Carboxylic Acids

AS · 11 min

Carboxylic acids contain the carboxyl group, −COOH\ce{-COOH}: a C=O and an O–H on the same carbon. Ethanoic acid gives vinegar its sharp taste; methanoic acid is in ant stings. They are the only common organic compounds that behave as typical (if weak) acids, fizzing with carbonates and reacting with metals and alkalis to form salts. This note covers their structure and properties, the three ways the syllabus makes them (oxidation, hydrolysis of nitriles, hydrolysis of esters), and their five reactions: with reactive metals, alkalis and carbonates, esterification and reduction. Expect them in synthesis routes, in "identify the functional group from these tests" questions, and in titration calculations.

Structure and naming

The carboxyl carbon is sp² hybridised and the −COOH\ce{-COOH} group is planar. It contains a polar C=O and a polar O–H.

The carboxyl carbon is always carbon 1, and it is counted in the stem:

formulaname
HCOOH\ce{HCOOH}methanoic acid
CHX3COOH\ce{CH3COOH}ethanoic acid
CHX3CHX2COOH\ce{CH3CH2COOH}propanoic acid
CHX3CHX2CHX2COOH\ce{CH3CH2CH2COOH}butanoic acid
(CHX3)X2CHCOOH\ce{(CH3)2CHCOOH}2-methylpropanoic acid
HOOCCOOH\ce{HOOCCOOH}ethanedioic acid

Salts are named after the metal and the acid with "-oate": CHX3COONa\ce{CH3COONa} is sodium ethanoate; (CHX3COO)X2Mg\ce{(CH3COO)2Mg} is magnesium ethanoate.

Physical properties

Carboxylic acids form strong hydrogen bonds: the O–H hydrogen of one molecule bonds to the C=O oxygen of another. In the pure liquid and in non-polar solvents, pairs of molecules hydrogen bond to each other twice, forming dimers.

compoundMrM_rboiling point / ∘C^\circ\text{C}
propan-1-ol6097
ethanoic acid60118
propanal5848
  • Carboxylic acids have higher boiling points than alcohols of similar MrM_r: each molecule has two hydrogen-bonding oxygens, and dimer formation effectively doubles the size of the particle that must be separated.
  • Short-chain acids (up to about four carbons) are completely miscible with water, because they hydrogen bond to water molecules. Solubility falls as the hydrocarbon chain lengthens.

Carboxylic acids as weak acids

In water, a carboxylic acid partially dissociates:

CHX3COOH(aq)⇌CHX3COOX−(aq)+HX+(aq)\ce{CH3COOH(aq) <=> CH3COO^-(aq) + H+(aq)}

The position of equilibrium lies well to the left: only a small fraction of molecules donate their proton, so carboxylic acids are weak acids (see Acids and bases). A 0.1 mol dm−30.1\ \text{mol dm}^{-3} solution of ethanoic acid has a pH of about 3, compared with 1 for hydrochloric acid of the same concentration.

Tip

Carboxylic acids are much more acidic than alcohols. In the carboxylate ion, the negative charge is spread (delocalised) over two oxygen atoms, which stabilises it; in an alkoxide ion the charge sits on one oxygen and is intensified by the alkyl group. The detailed comparison of acid strengths is an A Level topic.

Making carboxylic acids

Key result
starting materialreagents and conditionsexample
primary alcoholacidified KX2CrX2OX7\ce{K2Cr2O7} (or acidified KMnOX4\ce{KMnO4}), excess, heat under refluxCHX3CHX2OH+2 [O]→CHX3COOH+HX2O\ce{CH3CH2OH + 2[O] -> CH3COOH + H2O}
aldehydeacidified KX2CrX2OX7\ce{K2Cr2O7} (or KMnOX4\ce{KMnO4}), heat under refluxCHX3CHO+[O]→CHX3COOH\ce{CH3CHO + [O] -> CH3COOH}
nitriledilute acid (e.g. HCl), heat under reflux; or dilute alkali (NaOH), heat under reflux, then acidifyCHX3CHX2CN+2 HX2O+HCl→CHX3CHX2COOH+NHX4Cl\ce{CH3CH2CN + 2H2O + HCl -> CH3CH2COOH + NH4Cl}
esterdilute acid, heat under reflux; or dilute alkali, heat under reflux, then acidifyCHX3COOCHX3+HX2O⇌CHX3COOH+CHX3OH\ce{CH3COOCH3 + H2O <=> CH3COOH + CH3OH}

Hydrolysis of nitriles

The C≡N\ce{C#N} group is converted to −COOH\ce{-COOH}, and the nitrogen ends up as ammonium ions (in acid) or ammonia (in alkali).

With dilute acid:

CHX3CHX2CN+2 HX2O+HX+→CHX3CHX2COOH+NHX4X+\ce{CH3CH2CN + 2H2O + H+ -> CH3CH2COOH + NH4+}

With dilute alkali, the product is the carboxylate salt and ammonia gas is released:

CHX3CHX2CN+NaOH+HX2O→CHX3CHX2COONa+NHX3\ce{CH3CH2CN + NaOH + H2O -> CH3CH2COONa + NH3}

Adding a strong acid afterwards converts the salt into the free acid:

CHX3CHX2COONa+HCl→CHX3CHX2COOH+NaCl\ce{CH3CH2COONa + HCl -> CH3CH2COOH + NaCl}

The nitrile carbon becomes the carboxyl carbon, so the acid has the same number of carbons as the nitrile. Because nitriles are made from halogenoalkanes (KCN) or carbonyl compounds (HCN), this is a route that lengthens the chain by one carbon: bromoethane → propanenitrile → propanoic acid.

Hydrolysis of esters

Esters are split by water into the acid and the alcohol they were made from. Acid hydrolysis is reversible. Alkaline hydrolysis goes to completion and gives the carboxylate salt, which must be acidified to release the free acid. Details are in Esters.

Reactions of carboxylic acids

Key result
reactionreagentproductsexample
redox with reactive metalMg, Zn, Nasalt + hydrogen2 CHX3COOH+Mg→(CHX3COO)X2Mg+HX2\ce{2CH3COOH + Mg -> (CH3COO)2Mg + H2}
neutralisationalkali, e.g. NaOH(aq)salt + waterCHX3COOH+NaOH→CHX3COONa+HX2O\ce{CH3COOH + NaOH -> CH3COONa + H2O}
acid–carbonateNaX2COX3\ce{Na2CO3} or NaHCOX3\ce{NaHCO3}salt + water + carbon dioxide2 CHX3COOH+NaX2COX3→2 CHX3COONa+HX2O+COX2\ce{2CH3COOH + Na2CO3 -> 2CH3COONa + H2O + CO2}
esterification (condensation)alcohol, conc. HX2SOX4\ce{H2SO4} catalyst, heatester + waterCHX3COOH+CHX3OH⇌CHX3COOCHX3+HX2O\ce{CH3COOH + CH3OH <=> CH3COOCH3 + H2O}
reductionLiAlHX4\ce{LiAlH4} in dry etherprimary alcoholCHX3COOH+4 [H]→CHX3CHX2OH+HX2O\ce{CH3COOH + 4[H] -> CH3CH2OH + H2O}

Reactions as an acid

These are the same reactions as for any acid, but slower than with hydrochloric acid of the same concentration, because there are fewer HX+\ce{H+} ions.

  • Metals: magnesium fizzes and dissolves, giving hydrogen (squeaky pop with a lighted splint). This is a redox reaction: Mg is oxidised (0 to +2+2), H is reduced (+1+1 to 0).
  • Alkalis: neutralisation gives a salt and water. This is the basis of titrations of carboxylic acids with sodium hydroxide.
  • Carbonates and hydrogencarbonates: effervescence of carbon dioxide (turns limewater milky).
CHX3COOH+NaHCOX3→CHX3COONa+HX2O+COX2\ce{CH3COOH + NaHCO3 -> CH3COONa + H2O + CO2}
Key result

Test for a carboxylic acid: add solid sodium carbonate or sodium hydrogencarbonate (or the aqueous solution). Effervescence; the gas turns limewater milky. Alcohols do not react (they are far too weakly acidic), so this test distinguishes a carboxylic acid from an alcohol.

Esterification

Heating a carboxylic acid with an alcohol and a few drops of concentrated sulfuric acid gives an ester:

CHX3CHX2COOH+CHX3CHX2OH⇌CHX3CHX2COOCHX2CHX3+HX2O\ce{CH3CH2COOH + CH3CH2OH <=> CH3CH2COOCH2CH3 + H2O}

The reaction is reversible and slow; the acid catalyst speeds it up. See Esters.

Reduction

Only the powerful reducing agent lithium tetrahydridoaluminate(III) reduces carboxylic acids, all the way to the primary alcohol (the aldehyde intermediate is reduced faster than it can be isolated). NaBHX4\ce{NaBH4} does not reduce carboxylic acids.

CHX3CHX2COOH+4 [H]→CHX3CHX2CHX2OH+HX2O\ce{CH3CH2COOH + 4[H] -> CH3CH2CH2OH + H2O}

Worked examples

Routine: reactions as an acid

Write equations for the reactions of propanoic acid with (a) magnesium, (b) potassium hydroxide, (c) calcium carbonate. Give one observation for (a) and (c).

Solution

(a) 2 CHX3CHX2COOH+Mg→(CHX3CHX2COO)X2Mg+HX2\ce{2CH3CH2COOH + Mg -> (CH3CH2COO)2Mg + H2}. Effervescence; magnesium dissolves.

(b) CHX3CHX2COOH+KOH→CHX3CHX2COOK+HX2O\ce{CH3CH2COOH + KOH -> CH3CH2COOK + H2O}.

(c) 2 CHX3CHX2COOH+CaCOX3→(CHX3CHX2COO)X2Ca+HX2O+COX2\ce{2CH3CH2COOH + CaCO3 -> (CH3CH2COO)2Ca + H2O + CO2}. Effervescence; the solid dissolves; the gas turns limewater milky.

Routine: three routes to one acid

Give the reagents and conditions to make propanoic acid from (a) propan-1-ol, (b) propanenitrile, (c) methyl propanoate.

Solution

(a) Excess acidified potassium dichromate(VI), heat under reflux. CHX3CHX2CHX2OH+2 [O]→CHX3CHX2COOH+HX2O\ce{CH3CH2CH2OH + 2[O] -> CH3CH2COOH + H2O}

(b) Dilute hydrochloric acid, heat under reflux. CHX3CHX2CN+2 HX2O+HCl→CHX3CHX2COOH+NHX4Cl\ce{CH3CH2CN + 2H2O + HCl -> CH3CH2COOH + NH4Cl} (or NaOH(aq), reflux, then acidify).

(c) Dilute acid, heat under reflux (or NaOH(aq), reflux, then acidify). CHX3CHX2COOCHX3+HX2O⇌CHX3CHX2COOH+CHX3OH\ce{CH3CH2COOCH3 + H2O <=> CH3CH2COOH + CH3OH}

Standard: finding Mr by titration

0.185 g0.185\ \text{g} of a monocarboxylic acid, CXnHX2n+1COOH\ce{C_{n}H_{2n+1}COOH}, is dissolved in water and titrated with 0.100 mol dm−30.100\ \text{mol dm}^{-3} sodium hydroxide. 25.00 cm325.00\ \text{cm}^3 is needed for neutralisation. Find the MrM_r and identify the acid.

Solution

n(NaOH)=0.100×25.00/1000=2.50×10−3 moln(\ce{NaOH}) = 0.100 \times 25.00 / 1000 = 2.50 \times 10^{-3}\ \text{mol}. The acid has one COOH, so n(acid)=2.50×10−3 moln(\text{acid}) = 2.50 \times 10^{-3}\ \text{mol}.

Mr=0.185/2.50×10−3=74.0M_r = 0.185 / 2.50 \times 10^{-3} = 74.0.

CXnHX2n+1COOH\ce{C_{n}H_{2n+1}COOH}: 14n+1+45=7414n + 1 + 45 = 74, so 14n=2814n = 28 and n=2n = 2. The acid is CX2HX5COOH\ce{C2H5COOH}, propanoic acid.

Standard: distinguishing three compounds

Describe tests to distinguish ethanoic acid, ethanol and ethanal, giving observations.

Solution

Add sodium hydrogencarbonate to each: ethanoic acid fizzes (gas turns limewater milky); the others show no change.

Warm the remaining two with Tollens' reagent: ethanal gives a silver mirror; ethanol shows no change. (Alternatively, 2,4-DNPH gives an orange precipitate with ethanal only.)

Ethanol is the one left; it could be confirmed with acidified potassium dichromate(VI) (orange to green).

Exam-hard: lengthening the chain

Butanoic acid is made from 1-bromopropane in two steps.

(a) Give the reagents and conditions for each step, the intermediate, and the type of each reaction. (b) Write equations for the alkaline route for step 2, including the acidification. (c) 12.3 g12.3\ \text{g} of 1-bromopropane gives 5.28 g5.28\ \text{g} of butanoic acid. Calculate the overall percentage yield. (ArA_r: H 1.0, C 12.0, O 16.0, Br 79.9)

Solution

(a) Step 1: KCN in ethanol, heat under reflux; nucleophilic substitution. Intermediate: butanenitrile, CHX3CHX2CHX2CN\ce{CH3CH2CH2CN}. Step 2: dilute HCl, heat under reflux (or NaOH(aq), reflux, then acidify); hydrolysis.

(b) CHX3CHX2CHX2CN+NaOH+HX2O→CHX3CHX2CHX2COONa+NHX3\ce{CH3CH2CH2CN + NaOH + H2O -> CH3CH2CH2COONa + NH3}, then CHX3CHX2CHX2COONa+HCl→CHX3CHX2CHX2COOH+NaCl\ce{CH3CH2CH2COONa + HCl -> CH3CH2CH2COOH + NaCl}.

(c) Mr(CX3HX7Br)=36.0+7.0+79.9=122.9M_r(\ce{C3H7Br}) = 36.0 + 7.0 + 79.9 = 122.9; n=12.3/122.9=0.100 moln = 12.3 / 122.9 = 0.100\ \text{mol}. Mr(CX3HX7COOH)=88.0M_r(\ce{C3H7COOH}) = 88.0; theoretical mass =8.80 g= 8.80\ \text{g}. Yield =5.28/8.80×100=60.0%= 5.28 / 8.80 \times 100 = 60.0\%.

Exam-hard: a molecule with two acidic groups

Lactic acid, CHX3CH(OH)COOH\ce{CH3CH(OH)COOH} (MrM_r 90.0), is treated separately with excess sodium and with excess sodium hydrogencarbonate. For 0.900 g0.900\ \text{g} of lactic acid, calculate the volume of gas (at room temperature and pressure) produced in each case. Explain the difference. (Molar gas volume 24.0 dm3 mol−124.0\ \text{dm}^3\ \text{mol}^{-1})

Solution

n(lactic acid)=0.900/90.0=0.0100 moln(\text{lactic acid}) = 0.900 / 90.0 = 0.0100\ \text{mol}.

Sodium reacts with both O–H groups (the COOH and the alcohol OH):

CHX3CH(OH)COOH+2 Na→CHX3CH(ONa)COONa+HX2\ce{CH3CH(OH)COOH + 2Na -> CH3CH(ONa)COONa + H2}

n(HX2)=0.0100 moln(\ce{H2}) = 0.0100\ \text{mol}; volume =0.240 dm3= 0.240\ \text{dm}^3 (240 cm3240\ \text{cm}^3).

Sodium hydrogencarbonate reacts only with the carboxylic acid group; the alcohol is too weak an acid:

CHX3CH(OH)COOH+NaHCOX3→CHX3CH(OH)COONa+HX2O+COX2\ce{CH3CH(OH)COOH + NaHCO3 -> CH3CH(OH)COONa + H2O + CO2}

n(COX2)=0.0100 moln(\ce{CO2}) = 0.0100\ \text{mol}; volume =0.240 dm3= 0.240\ \text{dm}^3.

The volumes happen to be equal, but for different reasons: sodium releases half a mole of HX2\ce{H2} per OH group (two groups), while NaHCOX3\ce{NaHCO3} releases one mole of COX2\ce{CO2} per COOH group (one group).

Watch out
  • NaBHX4\ce{NaBH4} and acids. NaBHX4\ce{NaBH4} does not reduce carboxylic acids. Use LiAlHX4\ce{LiAlH4} in dry ether.
  • Forgetting to acidify. Alkaline hydrolysis of a nitrile or ester gives the salt (e.g. sodium propanoate). The free acid only forms after adding a strong acid.
  • Nitrile hydrolysis by-product. In acid the nitrogen becomes NHX4X+\ce{NH4+} (ammonium chloride); in alkali it is released as NHX3\ce{NH3}. Not NX2\ce{N2}, not HCN.
  • Counting carbons. Propanenitrile, CHX3CHX2CN\ce{CH3CH2CN}, gives propanoic acid, CHX3CHX2COOH\ce{CH3CH2COOH}: three carbons each. The chain-lengthening step is the one that made the nitrile.
  • Carbonate test with alcohols. Alcohols do not fizz with carbonates. Sodium metal fizzes with both acids and alcohols, so it does not distinguish them.
  • Salt formulae with Group 2. Magnesium ethanoate is (CHX3COO)X2Mg\ce{(CH3COO)2Mg}: two ethanoate ions for each MgX2+\ce{Mg^2+}.
Exam tip
  • "Give a chemical test to show that X is a carboxylic acid": sodium carbonate or hydrogencarbonate, effervescence, gas turns limewater milky. Give the reagent, the observation and the identity of the gas.
  • Reagents and conditions: oxidation needs "acidified potassium dichromate(VI)" and "heat under reflux"; nitrile hydrolysis needs "dilute acid, heat under reflux" (or alkali, then acidify).
  • Titration questions with carboxylic acids are straightforward mole calculations; check whether the acid is mono- or dicarboxylic before setting the mole ratio.
  • When a molecule has several functional groups, decide which ones each reagent reacts with: Na (all O–H), NaHCOX3\ce{NaHCO3} (COOH only), NaBHX4\ce{NaBH4} (C=O of aldehydes and ketones only), LiAlHX4\ce{LiAlH4} (aldehydes, ketones and COOH).
Summary
  • Carboxyl group −COOH\ce{-COOH}; the carboxyl carbon is C1 and counted in the name.
  • Hydrogen bonding (dimers) gives high boiling points; short-chain acids are soluble in water.
  • Weak acids: RCOOH⇌RCOOX−+HX+\ce{RCOOH <=> RCOO- + H+}, partially dissociated.
  • Made by: oxidation of primary alcohols or aldehydes (acidified KX2CrX2OX7\ce{K2Cr2O7} or KMnOX4\ce{KMnO4}, reflux); hydrolysis of nitriles (dilute acid, reflux; or alkali then acidify); hydrolysis of esters (dilute acid or alkali, heat, then acidify).
  • Reactions: metals → salt + HX2\ce{H2}; alkalis → salt + water; carbonates → salt + water + COX2\ce{CO2} (test); alcohols + conc. HX2SOX4\ce{H2SO4} → esters; LiAlHX4\ce{LiAlH4} → primary alcohols.

Practice

Question
  1. Name: (a) HCOOH\ce{HCOOH}, (b) CHX3CHX2CHX2COOH\ce{CH3CH2CH2COOH}, (c) (CHX3)X2CHCOOH\ce{(CH3)2CHCOOH}, (d) CHX3CHX2COONa\ce{CH3CH2COONa}.
  2. Explain why ethanoic acid (MrM_r 60) has a higher boiling point than propan-1-ol (MrM_r 60).
  3. Write equations for the reactions of ethanoic acid with (a) zinc, (b) sodium carbonate, (c) aqueous ammonia (forming ammonium ethanoate).
  4. Give the reagents and conditions for converting ethanol into ethanoic acid, and explain why the reaction is carried out under reflux.
  5. Write equations for the hydrolysis of ethanenitrile (a) with dilute hydrochloric acid, (b) with aqueous sodium hydroxide.
  6. Write an equation, using [H]\ce{[H]}, for the reduction of propanoic acid, and name the reducing agent.
  7. Describe how you would distinguish propanoic acid, propan-1-ol and propanal using chemical tests.
  8. 0.220 g0.220\ \text{g} of a monocarboxylic acid requires 25.0 cm325.0\ \text{cm}^3 of 0.100 mol dm−30.100\ \text{mol dm}^{-3} NaOH for neutralisation. Calculate its MrM_r and give the structures and names of the two acids that fit.
  9. Propanedioic acid, HOOCCHX2COOH\ce{HOOCCH2COOH}, reacts with excess sodium hydrogencarbonate. Write the equation and calculate the volume of carbon dioxide (room temperature and pressure) from 1.04 g1.04\ \text{g} of the acid. (ArA_r: H 1.0, C 12.0, O 16.0; molar gas volume 24.0 dm3 mol−124.0\ \text{dm}^3\ \text{mol}^{-1})
  10. Plan a three-step synthesis of propanoic acid from ethanol. For each step give the reagents and conditions, the product, and the type of reaction.
Answers
  1. (a) Methanoic acid. (b) Butanoic acid. (c) 2-methylpropanoic acid. (d) Sodium propanoate.
  2. Both form hydrogen bonds, but ethanoic acid molecules hydrogen bond in pairs (dimers) through both the C=O and the O–H, so the hydrogen bonding is more extensive and the effective particle is larger. More energy is needed to separate the molecules.
  3. (a) 2 CHX3COOH+Zn→(CHX3COO)X2Zn+HX2\ce{2CH3COOH + Zn -> (CH3COO)2Zn + H2} (b) 2 CHX3COOH+NaX2COX3→2 CHX3COONa+HX2O+COX2\ce{2CH3COOH + Na2CO3 -> 2CH3COONa + H2O + CO2} (c) CHX3COOH+NHX3→CHX3COONHX4\ce{CH3COOH + NH3 -> CH3COONH4}
  4. Excess acidified potassium dichromate(VI), heat under reflux. Under reflux the intermediate ethanal (and ethanol) vapour condenses and returns to the flask, so it stays in contact with the oxidising agent until it is fully oxidised to ethanoic acid, instead of escaping.
  5. (a) CHX3CN+2 HX2O+HCl→CHX3COOH+NHX4Cl\ce{CH3CN + 2H2O + HCl -> CH3COOH + NH4Cl} (b) CHX3CN+NaOH+HX2O→CHX3COONa+NHX3\ce{CH3CN + NaOH + H2O -> CH3COONa + NH3}
  6. CHX3CHX2COOH+4 [H]→CHX3CHX2CHX2OH+HX2O\ce{CH3CH2COOH + 4[H] -> CH3CH2CH2OH + H2O}; lithium tetrahydridoaluminate(III), LiAlHX4\ce{LiAlH4}, in dry ether.
  7. Sodium hydrogencarbonate: only propanoic acid fizzes (carbon dioxide turns limewater milky). Tollens' reagent, warm, on the other two: propanal gives a silver mirror; propan-1-ol does not. (Or 2,4-DNPH: orange precipitate with propanal only.)
  8. n(NaOH)=2.50×10−3 mol=n(acid)n(\ce{NaOH}) = 2.50 \times 10^{-3}\ \text{mol} = n(\text{acid}); Mr=0.220/2.50×10−3=88.0M_r = 0.220 / 2.50 \times 10^{-3} = 88.0. CXnHX2n+1COOH\ce{C_{n}H_{2n+1}COOH}: 14n+46=8814n + 46 = 88, n=3n = 3: CX3HX7COOH\ce{C3H7COOH}. Butanoic acid, CHX3CHX2CHX2COOH\ce{CH3CH2CH2COOH}, and 2-methylpropanoic acid, (CHX3)X2CHCOOH\ce{(CH3)2CHCOOH}.
  9. HOOCCHX2COOH+2 NaHCOX3→NaOOCCHX2COONa+2 HX2O+2 COX2\ce{HOOCCH2COOH + 2NaHCO3 -> NaOOCCH2COONa + 2H2O + 2CO2}. Mr=104.0M_r = 104.0; n=1.04/104.0=0.0100 moln = 1.04 / 104.0 = 0.0100\ \text{mol}; n(COX2)=0.0200 moln(\ce{CO2}) = 0.0200\ \text{mol}; volume =0.480 dm3= 0.480\ \text{dm}^3 (480 cm3480\ \text{cm}^3).
  10. Step 1: ethanol to bromoethane. KBr and concentrated HX2SOX4\ce{H2SO4}, heat (or HBr); nucleophilic substitution. Step 2: bromoethane to propanenitrile. KCN in ethanol, heat under reflux; nucleophilic substitution (the chain gains one carbon). Step 3: propanenitrile to propanoic acid. Dilute HCl, heat under reflux (or NaOH(aq), reflux, then acidify); hydrolysis.

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