Chapter 8 · Aldehydes, Ketones and Carboxylic Acids

8.2 Reactions of Aldehydes and Ketones

Nucleophiles add to the δ+ carbonyl carbon; reducing and oxidising agents move the carbonyl up or down the oxidation ladder; α-hydrogens open the door to aldol and Cannizzaro reactions. Mechanisms, reactivity, the tests that tell aldehydes from ketones, and ring substitution.

A silver mirror forms on the inside of a test tube when a few drops of ethanal are warmed with an ammoniacal silver solution; with propanone nothing happens. A pinch of an orange-red solid precipitates the moment 2,4-dinitrophenylhydrazine meets either of them. Bitter almonds and cassava roots release poisonous hydrogen cyanide from the breakdown of cyanohydrins, compounds formed when $\mathrm{HCN}$ adds to a carbonyl group. And when ethanal stands with a little dilute alkali, two molecules join to make a new carbon–carbon bond. This is the aldol reaction, which chemists and living cells both use to build larger molecules.

All these reactions start from the polarity of the carbonyl group (Rule 8.1). The $\delta+$ carbon is attacked by nucleophiles, giving nucleophilic addition, the characteristic reaction of aldehydes and ketones. Ammonia derivatives add and then lose water. The carbonyl group can be reduced to an alcohol or all the way to $\mathrm{CH_{2}}$, and aldehydes, unlike ketones, are easily oxidised, which gives simple test-tube tests. Finally, the carbonyl group makes the hydrogen atoms on the next carbon (the α-hydrogens) acidic, and this is the origin of the aldol and Cannizzaro reactions.

Nucleophilic Addition

📐 Rule 8.6 — Mechanism of nucleophilic addition
  1. The nucleophile $\mathrm{Nu^{-}}$ attacks the electrophilic carbonyl carbon from above or below the plane of the $sp^{2}$ carbonyl group, roughly at right angles to it. As the new $\mathrm{C{-}Nu}$ bond forms, the π electrons of $\mathrm{C{=}O}$ move onto oxygen.
  2. The carbon becomes $sp^{3}$ and tetrahedral: the product of this slow step is a tetrahedral alkoxide intermediate, with the negative charge on oxygen.
  3. The alkoxide ion takes a proton from the reaction medium (water, acid, or the reagent itself), giving the neutral addition product:
$$\mathrm{{>}C{=}O+Nu^{-}\xrightarrow{\text{slow}}{>}C(Nu){-}O^{-}\xrightarrow[\text{fast}]{H^{+}}{>}C(Nu){-}OH}$$

The overall change is addition of $\mathrm{H{-}Nu}$ across the $\mathrm{C{=}O}$ double bond: $\mathrm{H}$ goes to oxygen and $\mathrm{Nu}$ to carbon. Acid catalysis works the other way round: protonating the oxygen first makes the carbon even more electrophilic, so that weak nucleophiles such as alcohols can add.

📐 Rule 8.7 — Reactivity of aldehydes and ketones towards nucleophiles

Aldehydes are generally more reactive than ketones, for two reasons:

  • Steric: a ketone carbon carries two alkyl or aryl groups, which crowd the incoming nucleophile and crowd one another even more in the tetrahedral product; an aldehyde has a small hydrogen in place of one of them.
  • Electronic: alkyl groups release electrons (+I effect), which reduces the positive charge on the carbonyl carbon. Two alkyl groups (a ketone) reduce it more than one (an aldehyde); methanal has none.

Hence $\mathrm{HCHO}\gt\mathrm{CH_{3}CHO}\gt\mathrm{CH_{3}CH_{2}CHO}$, and $\mathrm{CH_{3}CHO}\gt\mathrm{CH_{3}COCH_{3}}\gt\mathrm{CH_{3}COCH_{2}CH_{3}}$. Aromatic aldehydes and ketones are less reactive than their aliphatic counterparts: in benzaldehyde the ring's π electrons are delocalised onto the carbonyl group (resonance), which makes the carbonyl carbon less electrophilic. So benzaldehyde is less reactive than propanal, and acetophenone less reactive than propanone. Electron-withdrawing groups (–I, –R) have the opposite effect: 4-nitrobenzaldehyde is more reactive than benzaldehyde.

📐 Rule 8.8 — Addition of HCN, sodium hydrogensulphite, Grignard reagents and alcohols

1. Hydrogen cyanide. In the presence of a little base, which generates the stronger nucleophile $\mathrm{CN^{-}}$ from weakly acidic $\mathrm{HCN}$, aldehydes and ketones give cyanohydrins:

$$\mathrm{HCN+OH^{-}\rightleftharpoons CN^{-}+H_{2}O},\qquad \mathrm{{>}C{=}O+HCN\xrightarrow{OH^{-}}{>}C(OH)CN}$$

Cyanohydrins are useful because the $\mathrm{CN}$ group can be hydrolysed to $\mathrm{COOH}$ (giving α-hydroxy acids) or reduced to $\mathrm{CH_{2}NH_{2}}$: the carbon chain grows by one carbon.

2. Sodium hydrogensulphite (sodium bisulphite) adds to give crystalline hydrogensulphite addition compounds, soluble in water but not in the organic layer:

$$\mathrm{{>}C{=}O+NaHSO_{3}\rightleftharpoons{>}C(OH)SO_{3}Na}$$

The equilibrium lies to the right for most aldehydes and for small, mainly methyl, ketones, but to the left for most other ketones (steric hindrance). Dilute acid or alkali reverses the addition and gives back the carbonyl compound, so the reaction is used to separate and purify aldehydes.

3. Grignard reagents (Chapter 6) add $\mathrm{R'^{-}}$ to the carbonyl carbon; hydrolysis of the magnesium alkoxide gives an alcohol. Methanal gives a primary alcohol, other aldehydes secondary alcohols, and ketones tertiary alcohols:

$$\mathrm{{>}C{=}O+R'MgX\rightarrow{>}C(R'){-}OMgX\xrightarrow{H_{2}O}{>}C(R'){-}OH+Mg(OH)X}$$

4. Alcohols. With one equivalent of an alcohol and dry $\mathrm{HCl}$, an aldehyde gives an unstable hemiacetal, which reacts with a second molecule of alcohol to give a stable acetal. Ketones react poorly with simple alcohols but readily with ethane-1,2-diol (ethylene glycol), giving a cyclic ketal:

$$\mathrm{RCHO\xrightarrow{R'OH,\ dry\ HCl}RCH(OH)OR'\xrightarrow{R'OH}RCH(OR')_{2}+H_{2}O}$$ $$\mathrm{R_{2}CO+HOCH_{2}CH_{2}OH\overset{H^{+}}{\rightleftharpoons}R_{2}C(OCH_{2})_{2}+H_{2}O}$$

Dry $\mathrm{HCl}$ protonates the carbonyl oxygen, making the carbon reactive enough for the weakly nucleophilic alcohol. Water is a product, so the steps are reversible: aqueous mineral acid hydrolyses acetals and ketals back to the carbonyl compound. Chemists therefore use ketal formation to protect a carbonyl group while they change another part of a molecule.

📘 Definition 8.4 — Cyanohydrin, hemiacetal and acetal
A cyanohydrin has an $\mathrm{OH}$ and a $\mathrm{CN}$ group on the same carbon, $\mathrm{{>}C(OH)CN}$. A hemiacetal has an $\mathrm{OH}$ and an $\mathrm{OR}$ group on the same carbon, $\mathrm{{>}C(OH)OR}$; an acetal has two $\mathrm{OR}$ groups on the same carbon, $\mathrm{{>}C(OR)_{2}}$. An acetal made from a ketone is often called a ketal. (Cyclic hemiacetals reappear in Chapter 10: the ring forms of glucose and fructose are hemiacetal and hemiketal.)
✏️ Example 1 — Ranking reactivity towards nucleophilic addition
Arrange in order of increasing reactivity towards $\mathrm{HCN}$, and explain: (a) ethanal, propanone, methanal, butan-2-one; (b) benzaldehyde, 4-nitrobenzaldehyde, 4-methylbenzaldehyde. (c) Which is more reactive, benzaldehyde or propanal?
(a) Count the alkyl groups on the carbonyl carbon (0, 1, 2) and their size. Butan-2-one < propanone < ethanal < methanal. Butan-2-one has a larger ethyl group in place of one methyl: more crowding and slightly more +I effect than propanone.
(b) The ring and the steric situation are the same; only the para substituent changes. $\mathrm{{-}CH_{3}}$ (+I, hyperconjugation) pushes electrons towards the carbonyl carbon and reduces its δ+ charge, while $\mathrm{{-}NO_{2}}$ (–I, –R) withdraws electrons and increases it: 4-methylbenzaldehyde < benzaldehyde < 4-nitrobenzaldehyde.
(c) Propanal. In benzaldehyde the ring's π electrons are delocalised onto the carbonyl group, which reduces the positive charge on the carbonyl carbon (and the resonance stabilisation is lost when the carbon becomes $sp^{3}$).
Check The rule is "more δ+ and less crowding = faster": each comparison above changes only one of these two factors, or changes both in the same direction ✓.
✏️ Example 2 — Products of nucleophilic addition
Give the structure and IUPAC name of the main organic product: (a) ethanal + $\mathrm{HCN}$ (trace of base); (b) propanone + $\mathrm{NaHSO_{3}}$; (c) propanone + $\mathrm{CH_{3}MgBr}$, then $\mathrm{H_{2}O}$; (d) methanal + $\mathrm{C_{2}H_{5}MgBr}$, then $\mathrm{H_{2}O}$; (e) ethanal + excess methanol, dry $\mathrm{HCl}$; (f) propanone + ethane-1,2-diol, $\mathrm{H^{+}}$.
(a) $\mathrm{CH_{3}CH(OH)CN}$: the nitrile carbon is C-1, so the name is 2-hydroxypropanenitrile (ethanal cyanohydrin). C-2 carries four different groups, so it is formed as a racemic mixture: $\mathrm{CN^{-}}$ attacks the planar carbonyl group equally from either face.
(b) $\mathrm{(CH_{3})_{2}C(OH)SO_{3}Na}$, the crystalline hydrogensulphite addition compound (sodium 2-hydroxypropane-2-sulphonate).
(c) $\mathrm{CH_{3}^{-}}$ (from the Grignard reagent) adds to the carbonyl carbon, giving $\mathrm{(CH_{3})_{3}C{-}OMgBr}$, then $\mathrm{(CH_{3})_{3}COH}$, 2-methylpropan-2-ol (a tertiary alcohol, as expected from a ketone).
(d) $\mathrm{HCHO+C_{2}H_{5}MgBr\rightarrow C_{2}H_{5}CH_{2}OMgBr}$, then hydrolysis gives $\mathrm{CH_{3}CH_{2}CH_{2}OH}$, propan-1-ol: methanal adds one carbon and gives a primary alcohol.
(e) Through the hemiacetal $\mathrm{CH_{3}CH(OH)OCH_{3}}$ to the acetal $\mathrm{CH_{3}CH(OCH_{3})_{2}}$, 1,1-dimethoxyethane, and water.
(f) The cyclic ketal: a five-membered ring $\mathrm{{-}O{-}CH_{2}{-}CH_{2}{-}O{-}}$ closed on the former carbonyl carbon, which keeps its two methyl groups: 2,2-dimethyl-1,3-dioxolane, $\mathrm{C_{5}H_{10}O_{2}}$.
Interactive 1 — Nucleophilic Addition, Step by Step

Choose a carbonyl compound and a nucleophile, then press Play (or drag the slider). The nucleophile approaches the planar $sp^{2}$ carbon from above the plane, the π electrons move onto oxygen, the carbon turns tetrahedral, and finally a proton is added to the oxygen. The bars compare the two factors of Rule 8.7 for the chosen compound: how crowded the carbonyl carbon is, and how much of its δ+ charge is left after the attached groups have pushed electrons into it. The bars are a qualitative picture, not measured values.

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stage—
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type of product—

Addition–Elimination with Ammonia Derivatives; Reduction

Ammonia and its derivatives $\mathrm{H_{2}N{-}Z}$ add to the carbonyl group like any nucleophile, through the nitrogen lone pair. The addition product, however, still has an $\mathrm{N{-}H}$ next to the $\mathrm{C{-}OH}$, and it loses a molecule of water to form a $\mathrm{C{=}N}$ double bond.

📐 Rule 8.9 — Reactions with ammonia derivatives (addition–elimination)
$$\mathrm{{>}C{=}O+H_{2}N{-}Z\rightleftharpoons\Big[{>}C(OH){-}NH{-}Z\Big]\rightleftharpoons{>}C{=}N{-}Z+H_{2}O}$$

The reaction is catalysed by acid and works best in a weakly acidic solution (pH roughly 3.5–4.5). A little acid protonates the carbonyl oxygen and speeds up the attack; too much acid protonates the nitrogen of $\mathrm{H_{2}N{-}Z}$, which then has no lone pair and is no longer a nucleophile.

$\mathrm{Z}$Reagent $\mathrm{H_{2}N{-}Z}$Product $\mathrm{{>}C{=}N{-}Z}$
$\mathrm{{-}OH}$hydroxylamineoxime
$\mathrm{{-}NH_{2}}$hydrazinehydrazone
$\mathrm{{-}NHC_{6}H_{5}}$phenylhydrazinephenylhydrazone
$\mathrm{{-}NHC_{6}H_{3}(NO_{2})_{2}}$2,4-dinitrophenylhydrazine (2,4-DNP)2,4-dinitrophenylhydrazone
$\mathrm{{-}NHCONH_{2}}$semicarbazidesemicarbazone

The products are usually crystalline solids with sharp melting points. The 2,4-DNP test, a yellow, orange or red precipitate with 2,4-dinitrophenylhydrazine, shows the presence of a carbonyl group in an aldehyde or ketone, and the melting point of the derivative identifies the compound. Primary amines $\mathrm{RNH_{2}}$ react in the same way to give imines (Schiff bases), $\mathrm{{>}C{=}N{-}R}$.

✏️ Example 3 — Derivatives of carbonyl compounds
Give the structure, molecular formula and molar mass of the product of (a) propanone + hydroxylamine; (b) ethanal + 2,4-DNP; (c) cyclohexanone + semicarbazide; (d) benzaldehyde + phenylhydrazine. (e) Which of the oximes of ethanal and propanone exists as two geometrical isomers? ($\mathrm{C}=12$, $\mathrm{H}=1$, $\mathrm{N}=14$, $\mathrm{O}=16$.)
Method Replace the $\mathrm{{=}O}$ of the carbonyl compound by $\mathrm{{=}N{-}Z}$ and lose one $\mathrm{H_{2}O}$: formula of product = carbonyl + $\mathrm{H_{2}NZ}$ − $\mathrm{H_{2}O}$.
(a), (b) (a) $\mathrm{(CH_{3})_{2}C{=}NOH}$, propanone oxime, $\mathrm{C_{3}H_{7}NO}$, $M=73\ \mathrm{g\,mol^{-1}}$. (b) $\mathrm{CH_{3}CH{=}NNHC_{6}H_{3}(NO_{2})_{2}}$, ethanal 2,4-dinitrophenylhydrazone, $\mathrm{C_{8}H_{8}N_{4}O_{4}}$, $M=224\ \mathrm{g\,mol^{-1}}$.
(c), (d) (c) $\mathrm{C_{6}H_{10}{=}NNHCONH_{2}}$, cyclohexanone semicarbazone, $\mathrm{C_{7}H_{13}N_{3}O}$, $M=155\ \mathrm{g\,mol^{-1}}$. (d) $\mathrm{C_{6}H_{5}CH{=}NNHC_{6}H_{5}}$, benzaldehyde phenylhydrazone, $\mathrm{C_{13}H_{12}N_{2}}$, $M=196\ \mathrm{g\,mol^{-1}}$.
(e) In $\mathrm{CH_{3}CH{=}NOH}$ the carbon carries two different groups ($\mathrm{CH_{3}}$ and $\mathrm{H}$) and the nitrogen carries $\mathrm{OH}$ and a lone pair, so the $\mathrm{OH}$ can be on the same side as $\mathrm{CH_{3}}$ or as $\mathrm{H}$: ethanal oxime has two geometrical isomers. In $\mathrm{(CH_{3})_{2}C{=}NOH}$ both groups on carbon are the same, so there is only one form.
📐 Rule 8.10 — Reduction to alcohols

Sodium borohydride ($\mathrm{NaBH_{4}}$), lithium aluminium hydride ($\mathrm{LiAlH_{4}}$) or hydrogen with a nickel, platinum or palladium catalyst reduce aldehydes to primary and ketones to secondary alcohols:

$$\mathrm{RCHO\xrightarrow{NaBH_{4}\ \text{or}\ H_{2}/Ni}RCH_{2}OH},\qquad \mathrm{R_{2}CO\xrightarrow{NaBH_{4}\ \text{or}\ H_{2}/Ni}R_{2}CHOH}$$

The hydride reagents act as sources of $\mathrm{H^{-}}$, a nucleophile, and do not reduce isolated $\mathrm{C{=}C}$ bonds; catalytic hydrogenation reduces $\mathrm{C{=}C}$ as well.

📘 Definition 8.5 — Clemmensen and Wolff–Kishner reductions

Both reductions turn a carbonyl group into a methylene group, $\mathrm{{>}C{=}O\rightarrow{>}CH_{2}}$, converting an aldehyde or ketone into a hydrocarbon.

The Clemmensen reduction uses zinc amalgam and concentrated hydrochloric acid:

$$\mathrm{{>}C{=}O\xrightarrow{Zn{-}Hg,\ conc.\ HCl}{>}CH_{2}+H_{2}O}$$

The Wolff–Kishner reduction first forms the hydrazone, which is then heated with potassium hydroxide in ethylene glycol; nitrogen gas is lost:

$$\mathrm{{>}C{=}O\xrightarrow[-H_{2}O]{NH_{2}NH_{2}}{>}C{=}NNH_{2}\xrightarrow[\text{ethylene glycol, heat}]{KOH}{>}CH_{2}+N_{2}}$$

Clemmensen is used for compounds that are stable to strong acid, Wolff–Kishner for those that are stable to strong base.

✏️ Example 4 — Choosing a reduction
Give the reagent(s) for: (a) butan-2-one → butan-2-ol; (b) butan-2-one → butane, by two methods; (c) acetophenone → ethylbenzene; (d) but-2-enal → but-2-en-1-ol; (e) but-2-enal → butan-1-ol. (f) A ketone also carries an $\mathrm{OH}$ group on a tertiary carbon, which is easily dehydrated by strong acid. Which carbonyl-to-$\mathrm{CH_{2}}$ method would you choose?
(a) $\mathrm{NaBH_{4}}$ (or $\mathrm{LiAlH_{4}}$, or $\mathrm{H_{2}/Ni}$): $\mathrm{CH_{3}COCH_{2}CH_{3}\rightarrow CH_{3}CH(OH)CH_{2}CH_{3}}$.
(b) Clemmensen ($\mathrm{Zn{-}Hg}$, conc. $\mathrm{HCl}$) or Wolff–Kishner ($\mathrm{NH_{2}NH_{2}}$, then $\mathrm{KOH}$, ethylene glycol, heat): $\mathrm{CH_{3}COCH_{2}CH_{3}\rightarrow CH_{3}CH_{2}CH_{2}CH_{3}}$.
(c) Either method: $\mathrm{C_{6}H_{5}COCH_{3}\rightarrow C_{6}H_{5}CH_{2}CH_{3}}$. (Friedel–Crafts acylation followed by Clemmensen reduction is the standard way to attach a straight alkyl chain to benzene.)
(d), (e) (d) $\mathrm{NaBH_{4}}$ reduces only the $\mathrm{C{=}O}$: $\mathrm{CH_{3}CH{=}CHCH_{2}OH}$. (e) Excess $\mathrm{H_{2}}$ over $\mathrm{Ni}$ reduces both the $\mathrm{C{=}C}$ and the $\mathrm{C{=}O}$: $\mathrm{CH_{3}CH_{2}CH_{2}CH_{2}OH}$.
(f) Wolff–Kishner: it works in base, so the acid-sensitive tertiary alcohol is not dehydrated, as it would be by the concentrated $\mathrm{HCl}$ of the Clemmensen reduction.

Oxidation: Telling Aldehydes from Ketones

📐 Rule 8.11 — Oxidation of aldehydes and ketones

The $\mathrm{C{-}H}$ bond of the $\mathrm{CHO}$ group makes aldehydes easy to oxidise. Common oxidising agents (nitric acid, $\mathrm{KMnO_{4}}$, $\mathrm{K_{2}Cr_{2}O_{7}}$), and even mild ones such as $\mathrm{Ag^{+}}$ and $\mathrm{Cu^{2+}}$ in alkaline solution, convert them into carboxylic acids with the same number of carbon atoms:

$$\mathrm{RCHO\xrightarrow{[O]}RCOOH}$$

Ketones have no such hydrogen. They resist mild oxidants, and strong oxidants at high temperature break a carbon–carbon bond next to the carbonyl group, giving a mixture of carboxylic acids with fewer carbon atoms than the ketone. Mild oxidising agents therefore distinguish aldehydes from ketones.

📘 Definition 8.6 — Tollens' and Fehling's tests

Tollens' reagent is a freshly prepared ammoniacal silver nitrate solution, containing $\mathrm{[Ag(NH_{3})_{2}]^{+}}$. Warmed with an aldehyde, it deposits metallic silver as a bright silver mirror on the glass (or a black precipitate):

$$\mathrm{RCHO+2[Ag(NH_{3})_{2}]^{+}+3OH^{-}\rightarrow RCOO^{-}+2Ag+2H_{2}O+4NH_{3}}$$

Fehling's solution is made just before use by mixing equal volumes of Fehling's solution A (aqueous copper(II) sulphate) and Fehling's solution B (alkaline sodium potassium tartrate, Rochelle salt, which holds $\mathrm{Cu^{2+}}$ in solution as a deep blue complex). Heated with an aliphatic aldehyde, it gives a red-brown precipitate of copper(I) oxide:

$$\mathrm{RCHO+2Cu^{2+}+5OH^{-}\rightarrow RCOO^{-}+Cu_{2}O+3H_{2}O}$$

Both aliphatic and aromatic aldehydes give the Tollens' test, but aromatic aldehydes do not give the Fehling's test. Ketones give neither.

📘 Definition 8.7 — The haloform (iodoform) reaction

Aldehydes and ketones with at least one methyl group on the carbonyl carbon, $\mathrm{CH_{3}CO{-}}$ (that is, methyl ketones and ethanal), are oxidised by sodium hypohalite ($\mathrm{NaOX}$, made from $\mathrm{X_{2}}$ and $\mathrm{NaOH}$). The three hydrogens of the methyl group are replaced by halogen, and the $\mathrm{CX_{3}}$ group is then split off as haloform, $\mathrm{CHX_{3}}$. With iodine, yellow iodoform, $\mathrm{CHI_{3}}$, precipitates; it has a characteristic smell:

$$\mathrm{RCOCH_{3}+3I_{2}+4NaOH\rightarrow RCOONa+CHI_{3}\downarrow+3NaI+3H_{2}O}$$

The iodoform test is given by all compounds containing $\mathrm{CH_{3}CO{-}}$ and also by alcohols containing $\mathrm{CH_{3}CH(OH){-}}$ (ethanol, propan-2-ol, butan-2-ol…), which the hypoiodite first oxidises to $\mathrm{CH_{3}CO{-}}$. Methanal, other aldehydes and ketones without a $\mathrm{CH_{3}CO}$ group do not give it. Because hypohalite does not attack $\mathrm{C{=}C}$, the reaction can also be used to convert a methyl ketone into an acid with one carbon fewer.

✏️ Example 5 — Distinguishing tests
Give one chemical test, with the observation, to distinguish each pair: (a) propanal and propanone; (b) benzaldehyde and ethanal; (c) pentan-2-one and pentan-3-one; (d) acetophenone and benzophenone; (e) methanal and ethanal; (f) propan-2-ol and propan-1-ol.
(a) Tollens': propanal (an aldehyde) gives a silver mirror; propanone does not. (Fehling's also works: red-brown $\mathrm{Cu_{2}O}$ with propanal only.)
(b) Fehling's: ethanal gives a red-brown precipitate; benzaldehyde, an aromatic aldehyde, does not. (Iodoform also works: ethanal is $\mathrm{CH_{3}CHO}$, which has $\mathrm{CH_{3}CO{-}}$ with R = H.)
(c) Iodoform ($\mathrm{I_{2}/NaOH}$, warm): pentan-2-one, $\mathrm{CH_{3}CO{-}C_{3}H_{7}}$, gives yellow $\mathrm{CHI_{3}}$; pentan-3-one, $\mathrm{C_{2}H_{5}COC_{2}H_{5}}$, has no methyl ketone group and does not.
(d) Iodoform: acetophenone, $\mathrm{C_{6}H_{5}COCH_{3}}$, gives a yellow precipitate (and sodium benzoate); benzophenone does not.
(e) Iodoform: ethanal gives it, methanal does not (no $\mathrm{CH_{3}}$ group). Both give Tollens' and Fehling's tests, so those do not help.
(f) Iodoform: propan-2-ol, $\mathrm{CH_{3}CH(OH)CH_{3}}$, is oxidised to propanone and gives $\mathrm{CHI_{3}}$; propan-1-ol does not.

Reactions Due to α-Hydrogen; Cannizzaro Reaction; Ring Substitution

📐 Rule 8.12 — Acidity of α-hydrogens

The carbon next to the carbonyl group is the α-carbon, and the hydrogens on it are α-hydrogens. They are far more acidic than ordinary $\mathrm{C{-}H}$ hydrogens (an approximate $\mathrm{p}K_{a}$ of 19–20 for propanone, compared with about 50 for an alkane), for two reasons: the electron-withdrawing carbonyl group polarises the $\mathrm{C{-}H}$ bond, and the anion formed, the enolate ion, is stabilised by resonance, with the negative charge shared by the α-carbon and the oxygen:

$$\mathrm{{-}\overset{\displaystyle\ominus}{C}H{-}CH{=}O\ \longleftrightarrow\ {-}CH{=}CH{-}O^{\ominus}}$$

A base can therefore remove an α-hydrogen, and the enolate ion, a carbon nucleophile, attacks the carbonyl carbon of another molecule. This is the basis of the aldol reaction. Aldehydes with no α-hydrogen, such as $\mathrm{HCHO}$, $\mathrm{C_{6}H_{5}CHO}$ and $\mathrm{(CH_{3})_{3}CCHO}$, cannot form an enolate and take part in the Cannizzaro reaction instead.

📘 Definition 8.8 — Aldol reaction and aldol condensation

Aldehydes and ketones with at least one α-hydrogen react in the presence of dilute alkali to give β-hydroxy aldehydes (aldols) or β-hydroxy ketones (ketols). On heating, these lose water easily to form α,β-unsaturated carbonyl compounds, the aldol condensation products:

$$\mathrm{2CH_{3}CHO\xrightarrow{dil.\ NaOH}CH_{3}CH(OH)CH_{2}CHO\xrightarrow[-H_{2}O]{\Delta}CH_{3}CH{=}CHCHO}$$

ethanal → 3-hydroxybutanal (aldol) → but-2-enal

$$\mathrm{2CH_{3}COCH_{3}\xrightarrow{Ba(OH)_{2}}(CH_{3})_{2}C(OH)CH_{2}COCH_{3}\xrightarrow[-H_{2}O]{\Delta}(CH_{3})_{2}C{=}CHCOCH_{3}}$$

propanone → 4-hydroxy-4-methylpentan-2-one (ketol) → 4-methylpent-3-en-2-one

The new carbon–carbon bond joins the α-carbon of one molecule to the carbonyl carbon of the other, so the $\mathrm{OH}$ ends up on the β-carbon. The name comes from aldehyde + alcohol. Even though the condensation step is a dehydration, the whole sequence is called aldol condensation.

Cross aldol condensation. If two different carbonyl compounds that both have α-hydrogens are mixed with base, each can act as the enolate (donor) and each as the carbonyl partner (acceptor), so four products form: two from self-condensation and two crossed. From ethanal and propanal (after dehydration): but-2-enal and 2-methylpent-2-enal (self), and pent-2-enal and 2-methylbut-2-enal (crossed). A cross aldol is useful only when one partner has no α-hydrogen, so that it can only accept. With benzaldehyde (the Claisen–Schmidt reaction) the products are aromatic α,β-unsaturated carbonyl compounds:

$$\mathrm{C_{6}H_{5}CHO+CH_{3}COCH_{3}\xrightarrow[\Delta]{dil.\ NaOH}C_{6}H_{5}CH{=}CHCOCH_{3}+H_{2}O}$$

4-phenylbut-3-en-2-one (benzalacetone); with ethanal instead, 3-phenylprop-2-enal (cinnamaldehyde)

📘 Definition 8.9 — Cannizzaro reaction

Aldehydes that have no α-hydrogen, heated with concentrated alkali, undergo self oxidation and reduction (disproportionation): one molecule is reduced to an alcohol and the other is oxidised to the salt of a carboxylic acid:

$$\mathrm{2HCHO+conc.\ KOH\rightarrow CH_{3}OH+HCOOK}$$ $$\mathrm{2C_{6}H_{5}CHO+conc.\ NaOH\rightarrow C_{6}H_{5}CH_{2}OH+C_{6}H_{5}COONa}$$

In a crossed Cannizzaro reaction between methanal and another aldehyde without α-hydrogen, methanal (the more reactive aldehyde) is the one oxidised: $\mathrm{C_{6}H_{5}CHO+HCHO+NaOH\rightarrow C_{6}H_{5}CH_{2}OH+HCOONa}$. This makes benzyl alcohol efficiently from benzaldehyde.

📐 Rule 8.13 — Electrophilic substitution in aromatic aldehydes and ketones
The $\mathrm{{-}CHO}$ and $\mathrm{{-}COR}$ groups withdraw electrons from the ring (–I and –R), so they deactivate it and direct electrophiles to the meta position, where the electron density is least reduced: $$\mathrm{C_{6}H_{5}CHO\xrightarrow[273{-}283\ K]{conc.\ HNO_{3}/conc.\ H_{2}SO_{4}}3{-}O_{2}NC_{6}H_{4}CHO}\quad(\text{3-nitrobenzaldehyde})$$

Uses. A $40\%$ solution of methanal (formalin) preserves biological specimens, and methanal is a raw material for synthetic resins and adhesives. Ethanal is used to make ethanoic acid, ethyl ethanoate and other chemicals. Benzaldehyde is used in perfumes and flavours and in the dye industry. Propanone and butan-2-one are widely used solvents (nail-polish remover, paint thinners). Many aldehydes and ketones, such as vanillin, cinnamaldehyde, camphor and butyraldehyde, are valued for their odours and flavours.

✏️ Example 6 — Aldol or Cannizzaro?
Predict the organic product(s): (a) propanal with dilute $\mathrm{NaOH}$, then heat; (b) 2,2-dimethylpropanal with concentrated $\mathrm{NaOH}$; (c) benzaldehyde and methanal with concentrated $\mathrm{NaOH}$; (d) benzaldehyde and ethanal with dilute $\mathrm{NaOH}$, then heat. (e) How many aldol products (before dehydration) can form from a mixture of ethanal and propanone with dilute alkali? Name them.
First Look for α-hydrogens: present → aldol (dilute alkali); absent → Cannizzaro (concentrated alkali).
(a) Propanal has two α-H. The α-carbon of one molecule bonds to the carbonyl carbon of another: $\mathrm{CH_{3}CH_{2}CH(OH)CH(CH_{3})CHO}$ (3-hydroxy-2-methylpentanal), which on heating gives $\mathrm{CH_{3}CH_{2}CH{=}C(CH_{3})CHO}$, 2-methylpent-2-enal.
(b) $\mathrm{(CH_{3})_{3}CCHO}$ has no α-H (the α-carbon is quaternary): Cannizzaro: $$\mathrm{2(CH_{3})_{3}CCHO+NaOH\rightarrow(CH_{3})_{3}CCH_{2}OH+(CH_{3})_{3}CCOONa}$$ giving 2,2-dimethylpropan-1-ol and sodium 2,2-dimethylpropanoate.
(c) Crossed Cannizzaro: benzyl alcohol (phenylmethanol) and sodium methanoate; methanal is oxidised.
(d) Benzaldehyde has no α-H, so it can only accept; the enolate of ethanal attacks it: $\mathrm{C_{6}H_{5}CH(OH)CH_{2}CHO\rightarrow C_{6}H_{5}CH{=}CHCHO}$, 3-phenylprop-2-enal (cinnamaldehyde). Some self-aldol of ethanal also occurs.
(e) Both have α-H: 2 donors × 2 acceptors = 4. Ethanal + ethanal: 3-hydroxybutanal. Propanone + propanone: 4-hydroxy-4-methylpentan-2-one. Ethanal enolate + propanone: $\mathrm{(CH_{3})_{2}C(OH)CH_{2}CHO}$, 3-hydroxy-3-methylbutanal. Propanone enolate + ethanal: $\mathrm{CH_{3}CH(OH)CH_{2}COCH_{3}}$, 4-hydroxypentan-2-one.
Interactive 2 — Aldol Builder: Who Gives, Who Takes?

Choose two carbonyl compounds (choose the same one twice for a self-condensation). Only a compound with an α-hydrogen can become an enolate and give its α-carbon; every carbonyl compound can take it at its carbonyl carbon. The builder lists every donor–acceptor pairing, draws the β-hydroxy carbonyl compound with the new C–C bond, and the α,β-unsaturated product after loss of water. Press Play to step through the pairings. If no partner has an α-hydrogen, it shows the Cannizzaro products instead.

α-H (1 / 2)—
number of aldol products—
this aldol product—
after dehydration—
⚠️ Common mistakes with the reactions of aldehydes and ketones

1. Aldol needs α-H on the donor only. Benzaldehyde and methanal cannot form enolates but are excellent acceptors. "No α-H → no reaction with base" is wrong: in dilute base they accept enolates; in concentrated base they undergo Cannizzaro.

2. Counting α-hydrogens. They are on the carbon next to $\mathrm{C{=}O}$, not on the carbonyl carbon itself. $\mathrm{(CH_{3})_{3}CCHO}$ has nine hydrogens on β-carbons but none on the α-carbon.

3. Tollens' versus Fehling's. Aromatic aldehydes give Tollens' but not Fehling's test; methanal gives both. Ketones give neither.

4. Iodoform needs $\mathrm{CH_{3}CO{-}}$ or $\mathrm{CH_{3}CH(OH){-}}$. Methanal and propanal do not give it; ethanal and ethanol do.

5. Clemmensen and Wolff–Kishner give hydrocarbons, not alcohols. For an alcohol, use $\mathrm{NaBH_{4}}$, $\mathrm{LiAlH_{4}}$ or $\mathrm{H_{2}/Ni}$.

🚀 Beyond the board syllabus (JEE Main · NEET · JEE Advanced) — the mechanisms of the aldol and Cannizzaro reactions

In JEE Main, NEET and JEE Advanced 2026 (aldol condensation and Cannizzaro reaction); not required by the CBSE syllabus.

Aldol. (1) $\mathrm{OH^{-}}$ removes an α-hydrogen: $\mathrm{CH_{3}CHO+OH^{-}\rightleftharpoons{}^{-}CH_{2}CHO+H_{2}O}$. (2) The enolate carbon attacks the carbonyl carbon of a second molecule (nucleophilic addition, Rule 8.6), giving an alkoxide. (3) The alkoxide takes a proton from water, giving 3-hydroxybutanal and regenerating $\mathrm{OH^{-}}$, so the base is a catalyst. On heating, base removes the now doubly activated α-hydrogen of the aldol and $\mathrm{OH^{-}}$ leaves, forming the conjugated $\mathrm{C{=}C{-}C{=}O}$ system, whose stability drives the dehydration.

Cannizzaro. (1) $\mathrm{OH^{-}}$ adds to the carbonyl carbon of one aldehyde molecule, giving $\mathrm{ArCH(OH)O^{-}}$. (2) This anion re-forms its $\mathrm{C{=}O}$ and pushes the hydrogen, with its bonding pair, as a hydride ion onto the carbonyl carbon of a second molecule: the first becomes $\mathrm{ArCOOH}$ (oxidised) and the second $\mathrm{ArCH_{2}O^{-}}$ (reduced). (3) Proton exchange gives $\mathrm{ArCOO^{-}}$ and $\mathrm{ArCH_{2}OH}$. The hydride-transfer step is rate-determining, and the rate law is third order overall (second order in aldehyde, first order in $\mathrm{OH^{-}}$) under typical conditions; in the crossed reaction the more reactive $\mathrm{HCHO}$ is attacked by $\mathrm{OH^{-}}$ and becomes the hydride donor, which is why it ends up as methanoate.

Exercises

I
Nucleophilic addition.
  • 1 Write the mechanism of the addition of $\mathrm{HCN}$ to ethanal. Why is a trace of base needed?
  • 2 Arrange in increasing order of reactivity towards nucleophilic addition: $\mathrm{CH_{3}CHO}$, $\mathrm{HCHO}$, $\mathrm{CH_{3}COCH_{3}}$, $\mathrm{C_{6}H_{5}COCH_{3}}$.
  • 3 Why is the reaction with sodium hydrogensulphite used to purify aldehydes? Which of propanone and pentan-3-one forms the addition compound more readily, and why?
  • 4 Give the alcohol formed (after hydrolysis) from: (a) $\mathrm{HCHO}$ + $\mathrm{CH_{3}MgBr}$; (b) $\mathrm{CH_{3}CHO}$ + $\mathrm{C_{2}H_{5}MgBr}$; (c) $\mathrm{CH_{3}COCH_{3}}$ + $\mathrm{C_{6}H_{5}MgBr}$; (d) $\mathrm{C_{6}H_{5}CHO}$ + $\mathrm{CH_{3}MgI}$. Classify each as 1°, 2° or 3°.
  • 5 Give two combinations of a carbonyl compound and a Grignard reagent that make 2-methylbutan-2-ol.
  • 6 Write the formation of the hemiacetal and the acetal from ethanal and ethanol. Why is dry $\mathrm{HCl}$ used, and how is the ethanal recovered from the acetal?
  • 7 Propanone cyanohydrin is hydrolysed by aqueous acid. Give the product, its molecular formula, and explain why this is a way of lengthening a carbon chain.
  • 8 Ethanal cyanohydrin contains a carbon with four different groups, yet the product of $\mathrm{HCN}$ addition to ethanal does not rotate plane-polarised light. Explain.
1 $\mathrm{HCN+OH^{-}\rightleftharpoons CN^{-}+H_{2}O}$; $\mathrm{CN^{-}}$ attacks the δ+ carbon of $\mathrm{CH_{3}CHO}$, the π electrons go to O, giving $\mathrm{CH_{3}CH(CN)O^{-}}$; this takes $\mathrm{H^{+}}$ from $\mathrm{HCN}$ or water, giving $\mathrm{CH_{3}CH(OH)CN}$. $\mathrm{HCN}$ is a weak acid and itself a poor nucleophile; the base makes the strong nucleophile $\mathrm{CN^{-}}$.
2 $\mathrm{C_{6}H_{5}COCH_{3}\lt CH_{3}COCH_{3}\lt CH_{3}CHO\lt HCHO}$ (resonance with the ring; two groups; one group; none).
3 The adduct is a crystalline solid, soluble in water but not in organic solvents, so it can be separated from non-carbonyl impurities; dilute acid or alkali then releases the pure aldehyde. Propanone reacts more readily: it has small methyl groups, while the two ethyl groups of pentan-3-one hinder the attack and crowd the tetrahedral product.
4 (a) $\mathrm{CH_{3}CH_{2}OH}$, ethanol (1°); (b) $\mathrm{CH_{3}CH(OH)CH_{2}CH_{3}}$, butan-2-ol (2°); (c) $\mathrm{(CH_{3})_{2}C(OH)C_{6}H_{5}}$, 2-phenylpropan-2-ol (3°); (d) $\mathrm{C_{6}H_{5}CH(OH)CH_{3}}$, 1-phenylethan-1-ol (2°).
5 2-Methylbutan-2-ol is $\mathrm{(CH_{3})_{2}C(OH)CH_{2}CH_{3}}$. Either propanone + $\mathrm{C_{2}H_{5}MgBr}$, or butan-2-one + $\mathrm{CH_{3}MgBr}$; then $\mathrm{H_{2}O}$.
6 $\mathrm{CH_{3}CHO+C_{2}H_{5}OH\rightleftharpoons CH_{3}CH(OH)OC_{2}H_{5}}$ (hemiacetal); $\mathrm{+\,C_{2}H_{5}OH\rightleftharpoons CH_{3}CH(OC_{2}H_{5})_{2}+H_{2}O}$ (acetal, 1,1-diethoxyethane). Dry $\mathrm{HCl}$ protonates the carbonyl oxygen, activating it, while the absence of water pushes the equilibrium towards the acetal. Warming the acetal with dilute aqueous acid hydrolyses it back to ethanal.
7 $\mathrm{(CH_{3})_{2}C(OH)CN\xrightarrow{H_{3}O^{+}}(CH_{3})_{2}C(OH)COOH}$, 2-hydroxy-2-methylpropanoic acid, $\mathrm{C_{4}H_{8}O_{3}}$. Propanone has 3 carbons; the $\mathrm{CN}$ carbon becomes the $\mathrm{COOH}$ carbon, so the acid has 4.
8 The carbonyl group is planar and $\mathrm{CN^{-}}$ attacks either face with equal probability, giving equal amounts of the two enantiomers: a racemic mixture, whose rotations cancel.
II
Ammonia derivatives and reduction. Use $\mathrm{C}=12$, $\mathrm{H}=1$, $\mathrm{N}=14$, $\mathrm{O}=16$.
  • 1 Give the products: (a) ethanal + hydroxylamine; (b) propanone + hydrazine; (c) benzaldehyde + semicarbazide.
  • 2 The reaction of a carbonyl compound with an ammonia derivative is slow both in neutral solution and in strongly acidic solution, and fastest at about pH 4. Explain.
  • 3 Find the molecular formula and molar mass of the 2,4-dinitrophenylhydrazone of propanone.
  • 4 Semicarbazide, $\mathrm{H_{2}N{-}NH{-}CO{-}NH_{2}}$, has two $\mathrm{NH_{2}}$ groups, but only one of them reacts with carbonyl compounds. Which one, and why?
  • 5 Give the products: (a) propanal + $\mathrm{NaBH_{4}}$; (b) cyclohexanone + $\mathrm{LiAlH_{4}}$, then $\mathrm{H_{2}O}$; (c) benzaldehyde + $\mathrm{Zn{-}Hg}$/conc. $\mathrm{HCl}$; (d) propanone + $\mathrm{NH_{2}NH_{2}}$, then $\mathrm{KOH}$/ethylene glycol, heat.
  • 6 Write the Clemmensen reduction of butanal.
  • 7 Convert benzene into propylbenzene in two steps.
  • 8 A compound $\mathrm{C_{4}H_{8}O}$ gives an orange precipitate with 2,4-DNP, does not reduce Tollens' reagent, and gives butane on Clemmensen reduction. Identify it.
1 (a) $\mathrm{CH_{3}CH{=}NOH}$, ethanal oxime; (b) $\mathrm{(CH_{3})_{2}C{=}NNH_{2}}$, propanone hydrazone; (c) $\mathrm{C_{6}H_{5}CH{=}NNHCONH_{2}}$, benzaldehyde semicarbazone (each with loss of $\mathrm{H_{2}O}$).
2 Acid protonates the carbonyl oxygen and catalyses both the addition and the loss of water, so neutral solution is slow. In strong acid the nitrogen of $\mathrm{H_{2}N{-}Z}$ is protonated to $\mathrm{H_{3}N^{+}{-}Z}$, which has no lone pair and cannot attack. Weakly acidic solution balances the two.
3 $\mathrm{C_{3}H_{6}O+C_{6}H_{6}N_{4}O_{4}-H_{2}O=C_{9}H_{10}N_{4}O_{4}}$, $M=108+10+56+64=238\ \mathrm{g\,mol^{-1}}$.
4 The $\mathrm{NH_{2}}$ on the hydrazine nitrogen ($\mathrm{H_{2}N{-}NH{-}}$). The other $\mathrm{NH_{2}}$ is attached to $\mathrm{C{=}O}$: its lone pair is delocalised onto the carbonyl oxygen (amide resonance), so it is a very poor nucleophile.
5 (a) $\mathrm{CH_{3}CH_{2}CH_{2}OH}$, propan-1-ol; (b) cyclohexanol; (c) $\mathrm{C_{6}H_{5}CH_{3}}$, methylbenzene; (d) $\mathrm{CH_{3}CH_{2}CH_{3}}$, propane, with $\mathrm{N_{2}}$.
6 $\mathrm{CH_{3}CH_{2}CH_{2}CHO\xrightarrow{Zn{-}Hg,\ conc.\ HCl}CH_{3}CH_{2}CH_{2}CH_{3}+H_{2}O}$.
7 Friedel–Crafts acylation with propanoyl chloride and anhydrous $\mathrm{AlCl_{3}}$ gives $\mathrm{C_{6}H_{5}COCH_{2}CH_{3}}$; Clemmensen (or Wolff–Kishner) reduction gives $\mathrm{C_{6}H_{5}CH_{2}CH_{2}CH_{3}}$.
8 Carbonyl compound (2,4-DNP), not an aldehyde (Tollens'), unbranched four-carbon skeleton (butane): butan-2-one, $\mathrm{CH_{3}COCH_{2}CH_{3}}$.
III
Oxidation and tests. Use $\mathrm{I}=127$.
  • 1 Write the ionic equation for the reaction of ethanal with Tollens' reagent. What is seen?
  • 2 Both ethanal and benzaldehyde reduce Tollens' reagent. Which of them reduces Fehling's solution? Describe what is seen.
  • 3 Which of these give a positive iodoform test? ethanal, propanal, propanone, butan-2-one, pentan-3-one, ethanol, methanol, acetophenone, propan-2-ol, benzaldehyde.
  • 4 What is the maximum mass of iodoform that can be obtained from $1.16\ \mathrm{g}$ of propanone?
  • 5 Give a test to distinguish ethanal from ethanol.
  • 6 Give the products of the haloform reaction of (a) acetophenone with $\mathrm{I_{2}/NaOH}$; (b) 4-methylpent-3-en-2-one with $\mathrm{NaOCl}$.
  • 7 Why are ketones oxidised only under vigorous conditions, and why do the acids formed have fewer carbon atoms than the ketone?
  • 8 A compound $\mathrm{C_{5}H_{10}O}$ gives a 2,4-DNP derivative and a positive iodoform test, but no silver mirror. Which two structures fit? How would Clemmensen reduction tell them apart?
1 $\mathrm{CH_{3}CHO+2[Ag(NH_{3})_{2}]^{+}+3OH^{-}\rightarrow CH_{3}COO^{-}+2Ag+2H_{2}O+4NH_{3}}$: a silver mirror (or grey-black precipitate) forms on warming.
2 Only ethanal: the blue solution gives a red-brown precipitate of $\mathrm{Cu_{2}O}$ on heating. Aromatic aldehydes do not reduce Fehling's solution.
3 Positive: ethanal, propanone, butan-2-one, acetophenone ($\mathrm{CH_{3}CO{-}}$), ethanol and propan-2-ol ($\mathrm{CH_{3}CH(OH){-}}$): six. Negative: propanal, pentan-3-one, methanol, benzaldehyde.
4 1 mol propanone gives 1 mol $\mathrm{CHI_{3}}$. $n=\dfrac{1.16}{58}=0.0200\ \mathrm{mol}$; $M(\mathrm{CHI_{3}})=12+1+3(127)=394\ \mathrm{g\,mol^{-1}}$; mass $=7.88\ \mathrm{g}$.
5 Tollens' reagent: ethanal gives a silver mirror; ethanol does not. (Both give iodoform, so that test does not distinguish them.)
6 (a) $\mathrm{C_{6}H_{5}COONa}$ (sodium benzoate) + $\mathrm{CHI_{3}}$; (b) $\mathrm{(CH_{3})_{2}C{=}CHCOONa}$, sodium 3-methylbut-2-enoate, + $\mathrm{CHCl_{3}}$; the $\mathrm{C{=}C}$ is not attacked.
7 A ketone has no $\mathrm{H}$ on the carbonyl carbon, so it can only be oxidised by breaking a strong $\mathrm{C{-}C}$ bond next to $\mathrm{C{=}O}$, which needs strong oxidants and heat; the chain is split, so each acid fragment is shorter.
8 A methyl ketone $\mathrm{CH_{3}CO{-}C_{3}H_{7}}$: pentan-2-one or 3-methylbutan-2-one. Clemmensen reduction gives pentane from the first and 2-methylbutane from the second (they could be told apart, e.g., by boiling point: $309\ \mathrm{K}$ and $301\ \mathrm{K}$).
IV
α-Hydrogen, aldol, Cannizzaro and ring substitution. Use $\mathrm{C}=12$, $\mathrm{H}=1$, $\mathrm{O}=16$.
  • 1 How many α-hydrogens are there in: ethanal, propanone, butan-2-one, benzaldehyde, 2-methylpropanal, 2,2-dimethylpropanal, acetophenone, cyclohexanone?
  • 2 Give the aldol and the aldol condensation product of butanal, with IUPAC names.
  • 3 Which of these undergo aldol condensation with dilute alkali on their own: methanal, 2-methylpropanal, benzaldehyde, propanone, 2,2-dimethylpropanal?
  • 4 Give the products of the Cannizzaro reaction of (a) methanal; (b) 4-chlorobenzaldehyde, each with concentrated $\mathrm{KOH}$.
  • 5 Propanal and butanal are warmed with dilute $\mathrm{NaOH}$. Give the four α,β-unsaturated aldehydes formed and name them.
  • 6 Give the product of benzaldehyde with acetophenone in dilute alkali on warming (Claisen–Schmidt), with its IUPAC name.
  • 7 Give the product of nitration of benzaldehyde and explain the orientation.
  • 8 Industry. 2-Ethylhexan-1-ol, used to make plasticisers, is manufactured from butanal by aldol condensation followed by complete hydrogenation. Write the steps and find the maximum mass of 2-ethylhexan-1-ol from $1000\ \mathrm{kg}$ of butanal.
1 Ethanal 3; propanone 6; butan-2-one 5 (3 + 2); benzaldehyde 0; 2-methylpropanal 1; 2,2-dimethylpropanal 0; acetophenone 3; cyclohexanone 4.
2 $\mathrm{CH_{3}CH_{2}CH_{2}CH(OH)CH(C_{2}H_{5})CHO}$, 2-ethyl-3-hydroxyhexanal; on heating, $\mathrm{CH_{3}CH_{2}CH_{2}CH{=}C(C_{2}H_{5})CHO}$, 2-ethylhex-2-enal.
3 Only those with α-H: 2-methylpropanal and propanone.
4 (a) $\mathrm{CH_{3}OH}$ and $\mathrm{HCOOK}$; (b) $\mathrm{ClC_{6}H_{4}CH_{2}OH}$ (4-chlorophenyl)methanol and $\mathrm{ClC_{6}H_{4}COOK}$ potassium 4-chlorobenzoate.
5 Self: $\mathrm{CH_{3}CH_{2}CH{=}C(CH_{3})CHO}$ 2-methylpent-2-enal (from propanal) and $\mathrm{CH_{3}CH_{2}CH_{2}CH{=}C(C_{2}H_{5})CHO}$ 2-ethylhex-2-enal (from butanal). Crossed: propanal enolate + butanal, $\mathrm{CH_{3}CH_{2}CH_{2}CH{=}C(CH_{3})CHO}$, 2-methylhex-2-enal; butanal enolate + propanal, $\mathrm{CH_{3}CH_{2}CH{=}C(C_{2}H_{5})CHO}$, 2-ethylpent-2-enal.
6 $\mathrm{C_{6}H_{5}CH{=}CHCOC_{6}H_{5}}$, 1,3-diphenylprop-2-en-1-one (chalcone): acetophenone gives the enolate, benzaldehyde accepts.
7 3-Nitrobenzaldehyde. $\mathrm{{-}CHO}$ withdraws electrons by –R and –I; resonance puts δ+ on the ortho and para carbons, so the meta positions are the least deactivated and are attacked.
8 $\mathrm{2C_{3}H_{7}CHO\xrightarrow{dil.\ NaOH,\ \Delta}C_{3}H_{7}CH{=}C(C_{2}H_{5})CHO+H_{2}O}$; $\ \mathrm{C_{8}H_{14}O+2H_{2}\xrightarrow{Ni}C_{8}H_{18}O}$. 2 mol butanal ($2\times72=144\ \mathrm{g}$) give 1 mol product ($130\ \mathrm{g}$): $1000\times\dfrac{130}{144}=903\ \mathrm{kg}$.
🎯 Exam corner BoardJEE MainJEE AdvancedNEET

E1. Board · 3 marks Give a simple chemical test to distinguish between each pair: (a) propanal and propanone; (b) acetophenone and benzophenone; (c) benzaldehyde and ethanal.

(a) Tollens' reagent: silver mirror with propanal, none with propanone. (1 mark)
(b) Iodoform test ($\mathrm{I_{2}/NaOH}$): yellow $\mathrm{CHI_{3}}$ with acetophenone ($\mathrm{C_{6}H_{5}COCH_{3}}$), none with benzophenone. (1 mark)
(c) Fehling's solution: red-brown $\mathrm{Cu_{2}O}$ with ethanal, none with benzaldehyde (or iodoform: positive for ethanal only). (1 mark)

E2. JEE Main Which compound does not give a positive iodoform test? (A) ethanal  (B) propan-2-ol  (C) pentan-3-one  (D) acetophenone

Idea Look for $\mathrm{CH_{3}CO{-}}$ or $\mathrm{CH_{3}CH(OH){-}}$. Ethanal and acetophenone have $\mathrm{CH_{3}CO}$; propan-2-ol is oxidised to propanone. Pentan-3-one, $\mathrm{C_{2}H_{5}COC_{2}H_{5}}$, has neither.
Answer (C).

E3. NEET Which aldehyde undergoes the Cannizzaro reaction with concentrated $\mathrm{NaOH}$? (A) ethanal  (B) propanal  (C) 2,2-dimethylpropanal  (D) 2-methylpropanal

Idea The Cannizzaro reaction needs an aldehyde with no α-hydrogen. In $\mathrm{(CH_{3})_{3}C{-}CHO}$ the α-carbon carries three methyl groups and no H; the others all have α-H and undergo aldol reactions.
Answer (C).

E4. JEE Advanced One or more correct. Compound X, $\mathrm{C_{8}H_{8}O}$, gives an orange precipitate with 2,4-DNP and a yellow precipitate with $\mathrm{I_{2}/NaOH}$, but does not reduce Tollens' reagent. Clemmensen reduction of X gives ethylbenzene. Which statements are correct? (A) X is acetophenone. (B) The iodoform reaction of X also gives sodium benzoate. (C) X undergoes the Cannizzaro reaction with concentrated $\mathrm{NaOH}$. (D) X can react with benzaldehyde in dilute alkali, acting as the enolate (donor).

Identify Carbonyl (2,4-DNP), not an aldehyde (Tollens'), $\mathrm{CH_{3}CO{-}}$ (iodoform), $\mathrm{C_{6}H_{5}CH_{2}CH_{3}}$ skeleton (Clemmensen): $\mathrm{C_{6}H_{5}COCH_{3}}$ ($\mathrm{C_{8}H_{8}O}$ ✓). (A) ✓.
(B) $\mathrm{C_{6}H_{5}COCH_{3}+3I_{2}+4NaOH\rightarrow C_{6}H_{5}COONa+CHI_{3}+3NaI+3H_{2}O}$ ✓.
(C), (D) (C) ✗: Cannizzaro is a reaction of aldehydes without α-H; X is a ketone with three α-H. (D) ✓: its $\mathrm{CH_{3}}$ group forms an enolate, which adds to benzaldehyde (Claisen–Schmidt), giving 1,3-diphenylprop-2-en-1-one.
Answer (A), (B), (D).

E5. JEE Main Numerical value. $1.45\ \mathrm{g}$ of propanone is warmed with excess iodine and sodium hydroxide. The mass of iodoform formed, assuming complete reaction, is (nearest integer, in g) ______. ($\mathrm{C}=12$, $\mathrm{H}=1$, $\mathrm{O}=16$, $\mathrm{I}=127$.)

Moles $\mathrm{CH_{3}COCH_{3}}$ → $\mathrm{CHI_{3}}$ is 1 : 1. $n=\dfrac{1.45}{58}=0.0250\ \mathrm{mol}$.
Answer $0.0250\times394=9.85\ \mathrm{g}$, nearest integer $\mathbf{10}$.

E6. Board · Case study Read the passage and answer. When ethanal is shaken with dilute sodium hydroxide, two molecules combine to form a compound with both an alcohol and an aldehyde group, which loses water on warming. Benzaldehyde does not react this way on its own, but with concentrated alkali it forms two different products. (a) Write the equations for the reaction of ethanal and name the two organic products. (2 marks) (b) Why does benzaldehyde not undergo this reaction on its own? (1 mark) (c) Name and write the reaction of benzaldehyde with concentrated $\mathrm{NaOH}$. (1 mark)

(a) $\mathrm{2CH_{3}CHO\xrightarrow{dil.\ NaOH}CH_{3}CH(OH)CH_{2}CHO}$, 3-hydroxybutanal (aldol); $\mathrm{CH_{3}CH(OH)CH_{2}CHO\xrightarrow{\Delta}CH_{3}CH{=}CHCHO+H_{2}O}$, but-2-enal.
(b) Benzaldehyde has no α-hydrogen (the $\mathrm{CHO}$ is attached directly to the ring), so it cannot form an enolate ion.
(c) Cannizzaro reaction: $\mathrm{2C_{6}H_{5}CHO+NaOH\rightarrow C_{6}H_{5}CH_{2}OH+C_{6}H_{5}COONa}$ (benzyl alcohol and sodium benzoate).
🔬 Research–Teaching NexusResearch-tutored · optional, about 30 minutes

Research question. Why must bitter cassava be grated, soaked and dried before it is safe to eat, and how thoroughly?

Why it matters. Cassava (tapioca) feeds hundreds of millions of people in Africa, South America and Asia, including Kerala and Tamil Nadu. Its roots store a cyanogenic glucoside, linamarin. When the tissue is crushed, an enzyme splits linamarin into glucose and acetone cyanohydrin, $\mathrm{(CH_{3})_{2}C(OH)CN}$, the cyanohydrin of propanone. In neutral or alkaline conditions, or on warming, the cyanohydrin falls apart into propanone and hydrogen cyanide: the reverse of the addition you met in Rule 8.8. $\mathrm{HCN}$ boils at about $299\ \mathrm{K}$, so it escapes from grated, fermented and sun-dried cassava. Poorly processed bitter cassava has caused cyanide poisoning and a paralytic disease called konzo.

What to do.

  1. Write the equation for the decomposition of acetone cyanohydrin, and explain with the mechanism of Rule 8.6 (run backwards) why base speeds it up.
  2. (Illustrative data, constructed for this exercise.) A batch of bitter cassava roots contains cyanogens equivalent to $300\ \mathrm{mg}$ of $\mathrm{HCN}$ per kg of fresh root. Processing removes $95\%$ of the cyanogens, and $1\ \mathrm{kg}$ of fresh root gives $0.30\ \mathrm{kg}$ of flour. Find the $\mathrm{HCN}$ content of the flour in $\mathrm{mg\,kg^{-1}}$.
  3. International food standards set a limit of $10\ \mathrm{mg}$ total $\mathrm{HCN}$ per kg of cassava flour. Is this flour safe? What percentage of the cyanogens must processing remove for the flour to meet the limit?
  4. How much propanone (in g) is released per kg of fresh root when all the acetone cyanohydrin decomposes? ($\mathrm{H}=1$, $\mathrm{C}=12$, $\mathrm{N}=14$, $\mathrm{O}=16$.)

Tools: A calculator; Rules 8.6 and 8.8; Interactive 1 (choose CN⁻ and imagine the film running backwards).

Hand in: The equations, your three calculations, and a short leaflet (about 80 words) for a village cooperative explaining why each processing step matters.

Going further. Bitter almonds and apricot kernels contain amygdalin, which releases the cyanohydrin of benzaldehyde (mandelonitrile). Find out how much of a risk these foods pose, and compare it with cassava.

Instructor note (the cyanogen content and processing figures are illustrative data, constructed for this exercise; the 10 mg/kg flour limit is the Codex Alimentarius figure for edible cassava flour): $\mathrm{(CH_{3})_{2}C(OH)CN\rightleftharpoons(CH_{3})_{2}CO+HCN}$; base removes the OH proton, and the alkoxide expels $\mathrm{CN^{-}}$ as $\mathrm{C{=}O}$ re-forms. Flour: $300\times0.05/0.30=50\ \mathrm{mg\,kg^{-1}}$, five times the limit, unsafe. Need $300(1-f)/0.30\le10$, so $f\ge99.0\%$. Propanone: $0.300/27=0.0111\ \mathrm{mol}$, $\times58=0.644\ \mathrm{g}$ per kg of fresh root.

🧪 Self-check quiz

0/5 correct

Q1. Which compound is the most reactive towards nucleophilic addition?

Q2. Which compound gives a red-brown precipitate when heated with Fehling's solution?

Q3. The reaction of a ketone with 2,4-dinitrophenylhydrazine is best described as

Q4. Clemmensen reduction of acetophenone gives

Q5. The aldol formed from two molecules of ethanal in dilute alkali is

← 8.1 Aldehydes and Ketones: Structure and Preparation8.3 Carboxylic Acids: Preparation, Acidity and Reactions →