Cambridge International AS & A Level Chemistry 9701 · A Level topic 33

Carboxylic acids and derivatives (A Level)

What this chapter covers33.1–33.3

Vinegar, the sting of an ant, the sourness of rhubarb and the preservative in many soft drinks are all due to carboxylic acids. The –COOH group makes these compounds acidic, but they are far weaker acids than hydrochloric acid, and far stronger than phenol or ethanol. Replace the –OH of the group with –Cl and the compound changes character completely: an acyl chloride fumes in moist air and reacts at room temperature with water, alcohols, phenols, ammonia and amines. Acyl chlorides are among the most useful reagents in organic synthesis, used to make esters that cannot be made directly and amides such as those in medicines.

This chapter builds on the AS work on carboxylic acids and esters in topic 19. It covers the preparation of benzoic acid, the two carboxylic acids that can be oxidised further, the conversion of acids into acyl chlorides, and a quantitative and structural account of why some acids are stronger than others. It ends with the reactions of acyl chlorides and the addition–elimination mechanism that all of them share.

Making benzoic acid33.1.1

Any alkylbenzene can be converted into benzoic acid by oxidising its side chain. Methylbenzene is heated under reflux with potassium manganate(VII) in alkaline solution. The purple manganate(VII) is reduced (a brown precipitate of manganese(IV) oxide forms), and the methyl group is oxidised to a carboxyl group. Because the solution is alkaline, the product is present as the benzoate ion, C₆H₅COO⁻; adding dilute acid (for example dilute sulfuric acid) at the end protonates it, and benzoic acid, which is only sparingly soluble in cold water, separates as white crystals.

The ring itself is not oxidised: its delocalised π system is stable. Every alkyl side chain, whatever its length, is cut back to a single –COOH group on the ring, so ethylbenzene and propylbenzene also give benzoic acid, and a ring with two alkyl side chains gives a dicarboxylic acid (for example, 1,3-dimethylbenzene gives benzene-1,3-dicarboxylic acid).

Making acyl chlorides33.1.2, 33.3.1

An acyl chloride contains the group –COCl: the –OH of a carboxylic acid has been replaced by –Cl. Ethanoyl chloride, CH₃COCl, is derived from ethanoic acid; benzoyl chloride, C₆H₅COCl, from benzoic acid. Acyl chlorides are named by replacing -oic acid with -oyl chloride. Three reagents make the conversion, and all of them must be used in the absence of water, because water reacts with the acyl chloride formed.

With SOCl₂ the gases escape from the mixture, leaving the acyl chloride, which can be purified by distillation; with PCl₅ the liquid POCl₃ has to be separated from it. Alcohols react with the same three reagents to give chloroalkanes (topic 16), so every –OH group in a molecule is replaced.

Carboxylic acids that can be oxidised33.1.3

The carbon of a –COOH group is already in a high oxidation state, so carboxylic acids are generally not oxidised by the usual laboratory oxidising agents — that is why oxidation of a primary alcohol stops at the acid. Two acids are exceptions, and each has a structural reason.

Methanoic acid

Methanoic acid, HCOOH, is the only carboxylic acid whose carboxyl carbon is bonded to hydrogen. The molecule therefore contains an H–C=O group, as an aldehyde does, and it behaves as a reducing agent in the same tests:

• warm with Tollens' reagent: a silver mirror forms;
• warm with Fehling's solution: the blue solution gives a brick-red precipitate of copper(I) oxide;
• warm with acidified KMnO₄: the purple colour is removed; with acidified K₂Cr₂O₇, orange turns green.

Ethanedioic acid

Ethanedioic acid (oxalic acid), HOOCCOOH, has no H–C=O group and does not react with Tollens' or Fehling's reagents. Its two carboxyl carbons are, however, bonded directly to each other, and warm acidified manganate(VII) oxidises it, breaking that C–C bond and releasing carbon dioxide. The purple solution is decolourised and bubbles of gas are seen:

Past-paper practice · Set 33A · Benzoic acid, acyl chlorides and oxidation of carboxylic acids

Examination questions on this part of the unit. Try each one on paper before opening the answer.

Question 33A.1[4]

Answer and marking guidance

(i) L is benzene-1,3-dicarboxylic acid: both CH₃ side chains are oxidised to –COOH, C₆H₄(COOH)₂ ✔. (ii) step 1: heat (under reflux) with acidified or alkaline KMnO₄ (then acidify) ✔; step 2: PCl₅, or SOCl₂, or heat with PCl₃ ✔. (iii) C₈H₆O₄ + 2PCl₅ → C₈H₄O₂Cl₂ + 2POCl₃ + 2HCl, or C₈H₆O₄ + 2SOCl₂ → C₈H₄O₂Cl₂ + 2SO₂ + 2HCl, or 3C₈H₆O₄ + 2PCl₃ → 3C₈H₄O₂Cl₂ + 2H₃PO₃ ✔. Examiner insight: (i) was usually correct. In (ii), common errors were leaving out one component of step 1 — acidic or alkaline, or heat — and suggesting HCl, or Cl₂ with AlCl₃, for step 2. (iii) was found very difficult: products were often wrong, and many did not see that two molecules of PCl₅ or SOCl₂ are needed because there are two –COOH groups.

Question 33A.2[2]

Answer and marking guidance

(i) reaction 1: POCl₃ and HCl; reaction 2: SO₂ and HCl — both needed ✔. (ii) all the by-products of reaction 2 (SO₂ and HCl) are gases, so no liquid by-product has to be separated from the acyl chloride ✔. Examiner insight: (i) discriminated well. In (ii) many saw the significance of SO₂ and HCl being gaseous; some wrongly referred to aqueous conditions — the synthesis of benzoyl chloride would not work with water present.

Question 33A.3[6]

Part (a) of the same question is 33B.4.

Answer and marking guidance

(b)(i) methanoic acid only ✔. (ii) methanoic acid and ethanedioic acid ✔. (c)(i) A is CH₃COCl ✔, ethanoyl chloride ✔. (ii) step 1: PCl₅, PCl₃ or SOCl₂ ✔; step 2: NH₃ ✔. Examiner insight: the reaction of methanoic acid with Fehling's reagent, and of methanoic and ethanedioic acids with acidified manganate(VII), were not well known. The structural formula of A was well known but the name less so — "ethyl chloride" and "chloroethanoic acid" were seen. For step 2, answers needed to show that any water present would also react with the ethanoyl chloride.

Question 33A.4[2]

Answer and marking guidance

Two of the three points for 1 mark, all three for 2 marks: type: oxidation; observation: the purple solution is decolourised (becomes colourless or pale pink), or bubbles of gas; equation: HOOCCOOH + [O] → 2CO₂ + H₂O, or 5HOOCCOOH + 2MnO₄⁻ + 6H⁺ → 10CO₂ + 8H₂O + 2Mn²⁺ ✔✔. Examiner insight: performance was good. The equation was the most challenging part: some used [H] even though they had called the reaction an oxidation.

Question 33A.5[3]

Answer and marking guidance

Test 1: Tollens' reagent, warm → silver mirror; or Fehling's solution, warm → (brick-)red precipitate ✔. Test 2: acidified MnO₄⁻, warm → decolourises (or bubbles) ✔. Both observations correct ✔. Examiner insight: the award of all three marks was very rare. Many suggested reagents that react with none of the three acids. Only methanoic acid has an H–C=O group that Tollens' or Fehling's reagent can oxidise; methanoic and ethanedioic acids are both oxidised by warm acidified manganate(VII).

Carboxylic acids, phenols and alcohols33.1.4

A Brønsted–Lowry acid donates a proton. For every compound in this section the proton comes from an O–H bond, and the equilibrium is of the same form:

The position of this equilibrium, measured by K_a or pK_a (pK_a = −log K_a; the smaller the pK_a, the stronger the acid), depends on what X is. Two linked ideas explain every comparison:

• the O–H bond: a group X that withdraws electron density from the oxygen weakens and polarises the O–H bond, so H⁺ is released more easily; a group that donates electron density strengthens it;
• the anion (conjugate base): a group that spreads out the negative charge of X–O⁻ stabilises the anion, so the equilibrium lies further to the right; a group that concentrates charge on the oxygen destabilises it.

Either explanation, applied correctly to each compound, earns credit; using both together gives the most complete answer.

Ethanoic acid and the carboxylate ion

In ethanoic acid the O–H group is attached to a carbonyl carbon. The C=O group is strongly electron-withdrawing (it has an electronegative oxygen atom and a polar double bond), so the O–H bond is weakened. When the proton is lost, the negative charge of the ethanoate ion is delocalised over the carbon and both oxygen atoms: a p orbital on each oxygen overlaps with the p orbital of the carbon, and the two C–O bonds become identical. The charge is shared by two electronegative atoms, which is a very effective way to stabilise it.

Phenol and the phenoxide ion

In phenol a lone pair on oxygen is delocalised into the ring. In the phenoxide ion the charge is spread over the oxygen and the ring carbons (chapter 32), stabilising the ion — but most of that charge is carried by carbon atoms, which are less electronegative than oxygen, so the stabilisation is smaller than in a carboxylate ion. Phenol (pK_a about 10) is a much weaker acid than ethanoic acid.

Alcohols and the alkoxide ion

In ethanol the O–H is attached to an alkyl group, which is electron-donating (a positive inductive effect). This strengthens the O–H bond and increases the charge density on the oxygen of the ethoxide ion, destabilising it. Ethanol is a weaker acid even than water.

The order of acid strength is therefore:

The evidence: reactions with sodium, sodium hydroxide and sodium carbonate

The difference in strength is seen in the reactions of the three classes with sodium compounds. Only a carboxylic acid is a strong enough acid to release CO₂ from a carbonate or hydrogencarbonate; phenol reacts with the strong base NaOH but not with carbonate; an alcohol reacts with neither.

Chlorine-substituted carboxylic acids33.1.5

Replacing a hydrogen atom on the carbon next to the –COOH group with chlorine makes the acid much stronger. Chlorine is more electronegative than carbon and hydrogen, and it draws electron density towards itself along the chain of σ bonds (a negative inductive effect). Electron density is drawn away from the carboxyl group, so:

• the O–H bond is further weakened and polarised, and H⁺ is released more easily;
• the negative charge of the carboxylate ion is further spread out (its charge density is reduced), so the anion is more stable.

Three rules follow, and all three are examined:

1. More chlorine atoms, stronger acid. Cl₃CCOOH > Cl₂CHCOOH > ClCH₂COOH > CH₃COOH.
2. Closer to –COOH, stronger acid. The inductive effect is transmitted through σ bonds and weakens rapidly with distance: 2-chloropropanoic acid > 3-chloropropanoic acid > propanoic acid.
3. Electron-donating alkyl groups weaken an acid. A longer alkyl group has a slightly larger positive inductive effect, so ethanoic acid is stronger than butanoic acid. The effect is small compared with that of a chlorine atom.

Dicarboxylic acids: interpreting data

A dicarboxylic acid ionises in two stages, each with its own pK_a. The data below come from an examination question, with ethanoic acid (pK_a 4.76) for comparison.

The same ideas explain the pattern. Every pK_a(1) is below 4.76 because the second –CO₂H group is electron-withdrawing, like a chlorine atom; its effect, like that of chlorine, falls off as the number of CH₂ groups between the two carboxyl groups increases. Every pK_a(2) is above 4.76: the second proton has to be removed from an ion that is already negatively charged, and the –CO₂⁻ group already present repels further negative charge and is electron-donating, so the second ionisation is harder. As n increases, the two charges are further apart and pK_a(2) approaches the ethanoic acid value.

Past-paper practice · Set 33B · Relative acidities

Examination questions on this part of the unit. Try each one on paper before opening the answer.

Question 33B.1[6]

Part (c) of the same question is 33A.2 and part (d) is 33C.1.

Answer and marking guidance

(a) M1 order: benzoic acid > phenol > phenylmethanol ✔; two of the following for M2 and M3 ✔✔: benzoic acid — the negative inductive effect of C=O weakens the O–H bond, or the negative charge of the carboxylate ion is delocalised by C=O / over two O atoms, stabilising it; phenol — a lone pair on oxygen is delocalised into the ring and the O–H bond is weakened; phenylmethanol — the positive inductive effect of the CH₂ group strengthens the O–H bond. (b) Na(s): ✓ ✓ ✓; NaOH(aq): ✓ ✗ ✓; Na₂CO₃(aq): ✓ ✗ ✗ (columns benzoic acid, phenylmethanol, phenol); three correct 1 mark, six correct 2, nine correct 3 ✔✔✔. Examiner insight: in (a) many referred accurately to the electronegative C=O group, lone-pair delocalisation in phenol and the electron-donating CH₂ group. Some did not say which compound they meant, or wrote "the oxygen atom" for benzoic acid without saying which of its two oxygens. The group in phenylmethanol can be called an alkyl or CH₂ group but not a methyl group. It is better to say the O–H bond is strengthened or weakened than "harder or easier to break", and "OH group" should not be written when the O–H bond is meant. In (b) many ignored the instruction to put a cross where there is no reaction; blank boxes could not be credited.

Question 33B.2[4]

Answer and marking guidance

M1 order: ethanoic acid > butanoic acid > water > ethanol ✔. M2: a reason for one compound in terms of an electron-donating or electron-withdrawing group strengthening or weakening the O–H bond, or stabilising the anion ✔. Two of ✔✔: ethanol — positive inductive (electron-donating) effect of the ethyl group; butanoic acid — positive inductive effect of the propyl group; the acids — negative inductive effect of C=O, or the negative charge delocalised over COO⁻. Examiner insight: a difficult question. The importance of electron-donating groups such as C₂H₅ and electron-withdrawing groups such as C=O to the strength of the O–H bond was not appreciated by all. "The electronegative oxygen atom" did not score, since every compound contains oxygen. Many thought ethanol more acidic than water.

Question 33B.3[4]

Answer and marking guidance

M1 order: chloroethanoic acid > ethanoic acid > phenol > ethanol ✔. M2: acidity linked to a weakened O–H bond or a stabilised anion ✔. Two explanations for one mark, four for two ✔✔: ClCH₂COOH > CH₃COOH — electron-withdrawing (negative inductive) effect of Cl; CH₃COOH > phenol — electron-withdrawing effect of C=O; phenol > ethanol — lone pair of oxygen delocalised into the ring; ethanol weakest — the alkyl group is electron-donating. Examiner insight: well understood, and most linked acidity to weakening of the O–H bond. The roles of electron-donating groups (C₂H₅) and electron-withdrawing groups (Cl and C=O) were not appreciated by all. "The electronegative oxygen atom" was not credited because all four compounds contain oxygen.

Question 33B.4[3]

Parts (b) and (c) of the same question are 33A.3.

Answer and marking guidance

Order: ethanamide < ethanoic acid < trichloroethanoic acid ✔. Two of ✔✔: ethanamide is neutral / not a proton donor; chlorine is electronegative / electron-withdrawing; the O–H bond is weakened / the anion is stabilised; a correct statement linking acid strength to H⁺ donation. Examiner insight: found challenging, but the central idea of acidity — the ability to donate H⁺ — was understood by many and used as the basis of their answers.

Question 33B.5[3]

Answer and marking guidance

M1 order: 2-chloropropanoic acid > 3-chloropropanoic acid > propanoic acid ✔. M2: both chloro acids are more acidic because the electronegative Cl weakens the O–H bond / stabilises the carboxylate anion ✔. M3: 2-chloropropanoic acid is more acidic than 3-chloropropanoic acid because its Cl is closer to the –CO₂H group, so weakens the O–H bond, or stabilises the anion, more ✔.

Question 33B.6[3]

Answer and marking guidance

Order: Cl₃CCO₂H > ClCH₂CO₂H > CH₃CO₂H ✔. Two of ✔✔: Cl is electronegative / electron-withdrawing and trichloroethanoic acid has three Cl; this weakens the O–H bond so it ionises more readily, or stabilises the anion by reducing the charge density on COO⁻; CH₃ is electron-donating, so the O–H bond of ethanoic acid is stronger and it ionises less. Examiner insight: many gave the correct trend and linked it to the electronegativity of chlorine, but only a small number explained that the O–H bond is weakened and so ionises more easily.

Question 33B.7[3]

Answer and marking guidance

Order: benzoic acid > 4-methylphenol > phenylmethanol ✔. The methylphenoxide ion is stabilised by delocalisation of an oxygen lone pair over the ring ✔. Benzoic acid has an extra electronegative oxygen, the electron-withdrawing C=O ✔. Examiner insight: many had the right order, though some thought methylphenol more acidic than benzoic acid. Few explained adequately that benzoic acid is more acidic because the negative charge of its anion is delocalised over two electronegative oxygen atoms, while methylphenol is more acidic than phenylmethanol because its anion's charge is delocalised over the ring.

Question 33B.8[2]

Parts (d)(i)–(ii) and (e) of this question are in chapter 32.

Answer and marking guidance

All three points for two marks ✔✔: the (CO)O–H bond is weaker, so K donates H⁺ more easily; because of the negative inductive (electron-withdrawing) effect of the C=O / COOH group; the carboxylate anion is stabilised (the phenoxide ion less so). Examiner insight: many described how the electronegative oxygen of C=O weakens the O–H bond, but some left out that it also stabilises the carboxylate anion.

Question 33B.9[5]

Answer and marking guidance

(i) CH₃CHClCOOH + H₂O ⇌ CH₃CHClCOO⁻ + H₃O⁺ (or without water, giving H⁺) ✔. (ii) M1 [H⁺] = 10^−1.51 = 0.0309 mol dm⁻³ ✔; M2 concentration of acid = 0.150 ÷ 0.250 = 0.600 mol dm⁻³; K_a = 0.0309² ÷ 0.600 = 1.59 × 10⁻³ mol dm⁻³; pK_a = −log(1.59 × 10⁻³) = 2.80 ✔. (iii) CH₃CHClCOOH is the stronger acid (it gives a higher [H⁺], so a lower pH) ✔; because the electron-withdrawing effect of the Cl substituent weakens the O–H bond / stabilises the carboxylate anion ✔. Examiner insight: (i) was answered well; the commonest error was an equation for hydrolysis of the C–Cl bond (CH₃CHClCOOH + H₂O ⇌ CH₃CH(OH)COOH + HCl). (ii) was generally answered well. In (iii) most identified that the electronegative chlorine weakens the O–H bond, but many did not link this to the difference in pH in terms of acid strength.

Why acyl chlorides are so reactive33.3.4

Open a bottle of ethanoyl chloride and white, steamy fumes appear at once: the liquid is reacting with water vapour in the air, releasing hydrogen chloride. The reactivity comes from the carbonyl carbon. It is bonded to two strongly electronegative atoms, oxygen and chlorine, both of which withdraw electron density from it. The carbon therefore carries a large δ+ charge and is readily attacked by any species with a lone pair — a nucleophile. Because the carbon is trigonal planar, the nucleophile can approach from above or below the plane without hindrance.

This explains the relative ease of hydrolysis met in chapter 31:

• Acyl chloride — carbon bonded to O and Cl, strongly δ+; hydrolysed rapidly by cold water.
• Alkyl chloride — carbon bonded to one electronegative atom only, and the alkyl group is electron-donating; hydrolysed only on heating with NaOH(aq).
• Aryl chloride — a lone pair of chlorine is delocalised into the ring, giving the C–Cl bond partial double-bond character; it is not hydrolysed.

The addition–elimination mechanism33.3.2, 33.3.3

All the reactions of acyl chlorides in this syllabus follow the same two-stage mechanism, called addition–elimination (or nucleophilic addition–elimination). Water is used as the example in Figure 33.6.

1. Addition. A lone pair on the oxygen of a water molecule (the electron-rich site) is attracted to the δ+ carbonyl carbon (the electron-deficient site). A curly arrow from that lone pair to the carbon shows the new O–C bond forming. At the same time, one pair of electrons of the C=O double bond moves onto the oxygen — the arrow starts on the C=O bond and ends on O. The result is a tetrahedral intermediate, in which the carbon is bonded to four groups: CH₃, O⁻, Cl and the oxygen of water, which now carries a positive charge.
2. Elimination. A lone pair on the O⁻ moves back to re-form the C=O double bond (arrow from the lone pair to the C–O bond), and the C–Cl bond breaks heterolytically, both electrons going to chlorine (arrow from the C–Cl bond to Cl). A chloride ion is eliminated.
3. Loss of H⁺. The positively charged oxygen loses a proton, which combines with Cl⁻: the overall products are ethanoic acid and HCl.

Reactions of acyl chlorides33.3.2

Every reaction takes place at room temperature and produces HCl. The nucleophile decides the product: an O–H compound gives an acid or an ester; an N–H compound gives an amide.

With nitrogen nucleophiles, the new bond is formed between the carbonyl carbon and nitrogen, and the amine loses one hydrogen from nitrogen. A primary amine RNH₂ gives RCONHR′; a secondary amine gives RCONR′₂, whose nitrogen carries no hydrogen. A tertiary amine has no N–H and cannot form an amide. Amides are neutral: the lone pair on nitrogen is delocalised onto the C=O group, so, unlike an amine, an amide is not basic (chapter 34).

Making esters, including esters of phenols33.2.1

At AS Level, esters were made by heating a carboxylic acid with an alcohol and concentrated sulfuric acid. That route is slow and reversible, and it does not work at all for phenols: the lone pair of the phenol oxygen is delocalised into the ring, which makes phenol too weak a nucleophile to attack a carboxylic acid. An acyl chloride is so much more reactive than the acid that it reacts with phenols too. The reaction is faster still when the phenol is first dissolved in sodium hydroxide solution, which converts it into the phenoxide ion, a much better nucleophile. The ester phenyl benzoate is made this way:

The ester is named from the phenol (phenyl) and the acyl chloride (benzoate). Similarly, ethyl ethanoate is made from ethanol and ethanoyl chloride.

Past-paper practice · Set 33C · Acyl chlorides, esters and amides

Examination questions on this part of the unit. Try each one on paper before opening the answer.

Question 33C.1[5]

Answer and marking guidance

(i) Left-hand side: lone pair on the O of water; arrow from that O to the carbonyl C; dipole on C=O (C^δ+, O^δ−); arrow from the C=O bond to O — two correct 1 mark, all four 2 marks ✔✔. Right-hand side: M3 the correct tetrahedral intermediate (O⁻ and H₂O⁺ on the same carbon, Cl still attached) ✔; M4 arrow from a lone pair on O⁻ to re-form C=O AND arrow from the C–Cl bond to Cl ✔. (ii) addition–elimination ✔. Examiner insight: a difficult and very precise task. Most made a start and some reproduced the whole mechanism to a high standard; it required careful drawing of every curly arrow. Single-headed (fish-hook) arrows, which show the movement of single electrons, are wrong here and gained no credit. The name addition–elimination was not well known.

Question 33C.2[7]

Answer and marking guidance

(a) HOCH₂COOH + 2SOCl₂ → ClCH₂COCl + 2SO₂ + 2HCl ✔ (both –OH groups are replaced by Cl). (b) to remove / neutralise the acid (H⁺, HCl, SO₂) produced, or to react with any unreacted W ✔. (c) M1 arrow from the lone pair on :NH₂ to the carbonyl C^δ+; M2 dipole on C=O AND arrow from the C=O bond to O; M3 correct intermediate with charges; M4 arrow from the lone pair on O⁻ to re-form C=O AND arrow from C–Cl to Cl ✔✔✔✔. (d) the N atom can donate its lone pair ✔. Examiner insight: (a) was found difficult; the commonest error was HOCH₂COOH + SOCl₂ → ClCH₂COCl + SO₂ + H₂O. In (b), "to act as a buffer" and "to remove chloride ions" were common. In (c), errors included omitting the C=O dipole, omitting the lone pair on N or on O⁻ in the intermediate, omitting the arrow on the C=O bond, directing the N lone-pair arrow to the wrong atom, and wrong intermediates. In (d) many stated that N has a lone pair but did not say that it is donated.

Question 33C.3[2]

Answer and marking guidance

CH₃COCl + CH₃NH₂ → CH₃CONHCH₃ + HCl. M1 correct formula of CH₃COCl or of CH₃CONHCH₃ ✔; M2 the rest of the equation ✔. Examiner insight: answered well by many. Common errors were a wrong formula for ethanoyl chloride, a primary amide as the product, and leaving out HCl.

Question 33C.4[1]

Answer and marking guidance

R is proline with an ethanoyl group on the ring nitrogen: the N–H is replaced by N–COCH₃, and the –COOH group is kept (C₇H₁₁NO₃) — skeletal formula only ✔. Examiner insight: found difficult; the most common mistake was a structure that was not skeletal. (The reagent for reaction 3, reducing –COOH to –CH₂OH, is LiAlH₄.)

Question 33C.5[3]

Answer and marking guidance

Step 1: PCl₅, or PCl₃ (+ heat), or SOCl₂, added to propanoic acid ✔. Product of step 1: propanoyl chloride, CH₃CH₂COCl ✔. Step 2: add the product of step 1 to phenol in NaOH(aq) ✔. Examiner insight: propanoyl chloride and the reagent for step 1 were usually correct. Many did not describe generating the phenoxide ion (phenol in NaOH) before adding the propanoyl chloride.

Question 33C.6[3]

Answer and marking guidance

Step 1: treat benzoic acid with SOCl₂ or PCl₅ to make the acyl chloride ✔; the intermediate is C₆H₅COCl ✔; step 2: dissolve the 4-methylphenol in NaOH(aq) (and shake with the benzoyl chloride) ✔. Examiner insight: most did not follow the instruction to give a two-step route, and simply heated the acid and the phenol with concentrated sulfuric acid. Phenols are not nucleophilic enough to be esterified by the normal (carboxylic acid) route.

Question 33C.7[3]

The same comparison is developed in chapter 31.

Answer and marking guidance

Order: acyl chloride > alkyl chloride > aryl chloride ✔. Two of ✔✔: acyl chlorides — the carbon of C–Cl is more electron-deficient because it is also attached to an oxygen atom (two electronegative atoms), or the C–Cl bond is weakened; aryl chlorides — the C–Cl bond is part of the delocalised system (a lone pair on Cl is delocalised into the π ring), so it has partial double-bond character and is stronger; alkyl chlorides — the carbon is less δ+, being attached to one electronegative atom only, or the C–Cl bond is strengthened by the electron-donating (positive inductive) effect of the alkyl group. Examiner insight: this discriminated well. Many had the correct trend, but the significance of the different strengths of the C–Cl bonds was less well understood.

Misconceptions and how the topic is assessed33.1–33.3

Self-test33.1–33.3

Ten questions on the whole chapter. Each gives its reason once you answer.

Definitions to learn33.1–33.3

Summary

• Benzoic acid is made from methylbenzene (or any alkylbenzene) by heating with alkaline KMnO₄, then adding dilute acid.
• Carboxylic acids give acyl chlorides with PCl₅ (→ POCl₃ + HCl), PCl₃ and heat (→ H₃PO₃) or SOCl₂ (→ SO₂ + HCl), under anhydrous conditions.
• Methanoic acid is oxidised to CO₂ and H₂O by Tollens', Fehling's, acidified KMnO₄ or acidified K₂Cr₂O₇; ethanedioic acid is oxidised to CO₂ by warm acidified KMnO₄.
• Acid strength: carboxylic acid > phenol > water > alcohol. Carboxylate charge is delocalised over two O atoms; phenoxide charge over O and the ring; alkyl groups destabilise alkoxide ions.
• Only carboxylic acids release CO₂ from carbonates; phenols react with NaOH but not carbonates; alcohols react with Na only.
• Electron-withdrawing Cl strengthens a carboxylic acid: more Cl, and Cl closer to –COOH, give a stronger acid.
• Acyl chlorides react at room temperature with water, alcohols, phenols, ammonia and primary or secondary amines to give acids, esters, amides and HCl, by addition–elimination.
• Esters of phenols (e.g. phenyl benzoate) are made from acyl chlorides, not from carboxylic acids.

Examination checklist

• Can I give reagents and conditions for making benzoic acid from methylbenzene?
• Can I write equations for making an acyl chloride with PCl₅, PCl₃ and SOCl₂, including molecules with two –OH groups?
• Can I describe tests that distinguish methanoic, ethanedioic and other carboxylic acids?
• Can I order and explain the acidities of carboxylic acids, phenols, water and alcohols, and of chlorine-substituted acids?
• Can I calculate K_a and pK_a from pH, and explain pH differences in terms of acid strength?
• Can I write equations for the reactions of acyl chlorides with water, alcohols, phenols, ammonia and amines, and name the products?
• Can I draw the addition–elimination mechanism with all lone pairs, dipoles, charges and curly arrows?
• Can I plan a two-step synthesis of an ester of a phenol or of an amide from a carboxylic acid?

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