SECTION A
1. (a) What type of bond fission leads to the formation of a carbonium ion in a reaction?
(b) What is the name of the reaction type in which two or more substances combine to give a single product?
© Give the structure of two (2) nitrogen-based nucleophiles.
(d) Give one (1) example of a Lewis acid. (5 marks)
2. Give the name of the class of organic compound that is formed when a haloalkane (R–X) reacts with: (5 marks)
(a) Hot water
(b) Sodium
© Alcoholic solution of ammonia
(d) Magnesium in dry ether
(e) Sodium alkoxide in alcohol
3. Give the class of organic compound that is formed when an alkanol (R–OH) reacts with: (5 marks)
(a) Thionyl chloride
(b) Hot concentrated tetraoxosulphate(VI) acid
© Alkanoic acid
(d) Reactive metal
(e) Excess potassium tetraoxomanganate(VII) acid
4. (a) The reaction of alkanoate with lithium tetrahydridoaluminate can be classified as?
(b) Give the molecular formula of the organic product(s) formed when lithium tetrahydridoaluminate reacts with:
(i) ethyl ethanoate
(ii) methyl propanoate (5 marks)
5. (a) Why are aromatic alkanoic acids weaker than aliphatic alkanoic acids?
(b) Give the name and structural formula of one example each of an aromatic monoalkanoic acid and dialkanoic acid. (5 marks)
6. Classify the following substances in order of increasing basic strength: (5 marks)
(a) Ammonium ion
(b) Dimethylamine
© 3-aminopentane
(d) Ammonia
(e) Triethylamine
7. (a) Depict the structural formula of aminoethanoic acid as a zwitterion.
(b) Name three (3) simple proteins. (5 marks)
8. Classify these anhydrides as either simple anhydride or mixed anhydride, and indicate the alkanoic acids from which the anhydrides are formed: (5 marks)
(a) CH₃CH₂COOCOCH₃
(b) CH₃CH₂COOCOCH₂CH₃
9. (a) Define condensation polymerization.
(b) State the names of the monomers of: (i) PVC (ii) Teflon (iii) PVAc (5 marks)
10. Name five (5) conversion processes used to improve the yield and quality of petroleum products after fractional distillation. (5 marks)
SECTION B
11. (a) Write chemical equations to illustrate the preparation of alkanamines from:
(i) Alkyl halides
(ii) Alkanamides
(iii) Alkyl nitro compounds
(iv) Alkyl nitriles (8 marks)
(b) Give the IUPAC name of the following organic compounds (structures i–vi as shown in diagrams): (12 marks)
© Protonation of alkanol A and subsequent loss of water produces the intermediate B:
A = CH₃CH₂–C(CH₃)(OH)–CH₂CH₃
B = CH₃CH₂–C(CH₃)=CH–CH₃ (carbocation intermediate)
(i) Name alkanol A.
(ii) What type of species is intermediate B?
(iii) Draw the structures of the two alkenes which can be formed from species B by removal of a proton.
(iv) The intermediate B is readily attacked by nucleophiles such as water. What is the essential feature of a nucleophile? (5 marks)
12. (a) Use chemical equations to illustrate the reaction between ethanol and:
(i) methanoic acid
(ii) hot concentrated tetraoxosulphate(VI) acid
(iii) phosphorus pentachloride
(iv) acidified potassium tetraoxomanganate(VII)
(v) iodine/sodium trioxocarbonate(IV) solution (10 marks)
(b) Give the structure of the possible organic product of these reactions:
(i) C₂H₅NH₂ + HCl
(ii) CH₃CH₂NH₂ + HNO₂
(iii) CH₃CH₂NH₂ + CH₃COCl
(iv) CH₂=CH₂NH₂ + alc. KOH/CHCl₃
(v) CH₃CH₂NH₂ + CH₃Br (10 marks)
© (i) Each homologous series has its own general formula. State three (3) other characteristics of a homologous series.
(ii) State what is meant by the term structural isomers. (5 marks)
13. (a) Write equations showing how chloroethane may be used in preparing: (10 marks)
(i) Ethane
(ii) Ethene
(iii) Butane
(iv) Ethanol
(v) Propenonitrile
(b) Give the structure of the organic products obtainable from the hydrolysis of the following compounds:
(i) Propyl methanoate
(ii) Methyl butanoate
(iii) Ethyl ethanoate
(iv) 2-methylpropyl propanoate
(v) Propyl ethanoate (10 marks)
© (i) But-2-ene can exist in two isomeric forms. Give the structures of these two isomers and name the type of isomerism.
(ii) Give the name of the process shown in the reaction equation:
nC₃H₄ → –[–C–C–]ₙ– (polymerisation diagram shown) (5 marks)
14. (a) Determine the empirical formula of a compound with molecular mass of 138 which has 60.87% carbon, 4.35% hydrogen and 34.78% oxygen. (5 marks)
(b) Give the structure and IUPAC name of the organic product of the reaction between: (15 marks)
(i) Methanal and ammoniacal silver trioxonitrate(V) solution
(ii) Ethanal and chlorine gas
(iii) Ethanal and cold dilute sodium hydroxide solution
(iv) Propanone and hydrazine solution
(v) Propanone and sodium tetrahydridoborate in methanol solution, followed by acid treatment.
© A naturally-occurring triester, shown below, was heated under reflux with an excess of aqueous sodium hydroxide and the mixture produced was then distilled. One of the products distilled off and the other was left in the distillation flask.
CH₃(CH₂)₁₄COOCH₂
CH₃(CH₂)₁₆COOCH
CH₃(CH₂)₁₄COOCH₂
(i) Draw the structure of the product distilled off and give its name.
(ii) Give the formula of the product left in the distillation flask and give a use for it.
(iii) What is the name of this type of reaction? (5 marks)
SOLUTIONS TO ALL
Q1
(a) Heterolytic fission — the bonding pair of electrons goes entirely to one atom, generating a carbocation (carbonium ion, R⁺) and a carbanion or leaving group.
(b) Addition reaction (also called a combination reaction)
© Two nitrogen-based nucleophiles:
H H H
| | |
H — N : and H — N — C₂H₅
|
H
(Ammonia, NH₃) (Ethylamine)
Both have a lone pair on nitrogen for nucleophilic attack.
(d) Example of a Lewis acid: AlCl₃ (aluminium trichloride) — it is an electron-pair acceptor (has an incomplete octet).
Q2 — Haloalkane (R–X) reactions
(a) Hot water → Alkanol (alcohol) — hydrolysis: R–X + H₂O → R–OH + HX
(b) Sodium → Alkane — Wurtz reaction: 2R–X + 2Na → R–R + 2NaX
© Alcoholic solution of ammonia → Alkanamine (amine) — nucleophilic substitution: R–X + NH₃ → R–NH₂ + HX
(d) Magnesium in dry ether → Grignard reagent (alkylmagnesium halide) — R–X + Mg → R–MgX
(e) Sodium alkoxide in alcohol → Ether (alkoxyalkane) — Williamson synthesis: R–X + NaOR’ → R–O–R’ + NaX
Q3 — Alkanol (R–OH) reactions
(a) Thionyl chloride (SOCl₂) → Haloalkane (chloroalkane): R–OH + SOCl₂ → R–Cl + SO₂ + HCl
(b) Hot concentrated H₂SO₄ → Alkene — dehydration/elimination: R–CH₂–CH₂–OH → R–CH=CH₂ + H₂O
© Alkanoic acid → Ester (alkanoate) — esterification: R–OH + R’COOH → R’COOR + H₂O
(d) Reactive metal (e.g., Na) → Metal alkoxide + hydrogen gas: 2R–OH + 2Na → 2R–ONa + H₂↑
(e) Excess acidified KMnO₄ → Alkanoic acid (carboxylic acid) — oxidation of primary alcohol: R–CH₂OH → R–COOH
Q4 — LiAlH₄ reduction of alkanoates
(a) The reaction of an alkanoate (ester) with LiAlH₄ is classified as a nucleophilic addition-reduction reaction (or reduction reaction).
(b) LiAlH₄ reduces esters to two alcohols:
RCOOR’ + LiAlH₄ → RCH₂OH + R’OH
(i) Ethyl ethanoate (CH₃COOC₂H₅):
CH₃COOC₂H₅ + LiAlH₄ → CH₃CH₂OH + C₂H₅OH
Products: ethanol (C₂H₅OH) and ethanol (C₂H₅OH)
(Both products are ethanol — molecular formula C₂H₅OH for each)
(ii) Methyl propanoate (CH₃CH₂COOCH₃):
CH₃CH₂COOCH₃ + LiAlH₄ → CH₃CH₂CH₂OH + CH₃OH
Products: propan-1-ol (C₃H₇OH) and methanol (CH₃OH)
Q5 — Aromatic vs Aliphatic Alkanoic Acids
(a) Aromatic alkanoic acids (e.g., phenylacetic acid, C₆H₅CH₂COOH) are weaker than aliphatic alkanoic acids because the aromatic ring withdraws electron density from the carboxylate group through inductive and resonance effects, but the key reason is that the phenyl ring donates electrons by hyperconjugation/resonance into the –CH₂– group, making the –COOH less able to stabilise the negative charge on the carboxylate ion (COO⁻) after dissociation, reducing the tendency to release H⁺.
(More precisely: benzene ring’s electron-donating resonance effect destabilises the carboxylate anion, opposing dissociation → weaker acid.)
(b) Aromatic monoalkanoic acid:
-
Name: Phenylmethanoic acid (benzoic acid)
-
Structure: C₆H₅–COOH
Aromatic dialkanoic acid:
-
Name: Benzene-1,4-dicarboxylic acid (terephthalic acid)
-
Structure: HOOC–C₆H₄–COOH (COOH groups at para positions)
Q6 — Increasing basic strength
Basic strength of amines depends on availability of the lone pair on nitrogen. More electron-donating groups → stronger base. Protonated species (NH₄⁺) is the conjugate acid, so it is weakest base.
Order of increasing basic strength:
NH₄⁺ < NH₃ < 3-aminopentane < Dimethylamine < Triethylamine
i.e.: (a) < (d) < © < (b) < (e)
Reasons:
-
NH₄⁺ — already protonated, no lone pair available → effectively no basic character
-
NH₃ — no alkyl groups, lone pair least available
-
3-aminopentane (C₂H₅CH(NH₂)C₂H₅) — secondary amine, two alkyl groups donate electrons
-
Dimethylamine (CH₃)₂NH — secondary amine, two methyl groups
-
Triethylamine (C₂H₅)₃N — tertiary amine, three ethyl groups maximally donate electron density to N → strongest base
Q7 — Amino acids
(a) Zwitterionic structure of aminoethanoic acid (glycine):
H₃N⁺ — CH₂ — COO⁻
The amino group is protonated (–NH₃⁺) and the carboxyl group is deprotonated (–COO⁻) simultaneously.
(b) Three simple proteins:
-
Albumin (found in egg white/blood plasma)
-
Globulin (found in blood)
-
Keratin (found in hair, nails)
(Others acceptable: fibrin, collagen, myosin)
Q8 — Anhydrides
(a) CH₃CH₂COOCOCH₃ — This has two different acyl groups (propanoyl and ethanoyl), so it is a mixed anhydride formed from propanoic acid (CH₃CH₂COOH) and ethanoic acid (CH₃COOH).
(b) CH₃CH₂COOCOCH₂CH₃ — Both acyl groups are the same (both propanoyl), so it is a simple anhydride formed from two molecules of propanoic acid (CH₃CH₂COOH). Name: propanoic anhydride.
Q9 — Polymerization
(a) Condensation polymerization: A type of polymerization in which monomers join together with the repeated elimination of small molecules (usually water or HCl) to form a polymer chain.
(b) Monomers:
(i) PVC (polyvinyl chloride) — monomer: chloroethene (vinyl chloride), CH₂=CHCl
(ii) Teflon (PTFE) — monomer: tetrafluoroethene, CF₂=CF₂
(iii) PVAc (polyvinyl acetate) — monomer: vinyl acetate (ethenyl ethanoate), CH₂=CHOOCCH₃
Q10 — Petroleum conversion processes
Five conversion processes after fractional distillation:
-
Cracking — breaking large hydrocarbon molecules into smaller, more useful ones (thermal or catalytic)
-
Reforming — converting straight-chain alkanes into branched-chain alkanes or aromatic compounds to improve octane rating
-
Isomerisation — converting straight-chain hydrocarbons into branched-chain isomers
-
Alkylation — combining small alkanes and alkenes to form higher-octane branched alkanes
-
Polymerisation — joining small alkene molecules to form larger molecules suitable as fuels
SECTION B SOLUTIONS
Q11(a) — Preparation of alkanamines
(i) From alkyl halides (nucleophilic substitution with ammonia):
R–X + 2NH₃ → R–NH₂ + NH₄X
e.g., CH₃Br + 2NH₃ → CH₃NH₂ + NH₄Br
(Methylamine)
(ii) From alkanamides (Hofmann degradation with Br₂/NaOH):
R–CONH₂ + Br₂ + 4NaOH → R–NH₂ + Na₂CO₃ + 2NaBr + 2H₂O
e.g., CH₃CONH₂ + Br₂ + 4NaOH → CH₃NH₂ + Na₂CO₃ + 2NaBr + 2H₂O
(iii) From alkyl nitro compounds (reduction):
R–NO₂ + 6[H] → R–NH₂ + 2H₂O
e.g., C₂H₅NO₂ + 6[H] → C₂H₅NH₂ + 2H₂O
(Using Fe/HCl or H₂/catalyst)
(iv) From alkyl nitriles (reduction):
R–C≡N + 4[H] → R–CH₂–NH₂
e.g., CH₃CN + 4[H] → CH₃CH₂NH₂
(Using LiAlH₄ or H₂/Ni catalyst)
Q11(b) — IUPAC names of structures (based on visible structural features in diagrams):
From the structural diagrams visible in the images:
(i) Structure with –C(=O)–Cl group on a chain: Ethanoyl chloride (acetyl chloride) — CH₃COCl
(ii) Structure CH₃–CH₂–C(=O)–O–CH₂–CH₃: Ethyl propanoate
(iii) Structure showing branched ester with –OCH(CH₃)– group: 1-methylethyl ethanoate (isopropyl acetate)
(iv) Tertiary amine structure with two ethyl groups on N: N,N-diethylamine (diethylamine)
(v) Sulphonamide-type structure with S=O: Ethanesulphonic acid or ethanamide depending on exact structure
(vi) Structure with Br on benzene ring and alkyl chain: 2-methylpropyl bromide or a substituted benzene amine
(Note: precise IUPAC names for (i)–(vi) depend on the exact bond connectivity in the diagrams which are partially unclear from the image quality — the above are best interpretations)
Q11© — Alkanol A and Carbocation B
Given:
-
A = CH₃CH₂–C(CH₃)(OH)–CH₂CH₃ → 3-methylpentan-3-ol
-
B = CH₃CH₂–C⁺(CH₃)–CH₂CH₃ → tertiary carbocation (after loss of water)
(i) Name of alkanol A: 3-methylpentan-3-ol
(ii) Type of species B: A carbocation (carbonium ion) — specifically a tertiary carbocation, since the positive carbon is attached to three carbon groups.
(iii) Two alkenes formed from B by loss of H⁺:
Alkene 1 (loss of H from C-2):
CH₃CH₂–C(CH₃)=CHCH₃ → 3-methylpent-2-ene
Alkene 2 (loss of H from C-4, the other side):
CH₃CH=C(CH₃)–CH₂CH₃ → 3-methylpent-2-ene (E/Z isomers possible)
Or more precisely:
- 2-ethyl-1-butene and 3-methylpent-2-ene depending on which β-hydrogen is eliminated.
(iv) Essential feature of a nucleophile: A nucleophile must possess a lone pair of electrons (or a negative charge/π electrons) that it can donate to an electron-deficient centre (electrophile). Water qualifies because oxygen has two lone pairs.
Q12(a) — Reactions of ethanol
(i) Ethanol + methanoic acid (esterification):
C₂H₅OH + HCOOH ⇌ HCOOC₂H₅ + H₂O
Product: ethyl methanoate
(ii) Ethanol + hot conc. H₂SO₄ (dehydration, above 170°C):
C₂H₅OH → CH₂=CH₂ + H₂O
Product: ethene
(At 140°C: 2C₂H₅OH → C₂H₅OC₂H₅ + H₂O → ethoxyethane)
(iii) Ethanol + PCl₅ (phosphorus pentachloride):
C₂H₅OH + PCl₅ → C₂H₅Cl + POCl₃ + HCl
Product: chloroethane
(iv) Ethanol + acidified KMnO₄ (oxidation):
C₂H₅OH + [O] → CH₃COOH
Primary alcohol → ethanoic acid
(With excess KMnO₄: full oxidation to CH₃COOH)
(v) Ethanol + I₂/Na₂CO₃ solution (iodoform reaction):
C₂H₅OH + I₂ + NaOH → CHI₃↓ + HCOONa + …
Product: triiodomethane (iodoform, CHI₃) — yellow precipitate, characteristic smell
Q12(b) — Organic products of amine reactions
(i) C₂H₅NH₂ + HCl:
C₂H₅NH₂ + HCl → C₂H₅NH₃⁺Cl⁻
Product: ethylammonium chloride (ethylamine hydrochloride)
CH₃CH₂–NH₃⁺ Cl⁻
(ii) CH₃CH₂NH₂ + HNO₂ (nitrous acid — diazotisation of primary aliphatic amine):
CH₃CH₂NH₂ + HNO₂ → CH₃CH₂OH + N₂↑ + H₂O
Product: ethanol (unstable aliphatic diazonium salt decomposes immediately)
(iii) CH₃CH₂NH₂ + CH₃COCl (acylation):
CH₃CH₂NH₂ + CH₃COCl → CH₃CONHCH₂CH₃ + HCl
Product: N-ethylethanamide (N-ethylacetamide)
CH₃–C(=O)–NH–CH₂CH₃
(iv) CH₂=CH₂NH₂ + alc. KOH/CHCl₃ (carbylamine reaction — isocyanide test):
R–NH₂ + CHCl₃ + 3KOH(alc.) → R–N≡C + 3KCl + 3H₂O
Product: ethyl isocyanide (ethylcarbylamine)
CH₃CH₂–N≡C:
(Gives a foul smell — positive carbylamine test for primary amines)
(v) CH₃CH₂NH₂ + CH₃Br (N-alkylation):
CH₃CH₂NH₂ + CH₃Br → CH₃CH₂NHCH₃ + HBr
Product: N-methylethanamine (ethylmethylamine)
CH₃–NH–CH₂CH₃
Q12©
(i) Three other characteristics of a homologous series (besides having the same general formula):
-
Each successive member differs by a –CH₂– unit (14 in molecular mass).
-
Members have similar chemical properties because they contain the same functional group.
-
Physical properties (boiling point, melting point, density) show a gradual and regular trend as the molecular mass increases.
(ii) Structural isomers are compounds that have the same molecular formula but different structural arrangements of atoms — i.e., different connectivity of atoms, which may include different carbon skeletons, different positions of functional groups, or different types of functional groups.
Q13(a) — Uses of chloroethane (C₂H₅Cl)
(i) Preparation of ethane (Wurtz reaction):
2C₂H₅Cl + 2Na → C₂H₅–C₂H₅ + 2NaCl
(Product: butane — but for ethane: CH₃Cl + C₂H₅Cl + 2Na → CH₃CH₃… )
More correctly, ethane from chloroethane:
C₂H₅Cl + [H] (Zn/HCl or LiAlH₄) → C₂H₆ + HCl
Ethane: C₂H₅Cl + Zn + HCl → C₂H₆ + ZnCl₂
(ii) Preparation of ethene (dehydrohalogenation — elimination):
C₂H₅Cl + KOH(alc.) → CH₂=CH₂ + KCl + H₂O
Product: ethene
(iii) Preparation of butane (Wurtz reaction):
2C₂H₅Cl + 2Na → CH₃CH₂CH₂CH₃ + 2NaCl
Product: butane
(iv) Preparation of ethanol (hydrolysis):
C₂H₅Cl + NaOH(aq) → C₂H₅OH + NaCl
Or: C₂H₅Cl + H₂O (steam, high T) → C₂H₅OH + HCl
Product: ethanol
(v) Preparation of propenonitrile (acrylonitrile) via first making ethyl cyanide, then elimination… More directly:
C₂H₅Cl + NaCN → C₂H₅CN + NaCl (propanenitrile)
Then C₂H₅CN + KOH(alc.) → CH₂=CHCN + KCl + H₂O
Product: propenonitrile (CH₂=CH–CN)
Q13(b) — Hydrolysis products of esters
General hydrolysis: RCOOR’ + H₂O (acid catalyst) → RCOOH + R’OH
(i) Propyl methanoate (HCOOC₃H₇):
→ methanoic acid (HCOOH) + propan-1-ol (C₃H₇OH)
(ii) Methyl butanoate (CH₃CH₂CH₂COOCH₃):
→ butanoic acid (CH₃CH₂CH₂COOH) + methanol (CH₃OH)
(iii) Ethyl ethanoate (CH₃COOC₂H₅):
→ ethanoic acid (CH₃COOH) + ethanol (C₂H₅OH)
(iv) 2-methylpropyl propanoate (CH₃CH₂COOCH₂CH(CH₃)₂):
→ propanoic acid (CH₃CH₂COOH) + 2-methylpropan-1-ol ((CH₃)₂CHCH₂OH)
(v) Propyl ethanoate (CH₃COOC₃H₇):
→ ethanoic acid (CH₃COOH) + propan-1-ol (C₃H₇OH)
Q13©
(i) But-2-ene geometric isomers:
But-2-ene (CH₃–CH=CH–CH₃) has restricted rotation about the C=C double bond:
cis-but-2-ene (Z-but-2-ene):
CH₃ CH₃
\ /
C=C
/ \
H H
Both CH₃ groups on the same side.
trans-but-2-ene (E-but-2-ene):
CH₃ H
\ /
C=C
/ \
H CH₃
CH₃ groups on opposite sides.
Type of isomerism: Geometrical (cis-trans) isomerism — a type of stereoisomerism.
(ii) Name of the process nC₃H₄ → –[–C–C–]ₙ– :
This is addition polymerisation (the monomer propadiene/allene undergoes addition polymerisation to form a polymer chain).
Q14(a) — Empirical formula
Given: Molecular mass = 138; C = 60.87%, H = 4.35%, O = 34.78%
Moles:
-
C: 60.87/12 = 5.07
-
H: 4.35/1 = 4.35
-
O: 34.78/16 = 2.17
Divide by smallest (2.17):
-
C: 5.07/2.17 = 2.34 ≈ 7/3 → multiply all by 3
-
H: 4.35/2.17 = 2.0
-
O: 2.17/2.17 = 1.0
Multiply by 3: C: 7, H: 6, O: 3 → Empirical formula = C₇H₆O₃
Check: Empirical formula mass = 7(12) + 6(1) + 3(16) = 84 + 6 + 48 = 138 ✓
This matches the molecular mass exactly, so:
Molecular formula = C₇H₆O₃ (same as empirical formula)
(This is the formula of aspirin — acetylsalicylic acid)
Q14(b) — Carbonyl compound reactions
(i) Methanal + ammoniacal AgNO₃ (Tollens’ reagent):
HCHO is oxidised (it is an aldehyde); silver mirror forms.
HCHO + 2[Ag(NH₃)₂]⁺ + 2OH⁻ → HCOO⁻ + 2Ag↓ + 4NH₃ + H₂O
Product: methanoate ion (HCOO⁻) and silver mirror
(Methanal is oxidised to methanoic acid/methanoate)
(ii) Ethanal + Cl₂:
In presence of OH⁻ (haloform): CH₃CHO + 3Cl₂ + 3NaOH → CCl₃CHO → CHCl₃ + HCOONa
Or direct halogenation: CH₃CHO + Cl₂ → CH₂ClCHO
Product (direct): chloroethanal (CH₂ClCHO)
With excess Cl₂/NaOH: trichloromethane (chloroform, CHCl₃)
(iii) Ethanal + cold dilute NaOH (aldol condensation):
2CH₃CHO → CH₃CH(OH)CH₂CHO
Product: 3-hydroxybutanal (aldol)
CH₃–CH(OH)–CH₂–CHO
IUPAC name: 3-hydroxybutanal
(iv) Propanone + hydrazine (H₂N–NH₂):
(CH₃)₂C=O + H₂N–NH₂ → (CH₃)₂C=N–NH₂ + H₂O
Product: propanone hydrazone
(CH₃)₂C=N–NH₂
IUPAC name: propan-2-ylidenehydrazine
(v) Propanone + NaBH₄ in methanol, then acid treatment (reduction):
(CH₃)₂C=O + NaBH₄ → [(CH₃)₂CHOB H₃Na] → H₃O⁺ → (CH₃)₂CHOH
Product: propan-2-ol
OH
|
CH₃–CH–CH₃
IUPAC name: propan-2-ol
Q14© — Saponification of triester
The triester shown is a triglyceride (fat/oil):
-
CH₃(CH₂)₁₄COOCH₂ — palmitic acid ester
-
CH₃(CH₂)₁₆COOCH — stearic acid ester
-
CH₃(CH₂)₁₄COOCH₂ — palmitic acid ester
Reaction with excess NaOH (saponification):
Triester + 3NaOH → glycerol + 3 soap molecules
(i) Product distilled off: Glycerol (propane-1,2,3-triol)
CH₂OH
|
CHOH
|
CH₂OH
Name: glycerol (propane-1,2,3-triol)
It is the smaller, more volatile (lower boiling point compared to sodium salts) product that distils off.
(ii) Product left in distillation flask:
The sodium salts of the fatty acids (soaps):
-
CH₃(CH₂)₁₄COONa — sodium palmitate
-
CH₃(CH₂)₁₆COONa — sodium stearate
Use: As soap — for washing and cleaning (the long non-polar tail is lipophilic; the –COONa head is hydrophilic, enabling emulsification of grease).
(iii) Name of the reaction: Saponification (alkaline hydrolysis of an ester/fat)