If you have just opened this unit and the page in front of you looks like a blur of Rosenmund, Cannizzaro, Wolff–Kishner and Hell–Volhard–Zelinsky, take a breath. Nothing here is beyond you. Aldehydes, ketones and carboxylic acids feel frightening because the chapter is long, not because it is difficult. Underneath the twenty-odd named reactions sits one small, very well-behaved piece of chemistry — a carbon joined to an oxygen by a double bond — and the moment you understand what that little group actually wants, most of the chapter stops being something to memorise and starts being something you can work out.
We are going to build everything from zero here: how to name these compounds, where they come from, why they react the way they do, and how to tell them apart in a test tube. Along the way you will get a full name reactions list with examples, a clean answer to the classic distinguish between aldehyde and ketone test question, and a worked set of aldehydes ketones and carboxylic acids class 12 important questions at the end, each with the reasoning written out rather than just an answer. If a section feels heavy on the first read, that is normal — read it once for the shape of the idea, then come back for the detail.
One promise before we start. This chapter has a spine, and the spine is only three ideas wide. Everything a carbonyl compound does happens through one of three doors: something attacks the carbonyl carbon, or something pulls off a hydrogen from the carbon next door, or the whole group is pushed up or down in oxidation state. Twenty-two reactions, three doors. Keep that in your head and the unit shrinks to a manageable size.
Meet Your Tutor
Carbonyl chemistry becomes manageable when you identify the functional group first and then ask what the reagent can do to it. I will help you compare aldehydes, ketones and carboxylic acids through structure, reactivity and conversion maps, with small product checks that make named reactions easier to recall.
What You’ll Learn
Your Game Plan
- Learn to name these compounds first. You cannot answer a question about a compound you cannot read.
- Spend real time on the nature of the C=O group. Ten minutes here saves ten hours later.
- Learn the preparation routes as a map, not as a list — six starting materials, all pointing at one destination.
- Master nucleophilic addition as a single mechanism, then treat HCN, NaHSO₃, alcohols and ammonia derivatives as four costumes on the same actor.
- Separate the alpha-hydrogen reactions (aldol, cross aldol, iodoform, HVZ) into their own mental drawer.
- Drill the identification tests until they are reflex — these are almost free marks.
- Finish with carboxylic acid strength, where the reasoning, not the memory, earns the mark.
- Do the worksheet at the end with the answers covered. Honestly covered.
Study Notes
Nomenclature Of Aldehydes And Ketones
An aldehyde carries its carbonyl group at the end of a chain, so the carbon always has a hydrogen attached: —CHO. A ketone carries its carbonyl group inside the chain, flanked by two carbons: —CO—. That single structural difference drives the whole chapter, so let it sink in before anything else. An aldehyde is the last carriage of a train; a ketone is a carriage in the middle.
IUPAC names for aldehydes. Take the longest chain that contains the —CHO carbon, drop the final -e of the alkane name and add -al. The CHO carbon is always carbon number 1 — you do not get a choice. So CH₃CHO is ethanal and CH₃CH₂CH₂CHO is butanal. When the CHO group is stuck onto a ring rather than a chain, the ring keeps its own name and we add -carbaldehyde: hence cyclohexanecarbaldehyde and benzenecarbaldehyde (which everyone actually calls benzaldehyde).
IUPAC names for ketones. Same idea, but the ending is -one and the carbonyl carbon has to be given the lowest possible number. CH₃COCH₃ is propanone; CH₃COCH₂CH₂CH₃ is pentan-2-one, not pentan-4-one. When there are two carbonyls in the chain, keep the -e and use -dione: CH₃COCH₂COCH₃ is pentane-2,4-dione.
Common names still appear in exam papers, so you need both. Aldehyde common names come from the common name of the acid they oxidise to: formic acid gives formaldehyde, acetic acid gives acetaldehyde, benzoic acid gives benzaldehyde. Ketone common names simply list the two groups alphabetically and add the word ketone: CH₃COCH₂CH₃ is ethyl methyl ketone. Two ketones break the pattern and are worth memorising outright — C₆H₅COCH₃ is acetophenone and C₆H₅COC₆H₅ is benzophenone.
Step 1. Find the longest chain through the CHO carbon: CHO—CH—CH₂—CH₃, four carbons, so the parent is butanal.
Step 2. Number from the CHO carbon: C1 = CHO, C2 = CH(CH₃), C3 = CH₂, C4 = CH₃.
Step 3. The methyl branch sits on C2.
Answer: 2-methylbutanal. Its common name is α-methylbutyraldehyde, and notice that α and C-2 point at exactly the same carbon — that is the check you should always run.
Step 1. Longest chain containing C=O: CH₃—CO—CH=C(CH₃)—CH₃, five carbons, so pentanone.
Step 2. The carbonyl must get the lowest number, so number from the right-hand methyl: C1 = CH₃, C2 = C=O, C3 = CH, C4 = C, C5 = CH₃.
Step 3. The double bond starts at C3; the extra methyl is on C4.
Answer: 4-methylpent-3-en-2-one. You will meet this compound again as the dehydration product of the acetone aldol, where it goes by the old name mesityl oxide.
Nature Of The Carbonyl Group
Here is the ten minutes that pays for the whole chapter. The carbonyl carbon is sp² hybridised. It makes three sigma bonds that all lie in one flat plane at roughly 120° to each other, and its leftover p orbital overlaps sideways with a p orbital on oxygen to make the pi bond. So the carbonyl group and the three atoms attached to it are planar, and there are two open faces — above the plane and below it — where an incoming reagent can land.
Now the important part. Oxygen is far more electronegative than carbon, and a pi bond is loosely held, so the shared electrons sit much closer to oxygen. The group is therefore strongly polar: carbon carries a partial positive charge (δ+) and oxygen a partial negative charge (δ−), and simple aldehydes and ketones have dipole moments in the region of 2.3 to 2.9 D — large for organic molecules. We can write the same idea as a resonance picture: the real molecule is a blend of a neutral structure with a full C=O double bond and a charge-separated structure with a positive carbon and a negative oxygen.
This polarity also explains why the hydrogen on the carbon next to the carbonyl — the alpha hydrogen — is unusually acidic. When that hydrogen leaves, the negative charge that stays behind is not stranded on carbon; it can slide onto the electronegative oxygen through resonance, forming a stabilised enolate. Ordinary alkane hydrogens have no such escape route, which is why they are not acidic at all. Hold on to this: it is the entire basis of the aldol reaction, the iodoform test and the HVZ reaction.
If the sp²/sp³ language feels shaky, it is worth ten minutes with our Alcohols, Phenols and Ethers notes, where the same hybridisation and polarity arguments are set up in a gentler setting — and where you will meet the alcohols that this chapter turns into carbonyls.
Preparation From Alcohols
Before we walk through the individual routes, look at the map below. Six different families of starting material all funnel into the same destination. Learning them as a picture rather than a list is the difference between recalling four routes under exam pressure and recalling all six.
Now the first road. An alcohol already has an oxygen on the right carbon; all we have to do is remove two hydrogens. There are two ways to do that, and the exam loves both.
Route A — oxidation. A primary alcohol loses two hydrogens to give an aldehyde; a secondary alcohol gives a ketone. There is a catch, though, and it is the reason PCC exists. Aldehydes are themselves easy to oxidise, so a strong oxidising agent such as acidified K₂Cr₂O₇ or KMnO₄ will not stop at the aldehyde — it charges straight on to the carboxylic acid. To halt the reaction at the aldehyde we use a mild, gentle reagent: pyridinium chlorochromate (PCC) in dichloromethane. Ketones have no hydrogen left on the carbonyl carbon, so once a secondary alcohol has become a ketone it simply stops; any ordinary oxidising agent works there.
Route B — catalytic dehydrogenation. Pass the alcohol vapour over hot copper at about 573 K and the alcohol hands over two hydrogen atoms as H₂ gas. A primary alcohol gives an aldehyde, a secondary alcohol gives a ketone. A tertiary alcohol has no hydrogen at all on its carbinol carbon, so it cannot dehydrogenate — instead it dehydrates and you get an alkene. That contrast is a favourite one-mark question.
Propan-1-ol (primary, two H on the carbinol carbon) → loses H₂ → propanal, CH₃CH₂CHO.
Propan-2-ol (secondary, one H on the carbinol carbon) → loses H₂ → propanone, CH₃COCH₃.
2-Methylpropan-2-ol (tertiary, no H on the carbinol carbon) → cannot dehydrogenate, so it dehydrates instead → 2-methylpropene, (CH₃)₂C=CH₂.
Why it works: dehydrogenation needs one hydrogen from the C—OH oxygen and one from the carbon holding it. A tertiary alcohol simply has no hydrogen on that carbon to give.
Preparation From Acyl Chlorides — Rosenmund Reduction
An acyl chloride, RCOCl, is one rung above an aldehyde in oxidation state, so to reach an aldehyde we have to come down by exactly one step. Hydrogen gas over a palladium catalyst will do the reduction — but ordinary palladium is far too enthusiastic and will keep going, turning the aldehyde into a primary alcohol. The trick, and it is a lovely one, is to deliberately make the catalyst worse at its job.
In the Rosenmund reduction the palladium is spread on barium sulphate, a support with a low surface area, and then poisoned with a little sulphur or quinoline. The poisoned catalyst is just active enough to deliver one dose of hydrogen and no more, so the reaction parks neatly at the aldehyde:
CH₃COCl + H₂ → CH₃CHO (ethanal) + HCl
C₆H₅COCl + H₂ → C₆H₅CHO (benzaldehyde) + HCl
Why it works: the chlorine is a good leaving group and hydrogen simply takes its place on the carbonyl carbon. The poison stops the catalyst before the newly made C=O can be attacked again. Without the poison you would isolate ethanol and benzyl alcohol instead — a common trap in “what if the catalyst were not poisoned” questions.
Preparation From Nitriles And Esters — Stephen Reaction And DIBAL-H
A nitrile, R—C≡N, is also one rung above an aldehyde, and once again the whole art is stopping half-way. There are two accepted ways.
The Stephen reaction. Treat the nitrile with stannous chloride (SnCl₂) and hydrochloric acid. The triple bond is reduced only as far as an imine, which is isolated as its hydrochloride salt, RCH=NH·HCl. Add water and the imine hydrolyses to the aldehyde. Think of the imine as the aldehyde wearing a nitrogen mask — water takes the mask off.
DIBAL-H. Diisobutylaluminium hydride, written AlH(i-Bu)₂, is a bulky, one-dose hydride donor. On a nitrile it delivers a single hydride, giving an imine intermediate that hydrolyses on work-up to the aldehyde. On an ester it does the same trick: one hydride, and acidic work-up releases the aldehyde rather than the alcohol. This is the only routine way in your syllabus to get from an ester down to an aldehyde in one operation.
| Starting material | DIBAL-H, then H3O+ | LiAlH4, then H3O+ |
|---|---|---|
| Nitrile, RCN | Aldehyde, RCHO | Primary amine, RCH2NH2 |
| Ester, RCOOR′ | Aldehyde, RCHO | Primary alcohol, RCH2OH |
| Acyl chloride, RCOCl | Aldehyde (Rosenmund is the standard route) | Primary alcohol, RCH2OH |
Method 1 (Stephen). CH₃CH₂CN + SnCl₂/HCl → CH₃CH₂CH=NH·HCl, then H₃O⁺ → CH₃CH₂CHO + NH₄⁺.
Method 2 (DIBAL-H). CH₃CH₂CN + AlH(i-Bu)₂ → imine intermediate, then H₃O⁺ → CH₃CH₂CHO.
Note the carbon count: the nitrile carbon becomes the aldehyde carbon, so the number of carbons does not change. Three carbons in, three carbons out. Students often lose a carbon here out of habit — count them every time.
Preparation From Hydrocarbons
Four routes live here, and they are best learned by asking what kind of hydrocarbon you are starting from.
1. Ozonolysis of alkenes. Ozone attacks the C=C double bond to give an unstable ozonide, which is then broken apart by zinc dust in water. The double bond snaps cleanly in half and each carbon of the old C=C walks away with the oxygen it earned: a carbon carrying an H becomes an aldehyde, a carbon carrying two alkyl groups becomes a ketone. The zinc is not decoration — hydrolysis of an ozonide also produces hydrogen peroxide, and without zinc to destroy it the H₂O₂ would oxidise your fresh aldehyde into a carboxylic acid.
2. Hydration of alkynes. Warm an alkyne with water, dilute sulphuric acid and mercury(II) sulphate at about 333 K. Water adds across the triple bond following Markovnikov’s rule to give an enol, which immediately tautomerises to the far more stable carbonyl compound. Ethyne is the special case: it is the only alkyne that gives an aldehyde (ethanal). Every other alkyne gives a ketone.
3. Friedel–Crafts acylation. To hang a ketone group on a benzene ring, treat the arene with an acyl chloride (or an anhydride) and anhydrous aluminium chloride. Benzene plus ethanoyl chloride gives acetophenone, C₆H₅COCH₃. This is how aromatic ketones are made.
4. Gattermann–Koch reaction. Aromatic aldehydes need a different trick, because formyl chloride (HCOCl) is too unstable to bottle. Instead we generate it in the flask: pass carbon monoxide and hydrogen chloride into benzene in the presence of anhydrous AlCl₃ together with cuprous chloride, and out comes benzaldehyde. Think of CO + HCl as formyl chloride assembled on the spot.
Your textbook also shows aromatic aldehydes made by oxidising the side chain of toluene — with chromyl chloride, CrO₂Cl₂, in the Etard reaction, or by chlorinating the methyl group to C₆H₅CHCl₂ and then hydrolysing it. Both stop at benzaldehyde rather than running on to benzoic acid, which is exactly why they are used.
(a) The alkene is (CH₃)₂C=CHCH₃. Cut the double bond in half and put an O on each end. The (CH₃)₂C— fragment has two alkyl groups, so it becomes propanone, (CH₃)₂C=O. The —CHCH₃ fragment carries a hydrogen, so it becomes ethanal, CH₃CHO.
(b) Work backwards. If propanone is the only product, then both halves of the original double bond must have been identical, and each half must have been the (CH₃)₂C— fragment. Rejoin two such fragments across a double bond: (CH₃)₂C=C(CH₃)₂. Count the atoms — six carbons and twelve hydrogens, C₆H₁₂, exactly as required. The alkene is 2,3-dimethylbut-2-ene.
Why it works: ozonolysis is a clean scissor cut, so the products are a photograph of the two halves of the double bond. A single product always means a symmetrical alkene.
(a) HC≡CH + H₂O → CH₂=CH—OH (ethenol, an enol) → tautomerises → CH₃CHO, ethanal.
(b) CH₃C≡CH + H₂O → OH adds to the more substituted carbon (Markovnikov) → CH₃C(OH)=CH₂ → tautomerises → CH₃COCH₃, propanone.
Why it works: the enol always flips to the keto form because a C=O pi bond is considerably stronger than a C=C pi bond. Ethyne is unique only because it is symmetrical and terminal at both ends, so the OH has nowhere to go except a carbon that still keeps a hydrogen — and a carbonyl carbon holding a hydrogen is, by definition, an aldehyde.
Physical Properties Of Aldehydes And Ketones
Methanal is a gas at room temperature and ethanal a very volatile liquid (boiling point about 294 K); everything above them is a liquid or a solid. The interesting question is where their boiling points sit relative to their neighbours, and the answer is beautifully logical.
A carbonyl compound is strongly polar, so its molecules attract each other by dipole–dipole forces. That makes them boil considerably higher than hydrocarbons or ethers of similar molecular mass, which only have weak dispersion forces. But a carbonyl compound has no O—H bond, so its molecules cannot hydrogen bond to each other. That makes them boil noticeably lower than alcohols of similar mass. Aldehydes and ketones therefore sit in the middle of the ladder.
| Compound | Molar mass (g mol−1) | Boiling point (K) | Strongest force between molecules |
|---|---|---|---|
| n-Butane, C4H10 | 58.1 | 273 | Dispersion only |
| Methoxyethane, C3H8O | 60.1 | 281 | Weak dipole–dipole |
| Propanal, C3H6O | 58.1 | 322 | Strong dipole–dipole |
| Propanone, C3H6O | 58.1 | 329 | Strong dipole–dipole |
| Propan-1-ol, C3H8O | 60.1 | 370 | Hydrogen bonding |
Solubility follows a different rule and catches people out. Even though carbonyl molecules cannot hydrogen bond to each other, the lone pairs on the carbonyl oxygen accept hydrogen bonds happily from water. So methanal, ethanal and propanone are completely miscible with water, and solubility then falls away rapidly as the greasy hydrocarbon tail grows. By about five or six carbons the molecule is essentially insoluble. All of them dissolve freely in organic solvents.
Mechanism Of Nucleophilic Addition
This is Door 1, and it is worth learning as one single story that you then retell four times with different characters.
Step 1 — the nucleophile lands. The nucleophile approaches the carbonyl carbon from above or below the flat plane (never edge-on, because the pi cloud is in the way) and forms a bond to it. As that bond forms, the two pi electrons are pushed entirely onto oxygen. The carbon, which was sp² and flat with three attachments, now has four attachments and becomes sp³ and tetrahedral. What we have is an alkoxide ion — a negatively charged oxygen sitting on a tetrahedral carbon.
Step 2 — the oxygen picks up a proton. The alkoxide grabs a proton from the solvent, from HCN, or from the acidic work-up, and you are left with the neutral addition product. Notice what has happened overall: the pi bond has vanished and two new sigma bonds have appeared, one to the nucleophile and one to hydrogen. That is why we call it addition and not substitution — nothing left the molecule.
Under acidic conditions there is a preliminary move: a proton attaches to the carbonyl oxygen first. That makes the carbon dramatically more positive and therefore far more tempting to a weak nucleophile. So acid does not change the story, it just raises the volume. If you enjoy thinking about how reaction steps determine rate, our Chemical Kinetics chapter covers the rate-determining-step idea that underlies why the first step here controls everything.
HCHO > CH₃CHO > CH₃COCH₃ > CH₃COC₂H₅
Aromatic carbonyls sit even lower, because the benzene ring feeds electron density into the C=O by resonance — a friend who also pays for the meal. Hence CH₃COCH₃ > C₆H₅COCH₃ > C₆H₅COC₆H₅, with benzophenone the least reactive of the lot.
Count the crowding on each carbonyl carbon.
Ethanal, CH₃CHO — one small methyl and one hydrogen. Least crowded.
Benzaldehyde, C₆H₅CHO — one hydrogen, but a ring that donates electrons by resonance.
Propanone, CH₃COCH₃ — two methyl groups, no ring.
Acetophenone, C₆H₅COCH₃ — a methyl and a donating ring. Most deactivated.
Answer: ethanal > propanone > benzaldehyde > acetophenone.
Why propanone beats benzaldehyde: two methyl groups give a modest +I push, but a benzene ring conjugated to the C=O delivers a much larger resonance donation. Resonance beats induction, so the aromatic aldehyde is the less reactive of the pair.
Nucleophilic Addition Reactions
Four nucleophiles, one mechanism. Retell the story each time.
1. Hydrogen cyanide → cyanohydrin. HCN itself is a poor nucleophile because it barely ionises, so we add a trace of base. The base generates CN⁻, which is an excellent nucleophile, and that is what actually attacks. The product is a cyanohydrin — a carbon carrying both an OH and a CN. These are prized in synthesis because the nitrile group can afterwards be hydrolysed to —COOH, giving you a hydroxy acid with one more carbon than you started with.
2. Sodium hydrogensulphite → the bisulphite adduct. Shake an aldehyde with saturated NaHSO₃ solution and a white crystalline solid separates out. The reaction is reversible: treat the crystals with dilute acid or dilute alkali and the original carbonyl compound comes straight back. That reversibility is the whole point — it lets us purify an aldehyde or separate it from a non-reacting impurity. Because the sulphur atom is bulky, only aldehydes, methyl ketones and small cyclic ketones form these adducts; a crowded ketone such as benzophenone simply will not.
3. Alcohols → hemiacetals and acetals. One molecule of a dry alcohol adds across the C=O in the presence of dry HCl gas to give a hemiacetal (one OH and one OR on the same carbon). A second molecule of alcohol then replaces the OH, and you have an acetal — a carbon carrying two OR groups. Ketones are more reluctant, so we use ethylene glycol with dry HCl, and because both OH groups belong to the same molecule the product is a neat five-membered cyclic ketal. Acetals and ketals shrug off bases and reducing agents but fall apart in aqueous acid, which makes them ideal protecting groups for a carbonyl you want to keep out of the way.
4. Ammonia derivatives → addition–elimination. Reagents of the general shape H₂N—Z first add across the C=O exactly as before, but then the intermediate loses a molecule of water and a C=N double bond appears. So this one is addition followed by elimination. Change Z and you change the name of the product:
| Reagent H2N—Z | Name of reagent | Product |
|---|---|---|
| H2N—OH | Hydroxylamine | Oxime |
| H2N—NH2 | Hydrazine | Hydrazone |
| H2N—NHC6H5 | Phenylhydrazine | Phenylhydrazone |
| H2N—NHC6H3(NO2)2 | 2,4-Dinitrophenylhydrazine | 2,4-Dinitrophenylhydrazone — orange-red solid |
| H2N—NHCONH2 | Semicarbazide | Semicarbazone |
Step 1. CH₃CHO + HCN, trace base → CH₃CH(OH)CN, 2-hydroxypropanenitrile.
Step 2. Hydrolyse the nitrile with dilute acid and warm → CH₃CH(OH)COOH, 2-hydroxypropanoic acid.
Count the carbons: ethanal has two, the product has three. The extra carbon came in as the cyanide ion. This is one of only a handful of chain-lengthening tools in your syllabus — the other big one is the Grignard plus CO₂ route later in this chapter. Whenever a synthesis question demands one more carbon, look for these two.
Too acidic: the nitrogen of H₂N—Z has a lone pair, and in strong acid that lone pair is protonated. Once it becomes —NH₃⁺ it has no lone pair left to donate, so it is no longer a nucleophile and step one cannot happen.
Too neutral or basic: the second half of the reaction needs a proton to convert the —OH into —OH₂⁺, a good leaving group, so that water can depart. With no acid available, the elimination stalls.
Answer: a weakly acidic medium is the compromise — enough acid to help water leave, not so much that the nucleophile is switched off. This is a classic three-mark question, and the marks are for naming both failures, not one.
Reactivity Of Alpha Hydrogen — Aldol And Cross Aldol Condensation
Welcome to Door 2. Nothing attacks the carbonyl carbon here; instead a base quietly removes a hydrogen from the carbon next to it. Remember why that hydrogen is loose: the negative charge left behind can delocalise onto the electronegative carbonyl oxygen, giving a stabilised enolate. An enolate is a carbon nucleophile, and once you have a carbon nucleophile in a flask full of carbonyl compounds, it is only a matter of time before it attacks one.
The aldol reaction. Treat an aldehyde or ketone that has at least one alpha hydrogen with dilute alkali. One molecule becomes an enolate; it attacks the carbonyl carbon of a second, untouched molecule; the resulting alkoxide picks up a proton. The product is a beta-hydroxy aldehyde (an aldol) or a beta-hydroxy ketone (a ketol) — a molecule with an OH exactly two carbons away from the C=O.
The condensation step. Warm that aldol and it loses a molecule of water, because the alkene that forms ends up conjugated with the C=O and conjugation is stabilising. The final product is an alpha,beta-unsaturated carbonyl compound. Strictly, “aldol reaction” means stopping at the beta-hydroxy compound and “aldol condensation” means going on to lose water — read the question carefully to see which one is wanted.
(a) Ethanal, CH₃CHO. The alpha carbon is the CH₃. Base removes one of its hydrogens; the resulting carbanion attacks the carbonyl carbon of a second ethanal.
Aldol: CH₃CH(OH)CH₂CHO — 3-hydroxybutanal, C₄H₈O₂.
On heating, water leaves from C2 and C3: but-2-enal, CH₃CH=CHCHO, C₄H₆O.
(b) Propanone, CH₃COCH₃. Either methyl can act as the alpha carbon.
Ketol: (CH₃)₂C(OH)CH₂COCH₃ — 4-hydroxy-4-methylpentan-2-one, C₆H₁₂O₂.
On heating: 4-methylpent-3-en-2-one, (CH₃)₂C=CHCOCH₃, C₆H₁₀O — the compound we named back in Example 2.
Check your work with a carbon count: the aldol always has exactly twice as many carbons as the starting material. Two carbons in, four out. Three in, six out. If your product does not double, you have gone wrong.
Cross aldol condensation is the same reaction with two different carbonyl compounds in the flask. If both of them carry alpha hydrogens, each can become an enolate and each can be attacked, so you get four products — two self-aldols and two crossed ones — and the mixture is useless in practice. The reaction only becomes synthetically valuable when one partner has no alpha hydrogen at all. That partner cannot form an enolate, so it can only ever be the one that gets attacked, and a single clean product results.
Benzaldehyde, C₆H₅CHO, has no alpha hydrogen — the carbon next to its C=O is part of the aromatic ring and carries no H that a base can remove. So benzaldehyde can only play the role of the electrophile.
(a) Ethanal supplies the enolate. Product after loss of water: C₆H₅CH=CHCHO, 3-phenylprop-2-enal — better known as cinnamaldehyde, the smell of cinnamon.
(b) Propanone supplies the enolate. Product: C₆H₅CH=CHCOCH₃, 4-phenylbut-3-en-2-one.
Why it works: with only one possible enolate and only one possible electrophile, there is only one possible crossed product. That is the entire design principle behind a useful cross aldol.
Oxidation — Tollens’ And Fehling’s Tests
Door 3 opens. An aldehyde still has a hydrogen sitting on its carbonyl carbon, and that hydrogen is easy to prise off and replace with an OH — which is exactly what oxidation to a carboxylic acid amounts to. A ketone has no such hydrogen. To oxidise a ketone you must break a carbon–carbon bond, which needs a strong oxidising agent and vigorous heating. That single structural difference is the basis of every test in this chapter.
Tollens’ test. Tollens’ reagent is ammoniacal silver nitrate, in which the silver exists as the diamminesilver(I) complex ion, [Ag(NH₃)₂]⁺. Warm it gently with an aldehyde and the silver(I) is reduced to metallic silver, which plates the inside of a clean test tube as a beautiful silver mirror:
Fehling’s test. Fehling’s solution comes in two bottles that are mixed just before use: Fehling A is aqueous copper(II) sulphate, Fehling B is alkaline sodium potassium tartrate. Mixed, they give a deep blue solution in which the tartrate keeps the copper(II) in solution as a complex. Heat with an aliphatic aldehyde and copper(I) oxide drops out as a reddish-brown precipitate:
Both of these reagents are, at heart, coordination complexes doing a redox job — the ammine ligands in Tollens’ and the tartrate ligands in Fehling’s exist purely to keep the metal ion dissolved in alkaline conditions where it would otherwise precipitate as the hydroxide. If ligands and complex ions still feel unfamiliar, our Coordination Compounds notes explain exactly this kind of stabilisation.
Oxidation of ketones. Under strong oxidising conditions a ketone does eventually give way, but the carbon chain is cleaved on either side of the carbonyl group and you get a mixture of two smaller carboxylic acids. Only symmetrical ketones give a clean single product, which is why this is rarely a useful preparation.
The iodoform reaction. Methyl ketones have their own special oxidation. Warm any compound containing a CH₃CO— group (or a CH₃CH(OH)— group, which is oxidised to it in situ) with iodine and sodium hydroxide, and a pale yellow precipitate of iodoform, CHI₃, with its distinctive antiseptic smell, appears. The rest of the molecule is left as a carboxylate salt with one carbon fewer.
Positive (yellow CHI₃): ethanal CH₃CHO ✓, propanone CH₃COCH₃ ✓, acetophenone C₆H₅COCH₃ ✓, ethanol CH₃CH₂OH ✓, propan-2-ol CH₃CH(OH)CH₃ ✓.
Negative: propanal CH₃CH₂CHO ✗, propan-1-ol CH₃CH₂CH₂OH ✗.
The rule behind it: the molecule must already contain CH₃CO—, or contain CH₃CH(OH)— which the iodine oxidises into CH₃CO— before the test proper begins. Propanal has CH₃CH₂CHO — the methyl is two carbons from the carbonyl, not attached to it, so it fails. Ethanol passes because it is CH₃CH(OH)H, which oxidises to ethanal.
This is why the iodoform test is the standard way to distinguish propanal from propanone — a question that appears in some form almost every year.
Reduction — Clemmensen And Wolff–Kishner
Reduction is Door 3 travelled in the opposite direction, and you get to choose how far down you go.
Stopping at the alcohol. Sodium borohydride (NaBH₄) or lithium aluminium hydride (LiAlH₄) delivers a hydride to the carbonyl carbon; work-up adds the proton to oxygen. Catalytic hydrogenation with H₂ over nickel, platinum or palladium does the same job. An aldehyde gives a primary alcohol; a ketone gives a secondary alcohol. One caution: catalytic hydrogenation is indiscriminate and will also reduce any C=C double bond in the molecule, whereas NaBH₄ leaves alkenes untouched. If a question hands you an alpha,beta-unsaturated carbonyl and asks you to keep the double bond, NaBH₄ is your answer.
Going all the way to CH₂. Sometimes we want the oxygen gone completely, leaving a plain methylene group. Two named reactions do this, and they differ in one important respect — the conditions.
| Feature | Clemmensen reduction | Wolff–Kishner reduction |
|---|---|---|
| Reagents | Zinc amalgam (Zn–Hg) with concentrated HCl | Hydrazine, NH2NH2, then KOH in ethylene glycol, heated |
| Medium | Strongly acidic | Strongly basic |
| Product from >C=O | >CH2 | >CH2 |
| Choose it when | The molecule contains base-sensitive groups | The molecule contains acid-sensitive groups |
The Cannizzaro Reaction
Here is the elegant one. Take an aldehyde that has no alpha hydrogen and treat it with concentrated alkali. It cannot form an enolate, so an aldol is impossible. Instead the molecules turn on each other: one is oxidised to the salt of a carboxylic acid, and another is reduced to a primary alcohol. A reaction in which the same substance is simultaneously oxidised and reduced is called a disproportionation.
2 C6H5CHO + NaOH (conc.) → C6H5CH2OH + C6H5COONa
Mechanically, a hydroxide ion adds to the carbonyl carbon of one molecule; the resulting alkoxide is so electron-rich that it hands over a hydride ion to the carbonyl carbon of a second molecule. The donor becomes the acid salt, the receiver becomes the alcohol. Once you picture it as a hydride handshake, the reaction stops feeling arbitrary.
Neither aldehyde has an alpha hydrogen, so both are eligible. The one that is oxidised is the one more readily attacked by hydroxide in the first step — and that is methanal, whose carbonyl carbon is the least hindered and least electron-rich of any carbonyl compound. Methanal therefore becomes the hydride donor.
Products: HCOONa (sodium methanoate, from the oxidised methanal) and C₆H₅CH₂OH (benzyl alcohol, from the reduced benzaldehyde).
The memorable version: in a crossed Cannizzaro, methanal always sacrifices itself. It gets oxidised so the other aldehyde can be reduced. Recall that alongside “Fewer Friends, Faster Feast” — methanal is the most reactive carbonyl, so it is attacked first, and being attacked first is what condemns it to oxidation.
Electrophilic Substitution In Aromatic Aldehydes And Ketones
When an aldehyde or ketone group is attached to a benzene ring, the ring can still undergo the usual electrophilic substitutions — nitration, halogenation, sulphonation — but two things change, and both flow from the same cause.
The —CHO and —COR groups are electron-withdrawing. They pull electron density out of the ring both inductively and by resonance, because the carbonyl carbon is already electron-poor and is happy to accept the ring’s electrons. The consequences are:
- The ring becomes deactivated — substitution is slower and needs harsher conditions than with benzene itself.
- The incoming group goes to the meta position. Withdrawing electrons out of the ortho and para positions leaves the meta position comparatively the richest, so that is where an electrophile attacks.
The electrophile is the nitronium ion, NO₂⁺. The —CHO group is deactivating and meta-directing, so the nitro group enters position 3.
Main product: 3-nitrobenzaldehyde.
Why it works: draw the resonance structures of the intermediate carbocation for ortho, meta and para attack. For ortho and para attack, one contributing structure places a positive charge on the ring carbon that already bears the electron-withdrawing —CHO group — a highly unfavourable arrangement of two positive centres side by side. Meta attack never produces that structure, so the meta intermediate is the most stable and forms fastest.
Uses Of Aldehydes And Ketones
- Methanal. Its 40 per cent aqueous solution, formalin, preserves biological specimens and acts as a disinfectant. Industrially it is the feedstock for bakelite, urea–formaldehyde resins and other polymers.
- Ethanal. A major industrial intermediate on the way to ethanoic acid, ethyl ethanoate and a range of other chemicals.
- Propanone and butanone. Excellent solvents — propanone is the acetone in nail-varnish remover and is used industrially to dissolve resins, cellulose acetate and paints.
- Benzaldehyde. Used in perfumery and as a starting material for dyes; it carries the smell of bitter almonds.
- Higher aldehydes and ketones. Many have pleasant fragrances and appear in perfumes and flavourings — cinnamaldehyde from Example 12 is the taste of cinnamon.
Nomenclature Of Carboxylic Acids
A carboxylic acid carries the —COOH group, which is a carbonyl and a hydroxyl fused onto the same carbon. That fusion changes everything, as we are about to see, but the naming is the easiest part of the chapter.
IUPAC. Take the longest chain containing the —COOH carbon, drop the final -e and add -oic acid. Like the aldehyde carbon, the carboxyl carbon is always C-1. So HCOOH is methanoic acid, CH₃COOH is ethanoic acid, CH₃CH₂COOH is propanoic acid. Two —COOH groups on a chain keep the -e and take -dioic acid: HOOC—COOH is ethanedioic acid. When the group is attached to a ring we use -carboxylic acid, giving benzenecarboxylic acid (universally called benzoic acid) and cyclohexanecarboxylic acid.
Common names come from where the acids were first found, and they are charming enough to be worth knowing. Methanoic acid is formic acid, from formica, the Latin for ant — it is what makes an ant bite sting. Ethanoic acid is acetic acid, from acetum, vinegar. Butanoic acid is butyric acid, from butyrum, butter, and it is responsible for the smell of rancid butter. Ethanedioic acid is oxalic acid, from the wood sorrel Oxalis. As with aldehydes, the common system numbers substituent positions with Greek letters starting at the carbon next to —COOH.
(a) Longest chain through COOH is three carbons: COOH—CH—CH₃, so propanoic acid, with a methyl on C2. 2-Methylpropanoic acid.
(b) Three carbons, COOH is C1, chlorine on C3. 3-Chloropropanoic acid (common name β-chloropropionic acid — and note that C-3 and β are the same carbon).
(c) Three carbons with a COOH at each end. Keep the -e and use -dioic: propanedioic acid (common name malonic acid).
Acidic Nature And The Effect Of Substituents
Alcohols have an O—H bond and are barely acidic at all. Carboxylic acids have an O—H bond and are genuinely acidic. Something about attaching a carbonyl group next door has changed the situation completely — and understanding what is the single most rewarding piece of reasoning in this unit.
The answer lies not in the acid but in what is left behind. When an alcohol loses its proton, the negative charge sits on one oxygen and stays there, with nowhere to go. When a carboxylic acid loses its proton, the resulting carboxylate ion has two resonance structures that are exactly equivalent — the charge is shared evenly between two oxygen atoms, and both carbon–oxygen bonds become identical, halfway between a single and a double bond. Spreading a negative charge over two electronegative atoms is enormously stabilising, and a stable conjugate base means a strong acid.
Phenol sits in between. Its phenoxide ion does delocalise the charge, but into the ring, onto carbon atoms — and carbon is far less willing to hold a negative charge than oxygen is. So the order is settled: carboxylic acid > phenol > alcohol. The ladder below is drawn from real pKa values so you can see how large the gaps actually are. Remember that pKa runs backwards — a smaller pKa means a stronger acid, and each whole unit is a factor of ten.
Now for substituents. The logic never changes: anything that helps spread out the negative charge of the carboxylate makes the acid stronger. Electron-withdrawing groups pull charge away from the —COO⁻ end and stabilise it. Electron-donating groups push charge towards it, which is the last thing an already-negative ion wants, so they weaken the acid. Four patterns follow from that one sentence.
- More withdrawing groups, stronger acid. Look at the ladder above: ethanoic 4.76, chloroethanoic 2.86, dichloroethanoic 1.29, trichloroethanoic 0.65. Each extra chlorine drags the pKa down by roughly one and a half units.
- Closer withdrawing group, stronger acid. The inductive effect fades fast down a chain. For chlorobutanoic acids the pKa values run 2.84 (chlorine on C-2), 4.06 (on C-3) and 4.52 (on C-4), against 4.82 for butanoic acid itself. By the fourth carbon the chlorine is barely felt.
- More electronegative group, stronger acid. Among the 2-halo acids: FCH₂COOH 2.59, ClCH₂COOH 2.86, BrCH₂COOH 2.90, ICH₂COOH 3.18 — the same order as the electronegativities.
- Alkyl groups weaken acids. Methanoic acid (3.75) has no alkyl group at all and is the strongest simple aliphatic acid. Adding a methyl gives ethanoic acid (4.76); adding another gives propanoic acid (4.88).
Aromatic acids. Benzoic acid (4.20) is a slightly stronger acid than ethanoic acid (4.76), which surprises people. The reason is hybridisation: the ring carbon attached to —COOH is sp², and an sp² carbon holds its electrons more tightly than the sp³ carbon of a methyl group. Relative to an alkyl group, therefore, a phenyl group is mildly electron-withdrawing. Substituents on the ring then shift things further — an electron-withdrawing group such as —NO₂ makes the acid stronger, an electron-donating group such as —OCH₃ or —CH₃ makes it weaker.
Reason it out one at a time, using pKa values.
Ethanoic acid (4.76) — one electron-donating methyl group, nothing withdrawing. Weakest.
3-Chloropropanoic acid (about 3.98) — it does have a chlorine, but two carbons away from the —COOH, so the inductive pull has largely faded.
Methanoic acid (3.75) — no alkyl group at all to donate electrons, and nothing between the H and the —COOH to weaken anything. It edges ahead of the distant chlorine.
Chloroethanoic acid (2.86) — the chlorine is now directly next to the —COOH, so the withdrawal is strong.
Trichloroethanoic acid (0.65) — three chlorines next door. Strongest by a wide margin.
Answer (weakest → strongest): ethanoic acid < 3-chloropropanoic acid < methanoic acid < chloroethanoic acid < trichloroethanoic acid.
The catch worth noticing: it is tempting to assume that any acid with a chlorine must beat any acid without one. It is not so — a chlorine three bonds away is worth less than the simple absence of an electron-donating alkyl group. Always ask how far, not just whether.
The mark scheme wants a reason, not just an order. Write: “an electron-withdrawing group disperses the negative charge on the carboxylate ion, stabilising it and so increasing acid strength; the effect grows with the number of such groups and falls off sharply with distance.” That one sentence answers every question of this type.
The phenol comparison in that box is developed much more fully in our Alcohols, Phenols and Ethers chapter, and reading the two acidity discussions side by side is one of the most efficient revision hours you can spend in organic chemistry.
Methods Of Preparation Of Carboxylic Acids
Six routes again, and they sort neatly into two groups: oxidise something up to the carboxyl level, or hydrolyse something down to it.
1. From primary alcohols and aldehydes (oxidation up). Warm a primary alcohol or an aldehyde with acidified KMnO₄ or acidified K₂Cr₂O₇ and you get the carboxylic acid. Aldehydes are so easily oxidised that even the mild reagents from earlier in this chapter — Tollens’ and Fehling’s — will do it, though they give the carboxylate salt rather than the free acid.
2. From alkylbenzenes (oxidation up, with a twist). Heat an alkylbenzene with alkaline KMnO₄ and then acidify, and the entire side chain, however long, is chopped back to a single —COOH group. Toluene, ethylbenzene and propylbenzene all give the same product: benzoic acid. There is one condition — the carbon attached to the ring must carry at least one hydrogen. tert-Butylbenzene, whose side-chain carbon has no hydrogen, survives the treatment unchanged.
3. From nitriles and amides (hydrolysis down). A nitrile hydrolyses through the amide to the acid. With dilute acid you get the free acid directly; with alkali you get the carboxylate salt and must acidify at the end to release the acid. Amides hydrolyse the same way, releasing ammonia or an amine.
4. From Grignard reagents (the carbon-adding route). Bubble the Grignard reagent onto solid carbon dioxide — dry ice — and the strongly nucleophilic carbon of R—MgX attacks the electrophilic carbon of CO₂. You get the halomagnesium salt of the acid, and acidic work-up frees the acid itself. This route lengthens the carbon chain by exactly one carbon, and that makes it one of the most useful conversions in the whole of organic chemistry.
5. From acyl halides and anhydrides (hydrolysis down). Acyl chlorides hydrolyse vigorously in water; more controllably, treat them with aqueous alkali and then acidify. Anhydrides hydrolyse more slowly, and one anhydride gives two molecules of acid.
6. From esters (hydrolysis down). Acid hydrolysis is reversible, so a large excess of water is needed to push it forwards. Alkaline hydrolysis — saponification — is not reversible, because the carboxylate salt that forms is removed from the equilibrium; acidify afterwards to obtain the free acid.
Step 1. CH₃CH₂Br + Mg, dry ether → CH₃CH₂MgBr (ethylmagnesium bromide).
Step 2. CH₃CH₂MgBr + CO₂ (dry ice) → CH₃CH₂COOMgBr.
Step 3. Add dilute acid → CH₃CH₂COOH, propanoic acid.
Count the carbons: two in, three out. The new carbon arrived as carbon dioxide. Compare this with the cyanohydrin route in Example 9 — both add exactly one carbon, and between them they cover almost every “increase the chain by one” question you will meet.
One warning: a Grignard reagent is destroyed instantly by water, alcohols or any acidic hydrogen, so the ether must be scrupulously dry and the acid is only added at the very end.
Toluene, C₆H₅CH₃ → benzoic acid.
Ethylbenzene, C₆H₅CH₂CH₃ → benzoic acid.
Propylbenzene, C₆H₅CH₂CH₂CH₃ → benzoic acid.
tert-Butylbenzene, C₆H₅C(CH₃)₃ → no reaction.
Why it works: the oxidation begins by attacking the benzylic carbon — the side-chain carbon joined directly to the ring — and it needs a hydrogen there to get started. Once it does, the whole chain is stripped back to —COOH. In tert-butylbenzene that carbon carries three methyl groups and no hydrogen at all, so the reaction never begins.
Physical Properties Of Carboxylic Acids
Aliphatic carboxylic acids up to about nine carbons are colourless liquids at room temperature with distinctly unpleasant smells — think rancid butter and stale sweat. Above that they are waxy, odourless solids, simply because they are no longer volatile enough to reach your nose.
Boiling points are remarkably high — higher than those of comparable alcohols, let alone aldehydes and ketones. Ethanoic acid (molar mass 60.05) boils at 391 K, while propan-1-ol (molar mass 60.10) boils at 370 K. The reason is that a carboxylic acid does not merely hydrogen bond; it forms a cyclic dimer, with two hydrogen bonds locking a pair of molecules together into what behaves like a single, much heavier particle. To boil the liquid you must break both bonds at once.
Solubility. The first four members — methanoic, ethanoic, propanoic and butanoic acid — are completely miscible with water, because the —COOH group both donates and accepts hydrogen bonds. Beyond that the greasy hydrocarbon tail dominates and solubility falls off sharply. Benzoic acid is only slightly soluble in cold water but dissolves well in hot water, which is exactly why it is purified by recrystallisation. All carboxylic acids dissolve readily in less polar organic solvents such as ether, benzene and alcohol.
Reactions Involving The Carboxyl Group
Five reactions here, and four of them are the same move in different clothes: something replaces the —OH of the carboxyl group. Learn them as a family and they stop being five separate facts.
1. Salt formation. Carboxylic acids react with sodium hydroxide, sodium carbonate and sodium hydrogencarbonate to give the carboxylate salt. Only the last of these releases a gas, and that is what makes it a test: brisk effervescence of CO₂ with NaHCO₃.
2. Anhydride formation. Heat the acid with a dehydrating agent such as phosphorus pentoxide, P₄O₁₀. One molecule loses —OH, the other loses —H, water leaves and the two acyl fragments join through an oxygen. Two molecules of ethanoic acid give ethanoic anhydride.
3. Ester formation. Warm the acid with an alcohol and a few drops of concentrated sulphuric acid (or pass in dry HCl gas). This is Fischer esterification, and the crucial feature is that it is reversible — the reverse reaction is simply acid hydrolysis of the ester. To drive it forwards you either use excess alcohol or distil off the water as it forms. The acid catalyst works by protonating the carbonyl oxygen, which makes the carboxyl carbon much more electrophilic and so much easier for the alcohol to attack.
4. Acid chloride formation. Phosphorus pentachloride (PCl₅), phosphorus trichloride (PCl₃) or thionyl chloride (SOCl₂) will all convert —COOH into —COCl. Thionyl chloride is much the best choice, and the reason is worth remembering: its by-products are sulphur dioxide and hydrogen chloride, both gases, which simply escape from the flask and leave a pure product behind. PCl₅ leaves POCl₃ mixed in, which then has to be separated.
5. Amide formation. Ammonia first gives the ammonium carboxylate salt, an ordinary acid–base reaction. Heat that salt strongly and it loses water to give the amide. Do not skip the salt — questions frequently ask for both steps.
(a) CH₃COONa + H₂O + CO₂↑ (the fizz test).
(b) (CH₃CO)₂O, ethanoic anhydride, plus water removed by the P₄O₁₀.
(c) CH₃COOC₂H₅, ethyl ethanoate, plus water — and the reaction is reversible.
(d) CH₃COCl, ethanoyl chloride, plus SO₂ and HCl as escaping gases.
(e) First CH₃COONH₄ (ammonium ethanoate), then on strong heating CH₃CONH₂, ethanamide, plus water.
The pattern: in (b), (c), (d) and (e) the —OH of the carboxyl group has been swapped for —OCOCH₃, —OC₂H₅, —Cl and —NH₂ respectively. One move, four costumes.
Reduction And Decarboxylation
Reduction. Carboxylic acids are stubborn towards reduction — the carboxylate ion that forms in the reaction mixture is already stabilised and resists attack. Sodium borohydride is simply not strong enough and will not touch a —COOH group. The two reagents that work are lithium aluminium hydride (LiAlH₄) and diborane (B₂H₆), and both take the acid all the way down to a primary alcohol. Diborane has a useful selectivity: it reduces —COOH while leaving ester, nitro and halogen groups elsewhere in the molecule alone.
Decarboxylation means throwing the carboxyl carbon away as carbon dioxide. Heat the sodium salt of the acid with soda lime (sodium hydroxide mixed with calcium oxide) and you get an alkane with one carbon fewer. Sodium ethanoate gives methane; sodium propanoate gives ethane.
Kolbe electrolysis does something more interesting. Electrolyse an aqueous solution of the sodium or potassium salt of a carboxylic acid and, at the anode, two alkyl fragments couple together while carbon dioxide is released. Sodium ethanoate therefore gives ethane, not methane — two CH₃ fragments joining up. Hydrogen is liberated at the cathode.
The electrode reasoning behind Kolbe electrolysis — why oxidation happens at the anode and what determines which species is discharged — is set out properly in our Electrochemistry chapter, which is worth a look if the anode/cathode logic here feels like something you are taking on trust.
HVZ Reaction And Ring Substitution
These two reactions leave the —COOH group alone and attack the hydrocarbon part instead.
The Hell–Volhard–Zelinsky reaction. Treat a carboxylic acid that has an alpha hydrogen with chlorine or bromine in the presence of a small amount of red phosphorus, and the halogen substitutes at the alpha carbon. Hydrolysis afterwards gives the alpha-halo acid. The phosphorus is a catalyst, not a reagent — it converts a little of the acid into the acyl halide, which enolises far more readily than the acid itself and so accepts the halogen. Note the strict requirement: no alpha hydrogen, no HVZ reaction. Methanoic acid (HCOOH), 2,2-dimethylpropanoic acid and benzoic acid are all immune.
Ring substitution. Aromatic carboxylic acids undergo the usual electrophilic substitutions, but the —COOH group is electron-withdrawing, so exactly as with —CHO the ring is deactivated and the incoming group is directed meta. Brominating benzoic acid with Br₂ and FeBr₃ gives 3-bromobenzoic acid; nitrating it gives 3-nitrobenzoic acid. One further point that examiners like: benzoic acid does not undergo Friedel–Crafts reactions at all, both because the ring is too deactivated and because the Lewis-acid catalyst binds to the carboxyl group and is taken out of action.
Count the alpha hydrogens on each.
HCOOH — the group next to —COOH is a hydrogen, not a carbon, so there is no alpha carbon at all. No reaction.
CH₃COOH — three alpha hydrogens. Reacts, giving BrCH₂COOH.
(CH₃)₂CHCOOH — one alpha hydrogen. Reacts, giving (CH₃)₂CBrCOOH.
(CH₃)₃CCOOH — the alpha carbon holds three methyls and no hydrogen. No reaction.
C₆H₅COOH — the alpha position is an aromatic ring carbon with no removable hydrogen there. No reaction.
Why it works: the mechanism goes through an enol, and you cannot form an enol without a hydrogen on the alpha carbon to move. This is the same alpha-hydrogen requirement that governed the aldol reaction — Door 2, once again.
Uses Of Carboxylic Acids
- Ethanoic acid is the acid of vinegar (a six to eight per cent solution), a widely used laboratory solvent, and the starting point for esters and for cellulose acetate used in rayon.
- Methanoic acid coagulates natural rubber latex, and is used in textile dyeing and finishing, in leather tanning and in some medicines.
- Hexanedioic acid (adipic acid) is one of the two monomers of nylon-6,6, and benzene-1,4-dicarboxylic acid (terephthalic acid) is a monomer of the polyester used in bottles and fabric.
- Higher fatty acids such as stearic and palmitic acid are converted into soaps and detergents.
- Sodium benzoate is a common food preservative, and esters of benzoic acid are used in perfumery.
- Salicylic acid is the raw material for aspirin.
Distinguishing Tests — The TIFF Bench
Almost every year a question hands you two or three unlabelled bottles and asks you to tell them apart. These marks are among the easiest in the paper, provided you have a system rather than a jumble of half-remembered colours. Here is the system I want you to carry into the exam hall: four bottles on a bench, in a fixed order, spelling TIFF.
- T — Tollens’ reagent. A silver mirror means an aldehyde. Any aldehyde, aliphatic or aromatic.
- I — Iodoform (I₂ with NaOH). A yellow precipitate means a CH₃CO— group (or a CH₃CH(OH)— group that becomes one).
- F — Fehling’s solution. A reddish-brown precipitate means an aliphatic aldehyde. Aromatic aldehydes fail this one.
- F — Fizz with NaHCO₃. Effervescence of carbon dioxide means a carboxylic acid.
Run them in that order on an unknown and you cannot fail to place it. The flowchart below shows exactly how the branches fall.
| Reagent | Positive result you see | What it proves is present | Notable failures |
|---|---|---|---|
| Tollens’ reagent | Silver mirror on the glass | Any aldehyde | All ketones |
| Fehling’s solution | Reddish-brown Cu2O precipitate | Aliphatic aldehyde | Aromatic aldehydes and all ketones |
| I2 with NaOH (iodoform) | Yellow CHI3 precipitate | CH3CO— or CH3CH(OH)— group | Propanal, propan-1-ol, benzaldehyde |
| Aqueous NaHCO3 | Brisk effervescence of CO2 | Carboxylic acid, —COOH | Phenols, alcohols, all carbonyls |
| 2,4-Dinitrophenylhydrazine | Orange-red precipitate | A carbonyl group — aldehyde or ketone | Cannot separate aldehyde from ketone |
| Saturated NaHSO3 | White crystalline solid | Aldehyde, methyl ketone or small cyclic ketone | Bulky ketones such as benzophenone |
Test 1 — add aqueous NaHCO₃ to all four. Only one fizzes. That bottle is ethanoic acid. Set it aside.
Test 2 — warm the remaining three with Tollens’ reagent. Two give a silver mirror (ethanal and benzaldehyde); one does not. The one that does not is propanone.
Test 3 — heat the last two with Fehling’s solution. One gives a reddish-brown precipitate and one does not. The one that does is the aliphatic aldehyde, ethanal; the one that does not is the aromatic aldehyde, benzaldehyde.
All four identified in three tests. Note how each test roughly halves the remaining possibilities — that is what makes TIFF efficient rather than a scattergun. A neat alternative for Test 3 is the iodoform test: ethanal gives a yellow precipitate, benzaldehyde gives none.
Aldehydes, Ketones and Carboxylic Acids Class 12 Important Questions with Solved Examples
Four question types account for most of the marks in this unit: multi-step conversions, identify-the-unknown puzzles, name-reaction recall, and “give a reason” questions. Here is one of each, fully worked, followed by the complete name reactions list with examples that you should be able to reproduce from memory.
(a) Methane — lose a carbon. Neutralise with NaOH to get CH₃COONa, then heat the dry salt with soda lime (NaOH + CaO). Decarboxylation gives CH₄.
(b) Ethane — keep both carbons but couple two fragments. Make the sodium salt again, then electrolyse its aqueous solution (Kolbe electrolysis). Two CH₃ fragments couple at the anode to give CH₃CH₃, with CO₂ released.
(c) Propanoic acid — gain a carbon. Three steps:
CH₃COOH —LiAlH₄, then H₃O⁺→ CH₃CH₂OH
CH₃CH₂OH —PBr₃→ CH₃CH₂Br
CH₃CH₂Br —alcoholic KCN→ CH₃CH₂CN —dilute acid, warm→ CH₃CH₂COOH
How to spot the route: count carbons in the question before you write anything. Two → one means decarboxylation. Two → two with coupling means Kolbe. Two → three means you need a carbon-adding step, and your only two options are a cyanide or a Grignard with CO₂.
Clue 1 — orange precipitate with 2,4-DNP: A contains a carbonyl group. It is an aldehyde or a ketone.
Clue 2 — no silver mirror with Tollens’: it is not an aldehyde. So A is a ketone.
Clue 3 — yellow precipitate with I₂/NaOH: A contains a CH₃CO— group, so it is a methyl ketone.
Clue 4 — the formula C₄H₈O: four carbons, one oxygen, one degree of unsaturation (the C=O).
The only four-carbon methyl ketone is CH₃COCH₂CH₃.
A is butan-2-one (ethyl methyl ketone), C₄H₈O.
Check: its iodoform reaction gives CHI₃ plus sodium propanoate, CH₃CH₂COONa — three carbons left over from four, exactly as expected when the CH₃ leaves as iodoform.
(a) In a carboxylic acid the —OH oxygen sits directly on the carbonyl carbon, and one of its lone pairs is delocalised into the C=O group. That donation feeds electron density back onto the carbonyl carbon, so its δ+ charge is markedly smaller than in an aldehyde or ketone. A weakly electrophilic carbon does not attract nucleophiles, so addition does not occur. Instead the acid reacts by losing its acidic proton or by substitution of the whole —OH group.
(b) The same delocalisation is the answer. Because the lone pair is shared into the C=O region, that bond acquires partial single-bond character, and a bond with less double-bond character is longer and weaker.
The transferable idea: one structural feature — resonance donation from the hydroxyl oxygen — explains the acidity, the lack of addition reactions and the bond length. When an exam asks three questions about the same molecule, look for the single cause that answers all three.
Finally, the recall table. If you can reproduce this from a blank sheet, you have the name-reaction marks in the bag.
| Name reaction | Reagents and conditions | Worked example |
|---|---|---|
| Rosenmund reduction | H2, Pd–BaSO4, poisoned with S or quinoline | CH3COCl → CH3CHO |
| Stephen reaction | SnCl2/HCl, then H3O+ | CH3CN → CH3CHO |
| Gattermann–Koch | CO + HCl, anhydrous AlCl3 with CuCl | C6H6 → C6H5CHO |
| Friedel–Crafts acylation | RCOCl, anhydrous AlCl3 | C6H6 → C6H5COCH3 |
| Etard reaction | CrO2Cl2 in CS2, then hydrolysis | C6H5CH3 → C6H5CHO |
| Aldol condensation | Dilute NaOH, then heat | 2 CH3CHO → CH3CH=CHCHO |
| Cross aldol condensation | Dilute NaOH; one partner must lack α-H | C6H5CHO + CH3CHO → C6H5CH=CHCHO |
| Cannizzaro reaction | Concentrated NaOH; aldehyde must lack α-H | 2 C6H5CHO → C6H5CH2OH + C6H5COONa |
| Clemmensen reduction | Zn–Hg, concentrated HCl (acidic) | CH3COCH3 → CH3CH2CH3 |
| Wolff–Kishner reduction | NH2NH2, then KOH in ethylene glycol, heat (basic) | CH3COCH3 → CH3CH2CH3 |
| Tollens’ test | Ammoniacal AgNO3, warm | CH3CHO → CH3COO− + Ag mirror |
| Fehling’s test | Fehling A + B, heat; aliphatic aldehydes only | CH3CHO → CH3COO− + Cu2O |
| Iodoform reaction | I2 with NaOH, warm | CH3COCH3 → CHI3 + CH3COO− |
| HVZ reaction | Cl2 or Br2 with red phosphorus; needs α-H | CH3COOH → BrCH2COOH |
| Kolbe electrolysis | Electrolysis of the aqueous sodium salt | 2 CH3COONa → CH3CH3 + 2CO2 |
| Decarboxylation | Sodium salt heated with soda lime (NaOH + CaO) | CH3COONa → CH4 |
Practice Worksheet
Ten questions, written fresh for this chapter. Cover the answers, attempt each one on paper, and only then open the accordion. Reading an answer you have not first struggled with teaches you almost nothing.
Q1. Give the IUPAC name of (a) CH₃CH₂CH₂CH(CH₃)CHO and (b) CH₃COCH₂CH(CH₃)CH₃.
Show Answer
(b) 4-Methylpentan-2-one. Numbering from the end nearer the carbonyl gives the C=O the lowest locant (2), and the extra methyl falls on C-4. Both compounds are C₆H₁₂O, molar mass 100.16 g mol⁻¹ — they are isomers.
Q2. How would you distinguish pentan-2-one from pentan-3-one by a single chemical test?
Show Answer
Pentan-2-one, CH₃COCH₂CH₂CH₃, contains a CH₃CO— group, so it gives a yellow precipitate of iodoform, CHI₃.
Pentan-3-one, CH₃CH₂COCH₂CH₃, has ethyl groups on both sides of the carbonyl and no CH₃CO— group, so it gives no precipitate.
Tollens’ and Fehling’s are useless here — both compounds are ketones and both would fail.
Q3. Arrange in increasing order of reactivity towards addition of HCN: benzaldehyde, ethanal, propanone, methanal. Justify your order.
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Methanal has no alkyl group, so its carbonyl carbon is the most electron-poor and the least hindered. Ethanal has one methyl, propanone has two — each one both donates electron density (+I) and blocks the approaching nucleophile. Benzaldehyde is the least reactive because the benzene ring feeds electron density into the C=O by resonance, and resonance donation outweighs the inductive donation of two methyl groups.
Q4. 2,2-Dimethylpropanal is warmed with concentrated sodium hydroxide. Name the reaction and give both products.
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Products: (CH₃)₃CCH₂OH, 2,2-dimethylpropan-1-ol (the reduced half), and (CH₃)₃CCOONa, sodium 2,2-dimethylpropanoate (the oxidised half).
Q5. Which of butanal, benzaldehyde, propanone and 2,2-dimethylpropanal undergo aldol condensation? Write the final product from butanal.
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From butanal: the aldol is CH₃CH₂CH₂CH(OH)CH(C₂H₅)CHO, that is 2-ethyl-3-hydroxyhexanal. On warming it loses water to give 2-ethylhex-2-enal, CH₃CH₂CH₂CH=C(C₂H₅)CHO.
Carbon check: four carbons in, eight carbons out — the aldol always doubles the carbon count.
Q6. Which is the stronger acid, 2-chlorobutanoic acid or 4-chlorobutanoic acid? Explain using their pKa values of 2.84 and 4.52.
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Reason: the chlorine withdraws electron density inductively and so helps disperse the negative charge on the carboxylate ion, stabilising it. But the inductive effect is transmitted through sigma bonds and weakens sharply with every bond it crosses. In the 2-isomer the chlorine is on the carbon next to —COOH; in the 4-isomer it is three carbons away and its influence is almost spent.
Q7. An alkene of molecular formula C₈H₁₆ gives butan-2-one as its only ozonolysis product. Identify the alkene and name it.
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Rejoin two of them: CH₃(C₂H₅)C=C(C₂H₅)CH₃.
Atom count: 8 carbons and 16 hydrogens — C₈H₁₆, as required.
The alkene is 3,4-dimethylhex-3-ene.
Q8. Explain why benzoic acid does not undergo Friedel–Crafts alkylation.
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1. The —COOH group is strongly electron-withdrawing, so the ring is heavily deactivated and no longer electron-rich enough to attack a carbocation electrophile.
2. The Lewis-acid catalyst, aluminium chloride, coordinates to the lone pairs of the carboxyl group. That ties up the catalyst and simultaneously makes the —COOH group an even stronger electron-withdrawer, deactivating the ring further still.
Q9. Identify A, B and C: CH₃CH₂COOH —SOCl₂→ A —H₂, Pd–BaSO₄/S→ B —dil. NaOH, then warm→ C.
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B = CH₃CH₂CHO, propanal. These are Rosenmund conditions — the poisoned catalyst stops the reduction at the aldehyde.
C = 2-methylpent-2-enal, CH₃CH₂CH=C(CH₃)CHO. Propanal has alpha hydrogens, so dilute alkali gives the aldol 3-hydroxy-2-methylpentanal, which loses water on warming.
Q10. Ethanoic acid dissolved in benzene gives an experimental molar mass of about 120 g mol⁻¹. Explain, and calculate the van’t Hoff factor and the degree of association.
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Calculation. Normal molar mass of CH₃COOH = 60.05 g mol⁻¹.
van’t Hoff factor i = normal molar mass ÷ observed molar mass = 60.05 ÷ 120.1 = 0.50.
For dimerisation, i = 1 − α/2, so 0.50 = 1 − α/2, giving α = 1.0, that is essentially 100 per cent association.
An i below 1 always signals association; an i above 1 signals dissociation.
If some of those went wrong, that is information, not failure. Note down which door each mistake belonged to — the carbonyl carbon, the alpha carbon, or the oxidation state — and you will usually find your errors clustering in one place rather than scattered everywhere. Fixing one cluster is far easier than revising a whole chapter.
A last thought before you close the tab. You do not have to master this chapter today. You only have to get one more question right than you did yesterday. Do that every day between now and your exam and the arithmetic takes care of itself — small, honest, repeated improvement will carry you further than any single heroic night of cramming ever could. Tomorrow, come back and beat today’s score by one.