Aromatic Hydrocarbons: Reactions of the Ring and the Side Chain – Derivatives of Benzene (Senior 6 Chemistry) | Ganzaa

Derivatives of Benzene

Lessons

Senior 6 · Chemistry

Derivatives of Benzene

5 lessons

Lesson 2 of 5

Aromatic Hydrocarbons: Reactions of the Ring and the Side Chain

22 min

Two parts, two kinds of chemistry

Methylbenzene, $\mathrm{C_6H_5CH_3}$ (toluene), is the common solvent in paints, varnishes and glues, and a component of petrol. Its molecule has two different parts: an aromatic ring, which reacts like benzene, and a methyl side chain, which reacts like an alkane. By choosing the conditions, a chemist can make a reagent attack one part and leave the other alone.

Definition

Aromatic hydrocarbon (arene): a hydrocarbon that contains one or more benzene rings. Examples: methylbenzene, ethylbenzene, phenylethene (styrene), naphthalene (two fused rings).

Side chain: an alkyl group (or other carbon chain) attached to an aromatic ring. Example: the $\mathrm{-CH_2CH_3}$ group of ethylbenzene.

Figure 2.1: Methylbenzene reacts at the ring (blue) or at the side chain (red), depending on the conditions.

Reactions of the ring

The methyl group is weakly activating and 2,4-directing (Lesson 1), so methylbenzene undergoes electrophilic substitution faster than benzene, under milder conditions, giving mainly 2- and 4-substituted products. With enough nitric acid, all three activated positions react: $\mathrm{C_6H_5CH_3 + 3HNO_3 \rightarrow C_6H_2(CH_3)(NO_2)_3 + 3H_2O}$ . The product, 2-methyl-1,3,5-trinitrobenzene, is the explosive TNT.

Reactions of the side chain

Free-radical substitution. In ultraviolet light, with no catalyst and at the boiling point, chlorine attacks the methyl group, just as it attacks methane (S5):

$\text{initiation: } \mathrm{Cl_2 \rightarrow 2Cl^{\bullet}}$

$\text{propagation: } \mathrm{C_6H_5CH_3 + Cl^{\bullet} \rightarrow C_6H_5CH_2^{\bullet} + HCl} \qquad \mathrm{C_6H_5CH_2^{\bullet} + Cl_2 \rightarrow C_6H_5CH_2Cl + Cl^{\bullet}}$

$\text{termination, e.g. } \mathrm{2C_6H_5CH_2^{\bullet} \rightarrow C_6H_5CH_2CH_2C_6H_5}$

With more chlorine, further hydrogens are replaced, giving $\mathrm{C_6H_5CHCl_2}$ and $\mathrm{C_6H_5CCl_3}$ .

Oxidation. The benzene ring resists oxidation, but a side chain does not. Heating under reflux with alkaline potassium manganate(VII), then acidifying, converts any alkyl side chain, whatever its length, into a $\mathrm{-COOH}$ group joined to the ring:

$\mathrm{C_6H_5CH_3 + 3[O] \rightarrow C_6H_5COOH + H_2O}$

$\mathrm{C_6H_5CH_2CH_3 + 6[O] \rightarrow C_6H_5COOH + CO_2 + 2H_2O}$

The purple colour of manganate(VII) disappears and brown manganese(IV) oxide forms; benzoic acid crystallises when the mixture is acidified.

Table 2.1: Chlorine and methylbenzene: the conditions decide the product

Conditions	Part attacked	Mechanism	Product
$\mathrm{Cl_2}$ with $\mathrm{AlCl_3}$ (or $\mathrm{FeCl_3}$ ), dark, room temperature	ring	electrophilic substitution	1-chloro-2-methylbenzene and 1-chloro-4-methylbenzene
$\mathrm{Cl_2}$ in ultraviolet light, boiling, no catalyst	side chain	free-radical substitution	(chloromethyl)benzene, $\mathrm{C_6H_5CH_2Cl}$

Phenylethene (styrene), $\mathrm{C_6H_5CH{=}CH_2}$ , has a C=C double bond in its side chain. The side chain behaves like an alkene: it decolourises bromine water, and it undergoes addition polymerisation to give poly(phenylethene), polystyrene:

$\mathrm{n\,C_6H_5CH{=}CH_2 \rightarrow {-}\!\left[CH(C_6H_5){-}CH_2\right]_n{-}}$

Worked example

Problem. 1,4-Dimethylbenzene is oxidised to benzene-1,4-dicarboxylic acid, which is used to make PET plastic bottles. (a) Write an equation using [O]. (b) 10.6 g of 1,4-dimethylbenzene ( $M_{r}$ = 106.2) gives 12.5 g of the acid ( $M_{r}$ = 166.1). Calculate the percentage yield.

Solution. (a) Each $\mathrm{-CH_3}$ becomes $\mathrm{-COOH}$ :

$\mathrm{C_6H_4(CH_3)_2 + 6[O] \rightarrow C_6H_4(COOH)_2 + 2H_2O}$

(b) $n = \frac{10.6}{106.2} = 0.09981\,\mathrm{mol} \qquad \text{theoretical mass} = 0.09981 \times 166.1 = 16.58\,\mathrm{g}$

$\text{yield} = \frac{12.5}{16.58} \times 100 = 75.4\,\%$

Answer: 75.4%.

Worked example

Problem. Compound X, $\mathrm{C_8H_{10}}$ , is aromatic. On oxidation with hot alkaline potassium manganate(VII) followed by acid, it gives benzoic acid. Identify X, and explain why its isomer 1,2-dimethylbenzene would give a different product.

Solution. Benzoic acid has only one $\mathrm{-COOH}$ group, so X has only one side chain, containing the two extra carbons: X is ethylbenzene, $\mathrm{C_6H_5CH_2CH_3}$ . The side chain is oxidised to $\mathrm{-COOH}$ and the extra carbon is lost as $\mathrm{CO_2}$ . 1,2-Dimethylbenzene has two side chains, so it gives benzene-1,2-dicarboxylic acid, with two $\mathrm{-COOH}$ groups.

Warning

Solvent safety. Methylbenzene is much less toxic than benzene, which it replaced as a laboratory solvent, but it is highly flammable and its vapour harms the nervous system. Use it only in a well-ventilated room or fume cupboard, away from flames. Deliberately breathing solvent vapour ("glue sniffing") can cause sudden death.

Tip

"Sun for the Side chain; Catalyst in the Cold for the Cycle (ring)." Ultraviolet light gives radicals, which attack the alkyl side chain; a halogen carrier in the dark gives an electrophile, which attacks the ring.

Leah4sci: Aromatic Side Chain Oxidation to Carboxylic Acid (5 min). Watch for why every alkyl side chain, however long, ends up as a single –COOH group on the ring.

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Practice

Give the reagents and conditions needed to convert methylbenzene into (a) (chloromethyl)benzene, (b) a mixture of 1-chloro-2-methylbenzene and 1-chloro-4-methylbenzene.
Write the propagation steps for the reaction of methylbenzene with chlorine in ultraviolet light.
Name the product of oxidising each compound with hot alkaline $\mathrm{KMnO_4}$ , then acid: (a) propylbenzene, (b) 1,3-dimethylbenzene.
Explain why methylbenzene is nitrated at a lower temperature than benzene.
Phenylethene decolourises bromine water, but ethylbenzene does not. Explain.
What mass of benzoic acid ( $M_{r}$ = 122.1) can be made from 23.0 g of methylbenzene ( $M_{r}$ = 92.1)?
A student heats methylbenzene with chlorine in sunlight and with a trace of iron from a rusty needle. Evaluate the likely products.

Show answersHide answers

(a) Chlorine in ultraviolet light, boiling, no catalyst. (b) Chlorine with $\mathrm{AlCl_3}$ (or $\mathrm{FeCl_3}$ ), in the dark at room temperature.
$\mathrm{C_6H_5CH_3 + Cl^{\bullet} \rightarrow C_6H_5CH_2^{\bullet} + HCl}$ ; $\mathrm{C_6H_5CH_2^{\bullet} + Cl_2 \rightarrow C_6H_5CH_2Cl + Cl^{\bullet}}$ .
(a) Benzoic acid (the extra carbons are lost as $\mathrm{CO_2}$ ). (b) Benzene-1,3-dicarboxylic acid.
The methyl group releases electrons into the ring, increasing its electron density and making it more attractive to the electrophile $\mathrm{NO_2^+}$ , so the activation energy is lower.
Phenylethene has a C=C double bond in its side chain, which adds bromine like an alkene. Ethylbenzene's side chain is saturated, and its ring does not react with bromine water.
$n = \frac{23.0}{92.1} = 0.2497$ mol; mass = $0.2497 \times 122.1 = 30.5$ g.
Both kinds of reaction can happen at once. Sunlight produces chlorine radicals, which substitute the side chain to give (chloromethyl)benzene and further products; the iron forms $\mathrm{FeCl_3}$ , a halogen carrier, which allows ring substitution at 2 and 4. The student would get a mixture that is hard to separate, so the conditions should be controlled to favour one reaction.

Sources and credits

Video: Aromatic Side Chain Oxidation to Carboxylic Acid, Leah4sci, embedded from YouTube.
The text, reaction map, table, worked examples and Practice items were written and drawn by Ganzaa.

1Two parts, two kinds of chemistry

2Reactions of the ring

3Reactions of the side chain

Two parts, two kinds of chemistry

Reactions of the ring

Reactions of the side chain