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Notes/Chemistry/Paper 4/Introduction to A Level Organic Chemistry
CAIEA2 Level9701§29

Introduction to A Level Organic Chemistry

The A2 organic toolkit: the new functional groups, naming aromatic compounds, the delocalised π system of benzene, the electrophilic-substitution and addition–elimination mechanisms, and optical isomerism in drugs.

240 min read 6 sub-topics
118
question parts
2021–2025 · 32 papers
6 marks
per paper
≈ 6% of the paper
1.7/3
avg difficulty
moderate
#10
most examined
of 15 topics by marks

Every A2 organic question — drugs, dyes, polymers, syntheses — opens by assuming this topic: spot the functional groups, name the aromatic compound, know why the benzene ring reacts the way it does. Over 2021–2025 it carried 118 examined parts across 32 papers, and its ideas are reused far beyond that count because examiners bolt them onto questions filed under later topics. The route through is: §01 the functional groups that are NEW at A Level — amides, nitriles, arenes — and classifying alcohols and amines, §02 naming aromatic and cyclic compounds, §03 benzene's textsp2\\text{sp}^2 framework and delocalised pi\\pi cloud, §04 electrophilic substitution, §05 addition–elimination, and §06 optical isomerism — the single most-repeated strand, appearing in 39 parts on its own.

Before you start you should be able to

  • Functional-group recognition and basic IUPAC naming of aliphatic compounds (this subject's own AS Intro to Organic Chemistry note)

  • σ and π bonds in the C=C double bond, and E/Z stereoisomerism from restricted rotation (this subject's own AS Hydrocarbons note)

  • The polar C=O bond (δ+\delta+ carbon, δ−\delta- oxygen) and nucleophilic addition to aldehydes/ketones (this subject's own AS Carbonyl Compounds note)

  • Nucleophilic substitution of halogenoalkanes, including curly-arrow conventions (this subject's own AS Halogen Compounds note)

  • Curly arrows, lone pairs and dipoles as mechanistic notation (all AS organic notes)

By the end of this page you can

  • Identify and name the functional groups newly examined at A Level — amide, nitrile, arene — alongside the AS set, and classify alcohols, amines and halides as primary/secondary/tertiary

  • Match functional groups to their characteristic tests: Na metal, Na₂CO₃(aq), 2,4-DNPH, I₂/OH⁻, Fehling's solution and Br₂(aq)

  • Name substituted arenes systematically using locants, including -CO₂H as benzoic acid/benzenecarboxylic acid and -NH₂ on the ring as phenylamine

  • Describe benzene's bonding completely: sp² hybridisation, 120° trigonal planar geometry, σ bonds from end-on-end overlap and a delocalised π system from sideways p-orbital overlap

  • Count sp, sp² and sp³ hybridised carbon atoms in any molecule, including rings, intermediates and drug structures

  • Draw the electrophilic substitution mechanism with curly arrows, the arenium intermediate and loss of H⁺, and track where each electron pair starts and ends

  • Draw the addition–elimination mechanism of acyl chlorides with water or amines, showing the tetrahedral intermediate and both collapses

  • Locate chiral centres in complex molecules, draw enantiomer pairs in wedge–dash 3D, and count possible optical isomers

  • Define optical activity, enantiomers and racemic mixtures, and predict what plane-polarised light does through pure and racemic samples

  • Discuss chirality in pharmaceutical synthesis: why single enantiomers are preferred, the costs of producing them, and how chiral catalysts or enzymes deliver them

01

Functional groups new at A Level — and classifying them

Syllabus requirement · §29

“

understand that the compounds in the table contain a functional group which dictates their physical and chemical properties · interpret and use the general, structural, displayed and skeletal formulas of those classes of compound

”

Every drug question opens here

Paper 4's organic questions nearly always begin the same way: a molecule you have never seen printed across the page — procaine, neotame, capsaicin, salbutamol — and part (a)(i), worth one or two marks, asks simply identify the functional groups. It looks like free money, and it is, provided you know exactly which labels exist at A Level and where examiners hide them. The same list then pays out again later in the question, because every synthesis route you are asked to plan runs on "which reagent attacks which group".

Three groups are NEW at A Level, and the syllabus table sets them beside their AS cousins:

  • arene — the benzene ring itself counts as a functional group; every printed hexagon deserves the label;
  • amide — −CONH2-\text{CONH}_2 / −CONR−-\text{CONR}-: a carbonyl whose carbon TOUCHES nitrogen directly;
  • nitrile — −C≡N-\text{C}\equiv\text{N}: its carbon COUNTS as a member of the parent chain when naming (§02 leans on this).

Beside them sit two near-neighbours from AS that A2 papers name constantly: phenol (an OH attached DIRECTLY to the ring) and acyl chloride (−COCl-\text{COCl}, the reactive partner in §05's mechanism).

Classification: always count CARBON GROUPS

Primary / secondary / tertiary is decided by counting how many carbon groups are attached to the atom that owns the functional group — never by counting hydrogen atoms:

familyclassify at this atomprimarysecondarytertiary
alcoholthe C bearing the −OH-\text{OH}ethanol CH3CH2OH\text{CH}_3\text{CH}_2\text{OH} (1 C neighbour)propan-2-ol (CH(OH)\text{CH}(\text{OH}) between 2 C)2-methylpropan-2-ol
aminethe NCH3NH2\text{CH}_3\text{NH}_2 (1 C on N)(CH3)2NH(\text{CH}_3)_2\text{NH} (2 C on N)(CH3)3N(\text{CH}_3)_3\text{N} (3 C on N)
halidethe C bearing the halogenCH3CH2Cl\text{CH}_3\text{CH}_2\text{Cl}2-chloropropane2-chloro-2-methylpropane

The amine row hides the classic trap: (CH3)2NH(\text{CH}_3)_2\text{NH} is secondary even though nitrogen still carries one H — the count is carbon groups, and it reaches two. Salbutamol questions (2025) set exactly this.

Notice how the marks scale in real questions. Neotame (2024): identify four functional groups — any two correct earned [1], all four [2]. So when a question offers two marks for identifications, keep hunting after your first answer; there is always another group hiding in a printed structure.

Why the look-alikes are different chemicals, not just different names

The labels exist because the chemistry differs, and knowing the chemistry makes the labelling automatic:

  • Amine vs amide: an amine's nitrogen owns a FREE lone pair, so it grabs H+\text{H}^+ from acids — amines are weak bases. In an amide that same lone pair is dragged into the neighbouring C=O\text{C}{=}\text{O} π system by delocalisation: the N goes planar, stops being basic, and the C–N bond shortens below a normal single bond. One atom rearrangement explains every amide/amine difference you will ever be asked.
  • Nitrile: the C≡N\text{C}\equiv\text{N} triple bond pulls electron density towards nitrogen, leaving the nitrile carbon δ+\delta+ and the molecule polar. Hydrolysing one converts it into a carboxylic acid — which is why its carbon belongs IN the parent chain when you name the acid it becomes.
  • Phenol vs alcohol: the benzene ring tugs on the oxygen's lone pair, polarising the O–H bond further than in an alcohol. Phenol therefore parts with H+\text{H}^+ far more readily — the reason the test table treats "OH compounds" in two separate rows.
new at A Level — and the look-alikes that decide the nameARENEthe ring itself is a functional groupNRNITRILEthe nitrile C counts in the chainORNAMIDEthe C=O carbon TOUCHES Na C=O group is named by what its carbon TOUCHES:RCOOHcarboxylic acidRCOOR′esterRCONH₂amideRNH₂(no C=O)amineclassification (primary / secondary / tertiary): count the CARBON GROUPS attachedalcohols — at the C bearing the OH · amines — at the N · halides — at the C bearing the halogen

The new-at-A-Level groups and their look-alikes. Top row: an ARENE is any benzene ring; a NITRILE's triple-bonded carbon counts as part of the chain; an AMIDE has its C=O carbon touching nitrogen directly. Bottom strip: carbonyl families are named by what the C=O carbon touches — O of an OH gives carboxylic acid, O of an OR gives ester, N gives amide, and an N with no carbonyl anywhere near it is just an amine. Classification (primary/secondary/tertiary) always means counting carbon groups.

reagent

a positive result tells you

Na metal (effervescence)

an O–H\text{O–H} group: alcohol OR carboxylic acid

Na2CO3\text{Na}_2\text{CO}_3(aq) (effervescence)

carboxylic acid ONLY (alcohols do not react)

2,4-DNPH (orange ppt)

carbonyl: aldehyde or ketone

Fehling's solution (red ppt)

aldehyde (ketones give nothing)

I2/OH−\text{I}_2/\text{OH}^- (yellow ppt)

CH3CO−\text{CH}_3\text{CO}- group or CH3CH(OH)−\text{CH}_3\text{CH(OH)}- group

Br2\text{Br}_2(aq) (decolourises)

alkene, C=C\text{C=C}

The six identification tests one 2025 paper matched for four marks (9701/44 M/J 2025 Q6(a)): any two rows [1], three [2], five [3], all six [4]. Learn them as reagent → conclusion pairs.

Identifying every functional group on a printed drug structure
  1. 1

    Sweep the structure left to right, atom by atom, marking every heteroatom (O, N, halogen) as you pass it.

    Random spotting misses groups; a fixed sweep does not. Missed groups in examiner reports were 'not seen', not 'misunderstood'.

  2. 2

    At every C=O, ask what its carbon TOUCHES: O of an OH → carboxylic acid · O of an OR → ester · N → amide · only C/H → aldehyde or ketone.

    This one question separates the whole carbonyl family — the exact discrimination procaine and neotame questions pay for.

  3. 3

    At every N, decide its neighbourhood: N with H sitting ON the ring → phenylamine · N between carbon groups with no adjacent C=O → amine (classify it) · N beside a C=O → part of an amide.

    Ring-NH₂ is credited specifically as 'phenylamine'; mislabelling an amide nitrogen as an amine is the most common error in this family.

  4. 4

    Collect the specials: hexagon = arene · OH directly on the hexagon = phenol · C≡N = nitrile · C=C = alkene.

    Arene and nitrile are the two most-forgotten labels precisely because they are drawn as background furniture.

  5. 5

    Re-read the marks before writing: 'any two [1]' means two suffice; 'all four [2]' means keep hunting until the list is complete.

    Writing early locks attention onto groups already found — finish the sweep first, THEN write.

The routine on invented molecules

4 marks

Identify the functional groups in each molecule.

(a) Molecule W: C6H5CH2CONHCH3\text{C}_6\text{H}_5\text{CH}_2\text{CONHCH}_3

(b) Molecule V: CH2=CHCN\text{CH}_2{=}\text{CHCN}

Show full working
  1. 1

    (a) Sweep W left to right: the C6H5\text{C}_6\text{H}_5 hexagon is an arene; next, the −CO–NH−-\text{CO–NH}- link has a carbonyl carbon touching nitrogen directly — an amide.

    The C=O question decides it: its carbon touches N, so the group is an amide — full stop.

  2. 2

    The −NH−-\text{NH}- inside that amide is NOT a separate amine: an amine needs its nitrogen clear of any adjacent C=O. W's list ends at arene + amide.

    The classic double-count is listing 'amide AND amine' for one nitrogen. One nitrogen carries one label — the amide absorbs it.

  3. 3

    (b) Sweep V: the C=C\text{C=C} makes it an alkene; the −C≡N-\text{C}\equiv\text{N} makes it a nitrile — two groups on a three-carbon molecule.

    Small molecules can carry several labels; V is exactly the lidocaine precursor examined in 2021, where both answers were required together for the marks.

Answer

(a) arene + amide (the −NH− belongs to the amide; no separate amine) · (b) alkene + nitrile

One nitrogen wearing a C=O hat is an amide only. Ask 'what does this carbonyl carbon touch?' every single time.

Procaine: three groups, one mark

9701/41 M/J 2022 Q7(a)(i)1 mark

Procaine is a local anaesthetic. Identify THREE functional groups in procaine.

Procaine as printed with the question: an aromatic ring carrying NH₂ at one end and an ester-linked diethylamino chain at the other.

Procaine as printed with the question: an aromatic ring carrying NH₂ at one end and an ester-linked diethylamino chain at the other.

Show full working
  1. 1

    Sweep: the −CO2C2H4−-\text{CO}_2\text{C}_2\text{H}_4- link is an ester — its carbonyl carbon touches O of an OR group.

    Ester before amine: the sweep hits the carbonyl first, and 'what does it touch' answers instantly.

  2. 2

    The NH2\text{NH}_2 attached directly to the ring is a phenylamine group; the −N(C2H5)2-\text{N}(\text{C}_2\text{H}_5)_2 at the far end is an amine.

    Mark scheme credits exactly: phenylamine AND amine AND ester. Two different nitrogen labels in one molecule — the ring one is named as phenylamine, not 'amino'.

Answer

ester · phenylamine · amine

'Three groups' with only two obvious candidates? The ring's NH₂ counts separately as PHENYLAMINE — that third label lives on the hexagon.

Neotame: four groups for two marks

9701/41 M/J 2024 Q8(a)(ii)2 marks

Neotame is an artificial sweetener. Identify FOUR functional groups present in neotame.

Neotame as printed with the question: two benzene rings joined through an amide-bearing backbone, with a methyl ester end and a free acid end.

Neotame as printed with the question: two benzene rings joined through an amide-bearing backbone, with a methyl ester end and a free acid end.

Show full working
  1. 1

    Carbonyls first: one −CO2CH3-\text{CO}_2\text{CH}_3 end is an ester; the other end, −CO2H-\text{CO}_2\text{H}, is a carboxylic acid — same carbon family, different neighbour (OR vs OH).

    'What does the C=O carbon touch' splits them in seconds; writing just 'carbonyl' twice earns nothing.

  2. 2

    Nitrogens next: the −CO–NH−-\text{CO–NH}- links along the backbone are amide groups; the lone N between two carbon chains (the 3,3-dimethylbutyl side) is an amine.

    Mark scheme: amide / amine / ester / carboxylic acid — any two [1], all four [2]. Four labels from four atoms, each decided by one neighbourhood question.

Answer

ester · carboxylic acid · amide · amine

Common mistakes

  • Listing an amide's −NH− as an amine as well

    Amide = carbonyl carbon touching N. That nitrogen is part of the amide; an AMINE nitrogen has no adjacent C=O.

    One nitrogen, one label. Neotame genuinely contains both — but on DIFFERENT nitrogens — which is exactly how the question catches the careless.

  • Classifying (CH3)2NH(\text{CH}_3)_2\text{NH} as primary because it still has an H on N

    Count carbon groups on N: two methyls → SECONDARY amine.

    The count is carbon groups, never hydrogens. Salbutamol's NH between two carbon chains is secondary for the same reason.

  • Forgetting the arene label (or writing 'benzene ring' but not naming it)

    The hexagon IS a functional group at A Level — write 'arene'. An OH sitting directly ON it is separately a phenol.

    Capsaicin's credited list was phenol + amide + C=C: the phenol mark exists only if you noticed the ring under the OH.

Your turn

Three real drug molecules — sweep each one completely before checking the solution.

  1. 19701/41 O/N 2023 Q8(a)1 mark

    Capsaicin is the pungent component of chilli peppers. Name each of the three functional groups labelled on its structure.

    Capsaicin as printed: a phenol ring bearing a long tail that carries an amide link and a terminal C=C.

    Capsaicin as printed: a phenol ring bearing a long tail that carries an amide link and a terminal C=C.

    Stuck? Show hint

    One label sits on the ring, one at a C=O touching N, one in the middle of the chain.

    Show solution
    1. 1

      The OH directly on the benzene ring = phenol. The −CO–NH−-\text{CO–NH}- of the tail = amide. The isolated C=C\text{C=C} in the chain = alkene.

      Mark scheme credits all three for the single mark: phenol, amide AND alkene/C=C. Three different neighbourhood questions, three labels.

    Answer

    phenol · amide · alkene (C=C)

  2. 29701/42 F/M 2021 Q7(c)(i)2 marks

    Two compounds used in a synthesis of lidocaine are

    CH2(CO2C2H5)2andCH2=CHCN\text{CH}_2(\text{CO}_2\text{C}_2\text{H}_5)_2 \quad\text{and}\quad \text{CH}_2{=}\text{CHCN}

    Identify THREE functional groups present in these two compounds.

    Stuck? Show hint

    Count what the first molecule's two C=O carbons touch, then read the second molecule left to right.

    Show solution
    1. 1

      CH2(CO2C2H5)2\text{CH}_2(\text{CO}_2\text{C}_2\text{H}_5)_2: both carbonyl carbons touch O of ethoxy groups → ester (two of them — 'diester' also credited).

      Same group twice still counts as one functional-group TYPE; the mark scheme accepts '(di)ester'.

    2. 2

      CH2=CHCN\text{CH}_2{=}\text{CHCN}: the C=C\text{C=C} is an alkene and the −C≡N-\text{C}\equiv\text{N} is a nitrile. All three types correct earned both marks.

      'All three correct for two marks' — miss the nitrile and you bank half. It hides behind the alkene precisely because it is drawn small.

    Answer

    ester · alkene · nitrile

  3. 39701/42 M/J 2025 Q6(f)(i)2 marks

    Salbutamol, an asthma drug, has the condensed structure

    HO–C6H3(CH2OH)−CH(OH)−CH2−NH−C(CH3)3\text{HO–C}_6\text{H}_3(\text{CH}_2\text{OH}){-}\text{CH(OH)}{-}\text{CH}_2{-}\text{NH}{-}\text{C}(\text{CH}_3)_3

    Classify EACH alcohol present, and classify the amine.

    Stuck? Show hint

    Alcohols classify at the carbon carrying the OH; the amine classifies by counting carbon groups on N.

    Show solution
    1. 1

      First alcohol: the −CH2OH-\text{CH}_2\text{OH} carbon touches only ONE other carbon → primary alcohol. Second: the −CH(OH)−-\text{CH(OH)}- carbon touches two carbons (ring and CH2\text{CH}_2) → secondary alcohol.

      Both alcohols must be classified SEPARATELY — 'contains alcohols' scores nothing; the paper wants primary AND secondary identified.

    2. 2

      The amine N sits between two carbon groups (CH2\text{CH}_2 and tert-butyl C) with one H remaining → secondary amine.

      The hydrogen-count trap from this section: one H does not make it primary. Mark scheme: any three of the six identifications [1], all six [2].

    Answer

    −CH₂OH: primary · −CH(OH)−: secondary · N between two carbon groups: secondary amine

Practise spotting and classifying functional groupsReal past-paper questions · New functional groups

The rest of this note

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Can you do all of these?

  • I can name every functional group on a given drug molecule, including amide vs amine and ester vs acid distinctions

  • I can classify an alcohol, amine or halide as primary, secondary or tertiary

  • I can state which functional group answers positive to each of Na, Na₂CO₃, 2,4-DNPH, I₂/OH⁻, Fehling's and Br₂(aq)

  • I can name di- and tri-substituted arenes with correct locants, and draw positional isomers

  • I can describe benzene's shape, hybridisation and orbital overlaps in the three-point form mark schemes credit

  • I can count sp/sp²/sp³ carbons in any structure, including intermediates and fused rings

  • I can draw the electrophilic substitution mechanism completely and say where each electron pair moves

  • I can draw the addition–elimination mechanism of an acyl chloride, intermediate included

  • I can asterisk every chiral centre in a complex molecule and draw a pair of enantiomers in 3D

  • I can define optical activity, enantiomers and racemic mixture in mark-scheme wording

  • I can predict optical rotation behaviour of pure enantiomers and racemates

  • I can argue the benefits and costs of single-enantiomer drugs and how chiral catalysts deliver them

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