Aromatic Hydrocarbons

(c) A mixture of conc. $\text{HNO}_3$ and conc. $\text{H}_2\text{SO}_4$

This mixture is called the nitrating mixture. The conc. $\text{H}_2\text{SO}_4$ protonates $\text{HNO}_3$, which then loses water to give the nitronium ion $\text{NO}_2^+$, the actual electrophile that attacks the benzene ring. Conc. $\text{HNO}_3$ alone cannot generate $\text{NO}_2^+$ efficiently.

Necessary conditions for aromaticity

A compound is aromatic only if it satisfies all of the following conditions (Hückel's rule):

1. Cyclic structure — the molecule must contain a ring of atoms.
2. Planarity — all the atoms of the ring must lie in one plane, so that the p-orbitals can overlap sideways.
3. Complete conjugation — every atom in the ring must possess a p-orbital (or an unshared pair in a p-orbital) that overlaps continuously all around the ring, forming a closed π-cloud.
4. Hückel's rule — the number of π-electrons in the delocalised ring must be $4n + 2$, where $n = 0, 1, 2, 3, \dots$ (i.e. 2, 6, 10, 14, … π-electrons).
5. Extra stability — the delocalisation must give the compound a resonance energy large enough to make it unusually stable and to make it undergo substitution rather than addition.

Why benzene is aromatic

• Benzene, $\mathrm{C_{6}H_{6}}$, is a six-membered planar ring of carbon atoms; all six C–C bond lengths are equal ($139\ \text{pm}$), intermediate between a single and a double bond.
• Each carbon is $sp^2$ hybridised and has one unhybridised p-orbital perpendicular to the ring. These six p-orbitals overlap sideways to give a continuous closed π-cloud above and below the plane.
• The number of π-electrons is $6$, i.e. $4n + 2$ with $n = 1$. So Hückel's rule is obeyed.
• This delocalisation gives benzene a resonance energy of about $150\ \text{kJ mol}^{-1}$, making it very stable. Hence benzene is aromatic.

Why cyclohexene is not aromatic

• Cyclohexene is a six-membered ring, but it is not fully conjugated: it contains only one double bond, and the remaining four carbon atoms are $sp^3$ hybridised with no p-orbitals.
• Because there is a break in conjugation, a continuous cyclic π-cloud cannot form; the π-electrons are localised over just two carbon atoms.
• Hence the conditions of complete conjugation and Hückel's rule are not satisfied, and cyclohexene shows no resonance energy of the aromatic type. It behaves as an ordinary alkene — it readily undergoes addition reactions (e.g. decolourises bromine water) rather than substitution. Therefore cyclohexene is not aromatic.

Oxidation of toluene (methylbenzene)

Both alkaline $\text{KMnO}_4$ and hot acidified $\text{K}_2\text{Cr}_2\text{O}_7$ are strong oxidising agents. They attack the side chain, not the ring, and oxidise the methyl group ($-\text{CH}_3$) to a carboxyl group ($-\text{COOH}$). The benzene ring resists oxidation under these conditions.

1. With alkaline $\text{KMnO}_4$ (heated)

The purple colour of permanganate is discharged and brown $\text{MnO}_2$ is formed. The methyl group becomes potassium benzoate:

$\mathrm{C_6H_5CH_3 + 2KMnO_4 \xrightarrow{\ \text{alkaline},\ \Delta\ } C_6H_5COOK + 2MnO_2 + KOH + H_2O}$

On acidifying with dilute $\text{HCl}$, benzoic acid is liberated:

$\mathrm{C_6H_5COOK + HCl \longrightarrow C_6H_5COOH + KCl}$

Product: benzoic acid, $\mathrm{C_6H_5COOH}$.

2. With hot acidified $\text{K}_2\text{Cr}_2\text{O}_7$

The orange dichromate turns green because of $\mathrm{Cr^{3+}}$ ions. In terms of nascent oxygen:

$\mathrm{C_6H_5CH_3 + 3[O] \xrightarrow{\ K_2Cr_2O_7/H_2SO_4,\ \Delta\ } C_6H_5COOH + H_2O}$

The balanced equation is

$\mathrm{C_6H_5CH_3 + K_2Cr_2O_7 + 4H_2SO_4 \longrightarrow C_6H_5COOH + K_2SO_4 + Cr_2(SO_4)_3 + 5H_2O}$

Product: benzoic acid, $\mathrm{C_6H_5COOH}$.

Use of benzoic acid: it is used as a food preservative (as sodium benzoate) to stop microbes growing in pickles, jams and soft drinks.

Electrophilic substitution reactions are reactions in which an electrophile (an electron-deficient species, e.g. $\text{NO}_2^+$, $\text{Cl}^+$, $\text{R}^+$, $\text{RCO}^+$) attacks an aromatic ring and replaces one of its hydrogen atoms. The aromatic ring, being electron-rich, acts as a nucleophile. The characteristic feature is that the aromaticity of the ring is temporarily lost in the intermediate but is restored in the final product.

Mechanism of chlorination of benzene in the presence of anhydrous $\text{AlCl}_3$

This is an example of electrophilic substitution in which $\text{AlCl}_3$ acts as a Lewis acid catalyst (a halogen carrier).

Step 1 — Generation of the electrophile: Anhydrous $\text{AlCl}_3$ is a strong Lewis acid. It accepts a chloride ion from $\text{Cl}_2$, forming a polarised complex that acts as the electrophile:

$\text{Cl}_2 + \text{AlCl}_3 \longrightarrow \text{Cl}^+\text{—AlCl}_4^- \quad (\text{electrophile} = \text{Cl}^+)$

Step 2 — Attack of the electrophile on benzene: The $\pi$-electrons of benzene attack the electrophile $\text{Cl}^+$, giving an unstable arenium ion (sigma complex) in which one carbon is $sp^3$ hybridised and the ring loses aromaticity. The positive charge is delocalised over three carbons:

$\text{C}_6\text{H}_6 + \text{Cl}^+ \longrightarrow \text{C}_6\text{H}_5\text{Cl}^+\text{H} \quad (\text{arenium ion})$

Step 3 — Loss of a proton: The base $\text{AlCl}_4^-$ removes the proton from the $sp^3$ carbon, restoring the aromatic $\pi$-system:

$\text{C}_6\text{H}_5\text{Cl}^+\text{H} + \text{AlCl}_4^- \longrightarrow \text{C}_6\text{H}_5\text{Cl} + \text{HCl} + \text{AlCl}_3$

The catalyst $\text{AlCl}_3$ is regenerated, so only a small amount is needed.

Overall equation:

$\text{C}_6\text{H}_6 + \text{Cl}_2 \xrightarrow{\text{anhyd. } \text{AlCl}_3} \text{C}_6\text{H}_5\text{Cl} + \text{HCl}$

Friedel–Crafts alkylation of benzene

Benzene reacts with an alkyl halide in the presence of anhydrous $\text{AlCl}_3$ to give an alkylbenzene:

$\text{C}_6\text{H}_6 + \text{CH}_3\text{Cl} \xrightarrow{\text{anhyd. } \text{AlCl}_3} \text{C}_6\text{H}_5\text{CH}_3 + \text{HCl}$

Friedel–Crafts acylation of benzene

Benzene reacts with an acyl halide in the presence of anhydrous $\text{AlCl}_3$ to give an aromatic ketone:

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

In both cases the electrophile is generated by the action of $\text{AlCl}_3$ on the alkyl/acyl halide ($\text{R}^+$ or $\text{RCO}^+$), which then attacks the ring by the same three-step mechanism as above.