Topic 20

Topic 20 on a page

Everything from this topic in one place, for the last read before an exam. If anything here is unfamiliar, follow the link back to the full page.

The one idea

Every organic question in this topic is evidence, then deduction, or reagent, conditions, product. Identify a compound from data, build a longer chain with a Grignard reagent or cyanide, plan up to four steps by working backwards, then choose a procedure that suits the properties of the substances.

Numbers and facts to know

Quantity or fact Value
n(C) from combustion mass of CO₂ ÷ 44.0
n(H) from combustion 2 × (mass of H₂O ÷ 18.0)
oxygen in a compound by difference
molecular formula n = MrM_\text{r} ÷ empirical mass; multiply the empirical formula by n
C=O in IR 1680–1750 cm⁻¹, strong and sharp
O–H in IR alcohol 3230–3550 (broad); acid 2500–3300 (very broad)
TMS δ = 0
number of ¹H lines n + 1
carbon chain lengthened by one carbon KCN; HCN; CO₂ with a Grignard reagent
aspirin melting temperature about 136 °C
aspirin MrM_\text{r} 180.0 (2-hydroxybenzoic acid 138.0)

Definitions to learn word for word

Tests for functional groups

Reagent Positive result Shows
bromine water orange to colourless C=C
PCl₅ steamy white fumes –OH
Na₂CO₃(aq) fizzing; limewater milky carboxylic acid
acidified K₂Cr₂O₇ orange to green primary or secondary alcohol, aldehyde
2,4-dinitrophenylhydrazine orange precipitate C=O
Tollens' reagent silver mirror aldehyde
iodine, NaOH(aq) pale yellow precipitate CH₃CO– or CH₃CH(OH)–
NaOH(aq), HNO₃, AgNO₃ white, cream or yellow precipitate C–Cl, C–Br, C–I

Tests for functional groups.

Spectra

Spectrum What it gives
infrared bonds present; look for C=O and O–H, say what is absent
mass M⁺ gives MrM_\text{r}; fragments are positive ions (15, 29, 43 ...)
¹³C NMR number of peaks = carbon environments; C=O above 160
¹H NMR peaks = H environments; area = ratio; n + 1 splitting
A decision tree for an unknown compound containing C, H and O. Add sodium carbonate: fizzing means a carboxylic acid. No fizzing: add 2,4-dinitrophenylhydrazine. An orange precipitate means a carbonyl compound, then Tollens' reagent separates an aldehyde, which gives a silver mirror, from a ketone, which does not. No precipitate: add PCl5. Steamy fumes mean an alcohol, then acidified dichromate separates a primary or secondary alcohol, which turns green, from a tertiary alcohol, which stays orange.
A route through the tests for a compound of C, H and O.

Infrared and mass spectra, NMR shifts and areas, NMR splitting and solving structures.

Grignard reagents

Grignard and carbonyls.

Planning a route

  1. Compare the carbon count and the functional groups of the start and the target.
  2. If the chain must grow, choose KCN, HCN, or a Grignard reagent.
  3. Work backwards from the target. Name every intermediate.
  4. Write each step as reagent, conditions, product.

The reaction map and Planning a route.

Techniques and checks

Technique Use Check
reflux slow reactions with volatile or flammable liquids water in at the bottom; top open
wash with Na₂CO₃(aq) removes acid impurities; releases CO₂ open the tap to vent
solvent extraction takes a product out of water several small portions
drying removes water clear liquid; solid no longer clumps
distillation separates liquids by boiling temperature thermometer bulb level with the side arm
steam distillation volatile, immiscible compound that decomposes when hot distillate is two layers
recrystallisation purifies a solid minimum hot solvent; wash ice-cold
melting temperature purity of a solid sharp and matches the data book
boiling temperature purity of a liquid steady and matches the data book

The ten traps

  1. Forgetting that H₂O contains two H, so n(H) = 2 × n(H₂O).
  2. Rounding a ratio such as 1 : 1.5 instead of multiplying by 2.
  3. Counting every carbon as a peak in ¹³C NMR, not every environment.
  4. Counting the hydrogens on the same carbon when using the n + 1 rule.
  5. Writing a mass-spectrum fragment without its +.
  6. "Ether" without dry, and leaving out the dilute acid step with a Grignard reagent.
  7. Using ammonia to add a carbon (use KCN), or inventing a step that is not on the maps.
  8. Sealing the top of a reflux condenser, or water in at the top.
  9. Saying that recrystallisation raises the yield. It raises the purity.
  10. A control measure that does not match the hazard: "be careful", or goggles for a toxic vapour.

Mixed test

Try these with no notes, then check the answers.

  1. 0.690 g of a compound containing C, H and O burns to give 1.32 g of CO₂ and 0.810 g of H₂O. Its MrM_\text{r} is 46. Find its molecular formula.
  2. A compound C₃H₆O gives an orange precipitate with 2,4-dinitrophenylhydrazine and no silver mirror with Tollens' reagent. Its proton NMR spectrum has one peak. Identify it.
  3. Predict the splitting and area of each peak in the proton NMR spectrum of propan-2-ol.
  4. Give the reagents and conditions to convert 1-bromopropane into butanoic acid, naming each intermediate.
  5. Plan a three-step synthesis of propylamine from ethene.
  6. A crude liquid product contains acid and water. List the steps to purify it, in order, with the reason for each.
  7. 4.14 g of 2-hydroxybenzoic acid gives 3.78 g of aspirin. Calculate the percentage yield.
  8. Ethoxyethane is highly flammable. Potassium cyanide is very toxic and releases HCN with acid. Give one control measure for each hazard.

Answers cover these until you have tried the questions

  1. n(CO₂) = 1.32 ÷ 44.0 = 0.0300 mol, so n(C) = 0.0300 mol (0.360 g). n(H₂O) = 0.810 ÷ 18.0 = 0.0450 mol, so n(H) = 0.0900 mol (0.0900 g). Mass of O = 0.690 − 0.360 − 0.0900 = 0.240 g, so n(O) = 0.0150 mol. Ratio C : H : O = 2 : 6 : 1, so the empirical formula is C₂H₆O (46.0). Since MrM_\text{r} = 46, the molecular formula is C₂H₆O.
  2. The orange precipitate shows a C=O. No silver mirror shows it is a ketone. One proton environment fits only propanone, CH₃COCH₃.
  3. CH₃ groups (6 H): doublet, one H on the neighbouring carbon. CH (1 H): septet, six H on the neighbouring carbons. OH (1 H): singlet. The area ratio is 6 : 1 : 1.
  4. Mg in dry ether gives CH₃CH₂CH₂MgBr (propylmagnesium bromide). CO₂ in dry ether gives CH₃CH₂CH₂COOMgBr. Dilute acid gives butanoic acid, CH₃CH₂CH₂COOH.
  5. Ethene with HBr at room temperature gives bromoethane. Bromoethane with KCN in ethanol, heated under reflux, gives propanenitrile. Propanenitrile with LiAlH₄ in dry ether gives propylamine.
  6. Wash with sodium carbonate solution in a separating funnel, venting the CO₂, to remove the acid. Wash with water to remove salts. Dry with an anhydrous salt until the liquid is clear and the solid no longer clumps, to remove the water. Distil, collecting the fraction at the product's boiling temperature, to separate it from other liquids.
  7. n(acid) = 4.14 ÷ 138.0 = 0.0300 mol. Theoretical mass of aspirin = 0.0300 × 180.0 = 5.40 g. Percentage yield = 3.78 ÷ 5.40 × 100 = 70.0%.
  8. Ethoxyethane: heat with a water bath or electric heater, with no naked flames. Potassium cyanide: use a fume cupboard and keep it away from acids.