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9476 / 2027
Organic Chemistry

Lesson 16 of 34

The ring and side-chain respond to different conditions

State the catalyst, light condition and temperature before predicting where substitution occurs.

A-Level 9476 (2026-2027)

  • I can describe the specified aromatic-ring reactions.Syllabus

    Syllabus 9476, 11.4(d). Describe the specified aromatic-ring reactions.

  • I can describe alkyl-side-chain chemistry using methylbenzene.Syllabus

    Syllabus 9476, 11.4(e). Describe alkyl-side-chain chemistry using methylbenzene.

  • I can predict ring versus side-chain halogenation.Syllabus

    Syllabus 9476, 11.4(f). Predict ring versus side-chain halogenation.

  • I can predict substitution positions in monosubstituted arenes.Syllabus

    Syllabus 9476, 11.4(g). Predict substitution positions in monosubstituted arenes.

The ring and side-chain respond to different conditions

State the catalyst, light condition and temperature before predicting where substitution occurs.

Reactions of benzene and methylbenzene
ReactionReagents and conditionsProducts and catalytic role
Ring chlorinationCl2 with anhydrous AlCl3; electrophilic substitution conditions.Benzene → chlorobenzene + HCl. AlCl3 is a Lewis acid catalyst.
Ring brominationBr2 with anhydrous AlBr3.Benzene → bromobenzene + HBr. AlBr3 is a Lewis acid catalyst.
Benzene nitrationConcentrated HNO3 + concentrated H2SO4, maintained at 50 °C.Nitrobenzene + water. H2SO4 acts as a Bronsted-Lowry acid catalyst in generating the electrophile.
Methylbenzene nitrationThe same concentrated-acid mixture, maintained at 30 °C.Mainly 2- and 4-nitromethylbenzene under mononitration conditions.
Friedel-Crafts alkylationHalogenoalkane with anhydrous AlCl3 or AlBr3, as appropriate.Benzene + CH3Cl → methylbenzene + HCl; the Lewis acid activates the halogenoalkane.
Methyl side-chain halogenationCl2 or Br2, UV light at room temperature.C6H5CH3 → C6H5CH2X initially; further substitution can occur.
Complete side-chain oxidationHot alkaline KMnO4 followed by dilute acid, or hot acidified KMnO4.Methylbenzene → benzoic acid. The aromatic ring is retained.

Concentrated sulfuric acid protonates nitric acid, which forms NO2+: HNO3 + H2SO4 ⇌ NO2+ + HSO4- + H2O. NO2+ is the electrophile. Nitration then follows electrophilic substitution, and the acid catalyst is made again. The syllabus states 30 °C for methylbenzene and 50 °C for benzene, so learn these temperatures as part of the conditions.

Positions relative to an existing substituent at carbon 1
Existing groupFavoured incoming positionsElectronic point
Alkyl, OH, NH22 and 4 (with position 6 equivalent to 2 in the simple monosubstituted case).Electron donation generally activates the ring and favours ortho/para substitution.
NO2, COOH, CHO, COR, CN3 (with position 5 equivalent).Electron withdrawal generally deactivates the ring and favours meta substitution.
Cl or Br directly on the ring2 and 4.Halogens are the useful distinction: overall deactivating by withdrawal, yet ortho/para directing through lone-pair donation.

Activation tells you how fast the ring reacts compared with benzene. Direction tells you where the new group goes. These are two different properties. A bulky group can block the positions next to it, so do not assume every allowed position forms the same amount of product.

Worked exampleSame starting compound, different reaction site

Compare methylbenzene with Br2/AlBr3 and with Br2/UV at room temperature.

Try it first. Then show the worked stepsHide the worked steps
  1. Lewis-acid conditions generate a strong electrophile for the aromatic pi system.
  2. The methyl group directs ring substitution mainly to positions 2 and 4.
  3. UV conditions initiate a radical chain and favour substitution in the alkyl side-chain.
Answer

Br2/AlBr3 gives mainly 2- and 4-bromomethylbenzene; Br2/UV gives C6H5CH2Br initially. Use structures to avoid confusing a ring bromine with a bromomethyl side-chain.

Strong oxidation of a side chain needs a hydrogen on the carbon joined directly to the ring. Methylbenzene has one: C6H5CH3 + 3[O] → C6H5COOH + H2O. In alkaline conditions the carboxylate ion forms first, so you must add acid to get the carboxylic acid.

Check your understanding

Methylbenzene reacts with chlorine. Which conditions put the chlorine on the methyl group, not the ring?

  1. Hot alkaline KMnO4, then acid, by oxidative substitution
  2. UV light at room temperature, by free-radical substitution
  3. Anhydrous AlCl3 in the dark, by electrophilic substitution
Show the answer

UV light at room temperature, by free-radical substitution

UV light splits Cl2 into radicals, which substitute H on the CH3 side-chain. A Lewis acid such as AlCl3 instead makes an electrophile that substitutes the ring.