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Notes/Chemistry/Paper 4/Halogen Compounds
CAIEA2 Level9701§31

Halogen Compounds

Two homes for a halogen — on the ring or beside it — why the ring C–X bond will not break, the AgNO₃ and NaOH probes that show it, the acyl > alkyl > aryl hydrolysis ladder, and how inertness becomes a synthesis tool.

115 min read 6 sub-topics
32
question parts
2021–2025 · 19 papers
3 marks
per paper
≈ 3% of the paper
2.3/3
avg difficulty
moderate
#15
most examined
of 15 topics by marks

Paper 4 asks the same opening move every time this topic appears: put a halogen somewhere near a benzene ring, then ask what happens. Whether the answer is "cream precipitate" or "nothing at all" depends entirely on ONE structural fact — whether the halogen is bonded to the ring or to the carbon next to it. Over 2021–2025 this topic carried 32 examined parts across 19 papers, nearly all built on that single contrast, and at a mean difficulty of 2.31 out of 3 it punches above its size. The route through: §01 the two homes for a halogen — halogenoarene versus benzylic halide, and the naming traps between them, §02 making halogenoarenes — the catalyst route and its conditions, §03 why the ring C–X bond refuses to break: lone-pair delocalisation and partial double-bond character, the explanation behind almost every mark, §04 watching the difference happen — the warm AgNO₃ and NaOH probes, their equations and observations, §05 the full hydrolysis ladder — acyl chloride > alkyl chloride > aryl chloride, with the three-part explanation, and §06 using inertness — halogenoarenes as indestructible handles inside long synthesis routes.

Before you start you should be able to

  • Nucleophilic substitution of halogenoalkanes — the SN1 and SN2 mechanisms, hydrolysis with OH⁻, and the warm AgNO₃(aq) hydrolysis test with white/cream/yellow precipitates (this subject's AS Halogen Compounds note, §§04–05)

  • Benzene's delocalised π system and the electrophilic-substitution mechanism, including halogenation with a Lewis-acid carrier (this subject's A2 Hydrocarbons note, §§01–02)

  • Directing effects — an alkyl group such as −CH₃ activates the ring and directs to the 2- and 4-positions (A2 Hydrocarbons note, §07)

  • Acyl chlorides: the electron-deficient carbonyl carbon and addition–elimination reactions (this subject's A2 Carboxylic Acids note)

  • Free-radical substitution of alkanes initiated by UV light (AS Hydrocarbons note)

By the end of this page you can

  • Decide instantly whether a halogen sits in a halogenoarene (bonded directly to the ring) or a benzylic halide (on the side-chain sp³ carbon), and name each correctly

  • Recall the production of halogenoarenes: dry Cl₂ or Br₂ with an AlCl₃ or AlBr₃ catalyst, forming chlorobenzene from benzene and the 2- plus 4-chloromethylbenzene mixture from methylbenzene, with HX as co-product

  • Explain the difference in reactivity between a halogenoalkane and a halogenoarene (chloroethane versus chlorobenzene): the halogen lone pair delocalises into the ring π system, giving the C–X bond partial double-bond character and extra strength

  • Predict observations for halogenoarenes, benzylic halides and halogenoalkanes with warm AgNO₃(aq), NaOH(aq) and water — including precipitate colours — and write the hydrolysis-plus-precipitation equation pair, writing NO equation for the unreactive aryl compound

  • Rank acyl chlorides > alkyl chlorides > aryl chlorides for ease of hydrolysis and justify each rung with electron deficiency, inductive donation and lone-pair delocalisation respectively

  • Use the inertness of the aryl C–X bond deliberately in synthesis: choosing where the halogen sits to protect or expose it, and building chains from benzylic halides via KCN and reduction

01

Two homes for a halogen

Syllabus requirement · §31

“

recall the reactions by which halogenoarenes can be produced: substitution of an arene with Cl₂ or Br₂ in the presence of a catalyst, AlCl₃ or AlBr₃ to form a halogenoarene, exemplified by benzene to form chlorobenzene and methylbenzene to form 2-chloromethylbenzene and 4-chloromethylbenzene · explain the difference in reactivity between a halogenoalkane and a halogenoarene as exemplified by chloroethane and chlorobenzene (syllabus 31, learning outcomes 1 and 2)

”

One atom, two addresses

Every question filed under this topic begins with a step nobody sees: the examiner chose WHERE to hang the halogen. Bromine on the ring makes bromobenzene, C6H5Br\text{C}_6\text{H}_5\text{Br} — famously inert. Bromine on the carbon next to the ring makes C6H5CH2Br\text{C}_6\text{H}_5\text{CH}_2\text{Br} — which hydrolyses happily, exactly like the bromoalkanes you met at AS. Two compounds in the same formula family, opposite behaviour.

That choice is the whole topic. Learn to spot the address first and every later question — reactivity, observations, synthesis planning — becomes bookkeeping. This section builds the classification; the sections after it explain and exploit it.

Halogenoarene versus benzylic halide

Home 1 — ON the ring: the halogenoarene (aryl halide). The halogen is bonded directly to a ring carbon — one of the flat, sp² carbons of the delocalised π system. Chlorobenzene C6H5Cl\text{C}_6\text{H}_5\text{Cl}, bromobenzene C6H5Br\text{C}_6\text{H}_5\text{Br}, and the 2- and 4-chloromethylbenzenes (chlorine on the ring of methylbenzene) all live here.

Home 2 — on the SIDE CHAIN: the benzylic (aralkyl) halide. The halogen is bonded to an sp³ carbon attached to the ring — the −CH2−-\text{CH}_2- group of (chloromethyl)benzene, C6H5CH2Cl\text{C}_6\text{H}_5\text{CH}_2\text{Cl}, or any carbon further out along a chain. Chemically this compound is a halogenoalkane: the ring is a spectator, and every AS halogenoalkane reaction — OH⁻ hydrolysis, KCN, NH₃, warm AgNO₃ giving a precipitate — works unchanged — and benzylic halides often outpace plain halogenoalkanes, because ionising the C–X bond gives a benzyl carbocation stabilised by resonance with the neighbouring ring. The π system is one carbon removed and never touches the C–X bond.

one glance test — WHERE does the halogen sit?HALOGENOARENE(chlorobenzene)Clsp² Caryl halide — X ON the ringrefuses substitution — no SN1, no SN2BENZYLIC HALIDE(chloromethyl)benzeneCH₂Clsp³ CX on the side-chain sp³ carbonreacts like a NORMAL halogenoalkanelook-alike formulas — an arene plus a Cl either way — yet OPPOSITE chemistry:always locate WHERE the X sits before answering

The two addresses side by side. Left: chlorobenzene — Cl bonded directly to a ring carbon of the delocalised π system; behaviour label: C–X bond locked, no substitution. Right: (chloromethyl)benzene — Cl on the side-chain sp³ carbon, one full bond away from the ring; behaviour label: reacts exactly like a halogenoalkane.

The naming trap — read locants like a chemist

Real papers use overlapping names, so pin these down now:

  • 2-chloromethylbenzene and 4-chloromethylbenzene: the number is a POSITION on the ring, counted from the methyl group. The chlorine is ON the ring — a halogenoarene. (Older books call these o- and p-chlorotoluene.)
  • (chloromethyl)benzene: the brackets weld "chloromethyl" together as one side-chain substituent, −CH2Cl-\text{CH}_2\text{Cl}, hanging off the ring. This is the benzylic home — also sold under the informal name benzyl chloride.

So "chloromethylbenzene" without a locant is genuinely ambiguous — papers slip between the readings. Protect yourself: use a locant when you mean the ring isomer, and brackets (or the C6H5CH2Cl\text{C}_6\text{H}_5\text{CH}_2\text{Cl} formula) when you mean the side chain.

Classifying any halogenated aromatic, in two questions
  1. 1

    Find the halogen atom in the drawn or formula-given structure, and put a finger on the carbon it is bonded to.

    Sounds trivial, but misreading a condensed formula such as C₆H₅CH₂Cl is where marks go — count the atoms before judging.

  2. 2

    Ask: is MY carbon itself one of the six ring carbons?

    Ring carbons are sp² and hold the delocalised π electrons; a side-chain sp³ carbon does not.

  3. 3

    YES → halogenoarene: expect an unusually STRONG C–X bond and resistance to substitution (§03). NO → benzylic or ordinary halogenoalkane: expect normal AS reactivity — hydrolysis, nucleophilic substitution, instant AgX precipitate.

    This prediction is precisely what the examiner pays for in §04's observation questions.

Worked demo — sort four structures before the theory

Classify each compound by its halogen's home, and predict what is seen when each is shaken separately with warm aqueous silver nitrate:

A=C6H5BrB=C6H5CH2BrC=4-bromomethylbenzeneD=C6H5CH(Br)CH3\text{A} = \text{C}_6\text{H}_5\text{Br} \qquad \text{B} = \text{C}_6\text{H}_5\text{CH}_2\text{Br} \qquad \text{C} = \text{4-bromomethylbenzene} \qquad \text{D} = \text{C}_6\text{H}_5\text{CH(Br)CH}_3

Show full working
  1. 1

    A: the Br is bonded to a ring carbon itself → halogenoarene.

    In the formula C₆H₅Br there is no carbon outside the ring at all — the subscript 5 already tells you one ring hydrogen has been replaced by Br.

  2. 2

    C: the name carries the locant 4-, so the Br sits on ring position 4 → also a halogenoarene.

    Same home as A even though a CH₃ rides along; the methyl group does not change which carbon owns the bromine.

  3. 3

    B: Br is on the −CH2−-\text{CH}_2- carbon OUTSIDE the ring → benzylic halide, chemically a primary bromoalkane.

    One sp³ carbon separates Br from the π system, so Br's orbitals never overlap the ring — treat it as RCH₂Br with R = C₆H₅.

  4. 4

    D: again an sp³ side-chain carbon carries the Br → benzylic, secondary this time.

    Whether the side-chain carbon has one alkyl neighbour or two changes nothing about the address question.

  5. 5

    Observations: A and C — no change. B and D — cream precipitate of AgBr forms (hydrolysis frees Br⁻, which Ag⁺ grabs).

    Warm AgNO₃(aq) is the standard probe; its logic is §04's job — here just connect address to behaviour.

Answer

A, C aryl (Br on ring) → no change · B, D benzylic (Br on side-chain sp³ C) → cream ppt of AgBr

Formula first, verdict second: trace the bond from the halogen before saying 'aryl' or 'benzylic'.

Your turn

Classification drills — invented, so the thinking is yours before the past papers arrive.

  1. 14 marks

    Sort these four chlorine-containing aromatics into halogenoarenes and non-halogenoarenes, then name the odd one out:

    (a) C6H5CH2CH2Cl\text{C}_6\text{H}_5\text{CH}_2\text{CH}_2\text{Cl} (b) 2-chloromethylbenzene (c) C6H5COCl\text{C}_6\text{H}_5\text{COCl} (d) C6H5CH(Cl)CH3\text{C}_6\text{H}_5\text{CH(Cl)CH}_3

    Which single compound would you expect to react MOST rapidly with cold water? Justify your choice in one sentence.

    Stuck? Show hint

    One of the four has its chlorine bonded to something that is neither ring nor side-chain carbon — look at the CO.

    Show solution
    1. 1

      (a) Cl on the SECOND carbon of the side chain — two bonds away from the ring → not a halogenoarene; an ordinary primary chloroalkane wearing a phenyl group.

      Distance along the chain does not matter: once Cl is off the ring carbons, the ring cannot influence the C–Cl bond.

    2. 2

      (b) The locant 2- puts Cl on ring position 2 → halogenoarene.

      Locants always count positions ON the ring, starting from the substituent already there.

    3. 3

      (c) Cl is bonded to a CARBONYL carbon (−COCl-\text{COCl}) → an acyl chloride, not a halogenoarene at all.

      'Contains a benzene ring and a chlorine' is not the definition — the BOND destination is. This compound belongs to the Carboxylic Acids topic, and it is the fastest-reacting of the four.

    4. 4

      (d) Cl on an sp³ side-chain carbon (secondary) → not a halogenoarene; a benzylic chloroalkane.

      Same verdict as (a) but one carbon closer to the ring — still outside the π system.

    5. 5

      Fastest with water: (c), the acyl chloride — its carbonyl carbon is strongly electron-deficient (an electronegative O pulls electron density away), so the C–Cl bond there is weakened and water attacks readily.

      You have just met rung one of §05's hydrolysis ladder early — the full ranking returns there with the examiner's wording.

    Answer

    (b) only is a halogenoarene · (a)/(d) benzylic chloroalkanes · (c) acyl chloride — fastest with water because the C=O oxygen makes that carbon strongly δ+ and weakens C–Cl

  2. 24 marks

    A student draws C6H5CH2Cl\text{C}_6\text{H}_5\text{CH}_2\text{Cl} and labels it "2-chloromethylbenzene".

    (a) Explain the error, and give a correct name for the student's structure.

    (b) Draw, or describe precisely, the structure that the name 4-chloromethylbenzene DOES refer to, and state how its chemistry differs from C6H5CH2Cl\text{C}_6\text{H}_5\text{CH}_2\text{Cl} towards warm aqueous AgNO3\text{AgNO}_3.

    Stuck? Show hint

    Ask what the locant 2- is counting, and where the student's structure keeps its chlorine.

    Show solution
    1. 1

      (a) The label is wrong because "2-" counts a position ON THE RING, but the student's structure has its Cl on the side chain — no ring position is involved.

      The number is not decoration; it asserts which carbon owns the chlorine.

    2. 2

      Correct names for the drawn structure: (chloromethyl)benzene (brackets binding the side-chain substituent) — informally, benzyl chloride.

      Bracketed substituent names are the exam-safe way to signal 'the halogen is on the chain'.

    3. 3

      (b) 4-chloromethylbenzene: the methylbenzene ring with Cl bonded DIRECTLY to ring carbon 4 (para to −CH₃); every side-chain carbon is an unsubstituted −CH₃.

      Drawing it mentally as toluene first, then replacing the ring H at position 4, prevents drifting back to the side chain.

    4. 4

      Chemistry difference: with warm AgNO3\text{AgNO}_3(aq) the side-chain compound gives a white precipitate (AgCl from freed Cl⁻), while 4-chloromethylbenzene gives no change — its ring C–Cl bond is strengthened by interaction with the ring's π system (§03).

      Same molecular family, opposite observations — exactly the contrast the topic examines.

    Answer

    (a) locants count ring positions; the structure is (chloromethyl)benzene/benzyl chloride · (b) Cl directly on ring C-4 of methylbenzene; no ppt with AgNO₃(aq) unlike the white ppt from the side-chain isomer

Practise classifying aryl versus benzylic halides and predicting their reactionsReal past-paper questions · Halogenoalkane vs halogenoarene reactivity

The rest of this note

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Can you do all of these?

  • I can look at any halogenated aromatic structure and say within seconds whether the halogen is on the ring (halogenoarene) or on the side chain (behaves like a halogenoalkane)

  • I can state reagents and conditions for making chlorobenzene from benzene — dry chlorine, AlCl₃ or FeCl₃, no UV — and write the overall equation including the HCl co-product

  • I can predict that methylbenzene plus Cl₂/AlCl₃ gives a mixture of the 2- and 4-chloromethylbenzene isomers (and explain why the 3-isomer is minor)

  • I can give the two-step explanation for aryl C–X inertness: lone pair/p-orbital delocalises into the π system, then C–X gains partial double-bond character and is stronger

  • I can explain why neither SN1 nor SN2 can operate on the ring carbon

  • I can predict observations with warm AgNO₃(aq) — no change for aryl, white/cream/yellow ppt for alkyl and benzylic — and write the hydrolysis and precipitation equations, writing none for the aryl compound

  • I can design a fair comparison of hydrolysis rates (same temperature, concentrations, volumes; time to first precipitate)

  • I can rank acyl > alkyl > aryl chlorides for hydrolysis and attach the right reason to each rung

  • I can explain why phenol, like chlorobenzene, resists cleavage of its C–O bond

  • I can plan syntheses that pass THROUGH a halogenoarene untouched, or functionalise the side chain via UV halogenation, KCN and reduction instead

Now do the questions
32 real Paper 4 parts from 2021–2025, sorted by difficulty, with mark schemes