Chapter 8 · Aldehydes, Ketones and Carboxylic Acids

8.1 Aldehydes and Ketones: Structure and Preparation

The carbonyl group C=O: its planar, polar structure, how aldehydes and ketones are named, and the reactions that make them, from alcohols, alkenes, alkynes, acyl chlorides, nitriles, esters and aromatic hydrocarbons, with their physical properties.

A perfumer wants the smell of cinnamon, a food company the flavour of vanilla, a plastics factory several thousand tonnes of methanal a day, and a pharmaceutical chemist a ketone in the middle of a ten-step synthesis of a drug. In every case the target contains a carbonyl group, and the job is to put it in exactly the right place without over-oxidising it to a carboxylic acid or reducing it to an alcohol. Chemists have collected a toolkit of reactions for this, many of them named after the chemists who found them.

This section first looks at the carbonyl group itself: why it is planar and polar, and how aldehydes and ketones are named. It then builds the toolkit: general methods that start from alcohols, alkenes and alkynes, special methods that stop at the aldehyde stage (Rosenmund, Stephen, DIBAL-H), ways of attaching a $\mathrm{CHO}$ group to a benzene ring (Etard, side-chain chlorination, Gattermann–Koch), and ways of making ketones (dialkylcadmium, Grignard reagents with nitriles, Friedel–Crafts acylation). It ends with the physical properties that follow from the polarity of $\mathrm{C{=}O}$.

The Carbonyl Group: Structure and Names

📘 Definition 8.1 — Aldehydes and ketones

The carbonyl group is a carbon atom joined to an oxygen atom by a double bond, $\mathrm{{>}C{=}O}$. Compounds in which it is the functional group are called carbonyl compounds.

  • In an aldehyde the carbonyl carbon is bonded to at least one hydrogen atom; the group $\mathrm{{-}CHO}$ is the formyl (aldehyde) group. General formula $\mathrm{R{-}CHO}$, where R is an alkyl or aryl group or, in methanal $\mathrm{HCHO}$, a hydrogen atom.
  • In a ketone the carbonyl carbon is bonded to two carbon atoms: $\mathrm{R{-}CO{-}R'}$. The ketone is simple (symmetrical) if $\mathrm{R=R'}$ and mixed (unsymmetrical) if $\mathrm{R\ne R'}$.
  • Aldehydes and ketones with the carbonyl group attached to a benzene ring, such as $\mathrm{C_{6}H_{5}CHO}$ and $\mathrm{C_{6}H_{5}COCH_{3}}$, are called aromatic; the rest are aliphatic.

Saturated open-chain aldehydes and ketones with one carbonyl group have the general formula $\mathrm{C_{n}H_{2n}O}$, so an aldehyde and a ketone with the same number of carbon atoms are functional isomers (propanal and propanone are both $\mathrm{C_{3}H_{6}O}$).

📐 Rule 8.1 — Structure and polarity of the carbonyl group
  • The carbonyl carbon is $sp^{2}$ hybridised. It forms three σ bonds (to O and to two other atoms) that lie in one plane at about $120^{\circ}$ to each other: the carbonyl group and the three atoms bonded to its carbon are trigonal planar.
  • The unhybridised $p$ orbital of carbon overlaps sideways with a $p$ orbital of oxygen to form a π bond above and below this plane. Oxygen also carries two lone pairs.
  • Oxygen is much more electronegative than carbon, so the bond is polar: $\mathrm{\overset{\delta+}{C}{=}\overset{\delta-}{O}}$. The polarity is larger than in a $\mathrm{C{-}O}$ single bond because the π electrons are easily pulled towards oxygen. It is described by resonance between a neutral and a dipolar structure, $\mathrm{{>}C{=}O\ \longleftrightarrow\ {>}\overset{+}{C}{-}\overset{-}{O}}$; carbonyl compounds have dipole moments of about $2.3$–$2.9\ \mathrm{D}$.
  • Consequences: the carbonyl carbon is an electrophilic centre (attacked by nucleophiles, Section 8.2), the oxygen is a nucleophilic, weakly basic centre (protonated by acids), and the polar molecules attract one another (physical properties, end of this section).

Common names. Many aldehydes are named after the carboxylic acid they give on oxidation, with "-ic acid" replaced by "-aldehyde": formic acid → formaldehyde, acetic acid → acetaldehyde. Positions along the chain are marked by Greek letters, starting with α for the carbon next to the carbonyl group. Common names of ketones list the two alkyl or aryl groups in alphabetical order followed by "ketone" (ethyl methyl ketone); $\mathrm{CH_{3}COCH_{3}}$ is acetone, and the aromatic ketones acetophenone and benzophenone keep their old names.

📐 Rule 8.2 — IUPAC names of aldehydes and ketones
  1. Choose the longest chain that contains the carbonyl carbon. Replace the final "-e" of the alkane name by -al for an aldehyde and -one for a ketone: ethanal, propanone.
  2. Number the chain so that the carbonyl carbon has the lowest possible number. In an aldehyde the $\mathrm{CHO}$ carbon is always C-1, so no locant is needed for it (butanal, not butan-1-al). In a ketone the locant goes before "-one": pentan-2-one, pentan-3-one (only propanone needs none).
  3. Name and number substituents as for alkanes: 3-methylbutanal, 4-methylpentan-2-one. A double bond is shown in the stem: but-2-enal, 4-methylpent-3-en-2-one. Two $\mathrm{CHO}$ groups give -dial (butanedial); two ketone groups, -dione (pentane-2,4-dione).
  4. When the $\mathrm{CHO}$ group is joined to a ring, the suffix -carbaldehyde is added to the ring name, and the ring carbon carrying it is C-1: cyclohexanecarbaldehyde, benzenecarbaldehyde (benzaldehyde, the IUPAC-accepted name). Aromatic ketones are named as phenyl-substituted alkanones: $\mathrm{C_{6}H_{5}COCH_{3}}$ is 1-phenylethan-1-one; $\mathrm{C_{6}H_{5}COC_{6}H_{5}}$ is diphenylmethanone.
StructureCommon nameIUPAC name
$\mathrm{HCHO}$formaldehydemethanal
$\mathrm{CH_{3}CHO}$acetaldehydeethanal
$\mathrm{(CH_{3})_{2}CHCHO}$isobutyraldehyde2-methylpropanal
$\mathrm{CH_{3}CH(Br)CH_{2}CHO}$β-bromobutyraldehyde3-bromobutanal
$\mathrm{CH_{3}CH{=}CHCHO}$crotonaldehydebut-2-enal
$\mathrm{C_{6}H_{5}CHO}$benzaldehydebenzenecarbaldehyde (benzaldehyde)
$\mathrm{CH_{3}COCH_{3}}$acetonepropanone
$\mathrm{CH_{3}COCH_{2}CH_{3}}$ethyl methyl ketonebutan-2-one
$\mathrm{(CH_{3})_{2}CHCOCH_{3}}$isopropyl methyl ketone3-methylbutan-2-one
$\mathrm{(CH_{3})_{2}C{=}CHCOCH_{3}}$mesityl oxide4-methylpent-3-en-2-one
$\mathrm{C_{6}H_{5}COCH_{3}}$acetophenone1-phenylethan-1-one
$\mathrm{C_{6}H_{5}COC_{6}H_{5}}$benzophenonediphenylmethanone
✏️ Example 1 — Naming carbonyl compounds
Give IUPAC names: (a) $\mathrm{CH_{3}CH_{2}CH(CH_{3})CHO}$; (b) $\mathrm{CH_{3}COCH_{2}CH(CH_{3})_{2}}$; (c) $\mathrm{CH_{3}CH(OH)CH_{2}CHO}$; (d) $\mathrm{CH_{3}CH{=}CHCOCH_{3}}$; (e) $\mathrm{CH_{3}CH_{2}COCH(CH_{3})_{2}}$; (f) $\mathrm{OHC{-}CH_{2}CH_{2}{-}CHO}$; (g) $\mathrm{C_{6}H_{5}CH_{2}CHO}$; (h) $\mathrm{4{-}CH_{3}C_{6}H_{4}CHO}$ (a methyl group para to $\mathrm{CHO}$).
(a) Four-carbon chain ending in $\mathrm{CHO}$ (C-1), methyl on C-2: 2-methylbutanal (common name α-methylbutyraldehyde).
(b) Longest chain through $\mathrm{C{=}O}$: five carbons. Numbering from the left gives the carbonyl C-2 (from the right it would be C-4), so the methyl is on C-4: 4-methylpentan-2-one.
(c) $\mathrm{CHO}$ is C-1, so $\mathrm{OH}$ is on C-3. The aldehyde is the principal group and $\mathrm{OH}$ becomes the prefix hydroxy: 3-hydroxybutanal (the "aldol" of Section 8.2).
(d) Five carbons; the carbonyl gets the lower number (C-2), so the double bond is between C-3 and C-4: pent-3-en-2-one.
(e) Carbonyl at C-3 from either end; the tie is broken by giving the methyl the lower number: 2-methylpentan-3-one (ethyl isopropyl ketone).
(f) Two $\mathrm{CHO}$ groups on a four-carbon chain: butanedial (the "e" of butane is kept before a consonant).
(g) Two-carbon chain with a phenyl group on C-2: 2-phenylethanal (phenylacetaldehyde).
(h) The ring carbon carrying $\mathrm{CHO}$ is C-1, the methyl is on C-4: 4-methylbenzaldehyde (4-methylbenzenecarbaldehyde; common name p-tolualdehyde).
✏️ Example 2 — All the carbonyl isomers of C₅H₁₀O
Draw condensed formulas and give IUPAC names for every aldehyde and ketone with molecular formula $\mathrm{C_{5}H_{10}O}$ (open chains only). How many are there? Which of them are chiral?
Aldehydes An aldehyde is $\mathrm{C_{4}H_{9}{-}CHO}$, and there are four butyl groups, so four aldehydes: $\mathrm{CH_{3}CH_{2}CH_{2}CH_{2}CHO}$ pentanal; $\mathrm{CH_{3}CH_{2}CH(CH_{3})CHO}$ 2-methylbutanal; $\mathrm{(CH_{3})_{2}CHCH_{2}CHO}$ 3-methylbutanal; $\mathrm{(CH_{3})_{3}CCHO}$ 2,2-dimethylpropanal.
Ketones A ketone is $\mathrm{R{-}CO{-}R'}$ with four carbons shared between R and R′: methyl + propyl (two propyl groups) or ethyl + ethyl. $\mathrm{CH_{3}COCH_{2}CH_{2}CH_{3}}$ pentan-2-one; $\mathrm{CH_{3}COCH(CH_{3})_{2}}$ 3-methylbutan-2-one; $\mathrm{CH_{3}CH_{2}COCH_{2}CH_{3}}$ pentan-3-one.
Total $4+3=\mathbf{7}$ carbonyl isomers. Only 2-methylbutanal has a carbon with four different groups ($\mathrm{H}$, $\mathrm{CH_{3}}$, $\mathrm{C_{2}H_{5}}$, $\mathrm{CHO}$), so it alone exists as a pair of enantiomers.
Check Each formula has 5 C, 10 H and 1 O ✓. A ketone $\mathrm{C{=}O}$ can never be on C-1 (that carbon would carry an H, making an aldehyde), and "pentan-4-one" is only pentan-2-one numbered from the wrong end.
Interactive 1 — Carbonyl Name Builder

Build an open-chain aldehyde or ketone: choose the length of the main chain, where the $\mathrm{C{=}O}$ goes, and (optionally) a methyl branch. The skeletal structure is drawn with the IUPAC numbering, the δ+ carbon and δ− oxygen, and the name is assembled by Rule 8.2. Carbon 1 is always the $\mathrm{CHO}$ carbon of an aldehyde; for a ketone, watch the numbers flip so that the carbonyl gets the lower locant. Press Play to cycle through some common examples.

IUPAC name—
common name—
molecular formula—
molar mass—

Preparing Aldehydes and Ketones: General Methods

Three general methods put a carbonyl group into a molecule: take two hydrogen atoms away from an alcohol, cut a carbon–carbon double bond, or add water to a triple bond.

📐 Rule 8.3 — Aldehydes and ketones from alcohols, alkenes and alkynes

1. Oxidation of alcohols (Chapter 7). Primary alcohols give aldehydes; secondary alcohols give ketones:

$$\mathrm{RCH_{2}OH\xrightarrow{[O]}RCHO},\qquad \mathrm{R_{2}CHOH\xrightarrow{[O]}R_{2}CO}$$

Ketones resist further oxidation, but aldehydes are easily oxidised to carboxylic acids. So for aldehydes a mild reagent is used, pyridinium chlorochromate (PCC) in $\mathrm{CH_{2}Cl_{2}}$, or $\mathrm{CrO_{3}}$ under anhydrous conditions; with acidified $\mathrm{K_{2}Cr_{2}O_{7}}$ the more volatile aldehyde is distilled out of the mixture as soon as it forms. For ketones, acidified $\mathrm{K_{2}Cr_{2}O_{7}}$ or $\mathrm{CrO_{3}}$ is used.

2. Dehydrogenation of alcohols. Alcohol vapour passed over hot copper (or silver) loses hydrogen; this is the industrial route for volatile alcohols:

$$\mathrm{RCH_{2}OH\xrightarrow{Cu,\ 573\ K}RCHO+H_{2}},\qquad \mathrm{R_{2}CHOH\xrightarrow{Cu,\ 573\ K}R_{2}CO+H_{2}}$$

3. Ozonolysis of alkenes. Ozone adds across $\mathrm{C{=}C}$ to give an ozonide, which is broken up by zinc dust and water. Each doubly bonded carbon becomes a carbonyl carbon:

$$\mathrm{R_{2}C{=}CR'_{2}\xrightarrow[\text{(ii) }Zn/H_{2}O]{\text{(i) }O_{3}}R_{2}C{=}O+O{=}CR'_{2}}$$

A carbon carrying a hydrogen becomes an aldehyde, a carbon carrying two alkyl groups becomes a ketone. Zinc removes the hydrogen peroxide formed, which would otherwise oxidise the aldehydes.

4. Hydration of alkynes. With dilute sulphuric acid and mercury(II) sulphate at $333\ \mathrm{K}$, water adds across $\mathrm{C{\equiv}C}$ by Markovnikov's rule. The first product, an enol ($\mathrm{OH}$ on a $\mathrm{C{=}C}$ carbon), rearranges at once to the carbonyl compound:

$$\mathrm{HC{\equiv}CH+H_{2}O\xrightarrow[\text{333 K}]{\mathrm{HgSO_{4},\ dil.\ H_{2}SO_{4}}}[CH_{2}{=}CHOH]\rightarrow CH_{3}CHO}$$ $$\mathrm{CH_{3}C{\equiv}CH+H_{2}O\xrightarrow[\text{333 K}]{\mathrm{HgSO_{4},\ dil.\ H_{2}SO_{4}}}[CH_{3}C(OH){=}CH_{2}]\rightarrow CH_{3}COCH_{3}}$$

Ethyne is the only alkyne that gives an aldehyde. Every other alkyne gives a ketone, because the $\mathrm{OH}$ ends up on the more substituted carbon.

✏️ Example 3 — Reading an ozonolysis
(a) Name the products of ozonolysis (followed by $\mathrm{Zn/H_{2}O}$) of 2-methylbut-2-ene. (b) An alkene $\mathrm{C_{6}H_{12}}$ gives propanone as the only product. Identify it. (c) An alkene $\mathrm{C_{7}H_{14}}$ gives propanal and butan-2-one. Give its structure and name. (d) What does cyclohexene give?
(a) $\mathrm{(CH_{3})_{2}C{=}CHCH_{3}}$: the left carbon has two methyls (→ ketone), the right one has H (→ aldehyde). Products: $\mathrm{(CH_{3})_{2}C{=}O}$ propanone and $\mathrm{CH_{3}CHO}$ ethanal.
(b) Work backwards Join the carbonyl carbons of two propanone molecules by a double bond: $\mathrm{(CH_{3})_{2}C{=}C(CH_{3})_{2}}$, 2,3-dimethylbut-2-ene ($\mathrm{C_{6}H_{12}}$ ✓).
(c) Join $\mathrm{CH_{3}CH_{2}CH{=}}$ (from propanal) to $\mathrm{{=}C(CH_{3})CH_{2}CH_{3}}$ (from butan-2-one): $\mathrm{CH_{3}CH_{2}CH{=}C(CH_{3})CH_{2}CH_{3}}$. Main chain six carbons, double bond at C-3 from either end; the methyl then gets the lower number: 3-methylhex-3-ene ($3+4=7$ carbons ✓).
(d) Cutting the ring double bond opens the ring and leaves both ends as $\mathrm{CHO}$: $\mathrm{OHC(CH_{2})_{4}CHO}$, hexanedial.
Check Ozonolysis never changes the number of carbon atoms: the carbons of the products always add up to those of the alkene ✓.
✏️ Example 4 — Hydration of alkynes
Give the major product of hydration ($\mathrm{HgSO_{4}}$, dilute $\mathrm{H_{2}SO_{4}}$, $333\ \mathrm{K}$) of (a) propyne, (b) but-1-yne, (c) but-2-yne, (d) pent-2-yne. (e) Which alkyne would you hydrate to make ethanal, and why can no other alkyne give an aldehyde?
(a), (b) Terminal alkynes: $\mathrm{H}$ goes to the $\mathrm{CH}$ end and $\mathrm{OH}$ to the inner carbon (Markovnikov), and the enol becomes a methyl ketone. Propyne → propanone; but-1-yne, $\mathrm{CH_{3}CH_{2}C{\equiv}CH}$ → $\mathrm{CH_{3}CH_{2}COCH_{3}}$, butan-2-one.
(c) But-2-yne is symmetrical, so either way round gives $\mathrm{CH_{3}COCH_{2}CH_{3}}$, butan-2-one.
(d) $\mathrm{CH_{3}C{\equiv}CCH_{2}CH_{3}}$ has two similar alkyl groups, so the $\mathrm{OH}$ goes to both carbons: a mixture of pentan-2-one and pentan-3-one. This is why internal unsymmetrical alkynes are poor starting materials.
(e) Ethyne, $\mathrm{HC{\equiv}CH}$ → $\mathrm{CH_{3}CHO}$. For an aldehyde, the carbon that receives the $\mathrm{OH}$ must also carry a hydrogen at the end, i.e. both triple-bond carbons must carry H. Only ethyne has that; in any other alkyne the $\mathrm{OH}$ lands on a carbon bonded to another carbon, giving a ketone.
Interactive 2 — Ozonolysis and Hydration Decoder

Pick an alkene or an alkyne. For an alkene, ozonolysis pulls the two halves of $\mathrm{C{=}C}$ apart and caps each with an oxygen; for an alkyne, hydration adds $\mathrm{H}$ and $\mathrm{OH}$ (Markovnikov) and the enol then rearranges to the carbonyl compound. Press Play to run the reaction, or drag the slider to step through it. Use it backwards too: from the products, can you rebuild the starting compound?

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reagent—
product(s)—
carbon count—

Stopping at the Aldehyde: Acyl Chlorides, Nitriles, Esters and Aromatic Hydrocarbons

An aldehyde sits halfway along the oxidation ladder: alcohol → aldehyde → carboxylic acid. Making one by reducing an acid derivative, or by oxidising a methyl group, therefore needs a reagent that stops halfway. The methods below were designed for exactly that.

📘 Definition 8.2 — Rosenmund reduction and Stephen reaction

The Rosenmund reduction is the hydrogenation of an acyl chloride to an aldehyde over palladium on barium sulphate. Barium sulphate (sometimes with a little sulphur or quinoline) partly poisons the catalyst, so it cannot reduce the aldehyde further to an alcohol:

$$\mathrm{R{-}COCl+H_{2}\xrightarrow{Pd{-}BaSO_{4}}R{-}CHO+HCl}\qquad\text{e.g.}\ \mathrm{C_{6}H_{5}COCl\rightarrow C_{6}H_{5}CHO}$$

The Stephen reaction reduces a nitrile with tin(II) chloride and hydrochloric acid to an imine (as its salt), which is then hydrolysed by water to the aldehyde:

$$\mathrm{R{-}C{\equiv}N+SnCl_{2}+HCl\rightarrow R{-}CH{=}NH\xrightarrow{H_{3}O^{+}}R{-}CHO}$$
📐 Rule 8.4 — DIBAL-H: aldehydes from nitriles and esters

Diisobutylaluminium hydride, $\mathrm{AlH(i{-}Bu)_{2}}$ (DIBAL-H), delivers one hydride ion to a nitrile or an ester; hydrolysis then gives the aldehyde. It does not touch carbon–carbon double bonds:

$$\mathrm{R{-}C{\equiv}N\xrightarrow[\text{(ii) }H_{2}O]{\text{(i) }AlH(i{-}Bu)_{2}}R{-}CHO},\qquad \mathrm{R{-}COOR'\xrightarrow[\text{(ii) }H_{2}O]{\text{(i) DIBAL{-}H}}R{-}CHO+R'OH}$$

For example, $\mathrm{CH_{3}CH{=}CHCH_{2}CN}$ gives $\mathrm{CH_{3}CH{=}CHCH_{2}CHO}$ (pent-3-enal) with the double bond intact, and ethyl butanoate gives butanal. A strong reducing agent such as $\mathrm{LiAlH_{4}}$ would carry both reductions on to the alcohol (or, for the nitrile, to the amine).

Aromatic aldehydes from hydrocarbons. A methyl group on a benzene ring can be oxidised to $\mathrm{CHO}$ if the aldehyde is trapped as a derivative that resists further oxidation, and a $\mathrm{CHO}$ group can be put straight onto benzene with carbon monoxide.

📘 Definition 8.3 — Etard reaction and Gattermann–Koch reaction

In the Etard reaction, methylbenzene (toluene) is oxidised by chromyl chloride, $\mathrm{CrO_{2}Cl_{2}}$, in $\mathrm{CS_{2}}$. A brown chromium complex precipitates and is hydrolysed to benzaldehyde:

$$\mathrm{C_{6}H_{5}CH_{3}\xrightarrow{CrO_{2}Cl_{2},\ CS_{2}}C_{6}H_{5}CH(OCrOHCl_{2})_{2}\xrightarrow{H_{3}O^{+}}C_{6}H_{5}CHO}$$

In the Gattermann–Koch reaction, benzene (or a substituted benzene) is treated with carbon monoxide and hydrogen chloride in the presence of anhydrous aluminium chloride or copper(I) chloride, which introduces a $\mathrm{CHO}$ group into the ring:

$$\mathrm{C_{6}H_{6}+CO+HCl\xrightarrow{\text{anhyd. }AlCl_{3}/CuCl}C_{6}H_{5}CHO+HCl}$$

Two more routes from toluene are common. Chromic oxide in ethanoic anhydride traps the aldehyde as benzylidene diacetate, which is then hydrolysed:

$$\mathrm{C_{6}H_{5}CH_{3}\xrightarrow[273{-}283\ K]{CrO_{3},\ (CH_{3}CO)_{2}O}C_{6}H_{5}CH(OCOCH_{3})_{2}\xrightarrow[\Delta]{H_{3}O^{+}}C_{6}H_{5}CHO}$$

In side-chain chlorination, chlorine in light replaces two hydrogens of the methyl group to give benzal chloride, which water hydrolyses at $373\ \mathrm{K}$. This is a commercial method:

$$\mathrm{C_{6}H_{5}CH_{3}+2Cl_{2}\xrightarrow{h\nu}C_{6}H_{5}CHCl_{2}+2HCl},\qquad \mathrm{C_{6}H_{5}CHCl_{2}+H_{2}O\xrightarrow{373\ K}C_{6}H_{5}CHO+2HCl}$$
✏️ Example 5 — Choosing a reagent that stops at the aldehyde
Give the reagent(s) for each conversion and name the method: (a) benzoyl chloride → benzaldehyde; (b) ethanenitrile → ethanal (two methods); (c) methylbenzene → benzaldehyde (three methods); (d) benzene → benzaldehyde in one step; (e) pent-3-enenitrile, $\mathrm{CH_{3}CH{=}CHCH_{2}CN}$ → pent-3-enal. (f) Why would $\mathrm{LiAlH_{4}}$ be a poor choice in (a)?
(a) $\mathrm{H_{2}}$, $\mathrm{Pd{-}BaSO_{4}}$: Rosenmund reduction. $\mathrm{C_{6}H_{5}COCl+H_{2}\rightarrow C_{6}H_{5}CHO+HCl}$.
(b) $\mathrm{SnCl_{2}/HCl}$ then $\mathrm{H_{3}O^{+}}$ (Stephen reaction), or (i) DIBAL-H, (ii) $\mathrm{H_{2}O}$. Both give $\mathrm{CH_{3}CHO}$ from $\mathrm{CH_{3}CN}$.
(c) (i) $\mathrm{CrO_{2}Cl_{2}}$ in $\mathrm{CS_{2}}$, then $\mathrm{H_{3}O^{+}}$ (Etard reaction); (ii) $\mathrm{CrO_{3}}$ in $\mathrm{(CH_{3}CO)_{2}O}$, then $\mathrm{H_{3}O^{+}}$; (iii) $\mathrm{Cl_{2}}$, $h\nu$ (side-chain chlorination to $\mathrm{C_{6}H_{5}CHCl_{2}}$), then $\mathrm{H_{2}O}$ at $373\ \mathrm{K}$.
(d) $\mathrm{CO}$, $\mathrm{HCl}$, anhydrous $\mathrm{AlCl_{3}/CuCl}$: Gattermann–Koch reaction.
(e) (i) DIBAL-H, (ii) $\mathrm{H_{2}O}$. DIBAL-H reduces $\mathrm{C{\equiv}N}$ but not $\mathrm{C{=}C}$; catalytic hydrogenation would reduce the double bond as well.
(f) $\mathrm{LiAlH_{4}}$ is too powerful: it reduces the acyl chloride past the aldehyde to benzyl alcohol, $\mathrm{C_{6}H_{5}CH_{2}OH}$. The point of each named method is that it stops at the aldehyde.

Preparing Ketones; Physical Properties

📐 Rule 8.5 — Ketones from acyl chlorides, nitriles and benzene

1. Acyl chlorides and dialkylcadmium. A Grignard reagent and cadmium chloride give a dialkylcadmium, which replaces the chlorine of an acyl chloride by an alkyl group:

$$\mathrm{2R{-}MgX+CdCl_{2}\rightarrow R_{2}Cd+2Mg(X)Cl}$$ $$\mathrm{2R'{-}COCl+R_{2}Cd\rightarrow2R'{-}CO{-}R+CdCl_{2}}$$

A Grignard reagent itself is not used, because it would go on to attack the ketone and give a tertiary alcohol; the dialkylcadmium is less reactive and stops at the ketone.

2. Nitriles and Grignard reagents. The Grignard reagent adds once across $\mathrm{C{\equiv}N}$; acid hydrolysis of the imine salt gives a ketone:

$$\mathrm{CH_{3}CH_{2}C{\equiv}N+C_{6}H_{5}MgBr\xrightarrow{\text{dry ether}}CH_{3}CH_{2}C({=}NMgBr)C_{6}H_{5}}$$ $$\mathrm{CH_{3}CH_{2}C({=}NMgBr)C_{6}H_{5}\xrightarrow{H_{3}O^{+}}CH_{3}CH_{2}COC_{6}H_{5}}$$

The product is 1-phenylpropan-1-one (propiophenone).

3. Friedel–Crafts acylation of benzene. An acyl chloride and anhydrous aluminium chloride attach an acyl group to the ring:

$$\mathrm{C_{6}H_{6}+R{-}COCl\xrightarrow{\text{anhyd. }AlCl_{3}}C_{6}H_{5}{-}CO{-}R+HCl}$$

With ethanoyl chloride ($\mathrm{R=CH_{3}}$) this gives acetophenone; with benzoyl chloride ($\mathrm{R=C_{6}H_{5}}$), benzophenone. The $\mathrm{C{=}O}$ group withdraws electrons from the ring, so the product does not react a second time. Ketones are also made by the general methods of Rule 8.3 (2° alcohols, ozonolysis, hydration of alkynes).

✏️ Example 6 — Planning ketone syntheses
Give reagents for making butan-2-one, $\mathrm{CH_{3}COCH_{2}CH_{3}}$, from (a) an acyl chloride, (b) a nitrile, (c) an alcohol, (d) an alkyne; and make (e) 1-phenylethan-1-one from benzene and (f) 1-phenylethan-1-one from benzonitrile, $\mathrm{C_{6}H_{5}CN}$. In each of (a) and (b) give both possible pairs of reagents.
Idea Butan-2-one is methyl + $\mathrm{C{=}O}$ + ethyl. Each disconnection puts one of the alkyl groups into the reagent and the other into the substrate.
(a) $$\mathrm{2CH_{3}COCl+(C_{2}H_{5})_{2}Cd\rightarrow2CH_{3}COC_{2}H_{5}+CdCl_{2}}$$ or $\mathrm{2CH_{3}CH_{2}COCl}$ + $\mathrm{(CH_{3})_{2}Cd}$, which gives the same ketone.
(b) $\mathrm{CH_{3}CN}$ + $\mathrm{CH_{3}CH_{2}MgBr}$, or $\mathrm{CH_{3}CH_{2}CN}$ + $\mathrm{CH_{3}MgBr}$, in dry ether; then $\mathrm{H_{3}O^{+}}$.
(c), (d) Butan-2-ol with acidified $\mathrm{K_{2}Cr_{2}O_{7}}$ (or Cu at $573\ \mathrm{K}$). But-1-yne or but-2-yne with $\mathrm{HgSO_{4}}$, dilute $\mathrm{H_{2}SO_{4}}$, $333\ \mathrm{K}$ (Example 4).
(e) Friedel–Crafts acylation: $$\mathrm{C_{6}H_{6}+CH_{3}COCl\xrightarrow{\text{anhyd. }AlCl_{3}}C_{6}H_{5}COCH_{3}+HCl}$$
(f) $\mathrm{C_{6}H_{5}CN}$ + $\mathrm{CH_{3}MgBr}$ in dry ether, then $\mathrm{H_{3}O^{+}}$: $\mathrm{C_{6}H_{5}C({=}NMgBr)CH_{3}\rightarrow C_{6}H_{5}COCH_{3}}$.
Check Count carbons: in (a) $2+2$ or $3+1$, in (b) $2+2$ or $3+1$, giving 4 carbons ✓; in (e) and (f) $6+2$ and $7+1$ give the 8 carbons of acetophenone ✓.

Physical properties. Methanal is a gas at room temperature, ethanal a very volatile liquid (boiling point $294\ \mathrm{K}$), and the other lower aldehydes and ketones are liquids. Because of the polar $\mathrm{C{=}O}$ group their molecules attract one another by dipole–dipole forces, so they boil higher than hydrocarbons and ethers of similar molar mass. They cannot form hydrogen bonds with one another (they have no $\mathrm{O{-}H}$), so they boil lower than alcohols of similar molar mass:

CompoundFormula$M$ / $\mathrm{g\,mol^{-1}}$b.p. / KStrongest attraction between molecules
butane$\mathrm{CH_{3}CH_{2}CH_{2}CH_{3}}$58273dispersion forces only
methoxyethane$\mathrm{CH_{3}OCH_{2}CH_{3}}$60281weak dipole–dipole
propanal$\mathrm{CH_{3}CH_{2}CHO}$58322dipole–dipole ($\mathrm{C{=}O}$)
propanone$\mathrm{CH_{3}COCH_{3}}$58329dipole–dipole ($\mathrm{C{=}O}$)
propan-1-ol$\mathrm{CH_{3}CH_{2}CH_{2}OH}$60370hydrogen bonding

The lower members (methanal, ethanal, propanone) mix with water in all proportions, because water molecules hydrogen-bond to the carbonyl oxygen. Solubility falls as the hydrocarbon part grows; all aldehydes and ketones dissolve in organic solvents such as benzene, ether and chloroform. The lower aldehydes have sharp, pungent smells, while many larger aldehydes and ketones smell pleasant and are used in perfumes and flavourings (vanillin, cinnamaldehyde, the citrus aldehyde citral, the musk ketone muscone).

⚠️ Common mistakes in naming and preparing carbonyl compounds

1. Locants for aldehydes. $\mathrm{CHO}$ is always C-1: "butan-1-al" and "1-butanal" are wrong; write butanal.

2. Counting the CHO carbon. $\mathrm{C_{6}H_{5}CH_{2}CHO}$ is 2-phenylethanal (two carbons in the chain), and $\mathrm{(CH_{3})_{2}CHCHO}$ is 2-methylpropanal, not "2-methylethanal".

3. Over-reduction and over-oxidation. $\mathrm{LiAlH_{4}}$ or $\mathrm{H_{2}/Ni}$ take acyl chlorides and esters right down to alcohols; acidified $\mathrm{KMnO_{4}}$ takes a primary alcohol or a ring methyl group right up to the carboxylic acid. Use PCC, Rosenmund, Stephen, DIBAL-H, Etard as the job requires.

4. Hydration of alkynes. Only ethyne gives an aldehyde. Propyne gives propanone, not propanal: Markovnikov addition puts the $\mathrm{OH}$ on the middle carbon.

5. Grignard reagent with an acyl chloride. To stop at the ketone, use the dialkylcadmium; a Grignard reagent continues to a tertiary alcohol.

🚀 Beyond the board syllabus (JEE Main · NEET · JEE Advanced) — why these reagents stop halfway

In JEE Main, NEET and JEE Advanced 2026 (preparation of aldehydes; JEE Advanced names aldehydes from esters, acid chlorides and nitriles); not required by the CBSE syllabus.

The trick in every "stop at the aldehyde" method is to keep the aldehyde from being formed while the reducing agent is still active. DIBAL-H is used at low temperature (about $195\ \mathrm{K}$, $-78\ ^{\circ}\mathrm{C}$): after one hydride adds to an ester, the aluminium holds the oxygen in a stable tetrahedral intermediate that does not lose $\mathrm{R'O^{-}}$ until water is added at the end, so a second hydride has nothing to attack. With a nitrile, the first addition gives an $\mathrm{N{-}Al}$ imine complex, which is reduced no further; water then hydrolyses the imine. In the Stephen reaction the imine is protected as its salt in the same way. In the Etard reaction and the $\mathrm{CrO_{3}/(CH_{3}CO)_{2}O}$ method, the aldehyde carbon is held as a chromium complex or a diacetate (an acylal), oxidation states that the oxidant cannot raise further. JEE questions often ask which reagent gives the aldehyde rather than the alcohol or the acid; the answer is always the reagent whose intermediate "parks" the molecule at the aldehyde level.

Exercises

I
Structure and names. Use $\mathrm{C}=12$, $\mathrm{H}=1$, $\mathrm{O}=16\ \mathrm{g\,mol^{-1}}$.
  • 1 State the hybridisation of the carbonyl carbon and the bond angles around it. Explain why this carbon is attacked by nucleophiles while the oxygen is attacked by $\mathrm{H^{+}}$.
  • 2 Give IUPAC names: (a) $\mathrm{CH_{3}CH_{2}CH_{2}CHO}$; (b) $\mathrm{(CH_{3})_{3}CCHO}$; (c) $\mathrm{CH_{3}COCH_{2}COCH_{3}}$; (d) $\mathrm{(CH_{3})_{2}CHCH_{2}CH_{2}CHO}$.
  • 3 Give IUPAC names: (a) $\mathrm{(CH_{3})_{2}C{=}CHCOCH_{3}}$; (b) $\mathrm{C_{6}H_{5}CH{=}CHCHO}$ (cinnamaldehyde).
  • 4 Write condensed structures: (a) 3-methylbutanal; (b) 4-chloropentan-2-one; (c) cyclohexanecarbaldehyde; (d) 1-phenylpropan-1-one.
  • 5 Give IUPAC names for these common names: (a) methyl isobutyl ketone; (b) γ-chlorobutyraldehyde; (c) di-tert-butyl ketone; (d) valeraldehyde.
  • 6 Draw and name all the aldehydes and ketones with molecular formula $\mathrm{C_{4}H_{8}O}$.
  • 7 Each name is wrong. Draw the structure it describes and give the correct name: (a) 2-ethylpropanal; (b) butan-3-one; (c) 1-pentanal; (d) 2-methylbutan-3-one.
  • 8 Find the molecular formula and molar mass of acetophenone and of benzophenone.
1 $sp^{2}$; about $120^{\circ}$, trigonal planar. Oxygen is more electronegative and pulls the π electrons towards itself ($\mathrm{\overset{\delta+}{C}{=}\overset{\delta-}{O}}$, dipolar resonance form $\mathrm{\overset{+}{C}{-}\overset{-}{O}}$): the carbon is electron-poor (electrophilic), while the oxygen has lone pairs and a partial negative charge (basic, so it takes $\mathrm{H^{+}}$).
2 (a) butanal; (b) 2,2-dimethylpropanal; (c) pentane-2,4-dione; (d) 4-methylpentanal.
3 (a) 4-methylpent-3-en-2-one (C=O gets the lower locant); (b) 3-phenylprop-2-enal.
4 (a) $\mathrm{(CH_{3})_{2}CHCH_{2}CHO}$; (b) $\mathrm{CH_{3}CH(Cl)CH_{2}COCH_{3}}$; (c) $\mathrm{C_{6}H_{11}CHO}$, a $\mathrm{CHO}$ group on a cyclohexane ring; (d) $\mathrm{C_{6}H_{5}COCH_{2}CH_{3}}$.
5 (a) $\mathrm{CH_{3}COCH_{2}CH(CH_{3})_{2}}$, 4-methylpentan-2-one; (b) γ is the third carbon from $\mathrm{CHO}$, i.e. C-4: $\mathrm{ClCH_{2}CH_{2}CH_{2}CHO}$, 4-chlorobutanal; (c) $\mathrm{(CH_{3})_{3}CCOC(CH_{3})_{3}}$, 2,2,4,4-tetramethylpentan-3-one; (d) $\mathrm{CH_{3}(CH_{2})_{3}CHO}$, pentanal.
6 Three: $\mathrm{CH_{3}CH_{2}CH_{2}CHO}$ butanal; $\mathrm{(CH_{3})_{2}CHCHO}$ 2-methylpropanal; $\mathrm{CH_{3}COCH_{2}CH_{3}}$ butan-2-one (two propyl groups for the aldehydes, and only methyl + ethyl for the ketone).
7 (a) $\mathrm{CH_{3}CH_{2}CH(CH_{3})CHO}$: the longest chain has four carbons, 2-methylbutanal; (b) $\mathrm{CH_{3}CH_{2}COCH_{3}}$ numbered from the wrong end: butan-2-one; (c) the C-1 locant is never written: pentanal; (d) $\mathrm{CH_{3}CH(CH_{3})COCH_{3}}$, carbonyl must get the lowest number: 3-methylbutan-2-one.
8 Acetophenone $\mathrm{C_{6}H_{5}COCH_{3}}$ = $\mathrm{C_{8}H_{8}O}$, $M=8(12)+8(1)+16=120\ \mathrm{g\,mol^{-1}}$. Benzophenone $\mathrm{C_{6}H_{5}COC_{6}H_{5}}$ = $\mathrm{C_{13}H_{10}O}$, $M=156+10+16=182\ \mathrm{g\,mol^{-1}}$.
II
From alcohols, alkenes and alkynes. Use $22.7\ \mathrm{L}$ for the molar volume of a gas at STP ($273.15\ \mathrm{K}$, $1\ \mathrm{bar}$).
  • 1 Give the product of PCC oxidation of (a) butan-1-ol, (b) 2-methylpropan-1-ol, (c) cyclohexanol.
  • 2 When propan-1-ol is oxidised with acidified $\mathrm{K_{2}Cr_{2}O_{7}}$, the mixture is warmed and the product distilled off as it forms. Why does this give propanal rather than propanoic acid? (Boiling points: propanal $322\ \mathrm{K}$, propan-1-ol $370\ \mathrm{K}$.)
  • 3 Write a balanced equation for the dehydrogenation of propan-2-ol over copper at $573\ \mathrm{K}$.
  • 4 Name the products of ozonolysis ($\mathrm{O_{3}}$, then $\mathrm{Zn/H_{2}O}$) of (a) but-1-ene, (b) 2-methylpropene, (c) 1-methylcyclohexene.
  • 5 Identify the alkene that gives on ozonolysis (a) ethanal only; (b) methanal and butan-2-one; (c) benzaldehyde and propanone.
  • 6 An alkene $\mathrm{C_{5}H_{10}}$ gives ethanal and propanal on ozonolysis. Name it.
  • 7 Give the product of hydration ($\mathrm{HgSO_{4}}$, dilute $\mathrm{H_{2}SO_{4}}$) of (a) 3-methylbut-1-yne, (b) phenylethyne.
  • 8 Industry. A plant dehydrogenates propan-2-ol over copper. Assuming complete conversion, what mass of propanone and what volume of hydrogen (at STP) does $1.00$ tonne of propan-2-ol give?
1 (a) butanal $\mathrm{CH_{3}CH_{2}CH_{2}CHO}$; (b) 2-methylpropanal $\mathrm{(CH_{3})_{2}CHCHO}$; (c) cyclohexanone (PCC also oxidises secondary alcohols, to ketones).
2 Propanal boils $48\ \mathrm{K}$ lower than the alcohol (no hydrogen bonding between its molecules), so at a suitable temperature it evaporates out of the flask as soon as it forms, before the oxidant can convert it to the acid.
3 $\mathrm{CH_{3}CH(OH)CH_{3}\xrightarrow{Cu,\ 573\ K}CH_{3}COCH_{3}+H_{2}}$ (C 3 = 3, H 8 = 6 + 2, O 1 = 1 ✓).
4 (a) $\mathrm{CH_{3}CH_{2}CH{=}CH_{2}}$ → propanal + methanal; (b) $\mathrm{(CH_{3})_{2}C{=}CH_{2}}$ → propanone + methanal; (c) the ring opens into one molecule with a ketone at the methyl-bearing carbon and an aldehyde at the other: $\mathrm{CH_{3}CO(CH_{2})_{4}CHO}$, 6-oxoheptanal.
5 (a) $\mathrm{CH_{3}CH{=}CHCH_{3}}$, but-2-ene; (b) $\mathrm{CH_{2}{=}C(CH_{3})CH_{2}CH_{3}}$, 2-methylbut-1-ene; (c) $\mathrm{C_{6}H_{5}CH{=}C(CH_{3})_{2}}$, 2-methyl-1-phenylprop-1-ene.
6 $\mathrm{CH_{3}CH{=}CHCH_{2}CH_{3}}$, pent-2-ene ($2+3=5$ carbons ✓).
7 (a) $\mathrm{(CH_{3})_{2}CHCOCH_{3}}$, 3-methylbutan-2-one; (b) $\mathrm{C_{6}H_{5}COCH_{3}}$, acetophenone (1-phenylethan-1-one). Both are Markovnikov products: the $\mathrm{OH}$ goes to the inner carbon.
8 $n=\dfrac{1.00\times10^{6}\ \mathrm{g}}{60\ \mathrm{g\,mol^{-1}}}=1.67\times10^{4}\ \mathrm{mol}$; propanone $=1.667\times10^{4}\times58=9.67\times10^{5}\ \mathrm{g}=967\ \mathrm{kg}$; hydrogen $=1.667\times10^{4}\times22.7\ \mathrm{L}=3.78\times10^{5}\ \mathrm{L}=378\ \mathrm{m^{3}}$ at STP.
III
Named methods.
  • 1 Write the Rosenmund reduction of ethanoyl chloride. What is the role of the barium sulphate?
  • 2 Write the Stephen reaction for propanenitrile, showing the intermediate.
  • 3 Give the products of treating (a) ethyl benzoate, (b) but-3-enenitrile $\mathrm{CH_{2}{=}CHCH_{2}CN}$ with (i) DIBAL-H, (ii) $\mathrm{H_{2}O}$.
  • 4 Which aldehyde is formed when 4-nitrotoluene undergoes the Etard reaction? Name the reagent and solvent.
  • 5 Methylbenzene undergoes the Gattermann–Koch reaction mainly at the position para to the methyl group. Give the product and the reagents.
  • 6 Give the ketones formed from (a) dimethylcadmium and benzoyl chloride; (b) diethylcadmium and propanoyl chloride.
  • 7 Give the ketone formed when each nitrile reacts with the Grignard reagent and the product is hydrolysed: (a) $\mathrm{CH_{3}CN}$ + $\mathrm{C_{6}H_{5}MgBr}$; (b) $\mathrm{C_{6}H_{5}CN}$ + $\mathrm{C_{2}H_{5}MgBr}$.
  • 8 Write the Friedel–Crafts reaction of benzene with propanoyl chloride. Why is a second acyl group not introduced?
1 $\mathrm{CH_{3}COCl+H_{2}\xrightarrow{Pd{-}BaSO_{4}}CH_{3}CHO+HCl}$. $\mathrm{BaSO_{4}}$ partly poisons the palladium, so the catalyst is too weak to hydrogenate the aldehyde to ethanol.
2 $\mathrm{CH_{3}CH_{2}CN+SnCl_{2}+HCl\rightarrow CH_{3}CH_{2}CH{=}NH}$ (imine, as its salt) $\mathrm{\xrightarrow{H_{3}O^{+}}CH_{3}CH_{2}CHO}$ (propanal) $+\ \mathrm{NH_{4}^{+}}$.
3 (a) $\mathrm{C_{6}H_{5}CHO}$ (benzaldehyde) and $\mathrm{C_{2}H_{5}OH}$; (b) $\mathrm{CH_{2}{=}CHCH_{2}CHO}$, but-3-enal; the $\mathrm{C{=}C}$ is untouched.
4 4-nitrobenzaldehyde, $\mathrm{O_{2}NC_{6}H_{4}CHO}$; chromyl chloride ($\mathrm{CrO_{2}Cl_{2}}$) in $\mathrm{CS_{2}}$, followed by hydrolysis.
5 4-methylbenzaldehyde, $\mathrm{CH_{3}C_{6}H_{4}CHO}$; $\mathrm{CO}$ and $\mathrm{HCl}$ with anhydrous $\mathrm{AlCl_{3}/CuCl}$.
6 (a) $\mathrm{2C_{6}H_{5}COCl+(CH_{3})_{2}Cd\rightarrow2C_{6}H_{5}COCH_{3}+CdCl_{2}}$: acetophenone; (b) $\mathrm{CH_{3}CH_{2}COCH_{2}CH_{3}}$, pentan-3-one.
7 (a) $\mathrm{CH_{3}COC_{6}H_{5}}$, acetophenone; (b) $\mathrm{C_{6}H_{5}COC_{2}H_{5}}$, 1-phenylpropan-1-one. The nitrile carbon becomes the carbonyl carbon; the Grignard alkyl or aryl group becomes its second substituent.
8 $\mathrm{C_{6}H_{6}+CH_{3}CH_{2}COCl\xrightarrow{\text{anhyd. }AlCl_{3}}C_{6}H_{5}COCH_{2}CH_{3}+HCl}$. The acyl group withdraws electrons from the ring (by resonance and induction), deactivating it towards a second electrophilic attack.
IV
Physical properties and planning.
  • 1 Arrange in increasing order of boiling point and explain: $\mathrm{CH_{3}CH_{2}CH_{2}CH_{3}}$, $\mathrm{CH_{3}COCH_{3}}$, $\mathrm{CH_{3}CH_{2}CH_{2}OH}$, $\mathrm{CH_{3}OCH_{2}CH_{3}}$.
  • 2 Why do methanal, ethanal and propanone mix with water in all proportions, while benzaldehyde is only slightly soluble?
  • 3 Arrange in increasing order of boiling point: $\mathrm{CH_{3}CHO}$, $\mathrm{CH_{3}CH_{2}OH}$, $\mathrm{CH_{3}OCH_{3}}$, $\mathrm{CH_{3}CH_{2}CH_{3}}$.
  • 4 Convert propene into propanone in two steps.
  • 5 Convert benzene into 1-phenylpropan-1-one in one step, and ethanenitrile into propanone.
  • 6 A compound $\mathrm{C_{3}H_{6}O}$ is made by the oxidation of a secondary alcohol, and also by the hydration of an alkyne. Identify it, the alcohol and the alkyne.
1 Butane ($273\ \mathrm{K}$) < methoxyethane ($281\ \mathrm{K}$) < propanone ($329\ \mathrm{K}$) < propan-1-ol ($370\ \mathrm{K}$). Similar molar masses (58–60), so the order follows the forces between molecules: dispersion only < weak dipole–dipole < strong $\mathrm{C{=}O}$ dipole–dipole < hydrogen bonding.
2 Water molecules hydrogen-bond to the lone pairs of the carbonyl oxygen. In the small molecules this outweighs the hydrophobic hydrocarbon part; in benzaldehyde the large benzene ring dominates.
3 $\mathrm{CH_{3}CH_{2}CH_{3}}$ ($231\ \mathrm{K}$) < $\mathrm{CH_{3}OCH_{3}}$ ($248\ \mathrm{K}$) < $\mathrm{CH_{3}CHO}$ ($294\ \mathrm{K}$) < $\mathrm{CH_{3}CH_{2}OH}$ ($351\ \mathrm{K}$), for the same reasons as in part 1 (molar masses 44–46).
4 $\mathrm{CH_{3}CH{=}CH_{2}\xrightarrow{H_{2}O,\ H^{+}}CH_{3}CH(OH)CH_{3}}$ (Markovnikov hydration) $\mathrm{\xrightarrow{K_{2}Cr_{2}O_{7}/H^{+}}CH_{3}COCH_{3}}$.
5 $\mathrm{C_{6}H_{6}+CH_{3}CH_{2}COCl\xrightarrow{\text{anhyd. }AlCl_{3}}C_{6}H_{5}COCH_{2}CH_{3}+HCl}$. $\mathrm{CH_{3}CN}$ + $\mathrm{CH_{3}MgBr}$ in dry ether, then $\mathrm{H_{3}O^{+}}$, gives $\mathrm{CH_{3}COCH_{3}}$.
6 Propanone, $\mathrm{CH_{3}COCH_{3}}$, from propan-2-ol (the only secondary $\mathrm{C_{3}}$ alcohol) and from propyne, $\mathrm{CH_{3}C{\equiv}CH}$. (Propanal, the other $\mathrm{C_{3}H_{6}O}$ carbonyl compound, cannot come from a secondary alcohol.)
🎯 Exam corner BoardJEE MainJEE AdvancedNEET

E1. Board · 2 marks Write the chemical equations for (a) the Rosenmund reduction and (b) the Etard reaction, naming the reagents and conditions.

(a) $\mathrm{C_{6}H_{5}COCl+H_{2}\xrightarrow{Pd{-}BaSO_{4}}C_{6}H_{5}CHO+HCl}$: hydrogen over palladium poisoned with barium sulphate. (1 mark)
(b) $\mathrm{C_{6}H_{5}CH_{3}\xrightarrow{CrO_{2}Cl_{2},\ CS_{2}}C_{6}H_{5}CH(OCrOHCl_{2})_{2}\xrightarrow{H_{3}O^{+}}C_{6}H_{5}CHO}$: chromyl chloride in carbon disulphide, then hydrolysis of the chromium complex. (1 mark)

E2. JEE Main Ethyl benzoate is converted into benzaldehyde in good yield by (A) $\mathrm{LiAlH_{4}}$, then $\mathrm{H_{2}O}$  (B) DIBAL-H at low temperature, then $\mathrm{H_{2}O}$  (C) $\mathrm{NaBH_{4}}$ in ethanol  (D) $\mathrm{H_{2}}$, $\mathrm{Pd{-}BaSO_{4}}$

Idea Only DIBAL-H delivers one hydride to an ester and holds the intermediate until water is added. $\mathrm{LiAlH_{4}}$ reduces the ester to benzyl alcohol; $\mathrm{NaBH_{4}}$ hardly reduces esters at all; Rosenmund conditions are for acyl chlorides.
Answer (B).

E3. NEET The main product of the reaction of but-1-yne with water in the presence of $\mathrm{HgSO_{4}}$ and dilute $\mathrm{H_{2}SO_{4}}$ is (A) butanal  (B) butan-2-one  (C) butan-2-ol  (D) but-1-en-2-ol

Idea Markovnikov addition puts $\mathrm{OH}$ on C-2, giving the enol but-1-en-2-ol (D), which is not isolated: it rearranges at once to the ketone.
Answer (B) $\mathrm{CH_{3}CH_{2}COCH_{3}}$.

E4. JEE Advanced Integer type. An alkene of molecular formula $\mathrm{C_{8}H_{16}}$ gives butan-2-one as the only product of ozonolysis. The number of possible structures for the alkene, counting stereoisomers, is ______.

Rebuild Join the carbonyl carbons of two butan-2-one molecules: $\mathrm{CH_{3}CH_{2}C(CH_{3}){=}C(CH_{3})CH_{2}CH_{3}}$, 3,4-dimethylhex-3-ene ($\mathrm{C_{8}H_{16}}$ ✓). Only one skeleton is possible, because each half must be exactly $\mathrm{(CH_{3})(C_{2}H_{5})C{=}}$.
Stereo Each alkene carbon carries two different groups (methyl and ethyl), so the alkene has a cis (Z) and a trans (E) form; both give the same ozonolysis product.
Answer $\mathbf{2}$.

E5. JEE Main Numerical value. Vapour of $15.0\ \mathrm{g}$ of propan-2-ol is passed over copper at $573\ \mathrm{K}$, and $80\%$ of it is converted into propanone. The mass of propanone formed, in grams (nearest integer), is ______. ($\mathrm{C}=12$, $\mathrm{H}=1$, $\mathrm{O}=16$.)

Moles $\mathrm{CH_{3}CH(OH)CH_{3}\rightarrow CH_{3}COCH_{3}+H_{2}}$ is 1 : 1. $n=\dfrac{15.0}{60}=0.250\ \mathrm{mol}$; $80\%$ gives $0.200\ \mathrm{mol}$ of propanone.
Answer $0.200\times58=11.6\ \mathrm{g}$, nearest integer $\mathbf{12}$.

E6. Board · Case study Read the passage and answer. Benzaldehyde, the almond-flavoured aldehyde, is made on a large scale from methylbenzene (toluene) and, in some plants, directly from benzene. The main problem is that the aldehyde is easily oxidised further, to benzoic acid. (a) Give two methods, with equations, for converting methylbenzene into benzaldehyde. (2 marks) (b) In the Etard reaction, why is the product benzaldehyde and not benzoic acid? (1 mark) (c) Name the reaction, and give the reagents, by which benzene is converted directly into benzaldehyde. (1 mark)

(a) Etard: $\mathrm{C_{6}H_{5}CH_{3}\xrightarrow{CrO_{2}Cl_{2},\ CS_{2}}}$ chromium complex $\mathrm{\xrightarrow{H_{3}O^{+}}C_{6}H_{5}CHO}$. Side-chain chlorination: $\mathrm{C_{6}H_{5}CH_{3}+2Cl_{2}\xrightarrow{h\nu}C_{6}H_{5}CHCl_{2}+2HCl}$; $\mathrm{C_{6}H_{5}CHCl_{2}+H_{2}O\xrightarrow{373\ K}C_{6}H_{5}CHO+2HCl}$. (The $\mathrm{CrO_{3}/(CH_{3}CO)_{2}O}$ method is also acceptable.)
(b) The aldehyde carbon is held in the insoluble chromium complex, which the chromyl chloride cannot oxidise further; the free aldehyde is released only when the complex is hydrolysed, after the oxidant has been used up.
(c) Gattermann–Koch reaction: $\mathrm{CO}$ and $\mathrm{HCl}$ with anhydrous $\mathrm{AlCl_{3}}$ (and $\mathrm{CuCl}$).
🔬 Research–Teaching NexusResearch-tutored · optional, about 30 minutes

Research question. Which route to benzaldehyde wastes the fewest atoms?

Why it matters. Green chemistry asks not only "what is the yield?" but "how much of what goes in ends up in the product?" The atom economy of a reaction is $\dfrac{\text{molar mass of the desired product}}{\text{total molar mass of the reactants}}\times100\%$, read from the balanced equation. Chemical companies in Europe, India, China and the USA compare routes this way, because every atom that ends up in a by-product such as $\mathrm{HCl}$ or a chromium salt must be paid for, handled and disposed of.

What to do.

  1. Using $\mathrm{C}=12$, $\mathrm{H}=1$, $\mathrm{O}=16$, $\mathrm{Cl}=35.5$, $\mathrm{Cr}=52$, write the overall balanced equation for side-chain chlorination followed by hydrolysis, $\mathrm{C_{6}H_{5}CH_{3}+2Cl_{2}+H_{2}O\rightarrow C_{6}H_{5}CHO+4HCl}$, and calculate its atom economy.
  2. Do the same for the Gattermann–Koch reaction (overall $\mathrm{C_{6}H_{6}+CO\rightarrow C_{6}H_{5}CHO}$; $\mathrm{HCl}$ and $\mathrm{AlCl_{3}}$ are not consumed) and for catalytic air oxidation of toluene, $\mathrm{C_{6}H_{5}CH_{3}+O_{2}\rightarrow C_{6}H_{5}CHO+H_{2}O}$.
  3. The Etard reaction uses $2$ mol of $\mathrm{CrO_{2}Cl_{2}}$ for each mole of toluene. Find an upper limit for its atom economy (count only toluene and chromyl chloride as reactants).
  4. Rank the four routes. Then list one practical drawback of the "best" route that atom economy does not show (think about the hazards of $\mathrm{CO}$, the need for a selective catalyst, the yield).

Tools: A calculator or spreadsheet; the equations of Definition 8.3 and Rule 8.3.

Hand in: A table of the four routes with their atom economies, your ranking, and a paragraph recommending one route for a new plant.

Going further. Natural benzaldehyde (from bitter almonds or apricot kernels) sells at a much higher price than synthetic benzaldehyde. Find out how "natural" benzaldehyde is defined by food regulators and why it costs more.

Instructor note: chlorination + hydrolysis: reactants $92+142+18=252$, product $106$, atom economy $42.1\%$. Gattermann–Koch: $106/(78+28)=100\%$. Air oxidation: $106/(92+32)=85.5\%$. Etard: $\mathrm{CrO_{2}Cl_{2}}=155$, so at most $106/(92+310)=26.4\%$. Ranking: Gattermann–Koch > air oxidation > chlorination > Etard. Drawbacks of Gattermann–Koch: toxic $\mathrm{CO}$ under pressure, anhydrous $\mathrm{AlCl_{3}}$ that must be quenched (creating aluminium waste in practice); air oxidation needs a selective catalyst and low conversion per pass to avoid benzoic acid.

🧪 Self-check quiz

0/5 correct

Q1. The carbonyl carbon of propanone is

Q2. The IUPAC name of $\mathrm{(CH_{3})_{2}CHCOCH_{3}}$ is

Q3. Which reagent converts an acyl chloride $\mathrm{RCOCl}$ into an aldehyde $\mathrm{RCHO}$?

Q4. Hydration of propyne with $\mathrm{HgSO_{4}}$ and dilute $\mathrm{H_{2}SO_{4}}$ gives

Q5. Propanone ($M=58$) boils at $329\ \mathrm{K}$, between butane ($273\ \mathrm{K}$) and propan-1-ol ($370\ \mathrm{K}$). The best explanation is that propanone molecules

← Chapter 8 — Aldehydes, Ketones and Carboxylic Acids8.2 Reactions of Aldehydes and Ketones →