Worked examples — Carboxylic acids — acidity, derivatives (acid chlorides, anhydrides, esters, amides), Hell-Volhard-Zelinsky, esterificat

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The scenario matrix

Before solving, let us list every case class this topic can produce. Each cell is a distinct "flavour" of problem; the examples below are tagged with the cell they cover.

Cell	Case class	What varies / the twist
A	Rank acidity by inductive strength	which substituent, how many, how far
B	Rank acidity — resonance vs no resonance	carboxylic acid vs phenol vs alcohol (degenerate: no anion stabilisation)
C	Compute $p{K}_a\leftrightarrow K_a\leftrightarrow$ percent dissociation	a number answer, log arithmetic
D	Order acid derivatives by reactivity	leaving group + lone-pair donation
E	HVZ product with α-H present	pick the α-carbon
F	HVZ degenerate case: no α-H	reaction fails — the "zero input"
G	Fischer esterification — name the ester, trace the oxygen	¹⁸O labelling twist
H	Le Chatelier — shift a reversible esterification	limiting/driving behaviour
I	Real-world word problem	aspirin / soap, a synthesis choice
J	Exam-style trap	"which route?" chloride vs Fischer

We cover cells A–H with one example each (Examples 1–8), then combine cells I and J into a single example (Example 9) — a real-world synthesis choice that is the exam trap. So ten cells, nine examples, because I and J are two views of the same question.

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Example 1 — Cell A (inductive, distance & count)

Steps:

1. Recall the rule. Electron-withdrawing groups (here $Cl$) pull charge away from the carboxylate anion, spreading it out → more stable anion → stronger acid → smaller $p{K}_a$. See Inductive effect. Why this step? Acidity is decided by how comfortable the conjugate base is, per the parent note's "stabilise the anion" theme.

2. Count the chlorines. (iv) has two $Cl$; the rest have one or zero. Two withdrawers beat one → (iv) is the strongest acid (smallest $p{K}_a$). Why this step? Each EWG adds its own pull; effects roughly stack.

3. Compare distance for the single-$Cl$ ones. The inductive effect fades with distance through bonds. In (ii) the $Cl$ sits on the α-carbon (next to $COOH$); in (iii) it sits on the β-carbon (one bond further). Closer $Cl$ = stronger pull → (ii) more acidic than (iii). Why this step? Inductive pull travels through the σ-bond framework and weakens each bond it crosses.

4. The bare alkyl acid. (i) has no withdrawer — only a mildly electron-donating alkyl chain, which destabilises the anion → weakest acid (largest $p{K}_a$).

Answer (increasing $p{K}_a$ = strongest → weakest): (iv) < (ii) < (iii) < (i).

Recall Verify by the logic checklist

More Cl → stronger. Closer Cl → stronger. No Cl → weakest. Order holds. ✓

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Example 2 — Cell B (resonance vs none; degenerate anion)

Steps:

1. Ethanol — the degenerate case. Its anion $C{H}_{3}C{H}_2{O}^{-}$ (ethoxide) has the charge stuck on one oxygen, no delocalisation (Resonance and delocalisation gives nothing here). This is the degenerate input — no stabilising feature at all. Least stable anion → weakest acid, $p{K}_a\approx 16$. Why this step? Zero anion-stabilisation is the "empty input" — always the weakest.

2. Phenol — resonance onto carbon. Phenoxide delocalises the charge, but onto ring carbon atoms, which are poor at holding negative charge (low electronegativity). Some help, not much: $p{K}_a\approx 10$. Why this step? Resonance quality depends on which atom holds the charge, not just how many structures you can draw.

3. Acetic acid — resonance onto oxygen. Carboxylate spreads the charge over two electronegative oxygens — the ideal home. Most stable anion → strongest, $p{K}_a\approx 4.76$.

Answer (increasing acidity): ethanol < phenol < acetic acid.

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Example 3 — Cell C (numbers: $p{K}_a\to K_a\to$ % dissociation)

Steps:

1. Undo the log. $K_a=10^{-p{K}_a}=10^{-2.86}\approx 1.38\times 10^{-3}$. Why this step? $p{K}_a=-{\log }_{10}{K}_a$, so $K_a=10^{-p{K}_a}$ — reversing the definition from the parent note.

2. Set up the equilibrium. Let $x$ = concentration dissociated. $K_a=\frac{x^2}{0.10-x}.$ Why this step? $[RCO{O}^{-}]=[H_3{O}^{+}]=x$; the leftover undissociated acid is $0.10-x$.

3. Solve. $x^2=1.38\times 10^{-3}(0.10-x)$ → $x^2+1.38\times 10^{-3}x-1.38\times 10^{-4}=0$. $x=\frac{-1.38\times 10^{-3}+\sqrt{(1.38\times 10^{-3}{)}^2+4(1.38\times 10^{-4})}}{2}\approx 0.0111\text{M}.$

4. Percent dissociation. $\frac{x}{0.10}\times 100\approx 11.1\%$.

Answer: $K_a\approx 1.38\times 10^{-3}$, about $11\%$ dissociated.

Recall Verify (units & sanity)

A "moderately strong" acid should dissociate more than a weak one (acetic acid $\approx 1.3\%$ here) but far less than $100\%$. $11\%$ sits sensibly in between. Concentrations in M, dimensionless percent. ✓

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Example 4 — Cell D (reactivity order)

Steps:

1. Leaving-group ability first. The reaction is Nucleophilic acyl substitution: attack the carbonyl, expel the leaving group $L^{-}$. Best leaving group → most reactive. $C{l}^{-}$ (stable, weak base) is a great $L^{-}$; ${}^{-}N{H}_2$ (strong base) is a terrible one. Why this step? The rate is set by how easily $L^{-}$ departs after the tetrahedral intermediate forms.

2. Lone-pair donation second (same direction). $N$ donates its lone pair into the carbonyl strongly, "satisfying" it → amide is sluggish. $Cl$ donates poorly → carbonyl stays "hungry" → chloride reacts fast.

3. Rank. Chloride > anhydride > ester > amide (the mnemonic ladder).

Answer: $C{H}_{3}COCl$ > $(C{H}_{3}CO{)}_{2}O$ > $C{H}_{3}COO{C}_2{H}_5$ > $C{H}_{3}CON{H}_2$.

Recall Verify

Matches the parent's ladder "Chlorides Are Extremely Active... amides Aren't." Acetyl chloride indeed fumes in moist air. ✓

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Example 5 — Cell E (HVZ with α-H present)

Steps:

1. Generate the reactive intermediate. Red P $+B{r}_2\to PB{r}_3$; then $PB{r}_3$ turns the acid into the acyl bromide $(C{H}_3{)}_{2}CH\text{-}COBr$. Why this step? The plain acid does not enolise well; the acyl bromide does — this is the whole reason a catalyst is needed (parent note).

2. Find the α-carbon. The α-carbon is the one directly bonded to the carbonyl. Here that is the central $CH$ of $(C{H}_3{)}_{2}CH$-. It carries one α-H. Why this step? Only the α-carbon can become the nucleophilic enol carbon (see Keto–enol tautomerism) that attacks $B{r}_2$.

3. Halogenate that single α-H. The enol's α-carbon attacks $B{r}_2$, replacing the α-H with $Br$; a final hand-off with fresh acid regenerates the catalyst and gives the α-bromo acid. Why this step? HVZ swaps exactly one α-H for one halogen per molecule (mono-halogenation under normal conditions).

4. Write the product. $(C{H}_3{)}_{2}CBr\text{-}COOH$ — 2-bromo-2-methylpropanoic acid.

Answer: $(C{H}_3{)}_{2}CBr\text{-}COOH$.

Recall Verify (atom count)

Reactant $C_4{H}_8{O}_2$; product replaces one H with Br → $C_4{H}_{7}Br{O}_2$. Molecular formula balances one-for-one. ✓

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Example 6 — Cell F (degenerate: no α-H, reaction fails)

Steps:

1. Locate the α-carbon. It is the quaternary $C$ of $(C{H}_3{)}_{3}C$-. Its four bonds go to three $C{H}_3$ groups and the $COOH$ — no hydrogen on it. Why this step? HVZ requires at least one α-H (parent definition) to make the enol, even after the red P / $PB{r}_3$ makes the acyl bromide.

2. Consequence. No α-H ⟹ no enol can form ⟹ no nucleophilic α-carbon ⟹ no HVZ product. The acid is recovered unchanged. This is the degenerate input for HVZ — the feature the reaction depends on is absent. Why this step? This is the "zero input" degenerate case — the reaction simply cannot proceed.

Answer: No reaction (no α-hydrogen). Starting acid recovered.

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Example 7 — Cell G (Fischer: name ester + trace the oxygen)

Steps:

1. Name the ester. Acid $+$ methanol → methyl ester: $C_6{H}_{5}COOC{H}_3$ (methyl benzoate) $+H_{2}O$. Why this step? Fischer esterification joins the acyl group of the acid to the $O{R}^{\prime }$ of the alcohol.

2. Trace the mechanism's leaving oxygen. In the PADPED sequence above, the oxygen that leaves as water comes from the acid's $-OH$, not the alcohol. The alcohol contributes its $O\text{-}C{H}_3$ unit intact. Why this step? The alcohol's oxygen becomes the bridging ester oxygen; the acid's $OH$ is protonated (step 4) and expelled as $H_{2}O$ (step 5).

3. Place the label. Since the ¹⁸O is on the alcohol, it stays in the ester ($C_6{H}_{5}CO{\text{-}}^{18}O\text{-}C{H}_3$). The water formed carries ordinary ¹⁶O from the acid.

Answer: Product = methyl benzoate with ¹⁸O in the ester linkage; water is unlabelled.

Recall Verify (classic experiment)

This is exactly the historical ¹⁸O labelling result confirming which C–O bond breaks: the acid loses $OH$, the alcohol keeps its O. Matches the parent's steel-man. ✓

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Example 8 — Cell H (Le Chatelier — drive the equilibrium)

Steps:

1. Set up. Let $x$ = mol ester (and water) formed. Then acid $=$ alcohol $=1-x$; ester $=$ water $=x$. Volume cancels because there are equal numbers of species on both sides. $K=\frac{[\text{ester}][\text{water}]}{[\text{acid}][\text{alcohol}]}=\frac{x\cdot x}{(1-x)(1-x)}={\left(\frac{x}{1-x}\right)}^2=\mathrm{4.0.}$ Why this step? All four species share the same volume, so the volume factors cancel — we can work directly in moles.

2. Take the square root. Since both sides are squares, $\frac{x}{1-x}=\sqrt{4.0}=2$. Why this step? $x$ and $1-x$ are both positive (between $0$ and $1$ mol), so we take the positive root only.

3. Solve the linear equation. $x=2(1-x)=2-2x$ → $3x=2$ → $x=\frac{2}{3}\approx 0.667\text{mol}$.

4. Push further (Le Chatelier's principle). (a) Add excess alcohol — adding a reactant shifts the equilibrium to the right (toward ester). (b) Remove water — distil it off or add a drying agent; removing a product also shifts right. Why this step? Le Chatelier: disturb an equilibrium and it responds so as to partly undo the disturbance — both moves increase ester.

Answer: $\approx 0.667\text{mol}$ ester at equilibrium; drive further with excess alcohol or by removing water.

Recall Verify (plug back)

With $x=2/3$: ${\left(\frac{2/3}{1/3}\right)}^2=2^2=4.0=K$. ✓

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Example 9 — Cells I + J combined (real-world word problem that IS the exam trap)

Steps:

1. Recognise the reaction type. Attaching an acetyl group to an $-OH$ is Nucleophilic acyl substitution — attack the acyl carbon, expel a leaving group. Why this step? It tells us the reactivity ladder (Cell D) decides the best reagent.

2. Judge the Fischer route (acetic acid). Fischer esterification is reversible and slow, needs an acid catalyst, and it makes water — which can hydrolyse the product back. Poor yield unless you aggressively remove water. Why this step? Reversibility (parent note) caps the yield; not ideal for a clean product.

3. Judge the anhydride route. Acetic anhydride sits high on the reactivity ladder. It reacts fast and essentially irreversibly, releasing acetic acid (a decent leaving group), with no equilibrium fight. Why this step? "Synthesis flows downhill" — use the more reactive derivative for a clean, high-yield acylation.

4. State the exam-trap fix (Cell J). Both routes make an ester bond, so a careless student calls them equivalent. They are not: the anhydride (and chloride) route is fast & irreversible; Fischer is slow & reversible. Always choose the reactive derivative for yield.

Answer: Use acetic anhydride — fast, near-irreversible, high yield. (For the reverse philosophy — cleaving esters with base — see Saponification.)

Recall Verify (industrial fact)

Real aspirin manufacture uses acetic anhydride with a trace acid catalyst — matching our reactivity-ladder reasoning. ✓

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Recall Matrix coverage check

A ✓(Ex 1) · B ✓(Ex 2) · C ✓(Ex 3) · D ✓(Ex 4) · E ✓(Ex 5) · F ✓(Ex 6) · G ✓(Ex 7) · H ✓(Ex 8) · I ✓(Ex 9) · J ✓(Ex 9). Every cell hit; I and J deliberately combined into one example.

Back to the parent: main topic note.