Sustainable Chemical Manufacture
Sem 1 2026 · Side 1 of 2
Whole-unit revision · all topics
0 · Revision Blueprintread first
This unit has two halves that share one mindset: make the molecule (synthesis + polymers) and prove what you made (spectroscopy) — both judged against the green-chemistry yardstick of doing it with the least waste, energy and hazard.
Side 1 = analysis & sustainability metrics. Side 2 = reactions & polymers. The exam reflex you need: read a structure, predict its spectra; read spectra, deduce the structure; and for any route, judge its atom economy.
Most-tested skills: calculate atom economy / E-factor; assign an IR + MS + NMR set to one structure; pick Sₙ1 vs Sₙ2; classify a polymerisation; name a greener alternative (solvent, catalyst, feedstock).
1 · Green Chemistry · 12 PrinciplesAnastas & Warner 1998
The design framework for the whole unit. Memorise the mnemonic "PRODUCTIVELY" idea — but really know the high-yield five (★).
- Prevent waste — better than treating/cleaning it up
- Atom economy — maximise atoms of reactant in product
- Less hazardous synthesis (low toxicity to people/environment)
- Design safer chemicals (function with minimal toxicity)
- Safer solvents/auxiliaries — avoid where possible
- Design for energy efficiency — ambient T & P
- Renewable feedstocks not depleting ones
- Reduce derivatives (protecting groups add waste steps)
- Catalysis > stoichiometric reagents
- Design for degradation — break down after use, no persistence
- Real-time analysis to prevent pollution
- Inherently safer chemistry (accident prevention)
Prevention > remediation is the spine: principles 1, 2, 9 do most of the exam work.
1b · Worked · Atom Economytwo routes
Target = ethanol (C₂H₆O, Mr 46):
Hydration CH₂=CH₂ + H₂O → C₂H₅OH
AE = 46/(28+18) = 100% (addition, no by-product).
Fermentation C₆H₁₂O₆ → 2 C₂H₅OH + 2 CO₂
AE = (2·46)/180 = 51% — CO₂ is lost mass, but the feedstock is renewable.
⇒ greenness trades AE vs feedstock vs energy. Quote all three, don't crown a winner on AE alone.
2 · Sustainability Metricscalculate these
Yield tells you nothing about waste. A 100% yield reaction can still be wasteful if half the reactant mass ends up as by-product. That's why we measure atoms and mass.
Atom economy Trost 1991
% atom economyAE = (Mr desired product / Σ Mr all products) × 100
= Mr product / Σ Mr reactants × 100
Theoretical (uses the balanced equation, ignores yield). Addition & rearrangement → 100% AE; substitution & especially elimination/condensation lose atoms.
E-factor Sheldon
Environmental factorE = mass of waste / mass of product
ideal E = 0 (zero waste)
| Industry | Tonnage | E-factor |
|---|---|---|
| Bulk chems | 10⁴–10⁶ | <1–5 |
| Fine chems | 10²–10⁴ | 5–50 |
| Pharma | 10–10³ | 25–>100 |
Smaller-tonnage, higher-value products are the dirtiest per kg — pharma is the big target.
Process Mass Intensity
PMI & RMEPMI = total mass in / mass product (PMI = E + 1)
RME = mass product / Σ mass reactants × 100
PMI includes solvent & water — usually the largest mass in a process, which AE ignores.
3 · Feedstocksrenew vs deplete
Non-renewable: crude oil, natural gas, coal → cracked to platform chemicals (ethene, propene, BTX = benzene/toluene/xylene). Finite, CO₂-emitting.
Renewable: biomass — sugars, cellulose/lignocellulose, plant oils, terpenes. Bio-platform molecules: ethanol, lactic acid, succinic acid, HMF, glycerol (biodiesel by-product).
Tension: 1st-gen (food crops) vs 2nd-gen (waste lignocellulose) — the latter avoids the food-vs-fuel problem. Renewable ≠ automatically green (land, energy, processing all count).
3b · CO₂ & C1 Feedstockswaste → resource
Use abundant one-carbon sources as raw material: CO₂ → urea, cyclic carbonates, salicylic acid, and (with H₂) methanol; syngas (CO + H₂) → methanol, Fischer–Tropsch hydrocarbons.
Drop-in bio-chemicals = chemically identical to the fossil version (bio-ethene) ⇒ slot straight into existing plants. Novel bio-chemicals = new structures (lactic acid → PLA).
4 · Catalysis · Fundamentalsprinciple 9
A catalyst lowers Eₐ by providing an alternative pathway, is regenerated, and is not consumed. It speeds both directions equally — so it does not shift the equilibrium position or change ΔG/K; it only changes the rate.
Why it's green: replaces stoichiometric reagents (less waste), enables milder conditions (less energy), and improves selectivity.
Selectivity — 4 types
- Chemoselective — one functional group over another
- Regioselective — one position/orientation
- Stereoselective — one diastereomer
- Enantioselective — one enantiomer (asymmetric catalysis)
Homogeneous vs heterogeneous
| Homo | Hetero | |
|---|---|---|
| Phase | same as reactants | different (usually solid) |
| Selectivity | high, tunable | lower |
| Separation | hard (energy cost) | easy, recyclable |
| Example | Wilkinson's, Pd | Ziegler–Natta, zeolites |
Activity metrics
TurnoverTON = mol product / mol catalyst
TOF = TON / time = catalyst productivity rate
High TON ⇒ a little catalyst does a lot ⇒ greener.
5 · Catalysis · Typesknow examples
- Acid/base — protonation/deprotonation steps
- Organometallic — Pd/Rh/Ru; oxidative addition → migratory insertion → reductive elimination cycle
- Biocatalysis — enzymes; water, mild T/pH, exquisite enantioselectivity
- Organocatalysis — small organic molecules, metal-free (List & MacMillan, Nobel 2021)
- Photoredox — light drives single-electron steps
Asymmetric hydrogenation (chiral metal complex) and enzyme catalysis are the classic "how to make one enantiomer cleanly" answers.
5b · Worked · TON / TOFhow green
0.001 mol catalyst makes 0.8 mol product in 2 h:
TON = 0.8 / 0.001 = 800
TOF = 800 / 2 = 400 h⁻¹
Industrial/enzyme catalysts reach TON 10⁶–10⁹. Real TON is capped by deactivation (poisoning, leaching, sintering) — recyclability is the green prize.
6 · Structure Elucidationthe workflow
Combine the techniques — each answers a different question:
- MS → molecular formula & mass (what's the size?)
- IR → functional groups (what's attached?)
- ¹H/¹³C NMR → carbon–hydrogen skeleton (how connected?)
Step 1 always: DBE
Degrees of unsaturation (IHD/DBE)DBE = (2C + 2 + N − H − X) / 2
(O is ignored). Ring or π-bond = 1 each.
DBE ≥ 4 ⇒ suspect a benzene ring (3 C=C + 1 ring). C=O = 1, C≡N = 2, C≡C = 2.
7 · IR Spectroscopycm⁻¹
Bond stretching frequency ∝ √(k/μ): stronger bond & lighter atoms ⇒ higher cm⁻¹. Read the diagnostic 4000–1500 region; 1500–500 is the "fingerprint".
| Bond / group | cm⁻¹ | Note |
|---|---|---|
| O–H alcohol | 3200–3550 | broad |
| O–H carb. acid | 2500–3300 | v. broad |
| N–H amine/amide | 3300–3500 | 1–2 bands |
| C–H sp³ / sp² | <3000 / >3000 | key split |
| ≡C–H alkyne | ~3300 | sharp |
| C≡N / C≡C | 2210–2260 | weak/sharp |
| C=O | 1670–1750 | strong ★ |
| C=C alkene | 1620–1680 | weak |
| aromatic C=C | 1450–1600 | |
| C–O | 1000–1300 | strong |
C=O fine print: amide ~1650 < acid ~1710 ≈ ketone 1715 < aldehyde 1725 < ester 1735 < acyl chloride 1800. Conjugation lowers C=O by ~30.
7b · Reading an IR Spectrum3 questions
- C=O? strong, sharp 1650–1750 ⇒ carbonyl present
- Broad O–H / N–H? 2500–3550 ⇒ acid/alcohol/amine
- C–H above or below 3000? ⇒ sp²/aromatic vs sp³
Tell-tales: aldehyde = C=O + twin C–H (Fermi) ~2720/2820; nitrile = sharp 2250; anhydride = two C=O bands.
7c · Distinguish by IRisomer trap
| Group | O–H/N–H | C=O |
|---|---|---|
| Carb. acid | broad 2500–3300 | ~1710 |
| Ester | none | 1735 + strong C–O |
| Ketone | none | ~1715 only |
| Amide | N–H ~3300 | ~1650 |
| Alcohol | broad ~3300 | none |
8 · Mass Spectrometrym/z
M⁺• molecular ion = relative molecular mass. Then bonds fragment to give a pattern of cations.
Diagnostic rules
- Nitrogen rule: odd M⁺ ⇒ odd number of N atoms
- M+1 ≈ 1.1% × (number of C) — counts carbons
- Cl: M:M+2 ≈ 3 : 1
- Br: M:M+2 ≈ 1 : 1
Common neutral losses
| Δm | Lost | Implies |
|---|---|---|
| 15 | •CH₃ | methyl |
| 18 | H₂O | alcohol |
| 28 | CO / C₂H₄ | carbonyl |
| 29 | CHO / C₂H₅ | aldehyde |
| 31 | •OCH₃ | methyl ester |
| 45 | COOH | carb. acid |
Key fragments: m/z 43 = CH₃CO⁺ (acylium), 91 = tropylium (benzylic), 77 = phenyl. α-cleavage & McLafferty (γ-H transfer, loses an alkene from carbonyls) are the named fragmentations.
9 · ¹H NMRδ / ppm
Three readings: shift δ (environment), integration (relative # H), multiplicity (neighbours).
Splittingn equivalent neighbours ⇒ n + 1 peaks
J (coupling constant, Hz) shared by coupled partners
| Proton | δ ppm |
|---|---|
| TMS (ref) | 0 |
| R–CH₃ / CH₂ / CH | 0.8–1.7 |
| C–H α to C=O | 2.0–2.6 |
| C–H next to N | 2.2–2.9 |
| C–H next to O / X | 3.3–4.5 |
| alkene =C–H | 4.5–6.5 |
| aromatic Ar–H | 6.5–8.0 |
| aldehyde CHO | 9.5–10 |
| carb. acid COOH | 10–12 |
OH/NH = broad, variable, exchange with D₂O (peak vanishes ⇒ confirms OH/NH).
9b · Coupling Patternsrecognise instantly
| Group | Pattern |
|---|---|
| –CH₂CH₃ (ethyl) | triplet 3H + quartet 2H |
| (CH₃)₂CH– (iPr) | doublet 6H + septet 1H |
| –OCH₃ / isolated CH₃ | singlet (no neighbours) |
| para-disubst. ring | 2 doublets (AA′BB′) |
Equivalent H (by symmetry) = one signal & don't split each other. n+1 counts H on adjacent carbons only.
10 · ¹³C NMR + DEPTδ / ppm
Counts unique carbons (no integration; ¹H-decoupled = singlets). Wide 0–220 ppm range.
| Carbon | δ ppm |
|---|---|
| alkyl C | 0–50 |
| C–N | 30–60 |
| C–O | 50–90 |
| alkyne C | 65–90 |
| alkene / aromatic | 100–150 |
| C=O acid/ester/amide | 160–185 |
| C=O ketone/aldehyde | 190–220 |
DEPT
Edits by # attached H: CH & CH₃ up, CH₂ down, quaternary C absent (the give-away for C=O & substituted aromatic).
Counting signals = symmetry
# of signals = # of chemically distinct environments. Benzene = 1 ¹³C; para-disubstituted ring = 4 (two pairs equivalent). Fewer signals than carbons ⇒ symmetry — a fast structural clue.
11 · Worked · C₄H₈Oput it together
MS M⁺ = 72. DBE = (2·4+2−8)/2 = 1 ⇒ one C=O or C=C/ring.
IR 1715 cm⁻¹ strong ⇒ ketone C=O (no broad O–H, no ~2720 aldehyde C–H, no 1735 ester).
¹H NMR triplet 1.0 (3H), quartet 2.4 (2H), singlet 2.1 (3H) ⇒ ethyl + isolated methyl on C=O.
⇒ butan-2-one, CH₃COCH₂CH₃. MS loss of 15 (→57) and 29 (→43, CH₃CO⁺) confirm α-cleavage either side of C=O.
11b · Worked · C₈H₈Ospot the ring
DBE = (2·8+2−8)/2 = 5 ⇒ benzene ring (4) + one more (a C=O).
IR 1685 (conjugated C=O). ¹H: 5H multiplet 7.4–8.0 (mono-substituted ring) + singlet 2.6 (3H, CH₃CO).
⇒ acetophenone, C₆H₅COCH₃. ¹³C ~198 (C=O); m/z 105 (PhCO⁺) & 77 (C₆H₅⁺) confirm.
Which Technique?recap
| Question | Use |
|---|---|
| Mass / formula | MS (M+1, isotopes) |
| Functional groups? | IR (C=O, O–H, N–H) |
| # unique C; C type | ¹³C + DEPT |
| H count & connectivity | ¹H (δ, integ, J) |
| Rings / π count | DBE from formula |
Formula Beltside 1
AE = Mr(prod)/ΣMr(react) ×100
E = waste / product · PMI = E+1
DBE = (2C+2+N−H−X)/2
splitting = n+1 · M+1 ≈ 1.1%·nC
Cl 3:1 · Br 1:1 (M:M+2)