$ (twice), $\alpha - 1.414\beta$ (twice), $\alpha - 2\beta$. Eight electrons: two in the lowest, four in the first pair, one in each orbital of the nonbonding pair.

**Exercise 16.5 ★★.**

Compute the [delocalisation energy](#def-b2-huckel-delocalisation) of butadiene and give it in $\mathrm{kJ}/\mathrm{mol}$ with $|\beta| = 75\,\mathrm{kJ}/\mathrm{mol}$. Compare with the $10\,\mathrm{kJ}/\mathrm{mol}$ of the hydrogenation data for cyclohexa-1,3-diene.

**Solution of Exercise 16.5.**

$4.472\beta - 4\beta = 0.472\beta$, that is $0.472 \times 75 = 35\,\mathrm{kJ}/\mathrm{mol}$, against about $10\,\mathrm{kJ}/\mathrm{mol}$ from the hydrogenation data: Hückel’s model, with $\beta$ fitted to benzene, overestimates the conjugation of open chains.

**Exercise 16.6 ★★.**

Compute the $\pi$ bond indices of butadiene and predict which of its C–C bonds is the shortest.

**Solution of Exercise 16.6.**

$p_{12} = p_{34} = 0.894$, $p_{23} = 0.447$: the end bonds C1–C2 and C3–C4 are the shortest, the central bond longer but shorter than a single bond.

**Exercise 16.7 ★★.**

The levels of a five-membered ring are $\alpha + 2\beta$, $\alpha + 0.618\beta$ (twice), $\alpha - 1.618\beta$ (twice). Fill them for the cyclopentadienyl anion and cation and compare.

**Solution of Exercise 16.7.**

Anion, six electrons: $2 \times 2 + 4 \times 0.618$, $E_\pi = 6\alpha + 6.472\beta$, a closed shell: [aromatic](#def-b2-huckel-aromatic). Cation, four: $2 \times 2 + 2 \times 0.618$, $4\alpha + 5.236\beta$, with one electron in each orbital of the degenerate pair: [antiaromatic](#def-b2-huckel-aromatic), very unstable.

**Exercise 16.8 ★★.**

Using the levels of a seven-membered ring, explain why cycloheptatrienyl bromide behaves as a salt, tropylium bromide.

**Solution of Exercise 16.8.**

Seven-membered ring: $\alpha + 2\beta$, $\alpha + 1.247\beta$ (twice), then antibonding levels. Six electrons, as in the cation $\ce{C7H7+}$, fill the bonding levels exactly: an [aromatic](#def-b2-huckel-aromatic) cation. The compound ionises readily into tropylium and bromide.

**Exercise 16.9 ★★.**

Check the trace rule on hexatriene, whose levels are $\alpha \pm 1.802\beta$, $\alpha \pm
1.247\beta$, $\alpha \pm 0.445\beta$.

**Solution of Exercise 16.9.**

$1.802 + 1.247 + 0.445 - 0.445 - 1.247 - 1.802 = 0$: the sum of the energies is $6\alpha$.

**Exercise 16.10 ★★★.**

Derive the levels of hexatriene from the closed form, and its [delocalisation energy](#def-b2-huckel-delocalisation).

**Solution of Exercise 16.10.**

$n = 6$: $m_k = 2\cos(k\pi/7) = 1.802$, $1.247$, $0.445$, $-0.445$, $-1.247$, $-1.802$. Six electrons: $E_\pi = 6\alpha + 2(1.802 + 1.247 + 0.445)\beta = 6\alpha + 6.988\beta$; delocalisation $0.988\beta$.

**Exercise 16.11 ★★★.**

Square cyclobutadiene would have two unpaired electrons. Explain why the real molecule is rectangular, with two short and two long bonds.

**Solution of Exercise 16.11.**

In the square, the two electrons of highest energy sit one in each orbital of a degenerate nonbonding pair. Stretching two opposite bonds lifts the degeneracy: one orbital goes down, the other up, and both electrons pair in the lower one, which lowers the energy. The molecule settles as a rectangle with two localised double bonds; it remains [antiaromatic](#def-b2-huckel-aromatic) and extremely reactive.

**Exercise 16.12 ★★★.**

In a two-atom $\pi$ system $\ce{C=X}$ with $\alpha_X = \alpha + \beta$ (model value, $h = 1$) and $\beta_{\ce{CX}} = \beta$, find the two levels and the $\pi$ charges on C and X for two electrons.

**Solution of Exercise 16.12.**

Matrix (in units of $\beta$, with $\alpha$ as origin) $\begin{pmatrix}0 & 1\\ 1 & 1\end{pmatrix}$: $m = 1.618$ and $-0.618$, $E = \alpha + 1.618\beta$ (bonding) and $\alpha - 0.618\beta$. For the [bonding orbital](https://one-course.com/books/chemistry/3/en/chapter/14-molecular-orbitals-of-diatomic-molecules#def-b2-diatomic-mos-bonding) $c_X = 1.618c_C$, so $c_C^2 = 0.276$, $c_X^2 = 0.724$; with two electrons, $q_C = 0.55$, $q_X = 1.45$: the $\pi$ density is drawn to X.

## 16.6 Problem: Benzene’s Missing Energy

**Problem 16.1.**

Weekend problem — the aromatic stabilisation of benzene from enthalpies of hydrogenation, the Hückel levels of benzene and its delocalisation energy, the value of $|\beta|$, and the comparison with hexatriene and cyclo-octatetraene

Data: enthalpies of formation of the gases ($\mathrm{kJ}/\mathrm{mol}$): benzene 82.9, cyclohexa-1,3-diene 104.6, cyclohexene $-4.3$, cyclohexane $-123.1$. Bond lengths (pm): benzene 139.7, ethene 133.9, ethane 153.6.

**Part I — Hydrogenation.**

1. Compute $\Delta_r H^\circ$ of the hydrogenation of cyclohexene to cyclohexane.
2. Same for cyclohexa-1,3-diene.
3. Same for benzene.
4. What would three independent double bonds release?
5. Deduce the stabilisation of benzene.
6. Compare with that of the conjugated diene, and comment.

**Part II — Hückel benzene.**

7. Write the Hückel matrix of benzene.
8. Give its levels from the ring formula.
9. Fill them with six $\pi$ electrons.
10. Compute $E_\pi$ .
11. Compute $E_\pi$ of three isolated double bonds.
12. Deduce the [delocalisation energy](#def-b2-huckel-delocalisation) .
13. Check the trace rule.

**Part III — The value of $|\beta|$.**

14. Equate the [delocalisation energy](#def-b2-huckel-delocalisation) with the stabilisation of Part I and find $|\beta|$ .
15. With that value, predict the [delocalisation energy](#def-b2-huckel-delocalisation) of butadiene, and compare with Part I.
16. Compute the $\pi$ bond indices of benzene.
17. Compare them with those of butadiene.
18. Explain why the bond lengths of benzene are equal and where they fall.

**Part IV — Other rings and chains.**

19. Compute the [delocalisation energy](#def-b2-huckel-delocalisation) of hexatriene and compare with benzene.
20. Give the levels of planar cyclo-octatetraene.
21. Fill them with eight electrons: what is wrong?
22. How does the real molecule escape?
23. What is the situation of cyclobutadiene?
24. State [Hückel’s rule](#thm-b2-huckel-huckel-rule) .
25. State the value of $|\beta|$ obtained by matching benzene’s [delocalisation energy](#def-b2-huckel-delocalisation) to its measured stabilisation.

**Solution of Problem 16.1.**

**1.** $-123.1 - (-4.3) = -118.8\,\mathrm{kJ}/\mathrm{mol}$. **2.** $-123.1 - 104.6 = -227.7\,\mathrm{kJ}/\mathrm{mol}$. **3.** $-123.1 - 82.9 = -206.0\,\mathrm{kJ}/\mathrm{mol}$. **4.** $3 \times (-118.8) = -356.4\,\mathrm{kJ}/\mathrm{mol}$. **5.** $356.4 - 206.0 = 150\,\mathrm{kJ}/\mathrm{mol}$. **6.** The diene releases $2 \times 118.8 - 227.7 = 10\,\mathrm{kJ}/\mathrm{mol}$ less than two isolated double bonds: conjugation of an open chain gives a small gain, the closed ring of benzene fifteen times more. **7.** A $6 \times 6$ matrix with 1 between each atom and its two neighbours (1–2, 2–3, …, 6–1) and 0 elsewhere. **8.** $m = 2\cos(2\pi k/6)$: $2$, $1$, $1$, $-1$, $-1$, $-2$: $E = \alpha + 2\beta$, $\alpha + \beta$ (twice), $\alpha - \beta$ (twice), $\alpha - 2\beta$. **9.** Two electrons in $\alpha + 2\beta$, four in the pair $\alpha + \beta$. **10.** $E_\pi = 6\alpha + 8\beta$. **11.** $3(2\alpha + 2\beta) = 6\alpha + 6\beta$. **12.** $2\beta$, that is $2|\beta|$ of stabilisation. **13.** $2 + 1 + 1 - 1 - 1 - 2 = 0$: the six levels add up to $6\alpha$. **14.** $2|\beta| = 150\,\mathrm{kJ}/\mathrm{mol}$: $|\beta| = 75\,\mathrm{kJ}/\mathrm{mol}$. **15.** $0.472 \times 75 = 35\,\mathrm{kJ}/\mathrm{mol}$, against $10\,\mathrm{kJ}/\mathrm{mol}$: the model fitted on benzene overestimates the open chain. **16.** The occupied orbitals can be taken as $\psi_1 = (1, 1, 1, 1, 1, 1)/\sqrt6$, $\psi_2 = (2, 1, -1, -2, -1, 1)/\sqrt{12}$ and $\psi_3 = (0, 1, 1, 0, -1, -1)/2$. For the bond 1–2: $p_{12} = 2(\frac16 + \frac{2}{12} + 0) = \frac23$; by symmetry every bond has $p = 2/3$. **17.** Between the end (0.894) and central (0.447) bonds of butadiene. **18.** The six bonds are equivalent by symmetry; their length, $139.7\,\mathrm{pm}$, lies between the double bond of ethene ($133.9\,\mathrm{pm}$) and the single bond of ethane ($153.6\,\mathrm{pm}$). **19.** $0.988|\beta| = 74\,\mathrm{kJ}/\mathrm{mol}$, half of benzene’s: closing the ring doubles the [delocalisation energy](#def-b2-huckel-delocalisation). **20.** $\alpha + 2\beta$, $\alpha + 1.414\beta$ (twice), $\alpha$ (twice), $\alpha - 1.414\beta$ (twice), $\alpha - 2\beta$. **21.** The last two electrons go one in each orbital of the nonbonding pair: an open shell, [antiaromatic](#def-b2-huckel-aromatic). **22.** It folds into a tub: the p orbitals of neighbouring double bonds are no longer parallel, the bonds alternate, and the molecule behaves as a polyene. **23.** Four electrons: the same half-filled pair, worse because the molecule cannot avoid planarity; it distorts into a rectangle and is extremely reactive. **24.** A planar monocyclic conjugated system is [aromatic](#def-b2-huckel-aromatic) with $4N + 2$ $\pi$ electrons, [antiaromatic](#def-b2-huckel-aromatic) with $4N$. **25.** $\boldsymbol{|\beta| \approx 75\,\mathrm{kJ}/\mathrm{mol}}$.