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

Level 1 — Recognition

Q1 (L1). Spot the carboxylic acid

Which of these contains a carboxyl group $- C (= O) - O - H$ ? (a) $C H_{3} C H_{2} O H$ (b) $C H_{3} C H O$ (c) $C H_{3} C O O H$ (d) $C H_{3} C O C H_{3}$

Recall Solution

A carboxyl group is a carbonyl ( $C = O$ ) and a hydroxyl ( $O - H$ ) on the same carbon.

(a) alcohol only — hydroxyl but no $C = O$ . ✗
(b) aldehyde — $C = O$ but the H is on carbon, no $O - H$ . ✗ (see Aldehydes and ketones)
(c) $C H_{3} C O O H$ — carbonyl + hydroxyl on one carbon. ✓
(d) ketone — $C = O$ flanked by two carbons, no $O - H$ . ✗

Answer: (c).

Q2 (L1). Name the derivative

For $R - C (= O) - L$ , match the leaving group $L$ to the derivative name: $- C l$ , $- O R^{'}$ , $- N H_{2}$ , $- O O C R$ .

Recall Solution

$L$	Derivative
$- C l$	acid chloride ( $R C O C l$ )
$- O O C R$	anhydride ( $(R C O)_{2} O$ )
$- O R^{'}$	ester ( $R C O O R^{'}$ )
$- N H_{2}$	amide ( $R C O N H_{2}$ )

Reactivity order: $chloride > anhydride > ester > amide$ . Why this order (not rote — two effects push the same way):

Leaving-group ability: substitution ends by kicking out $L^{-}$ . $C l^{-}$ is a stable, weak base → leaves easily → chloride most reactive. $^{-} N H_{2}$ is a terrible, high-energy leaving group → amide least reactive.
Lone-pair donation into $C = O$ (Resonance and delocalisation): $N$ donates its lone pair strongly into the carbonyl, making the amide carbon electron-rich and "satisfied" (unreactive). $C l$ donates poorly, so the chloride carbon stays hungry ( $δ +$ ).

Both effects rank the family the same way — hence the ladder.

Level 2 — Application

Q3 (L2). Rank by acidity

Rank from most to least acidic (smallest $p K_{a}$ first): $C H_{3} C O O H$ , $C l C H_{2} C O O H$ , $C l_{3} C C O O H$ , $C H_{3} C H_{2} C O O H$ .

Recall Solution

Use the Inductive effect: electron-withdrawing $C l$ pulls charge off the carboxylate, spreading the negative charge, stabilising the anion ⟹ stronger acid. More $C l$ = stronger. Alkyl groups donate electrons ⟹ weaker.

$C l_{3} C C O O H (0.65) > C l C H_{2} C O O H (2.86) > C H_{3} C O O H (4.76) > C H_{3} C H_{2} C O O H (4.87)$

Most → least acidic: $C l_{3} C C O O H > C l C H_{2} C O O H > C H_{3} C O O H > C H_{3} C H_{2} C O O H$ .

Q4 (L2). $K_{a}$ from $p K_{a}$

Acetic acid has $p K_{a} = 4.76$ . Find $K_{a}$ .

Recall Solution

By definition $p K_{a} = - lo g_{10} K_{a}$ , so invert it: $K_{a} = 1 0^{- p K_{a}} = 1 0^{- 4.76} \approx 1.7 \times 1 0^{- 5} .$ Answer: $K_{a} \approx 1.7 \times 1 0^{- 5}$ — small, so acetic acid is only partially ionised (weak acid).

Q5 (L2). Predict the HVZ product

$C H_{3} C H_{2} C H_{2} C O O H + B r_{2} cat. red P$ ?

Recall Solution

HVZ halogenates the α-carbon — the carbon directly attached to the carbonyl of $- C O O H$ (see the notation reminder at the top). Number the chain from the acid carbon: $C1 (C O O H) C O O H - C2 = α C H_{2} - C3 = β C H_{2} - C4 = γ C H_{3}$ The α-carbon is the first $C H_{2}$ (C2), the one touching $C O O H$ . Only it can become the nucleophilic enol carbon (Keto–enol tautomerism), so only it gets brominated. The $B r$ therefore goes on C2, giving: $Product: C H_{3} C H_{2} C H B r - C O O H (2-bromobutanoic acid) .$ Reading the product left-to-right: $C H_{3}$ (C4) — $C H_{2}$ (C3) — $C H B r$ (C2 = α) — $C O O H$ (C1). The bromine sits on the carbon adjacent to $C O O H$ , exactly as required.

Level 3 — Analysis

Q6 (L3). The ¹⁸O labelling experiment

In Fischer esterification $C H_{3} C O O H + C H_{3}^{18} O H H^{+} ?$ the alcohol's oxygen is the heavy isotope $^{18} O$ . Where does the $^{18} O$ end up — in the ester or in the water? Explain using the mechanism.

Recall Solution

The acid-catalysed mechanism runs in six steps, remembered as "PADPED" = Protonate, Add, Deprotonate, Protonate, Eliminate, Deprotonate:

Protonate the carbonyl $O$ → carbon becomes strongly $δ +$ (more electrophilic).
Add the alcohol $C H_{3}^{18} O H$ to that carbon → tetrahedral intermediate. The alcohol keeps its $O - C H_{3}$ bond intact — the red $^{18} O$ is now bonded to the central carbon.
Deprotonate the positively charged oxygen → neutral tetrahedral intermediate carrying two $- O H$ groups plus the red $-^{18} O C H_{3}$ .
Protonate one of the original ( $^{16} O$ ) $- O H$ groups → turns it into $- O H_{2}^{+}$ , a good leaving group.
Eliminate that $- O H_{2}^{+}$ as water → the $C = O$ re-forms. This departing oxygen is the one that came from the acid.
Deprotonate → neutral ester; catalyst $H^{+}$ regenerated.

The figure traces the red ( $^{18} O$ ) oxygen: it enters with the alcohol in step 2 and is still bonded in the final ester. The oxygen that leaves as water (step 5) is the black one from the acid's $- O H$ .

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

Figure: reactants (top) → product (bottom). The alcohol's oxygen is drawn in red $^{18} O$ ; follow it — it stays bonded all the way into the ester, while the acid's black oxygen departs as ordinary $H_{2}^{16} O$ .

The red oxygen ( $^{18} O$ , from the alcohol) stays bonded through the whole mechanism → it ends up in the ester.
The black oxygen (from the acid's $- O H$ ) leaves as water.

Answer: the $^{18} O$ appears in the ester, and the water is ordinary $H_{2}^{16} O$ . This proves the ester C–O bond comes from the alcohol, and it is the acid that loses its $- O H$ .

Q7 (L3). Why the reactivity order?

Explain, in terms of leaving-group ability and lone-pair donation, why an acid chloride reacts with water far faster than an amide.

Recall Solution

Two effects push the same way:

Leaving group: in the tetrahedral intermediate the $C = O$ re-forms by expelling $L^{-}$ . For the chloride $L = C l^{-}$ (stable, weak base, happy to leave); for the amide $L =^{-} N H_{2}$ (a terrible, high-energy leaving group). Better leaving group ⟹ faster substitution.
Lone-pair donation into $C = O$ (Resonance and delocalisation): nitrogen donates its lone pair strongly into the carbonyl (good overlap, low electronegativity). This makes the amide carbonyl electron-rich and "satisfied" — the carbon is barely $δ +$ , so nucleophiles ignore it. Chlorine donates poorly (mismatched orbital size), leaving the chloride carbonyl very $δ +$ and "hungry."

Answer: acid chloride ≫ amide because $C l^{-}$ leaves easily and the chloride carbonyl is far more electrophilic.

Level 4 — Synthesis

Q8 (L4). Two routes to ethyl acetate — compare

You must make $C H_{3} C O O C H_{2} C H_{3}$ . Write two routes: (i) Fischer, (ii) via the acid chloride. State catalyst, reversibility, and speed for each, then say which gives higher yield and why.

Recall Solution

Route (i) — Fischer: $C H_{3} C O O H + C H_{3} C H_{2} O H H^{+} ⇌ C H_{3} C O O C H_{2} C H_{3} + H_{2} O$ Needs conc. $H_{2} S O_{4}$ catalyst; reversible; slow. Yield limited by equilibrium — must push with excess alcohol or by removing water (Le Chatelier's principle).

Route (ii) — via acid chloride: $C H_{3} C O O H S O C l_{2} C H_{3} C O C l C H_{3} C H_{2} O H C H_{3} C O O C H_{2} C H_{3} + H C l$ No acid catalyst needed; fast; essentially irreversible because $C l^{-}$ is a superb leaving group and gaseous $H C l$ escapes.

Higher yield: Route (ii). The chloride is high up the reactivity ladder, so the reaction runs downhill to completion; Fischer is capped by its equilibrium.

Q9 (L4). Design a synthesis of an α-amino acid precursor

Starting from propanoic acid $C H_{3} C H_{2} C O O H$ , make 2-bromopropanoic acid, then convert it to its ethyl ester. Give reagents for each step.

Recall Solution

Step 1 — HVZ (install $B r$ at Cα): $C H_{3} C H_{2} C O O H B r_{2}, cat. red P C H_{3} C H B r - C O O H$ Step 2 — esterify (Fischer, or via chloride for a cleaner job): $C H_{3} C H B r - C O O H + C H_{3} C H_{2} O H H^{+} ⇌ C H_{3} C H B r - C O O C H_{2} C H_{3} + H_{2} O$ Product: ethyl 2-bromopropanoate, $C H_{3} C H B r - C O O C H_{2} C H_{3}$ . (The $B r$ at Cα could later be displaced by $N H_{3}$ to reach the amino acid alanine — but that step is beyond this exercise.)

Level 5 — Mastery

Q10 (L5). Drive the equilibrium quantitatively

For a Fischer esterification the equilibrium constant is $K = 4.0$ : $K = \frac{[ ester ] [ H _{2} O ]}{[ acid ] [ alcohol ]} .$ Start with 1.0 mol acid + 1.0 mol alcohol in a fixed 1.0 L vessel. Let $x$ mol of ester form. (a) Find $x$ at equilibrium. (b) What fraction of the acid is converted?

Recall Solution

At equilibrium: ester $= x$ , water $= x$ , acid $= 1 - x$ , alcohol $= 1 - x$ . Substitute: $K = \frac{x \cdot x}{( 1 - x ) ( 1 - x )} = (\frac{x}{1 - x})^{2} = 4.0.$ Take the (positive) square root: $\frac{x}{1 - x} = 2 \Rightarrow x = 2 (1 - x) \Rightarrow x = \frac{2}{3} \approx 0.667.$ (a) $x = 0.667$ mol ester. (b) Fraction converted $= x /1.0 = 0.667 = 66.7%$ . The rest stays as acid — this cap is exactly why we add excess alcohol or strip water (Le Chatelier's principle).

Q11 (L5). Push it with excess alcohol

Same $K = 4.0$ , but now start with 1.0 mol acid + 4.0 mol alcohol. Find $x$ and the new fraction of acid converted. Comment.

Recall Solution

Now acid $= 1 - x$ , alcohol $= 4 - x$ , ester $= x$ , water $= x$ : $K = \frac{x \cdot x}{( 1 - x ) ( 4 - x )} = 4.0.$ Clear the denominator and expand the product $(1 - x) (4 - x) = 4 - 5 x + x^{2}$ : $x^{2} = 4 (4 - 5 x + x^{2}) = 16 - 20 x + 4 x^{2} .$ Bring everything to one side (subtract $x^{2}$ from both sides): $0 = 16 - 20 x + 4 x^{2} - x^{2} = 3 x^{2} - 20 x + 16.$ Solve with the quadratic formula: $x = \frac{20 \pm ( - 20 ) ^{2} - 4 \cdot 3 \cdot 16}{2 \cdot 3} = \frac{20 \pm 400 - 192}{6} = \frac{20 \pm 208}{6} .$ $208 \approx 14.42$ , so the two roots are $x = \frac{20 - 14.42}{6} \approx 0.930 or x = \frac{20 + 14.42}{6} \approx 5.74.$ Discard $x \approx 5.74$ : we only started with $1.0$ mol of acid, so at most $1.0$ mol of ester can form; $x$ must satisfy $0 < x < 1$ . Only $x \approx 0.930$ is physical. $x \approx 0.930$ mol ester ⟹ $93%$ of the acid converted.

Comment: flooding with alcohol shifted equilibrium right (66.7% → 93%) exactly as Le Chatelier's principle predicts — no change to $K$ , just a change in the mixture.

Q12 (L5). Saponification vs Fischer

Ethyl acetate is hydrolysed (i) with dilute $H^{+}$ and (ii) with $N a O H$ (Saponification). One is reversible, one is not. Which, and why?

Recall Solution

(i) Acid hydrolysis is just Fischer run in reverse — every step is an equilibrium, so it is reversible. Water attacks; you get acid + alcohol back, but capped by $K$ .

(ii) Base hydrolysis (saponification) is irreversible. The final step deprotonates the carboxylic acid to the carboxylate $C H_{3} C O O^{-}$ , whose charge is delocalised over two oxygens (Resonance and delocalisation). A carboxylate is a lousy electrophile and its negative charge repels the incoming alcohol/alkoxide, so the reverse reaction is dead. The reaction runs to completion.

Answer: acid-catalysed hydrolysis is reversible; saponification is irreversible, because the carboxylate product is a stable, unreactive sink.

Recall One-line self-checks

Carboxyl group needs $O - H$ on the carbonyl carbon (not $C - H$ ) ::: True — that distinguishes it from aldehydes. HVZ halogenates the α-carbon only ::: because only it becomes the nucleophilic enol carbon. In Fischer esterification the water's oxygen comes from the ==acid's $- O H$ == ::: proven by the ¹⁸O experiment. Excess alcohol changes the equilibrium position but not ::: the value of $K$ . Saponification is irreversible because the product is the stable ::: carboxylate anion.