Chemistry: Atoms First 2e · Organic Chemistry
Aldehydes, Ketones, Carboxylic Acids, and Esters
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This topic covers four related families built on the Carbonyl group \(\mathrm{C{=}O}\) double bond Full entry →, \(\mathrm{C{=}O}\): aldehydes (\(\mathrm{R{-}CHO}\), carbonyl at the end of a chain), ketones (\(\mathrm{R{-}CO{-}R'}\), carbonyl in the middle), carboxylic acids (\(\mathrm{R{-}COOH}\), carbonyl bonded to an –OH), and esters (\(\mathrm{R{-}COO{-}R'}\), carbonyl bonded to an –OR group). Because all four share the polar, electron-rich C=O bond, they form a connected story: oxidation converts primary alcohols up the ladder to aldehydes then carboxylic acids, and acids plus alcohols condense into esters. Their physical properties climb with hydrogen bonding — aldehydes and ketones cannot donate hydrogen bonds, carboxylic acids form strong hydrogen-bonded pairs, and esters are the "fragrance" family. Formaldehyde, acetone, vinegar (acetic acid), and the flavor esters in fruits and perfumes are everyday examples of these four functional groups.
Why this matters
Carbonyl chemistry is the chemistry of biology and of the products around us. Formaldehyde preserves biological specimens and is a building block of resins; acetone is a solvent and nail-polish remover; vinegar is a dilute solution of acetic acid; esters give bananas, pineapples, and wintergreen their smells and flavors. In the body, glucose and fructose are aldehydes/ketones, fatty acids are carboxylic acids, and fats are esters of glycerol. Pharmaceuticals are saturated with these groups: aspirin is an Ester \(\mathrm{R{-}COO{-}R'}\) — carbonyl + –OR Full entry → of salicylic acid, and many pain relievers, anesthetics, and antibiotics carry carbonyl groups. In healthcare contexts, recognizing that "Aldehyde \(\mathrm{R{-}CHO}\) — carbonyl at chain end Full entry →" disinfectants (like glutaraldehyde) are chemically distinct from "alcohol" sanitizers matters for correct use, and carboxylic acids appear in medications (e.g., ibuprofen) whose acid group affects how they behave in the body. Understanding the carbonyl Oxidation ladder 1° alcohol → aldehyde → acid; 2° → ketone Full entry → also explains why wine exposed to air turns to vinegar — ethanol → acetaldehyde → acetic acid.
The college version
Core Concepts
The carbonyl group: the shared engine
The C=O double bond is strongly polar: oxygen pulls electron density away from carbon, leaving the carbon electron-poor (electrophilic) and the oxygen electron-rich (nucleophilic). This polarity drives nearly every reaction of these families. In aldehydes, the carbonyl carbon is bonded to one carbon group and one hydrogen (\(\mathrm{R{-}CHO}\)); in ketones, it is bonded to two carbon groups (\(\mathrm{R{-}CO{-}R'}\)). The difference is small structurally but changes naming, reactivity, and oxidation behavior.
Naming: -al, -one, -oic acid, -oate
- Aldehydes end in -al; because the carbonyl is always at the end, it is C1 and needs no locant (ethanal, propanal; common names: formaldehyde, acetaldehyde).
- Ketones end in -one with a locant for the carbonyl position (2-propanone is acetone; 2-butanone).
- Carboxylic acids end in -oic acid; the acid carbon is C1 (ethanoic acid = acetic acid; methanoic acid = formic acid).
- Esters are named as alkyl alkanoates: the group from the alcohol comes first as an alkyl name, then the acid part with -oate (ethyl ethanoate = ethyl acetate).
Oxidation states and the carbonyl ladder
The oxidation level of the carbonyl carbon tracks the number of bonds to oxygen. Methanol \(\mathrm{CH_3OH}\) → methanal (formaldehyde) \(\mathrm{HCHO}\) → methanoic acid \(\mathrm{HCOOH}\) → \(\mathrm{CO_2}\) is a rising ladder: primary alcohols oxidize to aldehydes and then acids; secondary alcohols oxidize to ketones (which resist further oxidation because the carbonyl carbon holds no hydrogen); tertiary alcohols do not oxidize. This ladder is a favorite exam map and explains real processes like wine turning to vinegar.
Physical properties: hydrogen bonding sets the order
Aldehydes and ketones are polar (the C=O dipole) but cannot donate hydrogen bonds, so they boil higher than alkanes of similar size but much lower than alcohols. Carboxylic acids can donate and accept hydrogen bonds, and two acid molecules lock together into stable Dimer Two molecules held together (here by two H-bonds) Full entry → pairs, giving acids unusually high boiling points (acetic acid: 118 °C vs. ethanol: 78 °C). Esters have no O–H, so they boil below acids and alcohols of similar size — and their lack of hydrogen bonding to water keeps them only slightly soluble. Small aldehydes and ketones (formaldehyde, acetaldehyde, acetone) are water-soluble because they accept hydrogen bonds from water.
Esterification and hydrolysis: the acid–alcohol condensation
A Carboxylic acid \(\mathrm{R{-}COOH}\) — carbonyl + hydroxyl Full entry → and an alcohol react in the presence of an acid catalyst to form an ester plus water — a condensation (Fischer esterification Acid-catalyzed acid + alcohol → ester + water Full entry →):
\[ \mathrm{RCOOH + R'OH \rightleftharpoons RCOOR' + H_2O} \]
The reaction is reversible: adding water (Hydrolysis Splitting a molecule with water) pushes it back toward the acid and alcohol. This equilibrium is why ester synthesis uses excess alcohol or removal of water, and why esters in foods slowly hydrolyze, changing flavor over time.
Common Confusions
| Do Not Confuse | With | Difference |
|---|---|---|
| Aldehyde (R–CHO) | Ketone (R–CO–R') | Aldehyde has an H on the carbonyl carbon and sits at the chain end; ketone has two carbon groups |
| Aldehyde | Carboxylic acid | Aldehydes oxidize to acids; the acid has an –OH on the carbonyl carbon |
| Carboxylic acid | Alcohol | The acid's –OH is attached to a carbonyl carbon — much more acidic than an alcohol's –OH |
| Ester (R–COO–R') | Ether (R–O–R') | Ester has a C=O next to the O; ether does not. Ethyl acetate ≠ diethyl ether |
| "Acetic acid" | "Acetone" | Acetic acid = \(\mathrm{CH_3COOH}\) (vinegar, an acid); acetone = \(\mathrm{CH_3COCH_3}\) (a ketone solvent) |
| Formaldehyde | Formic acid | Formaldehyde (methanal) is an aldehyde gas; formic acid (methanoic acid) is its oxidation product |
| Esterification "condensation" | Simple addition | Water is eliminated; it is a reversible equilibrium, not a one-way reaction |
| Aldehyde oxidation product | Ketone oxidation product | Aldehydes go on to acids; ketones stop — no H on the carbonyl carbon to lose |
Eli explains
The same idea, in plain words
Explain it like I’m 10
Imagine a "C=O" as a magnet pair — a carbon and an oxygen holding hands twice. If that pair sits at the end of a chain with a hydrogen beside it, it's an aldehyde (like the chemical that embalms specimens); if it sits in the middle between two carbons, it's a ketone (like nail-polish remover); if a hydrogen next to it is replaced by an –OH, it's a carboxylic acid (like vinegar); and if that –OH is replaced by an –OR group, it's an ester (the stuff that smells like bananas or pineapple). Same magnet, different neighbors, totally different smells and behaviors.
Worked example
Example 1: Following the oxidation ladder
Classify the transformation and name each product: \(\mathrm{CH_3CH_2OH}\) is oxidized first with one equivalent of oxidizing agent, then with a stronger one.
Step 1 — Identify the starting alcohol. Ethanol, \(\mathrm{CH_3CH_2OH}\), is a primary alcohol (the –OH carbon bonds to one other carbon).
Step 2 — First oxidation. Primary alcohols oxidize to aldehydes: ethanal (acetaldehyde), \(\mathrm{CH_3CHO}\). Name: ethanal.
Step 3 — Second oxidation. Aldehydes oxidize further to carboxylic acids: ethanoic acid (acetic acid), \(\mathrm{CH_3COOH}\). Name: ethanoic acid.
Step 4 — Verify with the general map. 1° → aldehyde → acid ✓. If the starting material had been 2-propanol (secondary), the product would stop at the ketone, propanone (acetone); a tertiary alcohol would not react. This single map answers most oxidation questions on exams.
Example 2: Esterification — how much ethyl acetate can you make?
Ethanoic acid reacts with ethanol to form ethyl ethanoate (ethyl acetate):
\[ \mathrm{CH_3COOH + CH_3CH_2OH \rightleftharpoons CH_3COOCH_2CH_3 + H_2O} \]
How many grams of ethyl acetate (theoretical) can be made from 30.0 g of ethanoic acid with excess ethanol? Molar masses: \(\mathrm{CH_3COOH}\) = 60.05 g/mol; \(\mathrm{CH_3COOCH_2CH_3}\) = 88.11 g/mol.
Step 1 — Grams of acid to moles:
\[ 30.0\ \text{g CH}_3\mathrm{COOH} \times \frac{1\ \text{mol CH}_3\mathrm{COOH}}{60.05\ \text{g}} = 0.4996\ \text{mol CH}_3\mathrm{COOH} \]
Step 2 — Mole ratio (1:1 from the balanced equation): 0.4996 mol acid → 0.4996 mol ester.
Step 3 — Moles of ester to grams:
\[ 0.4996\ \text{mol C}_4\mathrm{H}_8\mathrm{O}_2 \times \frac{88.11\ \text{g}}{1\ \text{mol}} = 44.0\ \text{g ethyl acetate} \]
Check: 88.11/60.05 ≈ 1.47, and 30.0 g × 1.47 ≈ 44.1 g — the ester is heavier than the acid because the ethyl group (29 g/mol) replaces the acid's H. Note this is the theoretical yield: because esterification is an equilibrium, real yields are lower unless water is removed or alcohol is in excess.
Example 3: Predicting boiling points by structure
Rank these four molecules by boiling point (all similar size): butane, butanal, 1-butanol, butanoic acid.
Step 1 — Identify intermolecular forces. Butane: dispersion only. Butanal (aldehyde): dispersion + dipole (polar C=O), no H-bond donation. 1-Butanol: hydrogen bonding (O–H). Butanoic acid: hydrogen bonding plus dimer formation (two H-bonds per pair).
Step 2 — Rank by force strength. Dispersion < dipole < H-bonding < H-bonding + dimer.
Step 3 — Conclusion. butane < butanal < 1-butanol < butanoic acid. Observed values (butane −0.5 °C, butanal 75 °C, 1-butanol 118 °C, butanoic acid 164 °C) follow this order — a pattern you can reproduce from structure alone, no memorized numbers required.
Key takeaways
- Carbonyl group \(\mathrm{C{=}O}\) is polar: C is electron-poor (electrophilic), O is electron-rich.
- Aldehyde \(\mathrm{R{-}CHO}\) = carbonyl at chain end; ketone \(\mathrm{R{-}CO{-}R'}\) = carbonyl in the middle.
- Naming suffixes: -al (aldehyde), -one (ketone), -oic acid (acid), -oate (ester: alkyl + oate).
- Oxidation ladder: 1° alcohol → aldehyde → carboxylic acid; 2° alcohol → ketone (stops there); 3° alcohol → no oxidation.
- Boiling points: carboxylic acids > alcohols > aldehydes/ketones > alkanes (of similar size), driven by hydrogen-bonding ability.
- Carboxylic acids form hydrogen-bonded dimers, which is why they boil unusually high.
- Esterification: \(\mathrm{RCOOH + R'OH \rightleftharpoons RCOOR' + H_2O}\) — reversible condensation; hydrolysis is the reverse.
- Esters are responsible for many fruit and flower fragrances; they are generally pleasant-smelling and only slightly water-soluble.
- Formaldehyde (methanal) and glutaraldehyde are used as disinfectants; aldehydes are more reactive than ketones toward nucleophiles.
- Acetic acid is a weak acid — it only partially ionizes in water: \(\mathrm{CH_3COOH(aq) \rightleftharpoons H^+(aq) + CH_3COO^-(aq)}\).
Check yourself
6 review questions from the chapter. Try each one, then open the answer.
What is the functional group shared by aldehydes, ketones, carboxylic acids, and esters?
Show answer
The carbonyl group, \(\mathrm{C{=}O}\).
Give the IUPAC names of \(\mathrm{CH_3CHO}\), \(\mathrm{CH_3COCH_3}\), and \(\mathrm{CH_3CH_2COOH}\).
Show answer
Ethanal (acetaldehyde); propanone (acetone); propanoic acid.
When 2-propanol is oxidized, what product forms and why does it not oxidize further?
Show answer
Propanone (acetone), a ketone. The carbonyl carbon of a ketone has no hydrogen attached, so there is no H to remove in further oxidation — oxidation stops at the ketone.
Why does butanoic acid boil far higher than butanal despite having the same number of carbons?
Show answer
Carboxylic acids form strong hydrogen bonds and pair into dimers (two molecules joined by two H-bonds). Aldehydes have only dipole–dipole forces because they cannot donate hydrogen bonds.
Write the equilibrium equation for making ethyl acetate from ethanoic acid and ethanol.
Show answer
\(\mathrm{CH_3COOH + CH_3CH_2OH \rightleftharpoons CH_3COOCH_2CH_3 + H_2O}\).
If 0.250 mol of ethanoic acid reacts with excess ethanol (100% yield), how many grams of ethyl acetate form?
Show answer
\(0.250\ \text{mol} \times 88.11\ \text{g/mol} = 22.0\ \text{g}\) (1:1 mole ratio; theoretical yield before equilibrium losses).
Study tools & related lessonsKey vocabulary · Related
Key vocabulary
- Carbonyl group
- \(\mathrm{C{=}O}\) double bond
- Aldehyde
- \(\mathrm{R{-}CHO}\) — carbonyl at chain end
- Ketone
- \(\mathrm{R{-}CO{-}R'}\) — carbonyl between carbons
- Carboxylic acid
- \(\mathrm{R{-}COOH}\) — carbonyl + hydroxyl
- Ester
- \(\mathrm{R{-}COO{-}R'}\) — carbonyl + –OR
- Fischer esterification
- Acid-catalyzed acid + alcohol → ester + water
- Hydrolysis
- Splitting a molecule with water
- Oxidation ladder
- 1° alcohol → aldehyde → acid; 2° → ketone
- Dimer
- Two molecules held together (here by two H-bonds)
Sources & references
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