Year 12

AQA A level Chemistry Analytical Organic Chemistry Notes

Analytical Organic Chemistry

— Chemistry · AS Level (7405) · Crash Course —
★ Exam Tip: Analytical = IR spectroscopy + mass spectrometry, usually as the last part of a long Paper 4-style/structured question. The IR table is printed in the paper — you just need to spot C=O (~1700) and O–H (2500–3600). In mass spec, learn the M+1 carbon-count formula and the M+2 patterns for Cl (3 : 1) and Br (1 : 1). Sir: “the examiner will give enough information and only ask what is doable — don’t worry.”

1. Infrared (IR) Spectroscopy

  • Different bonds absorb different infrared frequencies (they make the bond vibrate).
  • Shine a range of IR frequencies through a sample, put a sensor behind it, and measure how much of each frequency is transmitted (% transmittance).
  • Absorption = a trough (dip) in the plot; the position of the dip (wavenumber, cm⁻¹) tells you which bond is present — so the plot identifies the bonds in an unknown sample.
% Twavenumber (cm⁻¹), high → low4000300017001000O–H troughC=O troughpeaks point DOWN (absorption)
in IR spectra, wavenumber decreases left → right and each absorption is a downward trough
Exam note: The data table (“Characteristic infrared absorption frequencies”) will be given in the paper where needed — you do not have to memorise it. But you should recognise the starred ones: C=C, C=O, N–H, O–H.
Bond Functional groups Range / cm⁻¹ Peak
C–O alcohols, ethers, esters 1040–1300 strong
C=C aromatic compounds, alkenes 1500–1680 weak unless conjugated
C=O amides 1640–1690 · ketones & aldehydes 1670–1740 · esters 1710–1750 1640–1750 strong
C≡C alkynes 2150–2250 weak unless conjugated
C≡N nitriles 2200–2250 weak
C–H alkanes (–CH₂–H) 2850–2950 · alkenes/arenes (=C–H) 3000–3100 2850–3100 strong / weak
N–H amines, amides 3300–3500 weak
O–H carboxylic acids RCO₂–H 2500–3000 · H-bonded alcohol RO–H 3200–3600 · free alcohol 3580–3650 2500–3650 strong; carboxylic acid very broad; free alcohol sharp

Easily recognised groups — always look for these two first

  1. Focus at ≈1700 cm⁻¹ → C=O (carbonyl: aldehyde, ketone, carboxylic acid, ester, amide — a sharp, strong dip).
  2. Focus on 2500–3600 cm⁻¹ → O–H: carboxylic acid 2500–3000 (very broad, overlaps C–H); alcohol 3200–3600 (broad, centred ≈3300).
ethanol CH₃CH₂OHbig broad dip 3200–3600= O–H of an alcohol(no C=O dip at 1700)ethanoic acid CH₃CO₂Hvery broad 2500–3000 (O–H)+ sharp C=O ≈ 1700both present ⇒ acidethyl ethanoate (ester)C=O ≈ 1700, strongNO broad O–H ⇒ not acid/alcohol(C–O 1040–1300 also)propanone (ketone)C=O ≈ 1700 onlyno O–H(aldehyde looks the same)
how the four board spectra differ: look for O–H first, then C=O
Spectrum What you see Conclusion
2-hydroxypropanoic acid (lactic acid) very broad dip 2500–3500 = O–H of alcohol + O–H of carboxylic acid merged; C=O at 1700 has both –OH and –COOH
propan-1-ol broad O–H 3200–3600 (strong); C–H just below 3000; C–O ≈ 1050 alcohol only; no C=O
Practical tips: Sir: the board graphs are taken from the internet, so some look slightly conflicting (e.g. the O–H dip of the acid looks broader than expected). Don’t worry — if you forget the numbers, just remember: find C=O and O–H first. Also, a C–H absorption (2850–3100) is present in almost every organic compound, so ignore it unless asked. Questions often say “referring only to absorptions above 1500 cm⁻¹” so you avoid the busy fingerprint region.

2. Mass Spectrometry

From the mass spectrum of an organic sample you can find out: ① its Mr; ② the number of C atoms; ③ whether the compound contains Cl or Br; ④ fragments (clues to structure).

① Mr and number of carbon atoms

  • The molecular ion peak (M peak) — the peak with the highest m/z (rightmost main peak), M⁺ = the molecule minus one electron — gives Mr (e.g. hexane C₆H₁₄⁺: m/z = 86; octane: 114). Everything to its left are fragments.
  • The small peak one unit to the right is the M+1 peak: it is there because ~1.1 % of carbon is ¹³C. A molecule with n carbons has a 1.1 × n % chance of containing one ¹³C.
M+1 peak (%) / M peak (%) = 1.1 n / 100 ⇒ n = 100 × (M+1 abundance)1.1 × (M abundance)
Example Abundances Working Result
hexane (M = 86) M 22 : M+1 1.45 100 / 22 ⇒ 1.1 n = 1.45 × (100/22) = 6.6; n = 5.99 ≈ 6 C C₆H₁₄ ✓
octane (M = 114) M 23 : M+1 2 n = (2 × 100) / (23 × 1.1) = 7.9 ≈ 8 C C₈H₁₈ ✓
Board method (hexane):M : M+1 = 100 : 1.1n22 : 1.45 (cross-multiply)1.45 × 100 = 22 × 1.1n ⇒ n = 5.99round to nearest whole number ⇒ 6 C atoms
Remember: Always round to the nearest whole number (n = 5.99 → 6, 7.9 → 8). The M peak contains only ¹²C, ³⁵Cl (lightest isotopes) — so do Mr calculations on the M peak. Don’t use the M+1 peak for Mr.

② Does it contain Cl or Br? — the M+2 peak

Chlorine has isotopes ³⁵Cl and ³⁷Cl; bromine has ⁷⁹Br and ⁸¹Br. A compound with Cl or Br gives an M+2 peak in addition to M.

Halogen Isotopes Peak ratio
Cl present ³⁵Cl : ³⁷Cl ≈ 3 : 1 M : M+2 = 3 : 1
Br present ⁷⁹Br : ⁸¹Br ≈ 1 : 1 M : M+2 = 1 : 1 (two equal peaks)
MM+2Cl: 3 : 1MM+2Br: 1 : 1
chlorine: M is 3× the M+2 peak; bromine: the two peaks are equal

Worked example (board)

M peak 134 (20 %), peaks at 135 (1.54 %) and 136 (6.67 %)compound CxHyX, X = halogen — find the exact molecular formula
M : M+1 = 20 : 1.54 ⇒ 1.1n = (1.54 × 100)/20 = 7.7n = 7 C atomsM : M+2 = 20 : 6.67 ≈ 3 : 1 ⇒ Cl (³⁵Cl)Mr − Cl = 134 − 35 = 99H atoms = 99 − 7(12) = 15molecular formula = C₇H₁₅Cl
Method: Sir’s mathematical method: when you know the number of C atoms and the halogen, the number of hydrogens is simply (Mr − halogen − 12 × n). If the compound also contains oxygen, subtract 16 first.

③ Fragmentation

  • The molecular ion is unstable and can break into smaller pieces: M⁺ → X⁺ + Y•.
  • Upon fragmentation the compound breaks into a positive ion and a free radical: only the positive ion X⁺ is detected (produces a line); the radical Y• is neutral and does not produce a line.
  • Many fragmentations are possible, so there are many lines; the tallest line = the base peak (most stable fragment).
[M]⁺X⁺ → makes a lineY• → no line (neutral)
charged fragments are detected; radicals are not

Naming the fragments: m/z → formula

Pentane CH₃CH₂CH₂CH₂CH₃, Mr = 72m/z 29: 12 × 2 = 24; 29 − 24 = 5 H ⇒ [CH₃CH₂]⁺[CH₃CH₂CH₂CH₂CH₃]⁺ → [CH₃CH₂]⁺ + [CH₃CH₂CH₂]•m/z 43: 12 × 3 = 36; 43 − 36 = 7 H ⇒ [CH₃CH₂CH₂]⁺[CH₃CH₂CH₂CH₂CH₃]⁺ → [CH₃CH₂CH₂]⁺ + [CH₃CH₂]•m/z 57: 12 × 4 = 48; 57 − 48 = 9 H ⇒ [CH₃CH₂CH₂CH₂]⁺[CH₃CH₂CH₂CH₂CH₃]⁺ → [CH₃CH₂CH₂CH₂]⁺ + [CH₃]•
Pentan-3-one CH₃CH₂COCH₂CH₃ (Mr = 86)m/z 57: (12 × 3) + 5 + 16 = 57 ⇒ [CH₃CH₂C=O]⁺[CH₃CH₂COCH₂CH₃]⁺ → [CH₃CH₂CO]⁺ + [CH₂CH₃]•m/z 29: (12 × 2) + 5 = 29 ⇒ [CH₃CH₂]⁺[CH₃CH₂COCH₂CH₃]⁺ → [CH₃CH₂]⁺ + [CH₃CH₂CO]•
Exam checklist: Method: write the m/z, subtract 12 for each C you assume (and 16 for each O), the remainder is the number of H. Always give the + charge on the fragment (put it in square brackets with +) and a • on the radical. The examiner will give enough information — only ask doable fragments (C–C and C–CO cleavage), don’t worry.
29435772 (M)CH₃CH₂⁺CH₃CH₂CH₂⁺ (base peak)C₄H₉⁺C₅H₁₂⁺
pentane mass spectrum: M = 72; fragments 57, 43 (tallest), 29 — the fragments are the lines to the left of M

3. Using Spectra with Chemical Tests

Distinguish Reagent Positive result
Aldehyde vs ketone Tollens’ reagent silver mirror with the aldehyde
Methyl ketone vs other ketone I₂(aq) + NaOH(aq) yellow ppt (CHI₃) with the methyl ketone
Ketone vs ester 2,4-DNPH orange/yellow ppt with the ketone (C=O) only
Alkene vs tertiary alcohol aqueous Br₂ brown → colourless with the alkene (primary alcohol + alkene D2)
Carboxylic acid Na₂CO₃(aq) effervescence of CO₂
Aldehyde (Fehling’s) warm Fehling’s orange/brick-red ppt

4. Long CIE-style Questions (board)

CIE Q3 · distinguishing compounds + IR/MSTable 3.1 and spectra of E and F

(a)(i) A1 propanal / A2 propanone: Tollens’ reagent — silver mirror with A1. B1 (methyl ketone) / B2 pentan-3-one: I₂(aq) + NaOH(aq) — yellow ppt with B1. C1 ketone / C2 ester: 2,4-DNPH — yellow ppt with C1. D1 tertiary alcohol (1-methylcyclobutanol) / D2 but-3-en-1-ol: aqueous Br₂ — brown → colourless with D2.

(ii) structural isomers = compounds having the same molecular formula but different structural formula. (iii) B2 = pentan-3-one. (iv) D1 = C₅H₁₀O. (b)(i) D2 ⇒ addition polymer. (ii) one repeat unit: –CH₂–CH(CH₂CH₂OH)– (continuation bonds both ends).

(c) E has three C atoms, reacts with cold dilute acidified KMnO₄ to give one compound F, Mr = 154.9.

(i) E’s IR (above 1500 cm⁻¹): absorption between 1500–1680 ⇒ C=C(ii) F’s IR: absorption at 3200–3650 ⇒ alcohol (O–H)(iii) M and M+2 equal ⇒ Br (⁷⁹Br/⁸¹Br); m/z 120 and 122 ⇒ Mr(E) = 120(iv) C₃H₅Br: E = CH₂=CH–CH₂Br (3-bromoprop-1-ene)(v) CH₂=CHCH₂Br + [O] + H₂O → CH₂(OH)CH(OH)CH₂Br

(d) C2 (propyl propanoate) from A1 (propanal), a multi-step route: Step 1 reduce a sample of A1 with LiAlH₄ in dry ether ⇒ CH₃CH₂CH₂OH. Step 2 oxidise another sample of A1 by heating under reflux with acidified K₂Cr₂O₇ ⇒ CH₃CH₂COOH (distil off). Step 3 mix the alcohol and the acid with a few drops of conc. H₂SO₄ and warm gently ⇒ ester C2 (propyl propanoate).

CIE Q5 · lactonescyclic esters; mass spec of lactone Q

(a)(i) M (the ring alkene, 1-methylcyclopentene, as the formula of N implies — the board’s drawing of M is hard to read) + hot conc. acidified KMnO₄ ⇒ N, C₆H₁₀O₃ = HOOC–CH₂CH₂CH₂–CO–CH₃: the ring C=C breaks, one carbon becomes a ketone (C bonded to 2 alkyl groups) and the other a COOH. (ii) N + NaBH₄ → 5-hydroxyhexanoic acid: C₆H₁₀O₃ + 2[H] → C₆H₁₂O₃. (iii) reaction 2 is nucleophilic addition: the C=O of N is planar so the hydride can attack from either side ⇒ 2 optical isomers (racemic mixture). (iv) P = lactone: ring closure by loss of H₂O between OH and COOH — a 6-membered ring with a CH₃ on the carbon next to ring O.

(b) IR cannot distinguish N from 5-hydroxyhexanoic acid: except for the C=O and O–H (alcohol) absorptions, all other bonds are similar and the absorptions overlap (both have C=O, O–H of the acid, C–O, C–H).

(c) Q: M 72 (95.5), M+1 73 (3.15)M : M+1 = 100 : 1.1n ⇒ 95.5 : 3.15n = (3.15 × 100) / (95.5 × 1.1) = 2.99 ≈ 3 CMr 72, cyclic ester ⇒ C₃H₄O₂ = β-propiolactone (4-membered ring: C=O, O, CH₂, CH₂)
CIE Q4 · compound V (liquid)77.2 % C, 11.4 % H, 11.4 % O; Mr = 280
(a) C : H : O = 77.2/12 : 11.4/1 : 11.4/16= 6.433 : 11.4 : 0.7125÷ 0.7125 ⇒ 9 : 16 : 1 (empirical C₉H₁₆O, mass 140)n = 280/140 = 2 ⇒ C₁₈H₃₂O₂

(b) V has a carboxylic acid and an alkene. (i) test for –COOH: add Na₂CO₃(aq) ⇒ effervescence of CO₂ gas.

(ii) n(V) = 2.800 / 280 = 0.0100 moln(Br₂) = 3.196 / (2 × 79.9) = 0.0200 mol1 mol V reacts with 2 mol Br₂ ⇒ 2 alkene groups

(c) W, X, Y all C₅H₁₀O, all give an orange 2,4-DNPH ppt: common group = carbonyl. W + Fehling’s ⇒ orange-red ppt ⇒ aldehyde; X + alkaline I₂ ⇒ yellow ppt CHI₃ ⇒ methyl ketone, no Fehling’s; Y: ketone, not a methyl ketone (= pentan-3-one). Two skeletal isomers of W: pentanal CH₃(CH₂)₃CHO and 2-methylbutanal (chain isomerism).

(d) ketone Z, C₅H₁₀O, mass spectrum M = 86; peaks at 29 and 57: m/z 29 = [CH₃CH₂]⁺, m/z 57 = [CH₃CH₂CO]⁺ ⇒ Z = pentan-3-one.

✓ Quick Recap — Analytical
  • IR: bonds absorb characteristic frequencies (dips). Look first for C=O ≈ 1700 and O–H (alcohol 3200–3600, acid 2500–3000 very broad). C=C 1500–1680 (weak). Table given in the paper.
  • MS: M peak = Mr (rightmost main peak); M+1 from ¹³C: n = 100 × (M+1)/(1.1 × M); round to nearest whole number.
  • M+2: Cl ⇒ M : M+2 = 3 : 1; Br ⇒ 1 : 1. Use the M peak for formula work: H = Mr − halogen − 12n (− 16 per O).
  • Fragmentation: M⁺ → X⁺ + Y•; only X⁺ gives a line. m/z → formula: subtract 12 per C (16 per O), remainder = H. Write [ ]⁺ and •.
  • Combine spectra with chemical tests (Tollens’, Fehling’s, I₂/NaOH, 2,4-DNPH, Br₂, Na₂CO₃) to deduce structures; give formulae/isomer checks.