Lipids — fatty acids, triglycerides, phospholipids; saponification: Exercises

Before we start, one shared picture we will point back to all page long — the triglyceride skeleton:

Figure — Lipids — fatty acids, triglycerides, phospholipids; saponification

Look at it: on the left is glycerol (the 3-carbon backbone, magenta), and three ester bonds (violet) each tie a fatty-acid tail (orange) onto one of glycerol's three arms. Every number game on this page is really just "count the arms, count the bonds, count the atoms."

L1 — Recognition

Recall Solution 1.1

The second number in $C : D$ shorthand is $D$ = the count of $C=C$ double bonds. Zero double bonds = saturated; one or more = unsaturated.

(a) $18 : 0$ → $D = 0$ → saturated → straight chain packs tightly → solid.
(b) $18 : 1$ → $D = 1$ → unsaturated → one cis kink → liquid.
(c) $18 : 2$ → $D = 2$ → unsaturated (more kinked than oleic) → liquid. The tight-packing story is exactly the Van der Waals Forces argument: straight chains touch along their whole length, so more weak attractions add up to a stronger total.

Recall Solution 1.2

Each of glycerol's three $-OH$ groups meets one acid's $-OH$ ; together an $-OH$ and an $-H$ leave as one water. Three bonds formed → $n = 3$ . The reaction type is a condensation (dehydration), the same idea covered in Carboxylic Acids and Esterification.

L2 — Application

Recall Solution 2.1

Tripalmitin = glycerol backbone $C_{3} H_{5}$ carrying three palmitate esters: $C_{3} H_{5} (OOC-C_{15} H_{31})_{3} + 3 NaOH ⟶ 3 C_{15} H_{31} COONa + C_{3} H_{8} O_{3}$ This is alkaline hydrolysis (see Hydrolysis Reactions): 3 ester bonds break, base traps each fatty acid as its sodium salt. Products: sodium palmitate (the soap) + glycerol ( $C_{3} H_{8} O_{3}$ ).

Recall Solution 2.2

Still 3 ester bonds — the backbone always has exactly three arms. Because the three tails differ, you get 3 distinct soap salts: $A-COO^{-} Na^{+}$ , $B-COO^{-} Na^{+}$ , $C-COO^{-} Na^{+}$ , plus one glycerol. Number of bonds depends only on the backbone; variety of products depends on the tails.

L3 — Analysis

Recall Solution 3.1

Reaction stoichiometry: $1 tristearin + 3 NaOH \to 3 soap + 1 glycerol$ .

NaOH consumed: $3 \times 40 = 120 g$ .
Glycerol made: $1 \times 92 = 92 g$ .
Soap made: $3 \times 306 = 918 g$ . Mass check (conservation): $fat 890 + NaOH 120 = soap 918 + glycerol 92$ $1010 = 1010$ ✓ — total mass in equals total mass out.

Recall Solution 3.2

Same length, so chain-length contributions to Van der Waals Forces are equal — the only difference is the one double bond.

Stearic: straight → whole-length contact → many stacked attractions → high m.p.
Oleic (cis): one kink bends the tail like a hockey stick → chains splay → far less contact → low m.p.
Elaidic (trans): a trans double bond keeps the chain almost straight, so packing is nearly as good as saturated. Prediction: m.p. much closer to stearic than to oleic (real value $\approx 4 5^{\circ} C$ — between them, but far above oleic). Lesson: it is the cis geometry, not merely "a double bond," that lowers melting point.

L4 — Synthesis

Recall Solution 4.1

(a) It becomes a phospholipid. (b) Swapping one greasy tail for a charged phosphate head gives the molecule two personalities → it is now amphipathic (one water-loving end, two water-hating tails). (c) In water the tails hide from water while the heads face it. The stable compromise is a bilayer: tails point inward toward each other, heads face the water on both sides. That self-assembled sheet is the Cell Membrane Structure — no template needed, physics alone drives it. Compare: a full triglyceride keeps all three tails → fully hydrophobic → good for energy storage, useless for membranes. One swap flips the function.

Recall Solution 4.2

Both are amphipathic and both are driven by the same rule: hide the tails from water, expose the heads.

Soap has one tail and a big head → the molecule is roughly a cone (wide head, narrow tail). Cones pack into a sphere → a micelle, with all tails pointing to the centre (see Soaps and Detergents).
Phospholipid has two tails → it is roughly a cylinder (head about as wide as the paired tails). Cylinders pack into a flat sheet → a bilayer. So the deciding factor is tail count → molecular shape → curvature: one tail curves tightly (sphere), two tails stay flat (sheet).

L5 — Mastery

Recall Solution 5.1

Step 1 — moles of fat in 1 g: $\frac{1}{890} mol$ . Step 2 — moles of KOH needed: each fat needs 3 KOH, so $\frac{3}{890} mol$ . Step 3 — mass of KOH in mg: $\frac{3}{890} \times 56 \times 1000 \approx 188.8 mg KOH per g .$ Step 4 — why shorter tails → higher value: shorter tails mean a smaller molar mass, so 1 g contains more molecules, hence more ester bonds per gram, hence more KOH needed per gram → higher saponification value. (Saponification value is inversely related to average chain length.)

Recall Solution 5.2

(a) Fatty-acid carbons are highly reduced — mostly $C-H$ bonds with almost no oxygen already attached. Oxidation (adding O, removing H) is where energy is released, so the more $C-H$ there is to oxidise, the more energy comes out. Carbohydrate carbons already carry oxygen ( $-OH$ ), so they are "partly pre-oxidised" and release less. The stepwise removal of two-carbon units is Energy Metabolism — Beta Oxidation. (b) Fat is anhydrous — stored dry. Carbohydrate (glycogen) is stored hydrated, dragging roughly 2–3 g of water per gram of carbohydrate. So per gram of usable stored fuel, fat's real advantage over carbohydrate is even larger than the raw $9$ vs $4$ ratio suggests.

Recall Solution 5.3

Glycerol alone: it has no ester bonds — it is just an alcohol with three $-OH$ groups. There is nothing for hydrolysis to cleave, so no soap forms; NaOH cannot saponify what has no ester.
Free fatty acid $RCOOH$ : also has no ester bond. But it does have a $-COOH$ , so NaOH simply neutralises it directly: $RCOOH + NaOH \to RCOO^{-} Na^{+} + H_{2} O$ . You still get soap — but by a plain acid–base neutralisation, not by breaking an ester. The distinction matters: saponification requires an ester to hydrolyse. These edge cases pin down the exact definition: saponification = hydrolysis of an ester in base; remove the ester and the reaction is no longer saponification.

Recall Quick self-test — cover the answers

Saponification value of tristearin (KOH, 890 g/mol, 3 bonds)? ::: $\approx 188.8$ mg KOH per g Mass of NaOH to saponify 1.00 mol tristearin? ::: 120 g Soap mass from 1.00 mol tristearin (sodium stearate, 306 g/mol)? ::: 918 g Distinct soap salts from a triglyceride with 3 different tails? ::: 3 Why does trans fat melt high like saturated fat? ::: The trans double bond keeps the chain straight, so it still packs tightly.