Topic 20 · page 8 of 15
Unfamiliar compounds
A compound you have never met is a set of groups you have. Find each functional group, predict what each one does on its own, and then say what the compound will do in terms of those groups.
Read first: Tests for functional groups
Break the molecule into groups 20.3(i)
Exam questions often give a compound with a long name that you have not studied, for example a drug or a natural product. Nothing in it is new. Every functional group in it is in the specification, and each group behaves as it does in a simple molecule. Treat the compound as a collection of groups.
MethodPredicting the properties of an unfamiliar compound
- Circle every functional group and name it.
- For each reagent, ask which group reacts and what the product is.
- Write the observation or the property, and give the group as the reason.
- Check each group in turn. A reagent may react with more than one.
The groups at a glance 20.3(i)
| Group | What it does | Typical reagent and result |
|---|---|---|
| alkene, C=C | electrophilic addition | bromine water decolourised; HBr; H₂ with nickel; steam; addition polymers |
| halogenoalkane, C–X | nucleophilic substitution, elimination | NaOH(aq) gives an alcohol; KCN a nitrile; NH₃ an amine; KOH in ethanol an alkene |
| alcohol, –OH | oxidation, dehydration, esterification | dichromate turns green (primary, secondary); PCl₅ fumes; acid and H₂SO₄ gives an ester |
| aldehyde, ketone, C=O | nucleophilic addition, reduction | HCN gives a hydroxynitrile; LiAlH₄ gives an alcohol; 2,4-dinitrophenylhydrazine gives a precipitate; only aldehydes oxidise |
| carboxylic acid, –COOH | acid, esterification, reduction | Na₂CO₃ fizzes; alcohol and H₂SO₄ gives an ester; PCl₅ gives an acyl chloride |
| acyl chloride, –COCl | acylation | water, alcohol, ammonia and amines react at room temperature |
| ester, –COO– | hydrolysis | dilute acid or alkali, heat under reflux, gives acid and alcohol |
| nitrile, –C≡N | hydrolysis, reduction | dilute HCl and heat give an acid; LiAlH₄ gives an amine |
| amine, –NH₂ | base, nucleophile, ligand | accepts H⁺ to form a salt; attacks halogenoalkanes and acyl chlorides |
| amide, –CONH– | made from acyl chloride | not basic; part of polyamides and proteins |
| arene, benzene ring | electrophilic substitution | nitration, halogenation, Friedel–Crafts; does not decolourise bromine water |
| phenol, –OH on a ring | reacts with bromine water | white precipitate of tribromophenol |
A worked example 20.3(i)
Worked example2-hydroxypropanoic acid (lactic acid)
2-Hydroxypropanoic acid, CH₃CH(OH)COOH, has a secondary alcohol group, –CH(OH)–, and a carboxylic acid group, –COOH. Predict its reactions.
| Reagent | Prediction | Because |
|---|---|---|
| Na₂CO₃(aq) | effervescence; CO₂ forms | the –COOH group is acidic |
| PCl₅ | steamy fumes of HCl | it has –OH groups (two, so both are replaced by Cl) |
| acidified K₂Cr₂O₇, heat | orange to green; CH₃COCOOH forms | a secondary alcohol is oxidised to a ketone |
| ethanol and concentrated H₂SO₄, heat | an ester, CH₃CH(OH)COOCH₂CH₃, forms | –COOH reacts with an alcohol |
| LiAlH₄ in dry ether | propane-1,2-diol, CH₃CH(OH)CH₂OH | the –COOH is reduced to –CH₂OH |
| iodine and NaOH(aq), warm | pale yellow precipitate of CHI₃ | the compound contains CH₃CH(OH)– |
| water | very soluble | –OH and –COOH groups hydrogen bond to water |
Worked exampleA hydroxy-aldehyde
Illustrative compound. 3-Hydroxybutanal, CH₃CH(OH)CH₂CHO, has a secondary alcohol and an aldehyde. Predict four reactions.
- Tollens' reagent: a silver mirror, because of the aldehyde.
- 2,4-dinitrophenylhydrazine: an orange precipitate, because of the C=O.
- Acidified dichromate: orange to green. Both groups are oxidised: the aldehyde to a carboxylic acid and the secondary alcohol to a ketone.
- Warm with iodine and NaOH(aq): a pale yellow precipitate, because the compound has CH₃CH(OH)–.
Physical properties follow the groups too 20.3(i)
Physical properties can be predicted in the same way.
- Solubility in water. Groups that hydrogen bond to water, –OH, –COOH and –NH₂, make a compound more soluble. A long hydrocarbon chain makes it less soluble.
- Boiling temperature. Molecules that hydrogen bond to each other, through –OH, –COOH or N–H, boil at higher temperatures than molecules of similar size that cannot.
- Acid or base behaviour. –COOH is acidic. –NH₂ is basic. A compound with both is amphoteric.
Check yourself
- Predict two reactions of propenoic acid, CH₂=CHCOOH, and give the observation for each.
- Explain why 2-hydroxypropanoic acid is very soluble in water.
- 4-Hydroxybutanoic acid, HOCH₂CH₂CH₂COOH, is heated under reflux with acidified potassium dichromate. Predict and explain the observation and name the organic product.
- 2-Hydroxypropanenitrile, CH₃CH(OH)CN, is (a) heated under reflux with dilute hydrochloric acid and (b) treated with LiAlH₄ in dry ether. Predict each product.
- 4-Aminobenzoic acid, H₂N–C₆H₄–COOH, has an amine and a carboxylic acid group. Predict its reactions with (a) hydrochloric acid, (b) sodium carbonate solution and (c) ethanol with concentrated sulfuric acid.
- Explain why the compound in question 5 does not decolourise bromine water as an alkene does.
Answers cover these until you have tried the questions
- Bromine water is decolourised from orange to colourless, because of the C=C group (it adds across the double bond). Sodium carbonate solution fizzes, because the –COOH group is acidic. (Also: ethanol and concentrated H₂SO₄ gives an ester; addition polymerisation.)
- The molecule has an –OH and a –COOH group. Both form hydrogen bonds with water molecules, and these bonds release enough energy to separate the molecules.
- The solution turns orange to green. A primary alcohol is oxidised, on reflux with excess dichromate, to a carboxylic acid, so the product is butanedioic acid, HOOCCH₂CH₂COOH. The –COOH group is already oxidised.
- (a) The –CN is hydrolysed to –COOH, giving 2-hydroxypropanoic acid, CH₃CH(OH)COOH. (b) The nitrile is reduced to an amine, giving 1-aminopropan-2-ol, CH₃CH(OH)CH₂NH₂.
- (a) The –NH₂ group is basic, so it accepts H⁺ and gives a salt that dissolves, H₃N⁺–C₆H₄–COOH Cl⁻. (b) The –COOH group is acidic, so there is fizzing and CO₂ forms. (c) The –COOH group reacts with the alcohol to give an ester, ethyl 4-aminobenzoate, and water.
- The compound has a benzene ring, not a C=C. The π electrons in the ring are delocalised, so the ring does not add bromine in the way a localised C=C does.