Arenes (Aromatic Compounds)

Arenes are hydrocarbons containing one or more benzene rings. Benzene ($\mathrm{C}_6\mathrm{H}_6$) is the archetypal aromatic compound. Its structure and reactivity differ fundamentally from those of alkenes, despite both containing $\pi$ electrons. The key distinction is that benzene's $\pi$ electrons are delocalised over the entire ring, forming an aromatic system with exceptional thermodynamic stability.

Benzene Structure

Molecular Formula and Degree of Unsaturation

Benzene has the molecular formula $\mathrm{C}_6\mathrm{H}_6$, corresponding to four degrees of unsaturation. This immediately rules out an acyclic structure and indicates significant multiple bonding or ring formation.

The Kekule Structure

August Kekule proposed a cyclic structure with alternating single and double bonds in 1865. While this was a productive historical hypothesis, it fails to explain several key observations and is now known to be incorrect.

Evidence for Delocalisation

1. Bond lengths are identical. X-ray diffraction shows that all six C--C bonds in benzene have the same length: $139\,\mathrm{pm}$. A Kekule structure with alternating single ($154\,\mathrm{pm}$) and double ($134\,\mathrm{pm}$) bonds would show two distinct bond lengths. The experimental value is intermediate between single and double, consistent with bond order of 1.5.

2. No isomers of 1,2-disubstituted benzene. The Kekule structure predicts two distinct isomers of 1,2-dibromobenzene (bromines on a "single" bond vs on a "double" bond). Only one compound is observed.

3. Thermochemical evidence. The experimental enthalpy of hydrogenation of benzene ($-208\,\mathrm{kJ/mol}$) is significantly less exothermic than the theoretical value for three isolated double bonds ($3 \times -120 = -360\,\mathrm{kJ/mol}$). The difference of approximately $152\,\mathrm{kJ/mol}$ is the delocalisation energy (also called resonance energy or aromatic stabilisation energy).

4. Resistance to addition reactions. Unlike alkenes, benzene does not readily undergo addition reactions that would destroy the aromatic $\pi$ system. It favours substitution, which preserves the delocalised ring.

The Delocalised Model

In the modern model:

• Each carbon is $sp^2$ hybridised, forming three $\sigma$ bonds (two C--C and one C--H) at $120^\circ$ angles in a planar arrangement.
• Each carbon has one remaining electron in an unhybridised $p_z$ orbital, perpendicular to the ring plane.
• The six $p_z$ orbitals overlap laterally, forming a delocalised $\pi$ molecular orbital that extends over all six carbons, above and below the ring.
• The electron density is distributed uniformly (shown experimentally by X-ray and electron diffraction).
• The ring current induced by an external magnetic field explains the diamagnetic ring current (detectable by NMR spectroscopy: aromatic protons are deshielded and appear at $\delta \approx 7.2\,\mathrm{ppm}$).

Huckel's Rule

A planar, cyclic, fully conjugated system with $4n + 2$ $\pi$ electrons (where $n$ is a non-negative integer) is aromatic. Benzene has $6\,\pi$ electrons ($n = 1$), satisfying the rule. Cyclobutadiene ($4\,\pi$ electrons) and cyclooctatetraene ($8\,\pi$ electrons) are not aromatic because they have $4n\,\pi$ electrons (anti-aromatic or non-aromatic).

Electrophilic Aromatic Substitution

The defining reaction of arenes is electrophilic substitution, not addition. The mechanism involves two steps:

1. Attack: The electrophile attacks the delocalised $\pi$ system, forming a cyclohexadienyl cation intermediate (also called a sigma complex or arenium ion). This step is endothermic because it destroys the aromatic stabilisation.
2. Deprotonation: A base removes a proton from the arenium ion, restoring aromaticity. This step is exothermic.

The overall reaction is thermodynamically favourable because the aromatic system is regenerated. The rate-determining step is the initial electrophilic attack.

Nitration

Reagents: Concentrated nitric acid and concentrated sulphuric acid (the latter acts as a catalyst and dehydrating agent).

Conditions: $50$--$60^\circ\mathrm{C}$ (gentle heating). Higher temperatures produce di- and tri-nitration.

Generation of the electrophile (the nitronium ion, $\mathrm{NO}_2^+$):

$$\mathrm{HNO}_3 + \mathrm{H}_2\mathrm{SO}_4 \to \mathrm{NO}_2^+ + \mathrm{HSO}_4^- + \mathrm{H}_2\mathrm{O}$$

The sulphuric acid protonates nitric acid, which then loses water to form the nitronium ion. This is a strong electrophile because the positive charge is delocalised over three atoms ($\mathrm{O}=\mathrm{N}^+=\mathrm{O}$).

Overall reaction:

$$\mathrm{C}_6\mathrm{H}_6 + \mathrm{HNO}_3 \xrightarrow{\mathrm{H}_2\mathrm{SO}_4} \mathrm{C}_6\mathrm{H}_5\mathrm{NO}_2 + \mathrm{H}_2\mathrm{O}$$

Friedel-Crafts Acylation

Reagents: Acyl chloride ($\mathrm{RCOCl}$) and aluminium chloride ($\mathrm{AlCl}_3$) catalyst (anhydrous).

Electrophile: The acylium ion ($\mathrm{RCO}^+$), generated by:

$$\mathrm{RCOCl} + \mathrm{AlCl}_3 \to \mathrm{RCO}^+ + \mathrm{AlCl}_4^-$$

Overall reaction:

$$\mathrm{C}_6\mathrm{H}_6 + \mathrm{RCOCl} \xrightarrow{\mathrm{AlCl}_3} \mathrm{C}_6\mathrm{H}_5\mathrm{COR} + \mathrm{HCl}$$

The product is an aryl ketone. Friedel-Crafts acylation is preferred over Friedel-Crafts alkylation because:

• The acylium ion is a weaker electrophile, reducing over-alkylation.
• The ketone product is electron-withdrawing and deactivates the ring, preventing further substitution.
• No carbocation rearrangement occurs (the acylium ion is resonance-stabilised: $\mathrm{R}-\mathrm{C}\equiv\mathrm{O}^+ \leftrightarrow \mathrm{R}-\mathrm{C}^+=\mathrm{O}$).

Limitations: Does not work on deactivated rings (e.g. nitrobenzene, phenol derivatives with electron-withdrawing groups). The $\mathrm{AlCl}_3$ catalyst is destroyed by water and must be used under anhydrous conditions.

Halogenation

Reagents: Halogen ($\mathrm{Cl}_2$ or $\mathrm{Br}_2$) with a Lewis acid catalyst ($\mathrm{AlCl}_3$ or $\mathrm{FeCl}_3$ / $\mathrm{FeBr}_3$).

Electrophile: The halonium ion ($\mathrm{Cl}^+$ or $\mathrm{Br}^+$), generated by:

$$\mathrm{Cl}_2 + \mathrm{AlCl}_3 \to \mathrm{Cl}^+ + \mathrm{AlCl}_4^-$$

Overall reaction:

$$\mathrm{C}_6\mathrm{H}_6 + \mathrm{Cl}_2 \xrightarrow{\mathrm{AlCl}_3} \mathrm{C}_6\mathrm{H}_5\mathrm{Cl} + \mathrm{HCl}$$

Note: Benzene does not react with bromine water or chlorine water (no Lewis acid catalyst). This is a critical distinction from alkenes, which decolourise bromine water rapidly.

Phenol

Phenol ($\mathrm{C}_6\mathrm{H}_5\mathrm{OH}$) is benzene with a hydroxyl group directly attached to the ring. The $-\mathrm{OH}$ group donates electron density into the ring through resonance, making the ring much more reactive toward electrophilic substitution than benzene itself.

Increased Reactivity

The oxygen lone pair is delocalised into the ring, increasing electron density at the ortho and para positions. This has two consequences:

1. Faster reactions: Phenol reacts with bromine water without a catalyst (whereas benzene requires $\mathrm{FeBr}_3$).
2. Ortho/para directing: Substitution occurs preferentially at the 2- (ortho) and 4- (para) positions relative to the $-\mathrm{OH}$ group.

Bromination of Phenol

Phenol decolourises bromine water immediately at room temperature, producing 2,4,6-tribromophenol as a white precipitate:

$$\mathrm{C}_6\mathrm{H}_5\mathrm{OH} + 3\mathrm{Br}_2 \to \mathrm{C}_6\mathrm{H}_2\mathrm{Br}_3\mathrm{OH} + 3\mathrm{HBr}$$

The tribromo derivative precipitates because it is much less soluble in water than phenol.

Acidity of Phenol

Phenol is a weak acid ($\mathrm{p}K_a \approx 10$) compared with carboxylic acids ($\mathrm{p}K_a \approx 4\mathrm{--}5$) but is more acidic than alcohols ($\mathrm{p}K_a \approx 16$). The phenoxide ion ($\mathrm{C}_6\mathrm{H}_5\mathrm{O}^-$) is stabilised by delocalisation of the negative charge over the aromatic ring:

$$\mathrm{C}_6\mathrm{H}_5\mathrm{OH} \rightleftharpoons \mathrm{C}_6\mathrm{H}_5\mathrm{O}^- + \mathrm{H}^+$$

Phenol reacts with sodium hydroxide (but not sodium carbonate or sodium hydrogencarbonate) to form sodium phenoxide:

$$\mathrm{C}_6\mathrm{H}_5\mathrm{OH} + \mathrm{NaOH} \to \mathrm{C}_6\mathrm{H}_5\mathrm{ONa} + \mathrm{H}_2\mathrm{O}$$

Why Phenol Does Not React with $\mathrm{NaHCO}_3$

Phenol ($\mathrm{p}K_a \approx 10$) is a weaker acid than the $\mathrm{HCO}_3^-$ ion ($\mathrm{p}K_a \approx 6.4$). The equilibrium:

$$\mathrm{C}_6\mathrm{H}_5\mathrm{OH} + \mathrm{HCO}_3^- \rightleftharpoons \mathrm{C}_6\mathrm{H}_5\mathrm{O}^- + \mathrm{H}_2\mathrm{CO}_3$$

lies to the left because $\mathrm{H}_2\mathrm{CO}_3$ is a stronger acid than phenol. In contrast, carboxylic acids ($\mathrm{p}K_a \approx 4$) are stronger than $\mathrm{H}_2\mathrm{CO}_3$ and do react with $\mathrm{NaHCO}_3$, producing $\mathrm{CO}_2$. This distinction is a useful chemical test.

Directing Effects of Substituents

When a monosubstituted benzene undergoes further electrophilic substitution, the existing substituent determines both the rate and the position of the second substitution.

Activating Groups (Ortho/Para Directing)

These groups donate electron density into the ring, activating it toward electrophilic substitution and directing the electrophile to the ortho (2-, 6-) and para (4-) positions.

Group	Mechanism	Relative activating power
$-\mathrm{NH}_2$, $-\mathrm{NHR}$	Resonance donation (strong)	Very strong
$-\mathrm{OH}$	Resonance donation (strong)	Strong
$-\mathrm{OR}$	Resonance donation	Strong
$-\mathrm{R}$ (alkyl)	Inductive donation (weak)	Weak
$-\mathrm{C}_6\mathrm{H}_5$	Resonance donation (weak)	Weak

Why ortho/para? The resonance structures of the arenium ion intermediate show that ortho and para attack place the positive charge on the carbon bearing the electron-donating group, which can stabilise the arenium ion through resonance. Meta attack does not allow this stabilisation.

Deactivating Groups (Meta Directing)

These groups withdraw electron density from the ring, deactivating it toward electrophilic substitution and directing the electrophile to the meta (3-, 5-) position.

Group	Mechanism	Relative deactivating power
$-\mathrm{NO}_2$	Resonance withdrawal (strong)	Very strong
$-\mathrm{CN}$	Resonance withdrawal	Strong
$-\mathrm{COOH}$, $-\mathrm{COR}$	Resonance withdrawal	Strong
$-\mathrm{SO}_3\mathrm{H}$	Resonance withdrawal	Strong
$-\mathrm{CF}_3$	Inductive withdrawal	Moderate
Halogens ($-\mathrm{Cl}$, $-\mathrm{Br}$)	Inductive withdrawal, resonance donation	Weak deactivation

Why meta? The resonance structures of the arenium ion for ortho or para attack place the positive charge on the carbon bearing the electron-withdrawing group, which further destabilises the already electron-deficient intermediate. Meta attack avoids placing the positive charge adjacent to the electron-withdrawing group.

Halogens: A Special Case

Halogens are deactivating (they withdraw electrons inductively because they are electronegative) but ortho/para directing (they donate electrons by resonance through their lone pairs). The resonance donation is weaker than the inductive withdrawal (net: deactivating), but the resonance effect controls the orientation (ortho/para directing).

Worked Example: Predict the Products

Nitration of methylbenzene (toluene): The $-\mathrm{CH}_3$ group is activating and ortho/para directing. Nitration produces a mixture of 2-nitrotoluene (ortho, $\approx 60\%$) and 4-nitrotoluene (para, $\approx 40\%$), with very little 3-nitrotoluene (meta, trace).

Nitration of nitrobenzene: The $-\mathrm{NO}_2$ group is deactivating and meta directing. Nitration requires more vigorous conditions (concentrated $\mathrm{HNO}_3 / \mathrm{H}_2\mathrm{SO}_4$ at higher temperature) and produces predominantly 1,3-dinitrobenzene.

Summary Table

Group type	Effect on rate	Directing
$-\mathrm{NH}_2$, $-\mathrm{OH}$, $-\mathrm{OR}$	Activates strongly	Ortho/para
Alkyl ($-\mathrm{R}$)	Activates weakly	Ortho/para
Halogens	Deactivates weakly	Ortho/para
$-\mathrm{NO}_2$, $-\mathrm{CN}$, $-\mathrm{COR}$, $-\mathrm{COOH}$	Deactivates strongly	Meta

Common Pitfalls

1. Drawing the Kekule structure. The Kekule structure (alternating single/double bonds) is incorrect and should not be used. Draw a hexagon with a circle inside to represent the delocalised $\pi$ system.

2. Confusing addition and substitution for benzene. Benzene undergoes electrophilic substitution (preserving the aromatic ring), not addition (which would destroy the delocalisation energy). Alkenes undergo electrophilic addition.

3. Assuming benzene reacts with bromine water. Benzene does not react with bromine water. It requires $\mathrm{Br}_2$ with a Lewis acid catalyst ($\mathrm{FeBr}_3$). Phenol does react with bromine water.

4. Forgetting the catalyst in Friedel-Crafts reactions. $\mathrm{AlCl}_3$ (or equivalent Lewis acid) is essential for generating the electrophile. Without it, the reaction does not proceed.

5. Misunderstanding the role of $\mathrm{H}_2\mathrm{SO}_4$ in nitration. Concentrated $\mathrm{H}_2\mathrm{SO}_4$ generates the nitronium ion and is not simply a "catalyst" in the usual sense -- it is consumed stoichiometrically in generating the electrophile (one mole per mole of $\mathrm{NO}_2^+$), though it is regenerated when the arenium ion is deprotonated.

Friedel-Crafts Reactions in Detail

Friedel-Crafts Alkylation

An alkyl group is introduced onto the benzene ring using an alkyl halide and a Lewis acid catalyst ($\mathrm{AlCl}_3$):

$$\mathrm{C}_6\mathrm{H}_6 + \mathrm{CH}_3\mathrm{Cl} \xrightarrow{\mathrm{AlCl}_3} \mathrm{C}_6\mathrm{H}_5\mathrm{CH}_3 + \mathrm{HCl}$$

Mechanism:

1. $\mathrm{AlCl}_3$ coordinates to the chlorine of $\mathrm{CH}_3\mathrm{Cl}$, generating a carbocation-like electrophile ($\mathrm{CH}_3^+$) or a polarised complex.
2. The electrophile attacks the benzene ring, forming a sigma complex (arenium ion).
3. The $\mathrm{H}^+$ is removed by $\mathrm{AlCl}_4^-$, restoring aromaticity.

Limitations:

• The product is more reactive than the starting material (alkyl groups are activating), leading to polyalkylation. This makes Friedel-Crafts alkylation difficult to control for monosubstitution.
• Only primary alkyl halides can be used directly. Secondary and tertiary halides may undergo rearrangement via carbocation rearrangements (hydride or alkyl shifts) before attacking the ring.
• Strongly deactivated rings do not react.

Friedel-Crafts Acylation

An acyl group is introduced using an acyl chloride and $\mathrm{AlCl}_3$:

$$\mathrm{C}_6\mathrm{H}_6 + \mathrm{CH}_3\mathrm{COCl} \xrightarrow{\mathrm{AlCl}_3} \mathrm{C}_6\mathrm{H}_5\mathrm{COCH}_3 + \mathrm{HCl}$$

Advantages over alkylation:

• The acyl group is electron-withdrawing, so the product (aryl ketone) is less reactive than the starting material. Polyacylation does not occur.
• No carbocation rearrangements occur because the acylium ion ($\mathrm{RCO}^+$) is resonance-stabilised.
• The ketone product can be reduced to an alkyl group (Clemmensen or Wolff-Kishner reduction), providing an indirect route to unbranched alkylbenzenes.

Electrophilic Substitution Mechanisms in Detail

Nitration Mechanism

1. Generation of the nitronium ion:

$$\mathrm{HNO}_3 + 2\mathrm{H}_2\mathrm{SO}_4 \to \mathrm{NO}_2^+ + \mathrm{H}_3\mathrm{O}^+ + 2\mathrm{HSO}_4^-$$

2. Electrophilic attack: the $\mathrm{NO}_2^+$ ion attacks the benzene ring, forming a sigma complex.

3. Deprotonation: $\mathrm{HSO}_4^-$ removes $\mathrm{H}^+$ from the sigma complex, restoring aromaticity and forming nitrobenzene.

Sulphonation Mechanism

Sulphonation is reversible:

$$\mathrm{C}_6\mathrm{H}_6 + \mathrm{H}_2\mathrm{SO}_4 \rightleftharpoons \mathrm{C}_6\mathrm{H}_5\mathrm{SO}_3\mathrm{H} + \mathrm{H}_2\mathrm{O}$$

The electrophile is $\mathrm{SO}_3$ (generated from $\mathrm{H}_2\mathrm{SO}_4$), not $\mathrm{SO}_3\mathrm{H}^+$. Sulphonic acid groups are used in detergents (they make the organic compound water-soluble as the sulphonate salt) and in directing strategies (the $-\mathrm{SO}_3\mathrm{H}$ group is bulky and can block positions).

Disubstituted Benzenes: Positional Effects

When a benzene ring already has one substituent and a second electrophilic substitution is performed, the existing substituent controls the position of the new group:

Ortho/Para Directors (Activating)

These groups donate electron density to the ring, activating it and directing substitution to the ortho (positions 2 and 6) and para (position 4) positions:

• $-\mathrm{OH}$, $-\mathrm{OR}$ (strong activators)
• $-\mathrm{NH}_2$, $-\mathrm{NHR}$, $-\mathrm{NR}_2$ (strong activators)
• $-\mathrm{CH}_3$, $-\mathrm{R}$ (moderate activators)
• Halogens: $-\mathrm{F}$, $-\mathrm{Cl}$, $-\mathrm{Br}$, $-\mathrm{I}$ (deactivators but ortho/para directors)

Meta Directors (Deactivating)

These groups withdraw electron density from the ring, deactivating it and directing substitution to the meta position (position 3 and 5):

• $-\mathrm{NO}_2$ (very strong deactivator)
• $-\mathrm{CF}_3$, $-\mathrm{CCl}_3$
• $-\mathrm{CN}$
• $-\mathrm{SO}_3\mathrm{H}$
• $-\mathrm{CHO}$, $-\mathrm{COR}$
• $-\mathrm{COOH}$, $-\mathrm{COOR}$

Worked Examples

Example 1: Nitration of methylbenzene (toluene).

Methyl is an ortho/para director and activator. Nitration gives a mixture of ortho-nitrotoluene and para-nitrotoluene, with the para product predominating for steric reasons. The reaction is faster than nitration of benzene.

Example 2: Nitration of nitrobenzene.

$-\mathrm{NO}_2$ is a meta director and strong deactivator. Nitration gives meta-dinitrobenzene. The reaction requires more vigorous conditions (fuming $\mathrm{HNO}_3$/$\mathrm{H}_2\mathrm{SO}_4$, higher temperature) and is slower than nitration of benzene.

Phenol: Special Reactivity

Phenol is so strongly activated that it reacts with bromine water (dilute aqueous $\mathrm{Br}_2$) at room temperature without a catalyst, producing 2,4,6-tribromophenol (white precipitate). This is in stark contrast to benzene, which requires $\mathrm{Br}_2$ with $\mathrm{FeBr}_3$ catalyst and gives only monobromobenzene.

$$\mathrm{C}_6\mathrm{H}_5\mathrm{OH} + 3\mathrm{Br}_2 \to \mathrm{C}_6\mathrm{H}_2\mathrm{Br}_3\mathrm{OH} + 3\mathrm{HBr}$$

Phenol vs Toluene: Acidity Distinction

Phenol ($\mathrm{p}K_a \approx 10$) is a weak acid and reacts with $\mathrm{NaOH}$ to form sodium phenoxide, but does not react with $\mathrm{NaHCO}_3$ ($\mathrm{p}K_a \approx 6.4$). Carboxylic acids ($\mathrm{p}K_a \approx 4$--$5$) do react with $\mathrm{NaHCO}_3$, producing $\mathrm{CO}_2$. This provides a clean experimental test to distinguish phenols from carboxylic acids.

Practice Problems

Problem 1

Calculate the delocalisation energy of benzene using the following hydrogenation data:

• $\Delta H$ for $\mathrm{C}_6\mathrm{H}_6(g) + 3\mathrm{H}_2(g) \to \mathrm{C}_6\mathrm{H}_{12}(g)$: $-208\,\mathrm{kJ/mol}$
• $\Delta H$ for $\mathrm{C}_6\mathrm{H}_{12}(g)$: cyclohexane
• Mean C=C bond enthalpy: $612\,\mathrm{kJ/mol}$ (note: this is for a double bond, consisting of one $\sigma$ and one $\pi$ component)
• Mean C--C bond enthalpy: $348\,\mathrm{kJ/mol}$
• Mean C--H bond enthalpy: $413\,\mathrm{kJ/mol}$
• H--H bond enthalpy: $436\,\mathrm{kJ/mol}$

Solution:

Benzene has 6 C--C bonds (effectively bond order 1.5), 6 C--H bonds, and we break 3 H--H bonds. We form cyclohexane with 6 C--C bonds (single) and 12 C--H bonds.

Expected $\Delta H$ if benzene had 3 isolated C=C bonds:

Bonds broken: 3 C=C ($3 \times 612 = 1836$), 3 C--C ($3 \times 348 = 1044$), 6 C--H ($6 \times 413 = 2478$), 3 H--H ($3 \times 436 = 1308$). Total broken = $6666\,\mathrm{kJ/mol}$.

Bonds formed: 6 C--C ($6 \times 348 = 2088$), 12 C--H ($12 \times 413 = 4956$). Total formed = $7044\,\mathrm{kJ/mol}$.

Expected $\Delta H = 6666 - 7044 = -378\,\mathrm{kJ/mol}$.

Delocalisation energy = expected $-$ actual = $-378 - (-208) = -170\,\mathrm{kJ/mol}$.

The benzene ring is stabilised by approximately $170\,\mathrm{kJ/mol}$ relative to the Kekule structure with three isolated double bonds.

Problem 2

Explain why Friedel-Crafts acylation of nitrobenzene fails, while Friedel-Crafts acylation of phenol proceeds readily.

Solution:

The $-\mathrm{NO}_2$ group is strongly electron-withdrawing through both the inductive effect (the nitrogen is electron-deficient) and resonance (the $\pi$ electrons of the ring are delocalised onto the oxygen atoms of the nitro group). This deactivates the ring toward electrophilic attack by reducing the electron density in the $\pi$ system. Additionally, the $-\mathrm{NO}_2$ group makes the ring less nucleophilic, so the electrophilic attack step (already the rate-determining step) becomes prohibitively slow.

The $-\mathrm{OH}$ group on phenol is electron-donating through resonance (the oxygen lone pair delocalises into the ring, increasing electron density at ortho and para positions). This activates the ring toward electrophilic substitution, making the reaction much faster than for unsubstituted benzene.

Problem 3

Propose a synthesis of 4-nitrobenzaldehyde from benzene, explaining the order of steps and the choice of reagents.

Solution:

The key challenge is that $-\mathrm{CHO}$ is meta-directing, so nitration after introducing the aldehyde would give 3-nitrobenzaldehyde. We need the aldehyde in the para position relative to the nitro group, so we must use a group that directs para and can later be converted to $-\mathrm{CHO}$.

Step 1: Friedel-Crafts alkylation with $\mathrm{CH}_3\mathrm{Cl}/\mathrm{AlCl}_3$ to give methylbenzene.

$$\mathrm{C}_6\mathrm{H}_6 \xrightarrow{\mathrm{CH}_3\mathrm{Cl},\,\mathrm{AlCl}_3} \mathrm{C}_6\mathrm{H}_5\mathrm{CH}_3$$

Step 2: Nitration (methyl is ortho/para directing; para product is major):

\mathrm{C}_6\mathrm{H}_5\mathrm{CH}_3 \xrightarrow{\mathrm{HNO}_3/\mathrm{H}_2\mathrm{SO}_4}} 4\text{-}\mathrm{C}_6\mathrm{H}_4(\mathrm{CH}_3)(\mathrm{NO}_2)

Step 3: Oxidation of the methyl group to carboxylic acid:

$$4\text{-}\mathrm{C}_6\mathrm{H}_4(\mathrm{CH}_3)(\mathrm{NO}_2) \xrightarrow{\mathrm{KMnO}_4,\,\Delta} 4\text{-}\mathrm{C}_6\mathrm{H}_4(\mathrm{COOH})(\mathrm{NO}_2)$$

Step 4: Conversion of carboxylic acid to aldehyde. This is tricky as direct reduction of an aromatic carboxylic acid typically gives the primary alcohol. A suitable route is to reduce the acid to alcohol ($\mathrm{LiAlH}_4$), then oxidise to the aldehyde (distillation with $\mathrm{PCC}$ or acidified $\mathrm{K}_2\mathrm{Cr}_2\mathrm{O}_7$):

$$4\text{-}\mathrm{C}_6\mathrm{H}_4(\mathrm{COOH})(\mathrm{NO}_2) \xrightarrow{\mathrm{LiAlH}_4} 4\text{-}\mathrm{C}_6\mathrm{H}_4(\mathrm{CH}_2\mathrm{OH})(\mathrm{NO}_2) \xrightarrow{[\mathrm{O}],\,\text{distillation}} 4\text{-}\mathrm{C}_6\mathrm{H}_4(\mathrm{CHO})(\mathrm{NO}_2)$$

The directing group strategy ensures the nitro group ends up para to the substituent that will become the aldehyde.

Worked Examples: Arenes in Depth

Example 1: Nitration Mechanism with Curly Arrows

Complete mechanism for the nitration of benzene:

Step 1: Generation of the electrophile ($\mathrm{NO}_2^+$).

The nitronium ion is formed by the reaction of concentrated nitric acid with concentrated sulphuric acid:

$\mathrm{HNO}_3 + \mathrm{H}_2\mathrm{SO}_4 \to \mathrm{NO}_2^+ + \mathrm{HSO}_4^- + \mathrm{H}_2\mathrm{O}$

Electron flow: The lone pair on the oxygen of $\mathrm{HNO}_3$ attacks the hydrogen of $\mathrm{H}_2\mathrm{SO}_4$, forming $\mathrm{H}_2\mathrm{NO}_3^+$. Water is eliminated, leaving $\mathrm{NO}_2^+$.

Step 2: Electrophilic attack on benzene.

The $\pi$ electrons of benzene attack the nitrogen of $\mathrm{NO}_2^+$, forming the sigma complex (arenium ion). The positive charge is delocalised over three carbon atoms (the carbon bearing the nitro group and the two ortho carbons).

Step 3: Deprotonation.

$\mathrm{HSO}_4^-$ removes the proton from the carbon bearing the $\mathrm{NO}_2$ group, restoring aromaticity. The sulphuric acid is regenerated.

Example 2: Friedel-Crafts Acylation Mechanism

Acylation of benzene with ethanoyl chloride:

$\mathrm{C}_6\mathrm{H}_6 + \mathrm{CH}_3\mathrm{COCl} \xrightarrow{\mathrm{AlCl}_3} \mathrm{C}_6\mathrm{H}_5\mathrm{COCH}_3 + \mathrm{HCl}$

Step 1: Formation of the electrophile.

$\mathrm{AlCl}_3$ (a Lewis acid) coordinates to the chlorine of ethanoyl chloride, polarising the C--Cl bond and forming the acylium ion:

$\mathrm{CH}_3\mathrm{COCl} + \mathrm{AlCl}_3 \to \mathrm{CH}_3\mathrm{CO}^+ + \mathrm{AlCl}_4^-$

The acylium ion is resonance-stabilised: $\mathrm{CH}_3\mathrm{C}\equiv\mathrm{O}^+ \leftrightarrow \mathrm{CH}_3\mathrm{C}^+=\mathrm{O}$

Step 2: Electrophilic attack.

The $\pi$ electrons of benzene attack the carbonyl carbon of the acylium ion, forming the sigma complex.

Step 3: Deprotonation.

$\mathrm{AlCl}_4^-$ removes a proton, restoring aromaticity and regenerating $\mathrm{AlCl}_3$ and $\mathrm{HCl}$.

Example 3: Sequential Electrophilic Substitutions

Synthesis of 4-nitro-2-methylbenzoic acid from toluene.

Step 1: Nitration of toluene. The $-\mathrm{CH}_3$ group is ortho/para directing. Nitration gives a mixture of 2-nitrotoluene (ortho) and 4-nitrotoluene (para). Isolate 4-nitrotoluene by fractional distillation.

$\mathrm{C}_6\mathrm{H}_5\mathrm{CH}_3 \xrightarrow{\mathrm{HNO}_3/\mathrm{H}_2\mathrm{SO}_4} 4\text{-}\mathrm{O}_2\mathrm{NC}_6\mathrm{H}_4\mathrm{CH}_3 \text{ (major)}$

Step 2: Oxidation of the methyl group to carboxylic acid:

$4\text{-}\mathrm{O}_2\mathrm{NC}_6\mathrm{H}_4\mathrm{CH}_3 \xrightarrow{\mathrm{KMnO}_4,\,\Delta} 4\text{-}\mathrm{O}_2\mathrm{NC}_6\mathrm{H}_4\mathrm{COOH}$

Step 3: A second nitration would place the nitro group meta to the $-\mathrm{COOH}$ group (meta director). But we need it ortho to the $-\mathrm{COOH}$.

This synthesis requires careful ordering: the methyl group directs the first nitration, and after oxidation the $-\mathrm{COOH}$ directs any further substitution meta. To get the 2-nitro isomer, a different strategy is needed (e.g., reduce the nitro group, protect as amide, oxidise, deprotect, then nitrate).

Example 4: Delocalisation Energy Calculation

Calculate the delocalisation energy of benzene from hydrogenation data.

Hydrogenation of cyclohexene: $\mathrm{C}_6\mathrm{H}_{10} + \mathrm{H}_2 \to \mathrm{C}_6\mathrm{H}_{12}$, $\Delta H = -120\,\mathrm{kJ/mol}$

If benzene had three isolated C=C bonds, expected $\Delta H_\mathrm{hydrogenation} = 3 \times (-120) = -360\,\mathrm{kJ/mol}$

Experimental: $\mathrm{C}_6\mathrm{H}_6 + 3\mathrm{H}_2 \to \mathrm{C}_6\mathrm{H}_{12}$, $\Delta H = -208\,\mathrm{kJ/mol}$

$\text{Delocalisation energy} = -360 - (-208) = -152\,\mathrm{kJ/mol}$

The negative sign indicates that benzene is stabilised by $152\,\mathrm{kJ/mol}$ relative to the hypothetical Kekule structure with three isolated double bonds.

Example 5: Bromination of Phenol -- Detailed Mechanism

Reaction: $\mathrm{C}_6\mathrm{H}_5\mathrm{OH} + 3\mathrm{Br}_2 \to \mathrm{C}_6\mathrm{H}_2\mathrm{Br}_3\mathrm{OH} + 3\mathrm{HBr}$

Phenol is so activated that it reacts with bromine water (dilute aqueous $\mathrm{Br}_2$) without a Lewis acid catalyst.

Mechanism:

1. In aqueous solution, the $-\mathrm{OH}$ group is partially deprotonated to give the phenoxide ion ($\mathrm{C}_6\mathrm{H}_5\mathrm{O}^-$), which is even more electron-rich.

2. The phenoxide ion donates electron density into the ring through resonance. The oxygen lone pair can delocalise into the ring at the ortho and para positions.

3. $\mathrm{Br}_2$ acts as an electrophile (polarised by the electron-rich ring). The first bromination occurs at the position para to $-\mathrm{OH}$ (or ortho, since both are activated).

4. After the first bromination, the product is more deactivated, but still reactive enough for bromination at the second ortho position.

5. The third bromination occurs at the remaining ortho position, giving 2,4,6-tribromophenol.

Observation: A white precipitate forms immediately (2,4,6-tribromophenol is insoluble in water) and the bromine water is decolourised.

Example 6: Comparing the Acidity of Phenol and Ethanol

Phenol: $\mathrm{p}K_a = 10.0$. Ethanol: $\mathrm{p}K_a = 16.0$.

Phenol is approximately one million times more acidic than ethanol. The phenoxide ion is stabilised by resonance delocalisation of the negative charge over the aromatic ring, whereas the ethoxide ion ($\mathrm{CH}_3\mathrm{CH}_2\mathrm{O}^-$) has no such stabilisation.

Resonance structures of the phenoxide ion:

$\mathrm{C}_6\mathrm{H}_5\mathrm{O}^- \leftrightarrow \overset◆LB◆-◆RB◆◆LB◆\mathrm{C}◆RB◆_6\mathrm{H}_4 = \mathrm{O} \leftrightarrow \mathrm{C}_6\mathrm{H}_4 = \overset◆LB◆+◆RB◆◆LB◆\mathrm{O}◆RB◆ \leftrightarrow \dots$

The negative charge is distributed over the ortho and para carbons of the ring, making the phenoxide ion significantly more stable than the ethoxide ion.

Example 7: Synthesis of 4-Bromophenol from Benzene

The challenge: $-\mathrm{OH}$ is ortho/para directing and strongly activating, but $-\mathrm{Br}$ is deactivating. We need to introduce bromine para to the $-\mathrm{OH}$ group.

Route:

Step 1: Nitration of benzene: $\mathrm{C}_6\mathrm{H}_6 \to \mathrm{C}_6\mathrm{H}_5\mathrm{NO}_2$

Step 2: Reduction: $\mathrm{C}_6\mathrm{H}_5\mathrm{NO}_2 \to \mathrm{C}_6\mathrm{H}_5\mathrm{NH}_2$

Step 3: Diazotisation: $\mathrm{C}_6\mathrm{H}_5\mathrm{NH}_2 \xrightarrow{\mathrm{NaNO}_2/\mathrm{HCl}} \mathrm{C}_6\mathrm{H}_5\mathrm{N}_2^+\mathrm{Cl}^-$

Step 4: Hydrolysis of the diazonium salt to phenol:

$\mathrm{C}_6\mathrm{H}_5\mathrm{N}_2^+ + \mathrm{H}_2\mathrm{O} \xrightarrow{\Delta} \mathrm{C}_6\mathrm{H}_5\mathrm{OH} + \mathrm{N}_2 + \mathrm{H}^+$

Step 5: Bromination of phenol with bromine water (no catalyst needed):

$\mathrm{C}_6\mathrm{H}_5\mathrm{OH} + \mathrm{Br}_2(aq) \to 4\text{-}\mathrm{BrC}_6\mathrm{H}_4\mathrm{OH} \text{ (mono-bromination at para)}$

Note: Controlling mono-bromination of phenol requires careful control of bromine stoichiometry (use 1 equivalent of $\mathrm{Br}_2$ in a non-aqueous solvent such as $\mathrm{CS}_2$ at low temperature). In aqueous bromine water, tri-bromination occurs. For mono-bromination, use bromine in a non-polar solvent.

Practical Techniques for Aromatic Chemistry

Required Practical: Nitration of Methylbenzene (AQA RP 12)

Objective: To prepare methyl 3-nitrobenzoate by nitration of methyl benzoate (or nitration of methylbenzene).

Safety: Concentrated $\mathrm{HNO}_3$ and $\mathrm{H}_2\mathrm{SO}_4$ are highly corrosive. Nitrobenzene and its derivatives are toxic. Wear gloves, eye protection, and work in a fume cupboard.

Procedure (nitration of methylbenzene):

1. In a fume cupboard, add $10\,\mathrm{cm}^3$ of concentrated $\mathrm{HNO}_3$ to a conical flask and cool in an ice bath.
2. Slowly add $10\,\mathrm{cm}^3$ of concentrated $\mathrm{H}_2\mathrm{SO}_4$, with stirring, maintaining the temperature below $20^\circ\mathrm{C}$.
3. Cool the nitrating mixture to $0$--$5^\circ\mathrm{C}$.
4. Add $5\,\mathrm{cm}^3$ of methylbenzene dropwise, with vigorous stirring, keeping the temperature below $10^\circ\mathrm{C}$.
5. After addition, allow the mixture to stand at room temperature for 15 minutes, then pour onto crushed ice.
6. Filter the yellow solid (mixture of 2-nitrotoluene and 4-nitrotoluene) under reduced pressure.
7. Wash with cold water and recrystallise from ethanol.

Key control: Temperature must be kept low to prevent di-nitration and oxidation side reactions.

Exam-Style Questions with Full Mark Schemes

Q1 (6 marks)

Benzene does not react with bromine water under normal conditions, but phenol decolourises bromine water immediately. Explain this difference.

Mark Scheme:

6 marks:

• Benzene undergoes electrophilic substitution, which requires a strong electrophile (1 mark). Bromine water does not provide a sufficiently strong electrophile without a Lewis acid catalyst ($\mathrm{FeBr}_3$) (1 mark).
• Phenol has an $-\mathrm{OH}$ group that donates electron density into the ring through resonance, increasing the electron density at the ortho and para positions (1 mark).
• This makes the ring much more nucleophilic and more reactive towards electrophiles (1 mark).
• The increased electron density polarises the $\mathrm{Br}_2$ molecule sufficiently for it to act as an electrophile without a catalyst (1 mark).
• Phenol is so activated that it undergoes tri-bromination, giving 2,4,6-tribromophenol as a white precipitate (1 mark).

Q2 (5 marks)

Describe the mechanism for the nitration of benzene. Include the formation of the electrophile and the structure of the intermediate.

Mark Scheme:

5 marks:

• Formation of $\mathrm{NO}_2^+$: $\mathrm{HNO}_3 + \mathrm{H}_2\mathrm{SO}_4 \to \mathrm{NO}_2^+ + \mathrm{HSO}_4^- + \mathrm{H}_2\mathrm{O}$ (1 mark).
• Electrophilic attack: $\pi$ electrons attack $\mathrm{NO}_2^+$, forming the sigma complex (arenium ion) with the positive charge delocalised over three carbons (1 mark).
• Deprotonation: $\mathrm{HSO}_4^-$ removes $\mathrm{H}^+$ from the sigma complex, restoring aromaticity (1 mark).
• Overall equation: $\mathrm{C}_6\mathrm{H}_6 + \mathrm{HNO}_3 \to \mathrm{C}_6\mathrm{H}_5\mathrm{NO}_2 + \mathrm{H}_2\mathrm{O}$ (1 mark).
• Curly arrow diagram showing electron flow for the electrophilic attack and deprotonation steps (1 mark).

Q3 (4 marks)

Explain why Friedel-Crafts acylation is preferred over Friedel-Crafts alkylation for introducing alkyl groups onto the benzene ring.

Mark Scheme:

4 marks:

• Friedel-Crafts alkylation produces an alkylbenzene that is more reactive than the starting material (alkyl groups are activating), leading to polyalkylation (1 mark).
• Friedel-Crafts acylation produces an aryl ketone, and the acyl group is electron-withdrawing (deactivating), preventing further substitution (1 mark).
• The acylium ion ($\mathrm{RCO}^+$) is resonance-stabilised, so no carbocation rearrangement occurs (1 mark).
• The ketone product can subsequently be reduced to the alkyl group (e.g. Clemmensen reduction), providing a controlled route to linear alkylbenzenes (1 mark).

Q4 (6 marks)

Starting from benzene, propose a synthesis of 3-bromonitrobenzene. Explain the order of steps and justify your choice of reagents.

Mark Scheme:

6 marks:

Step 1: Nitration of benzene (1 mark): \mathrm{C}_6\mathrm{H}_6 \xrightarrow{\mathrm{HNO}_3/\mathrm{H}_2\mathrm{SO}_4}} \mathrm{C}_6\mathrm{H}_5\mathrm{NO}_2 (1 mark).

Step 2: Bromination of nitrobenzene (1 mark): \mathrm{C}_6\mathrm{H}_5\mathrm{NO}_2 \xrightarrow{\mathrm{Br}_2/\mathrm{FeBr}_3}} 3\text{-}\mathrm{BrC}_6\mathrm{H}_4\mathrm{NO}_2 (1 mark).

Justification (2 marks):

• The $-\mathrm{NO}_2$ group is meta-directing (1 mark), so bromination after nitration places bromine at the meta position (1 mark).
• If bromination were done first, the $-\mathrm{Br}$ group (ortho/para director) would place the nitro group at the ortho/para positions, giving the wrong isomer.

Q5 (4 marks)

Explain the term delocalisation energy as applied to benzene. State the experimental evidence for the delocalised model of benzene.

Mark Scheme:

4 marks:

• Delocalisation energy is the extra stability of benzene compared with the hypothetical Kekule structure with three isolated double bonds (1 mark).
• Evidence 1: All C--C bond lengths in benzene are equal ($139\,\mathrm{pm}$), intermediate between single ($154\,\mathrm{pm}$) and double ($134\,\mathrm{pm}$) bonds (1 mark).
• Evidence 2: The enthalpy of hydrogenation of benzene ($-208\,\mathrm{kJ/mol}$) is less exothermic than three times the enthalpy of hydrogenation of cyclohexene ($3 \times -120 = -360\,\mathrm{kJ/mol}$), giving a delocalisation energy of $152\,\mathrm{kJ/mol}$ (1 mark).
• Evidence 3: Benzene undergoes substitution (preserving the ring) rather than addition (which would destroy the aromatic system), indicating the ring is unusually stable (1 mark).

Retrosynthesis with Arenes

Key Disconnections

Target	Disconnection	Forward reagents
Nitroarene	Electrophilic nitration	$\mathrm{HNO}_3/\mathrm{H}_2\mathrm{SO}_4$
Aryl ketone	Friedel-Crafts acylation	$\mathrm{RCOCl}/\mathrm{AlCl}_3$
Phenol	Diazonium salt hydrolysis	$\mathrm{ArN}_2^+ + \mathrm{H}_2\mathrm{O},\,\Delta$
Haloarene	Electrophilic halogenation	$\mathrm{Br}_2/\mathrm{FeBr}_3$ or Sandmeyer
Azo dye	Diazonium coupling	$\mathrm{ArN}_2^+ + \text{phenol/amine}$

Directing Group Strategy in Multi-Step Synthesis

When planning the synthesis of polysubstituted benzenes, the order of introducing substituents is critical. The directing effects must be considered:

1. Introduce activating groups first (they make subsequent reactions easier).
2. If a meta-directing group is needed, introduce it last (or protect activating groups during steps requiring meta substitution).
3. Remember that strongly deactivating groups (e.g. $-\mathrm{NO}_2$) essentially stop further electrophilic substitution.

Worked example: Synthesis of 3-amino-4-methylbenzoic acid from toluene.

Step 1: Nitration of toluene (ortho/para directing). The major product is 4-nitrotoluene.

Step 2: Oxidation of methyl to carboxylic acid: 4-nitrotoluene $\to$ 4-nitrobenzoic acid.

Step 3: Reduction of nitro group: 4-nitrobenzoic acid $\to$ 4-aminobenzoic acid.

Wait -- we need 3-amino, not 4-amino. The correct approach:

Step 1: Nitration of toluene gives 2-nitrotoluene (ortho) and 4-nitrotoluene (para). We need 2-nitrotoluene.

Step 2: Oxidation: 2-nitrotoluene $\to$ 2-nitrobenzoic acid.

Step 3: Reduction: 2-nitrobenzoic acid $\to$ 2-aminobenzoic acid.

But we need 3-amino-4-methylbenzoic acid, which has three substituents. This requires a different approach entirely, demonstrating the complexity of multi-substituted aromatic synthesis.

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tip

Diagnostic Test Ready to test your understanding of Arenes? The diagnostic test contains the hardest questions within the A-Level specification for this topic, each with a full worked solution.

Unit tests probe edge cases and common misconceptions. Integration tests combine Arenes with other chemistry topics to test synthesis under exam conditions.

See Diagnostic Guide for instructions on self-marking and building a personal test matrix.