CHEM1011 · Fundamentals Of Chemistry 1a
Functional Groups, Fats and Polymers
This chapter closes the organic strand of CHEM1011 by showing how water breaks the carbonyl derivatives (esters, amides, acid chlorides, thioesters) and how that one reaction connects fats, oils and soap. You will explain melting points of fatty acids from chain shape, account for soap as an amphiphile that builds micelles, rank boiling points from the intermolecular-force ladder, and tell addition from condensation polymers. It is high-yield short-answer territory: markers reward the causal chain (structure to force to physical property), not just the final answer.
What this chapter covers
- 011. Hydrolysis of acid derivatives: ester/amide/acid-chloride/thioester + H₂O, acid (H⁺) vs base (OH⁻)
- 022. Triglycerides: glycerol + 3 fatty acids = a triester (fat/oil)
- 033. Saturated vs cis-unsaturated fatty acids: chain shape, packing and melting point
- 044. Saponification: base hydrolysis of a fat to glycerol + fatty-acid sodium salts (soap)
- 055. Amphiphiles, hydrophobic effect and micelle/bilayer self-assembly
- 066. Intermolecular forces: London < dipole-dipole < H-bond (< ion-dipole)
- 077. The boiling-point rule: stronger IMF means higher b.p.
- 088. Addition vs condensation polymers; amino acids, zwitterions and the peptide bond
Rank boiling points by intermolecular force
- +1Propane (CH₃CH₂CH₃): a non-polar hydrocarbon with no net dipole and no H on N/O/F, so the only force is London dispersion — the weakest type.
- +1Ethanal (CH₃CHO): the polar C=O gives a permanent dipole, but the H atoms sit on carbon (not N/O/F), so it cannot H-bond — its strongest force is dipole–dipole, stronger than propane.
- +1Ethanol (CH₃CH₂OH): the O–H group lets it act as a hydrogen-bond donor and acceptor, so it shows hydrogen bonding — the strongest force of the three.
- +1Apply the boiling-point rule (stronger IMF ⇒ higher b.p.). Because the masses are similar, London dispersion is roughly equal for all three, so the force type decides the order.
Key terms
- Hydrolysis
- The splitting of a carboxylic-acid derivative by water ("hydro" = water, "lysis" = breaking): the C=O group is attacked and the molecule cleaves into a carboxylic acid plus the leaving group, e.g. R–COO–R′ + H₂O ⇌ R–COOH + R′–OH.
- Saponification
- Base hydrolysis of a triglyceride: triglyceride + 3 NaOH → glycerol + 3 fatty-acid sodium salts (soap). Because the acid product is deprotonated to its carboxylate salt, the reaction goes to completion.
- Amphiphile (surfactant)
- A molecule with a polar/hydrophilic head (e.g. –COO⁻ Na⁺) and a non-polar/hydrophobic tail. In water it self-assembles tails-in, heads-out into a micelle (single tail) or a bilayer (double tail).
- Hydrogen bonding
- A strong dipole interaction occurring when an H atom bonded to N, O or F is attracted to a lone pair on another N, O or F. It is the strongest of the common neutral-molecule IMFs and far weaker than a covalent bond.
- Addition polymer
- A polymer formed when monomers containing a C=C double bond add together with no atoms lost (no by-product), e.g. ethene → polyethene. The repeating unit is –CH₂–CH₂–.
- Condensation polymer
- A polymer formed when bifunctional monomers join and expel a small molecule (usually H₂O) at each linkage, e.g. polyesters, polyamides (nylon) and peptides; counting the waters lost counts the links formed.
Functional Groups, Fats and Polymers FAQ
Why does an alcohol boil higher than an aldehyde or alkane of the same mass?
Because boiling has to overcome intermolecular forces, and the alcohol's O–H lets it hydrogen-bond — the strongest of the three forces. The aldehyde only has dipole–dipole (its H is on carbon, not O), and the alkane has London dispersion alone. At similar mass the dispersion is comparable, so the force type decides: H-bond > dipole–dipole > London, and the boiling points follow the same order.
Why is animal fat solid but plant oil liquid at room temperature?
Saturated fatty acids (only C–C single bonds) are straight chains that pack tightly, giving strong London dispersion between the tails and a high melting point — solid. Cis-unsaturated chains have a kink at each C=C, so they pack poorly, the forces are weaker, and the melting point is lower — liquid. Melting point drops as the number of cis C=C bonds rises.
What is the difference between acid and base hydrolysis of an ester?
Acid (H⁺) catalysis only speeds up a reversible equilibrium, so the ester can re-form and you get back the carboxylic acid plus alcohol. Base (OH⁻) hydrolysis deprotonates the acid product to its carboxylate salt, removing it from the system; by Le Châtelier the reaction is pulled all the way to completion. That irreversibility is why soap-making uses NaOH/KOH, not dilute acid.
How do I tell an addition polymer from a condensation polymer?
Look at the monomer and the by-product. If the monomer has a C=C that opens up and units add with no atoms lost, it is an addition polymer (polyethene, polystyrene). If the monomers have two functional groups that react and expel a small molecule — usually water — at each linkage, it is a condensation polymer (polyester, nylon, peptides).
What is on the exam for this topic?
Expect short-answer items that test the causal chain: rank boiling or melting points and justify them by naming the dominant intermolecular force; write or explain ester hydrolysis and saponification; explain how a soap's two ends let it clean (amphiphile → micelle); and classify or draw a polymer as addition vs condensation, including the amino-acid zwitterion and the peptide (amide) bond as a condensation.
Exam move
Memorise the structures and schemes the datasheet will NOT give you (ester hydrolysis, glycerol + 3 fatty acids = a fat, saponification → soap, the amphiphile/micelle picture, the addition-vs-condensation table, the amino-acid zwitterion). Then drill the one move markers reward in every short-answer: argue from structure to physical property. For fatty acids run shape → packing → London force → melting point; for boiling points name the strongest IMF first (London < dipole–dipole < H-bond), and use mass only as a tie-breaker within the same force type. Hold one variable fixed when ranking (equal chain length, or equal saturation), and never claim that boiling water breaks covalent O–H bonds — it breaks the H-bonds between the molecules.
