This MCQ module is based on: Properties Reactions
Properties Reactions
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Properties Reactions
8.3 Physical Properties of Aldehydes & Ketones L2
Methanal is a gas at room temperature; ethanal boils at 294 K and the next members are volatile liquids. Aldehydes and ketones have higher boiling points than the corresponding hydrocarbons or ethers of similar molar mass — but lower than alcohols or carboxylic acids of similar molar mass. The reason is the polar C=O group: dipole-dipole attractions raise b.p., but absence of OH means there is no intermolecular hydrogen bonding among aldehyde/ketone molecules.
| Compound | Class | M (g mol⁻¹) | B.p. (K) |
|---|---|---|---|
| n-Butane | Alkane | 58 | 273 |
| Diethyl ether | Ether | 74 | 308 |
| Propanal | Aldehyde | 58 | 322 |
| Acetone (Propan-2-one) | Ketone | 58 | 329 |
| 1-Propanol | Alcohol | 60 | 370 |
| Acetic acid | Acid | 60 | 391 |
8.4 Chemical Reactions of Aldehydes & Ketones L3
8.4.1 Nucleophilic Addition (the keystone reaction)
A nucleophile attacks the electrophilic carbon; the C=O π electrons shift to oxygen, generating a tetrahedral alkoxide intermediate that picks up H⁺.
(a) Addition of HCN — cyanohydrins
Reaction is slow with HCN alone; catalysed by base which generates the more nucleophilic CN⁻.
(b) Addition of sodium hydrogensulphite NaHSO3
White crystalline product separates from solution — handy for purifying aldehydes/methyl ketones; reversed by dilute acid or alkali to regenerate the carbonyl.
(c) Addition of alcohols — hemiacetals & acetals
In presence of dry HCl, an aldehyde adds an alcohol to give a hemiacetal which adds a second mole of alcohol to give an acetal (gem-dialkoxy compound).
Ketones react sluggishly with monohydric alcohols, but with ethylene glycol they form a 5-membered cyclic ketal.
(d) Addition of ammonia & its derivatives
The nitrogen lone pair attacks C=O. After loss of water the product is a C=N compound. Hydroxylamine gives oximes; hydrazine gives hydrazones; phenylhydrazine gives phenylhydrazones; 2,4-DNP gives the famous yellow-orange 2,4-dinitrophenylhydrazones (Brady's test); semicarbazide gives semicarbazones.
| Reagent | Product type | Use |
|---|---|---|
| NH3 | R2C=NH (imine) | Unstable; not used for ID |
| H2N–OH | R2C=N–OH (oxime) | Used for identification, m.p. |
| H2N–NH2 | R2C=N–NH2 (hydrazone) | Wolff-Kishner intermediate |
| C6H5NH–NH2 | R2C=N–NHC6H5 (phenylhydrazone) | Sugars; ID |
| 2,4-(NO2)2C6H3–NHNH2 | 2,4-DNP-hydrazone (yellow-red) | Brady's test for C=O |
| NH2–CO–NHNH2 | R2C=N–NH–CONH2 (semicarbazone) | ID and purification |
8.4.2 Reduction
(a) To alcohols: NaBH4 or LiAlH4 reduces aldehydes to 1° alcohols and ketones to 2° alcohols. H2/Ni, Pt or Pd also work.
(b) To hydrocarbons (C=O → CH2):
- Clemmensen reduction — Zn-Hg / conc. HCl (acidic).
- Wolff-Kishner reduction — NH2NH2, then KOH / ethylene glycol, heat (basic).
8.4.3 Oxidation — the test bench
Aldehydes are oxidised readily even by mild oxidants — the aldehydic C-H makes the difference. Ketones resist mild oxidation; strong oxidants (KMnO4) cleave the C-C bond next to C=O.
(a) Tollens' test — "silver mirror"
A bright silver mirror coats the test tube. Tollens reagent distinguishes aldehydes from ketones.
(b) Fehling's & Benedict's test — "red precipitate"
Fehling A (CuSO4) + Fehling B (alkaline Na-K tartrate) mixed give a deep-blue Cu(II) complex that turns into a brick-red Cu2O precipitate on heating with an aldehyde. Aromatic aldehydes do not give Fehling's test.
(c) Iodoform test — methyl ketones & CH3CH(OH) compounds
Aldehydes/ketones containing a CH3CO– group, and alcohols of the type CH3CH(OH)–, give a yellow precipitate of CHI3 (iodoform) when treated with NaOH + I2.
8.4.4 Reactions due to α-Hydrogen
α-Hydrogens (on the carbon next to C=O) are mildly acidic (pKa ~ 20) because the resulting carbanion is stabilised by resonance into the C=O. This makes possible the aldol condensation.
Aldol Condensation
Two molecules of an aldehyde (or ketone) bearing an α-H combine in dilute alkali to give a β-hydroxy aldehyde (or β-hydroxy ketone) — the aldol. On warming it loses water to give an α,β-unsaturated carbonyl.
aldol but-2-enal
Crossed aldol: Two different carbonyls give four products in general; useful only when one component has no α-H (e.g., HCHO, C6H5CHO) — it can only act as electrophile. Example: PhCHO + CH3COCH3 → PhCH=CH-COCH3.
Cannizzaro reaction
Aldehydes without an α-H (HCHO, C6H5CHO, (CH3)3CCHO) undergo disproportionation with concentrated alkali: one molecule is oxidised, another reduced.
8.4.5 Electrophilic Substitution in Aromatic Aldehydes/Ketones
–CHO and –COR are meta-directing, deactivating groups. Nitration of benzaldehyde or acetophenone gives the meta-isomer as the major product.
Setup: A reaction flask contains ethanal and propanal in dilute NaOH at 0 °C. Both have α-hydrogens.
Four products are possible — the two self-aldols and two crossed aldols:
- Self of ethanal: CH3CH(OH)CH2CHO (3-hydroxybutanal)
- Self of propanal: CH3CH2CH(OH)CH(CH3)CHO (3-hydroxy-2-methylpentanal)
- Crossed (ethanal α → propanal C=O): CH3CH2CH(OH)CH2CHO
- Crossed (propanal α → ethanal C=O): CH3CH(OH)CH(CH3)CHO
The poor selectivity is why crossed aldols are typically useful only when one partner lacks α-H.
Interactive: Carbonyl Test Identifier
Pick a compound; see whether Tollens, Fehling, NaHSO3, 2,4-DNP and iodoform tests are positive.
What product(s) form when (a) HCHO is heated with conc. NaOH, (b) CH3CHO is treated with dilute NaOH?
(a) HCHO has no α-H, so it undergoes Cannizzaro disproportionation: HCOONa + CH3OH.
(b) CH3CHO has α-H. With dilute NaOH it undergoes aldol condensation: 3-hydroxybutanal (and on warming, but-2-enal + H2O).
You need to convert 4-nitroacetophenone to 4-nitroethylbenzene. Which reduction will you use — Clemmensen or Wolff-Kishner? Justify.
Clemmensen uses Zn-Hg / conc. HCl. The strongly acidic medium would protonate / reduce the nitro group. Wolff-Kishner uses basic NaOH/glycol, which leaves the –NO2 intact while reducing C=O → CH2. Choose Wolff-Kishner.
Intext Practice L3
How will you distinguish between (i) propanal and propan-2-one, (ii) acetophenone and benzophenone, (iii) benzaldehyde and acetophenone?
(i) Propanal gives a positive Tollens / Fehling test (silver mirror / brick-red ppt); propan-2-one does not (it is a ketone).
(ii) Acetophenone (CH3COC6H5) is a methyl ketone, gives iodoform with NaOH + I2; benzophenone (C6H5COC6H5) does not.
(iii) Benzaldehyde gives a positive Tollens test (silver mirror); acetophenone does not. Conversely, acetophenone gives the iodoform test, benzaldehyde does not.
Competency-Based Questions
Assertion–Reason Questions
Options: (A) Both A & R true; R correct explanation of A. (B) Both true; R not correct explanation. (C) A true, R false. (D) A false, R true.
A1. Aldehydes are more reactive than ketones in nucleophilic addition.
R1. Alkyl groups in ketones donate electrons by +I and also crowd the carbonyl carbon, reducing both electrophilicity and accessibility.
A2. Benzaldehyde gives Cannizzaro reaction with conc. NaOH.
R2. Benzaldehyde does not possess an α-hydrogen.
A3. Wolff-Kishner reduction is preferred over Clemmensen for nitro-substituted ketones.
R3. Acidic conditions of Clemmensen would protonate / reduce the nitro group.
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