Chemistry Class 12 Part II…
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Properties Reactions

🎓 Class 12 Chemistry CBSE Theory Ch 8 – Aldehydes, Ketones and Carboxylic Acids ⏱ ~14 min
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આ MCQ મોડ્યુલ આના પર આધારિત છે: Properties Reactions

આ મૂલ્યાંકન આના પર આધારિત હશે: Properties Reactions

મૂલ્યાંકન બનાવવામાં તેમની સામગ્રી સામેલ કરવા ચિત્રો, PDF અથવા Word દસ્તાવેજ અપલોડ કરો.

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.

Solubility: The lower aldehydes and ketones (≤4 C) are miscible with water because the carbonyl O accepts H-bonds from water. Solubility drops sharply beyond four carbons because the hydrophobic alkyl chain dominates. All aldehydes and ketones dissolve in common organic solvents.
Boiling-point comparison (≈ same molar mass ~58–60 g mol⁻¹)
CompoundClassM (g mol⁻¹)B.p. (K)
n-ButaneAlkane58273
Diethyl etherEther74308
PropanalAldehyde58322
Acetone (Propan-2-one)Ketone58329
1-PropanolAlcohol60370
Acetic acidAcid60391

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⁺.

R R' C O δ⁺ δ⁻ Nu:⁻ → R R' C O⁻ Nu (tetrahedral)
Fig. 8.4: Nucleophilic addition — Nu attacks δ⁺ carbon, O gains the π electrons.
Reactivity order (toward Nu): HCHO > CH3CHO > CH3COCH3. Alkyl groups donate electrons (+I) reducing δ⁺ at C, and they also block the approach of nucleophiles (steric hindrance). Aromatic carbonyls are even less reactive due to ring resonance donation.

(a) Addition of HCN — cyanohydrins

R2C=O + HCN  ⇌  R2C(OH)CN  (cyanohydrin)

Reaction is slow with HCN alone; catalysed by base which generates the more nucleophilic CN⁻.

(b) Addition of sodium hydrogensulphite NaHSO3

R2C=O + NaHSO3  ⇌  R2C(OH)SO3Na  (bisulphite addition product)

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).

R-CHO  R'OH, dry HCl→  R-CH(OH)(OR')  R'OH, dry HCl→  R-CH(OR')2 + H2O

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.

Common ammonia-derivative products of R2C=O
ReagentProduct typeUse
NH3R2C=NH (imine)Unstable; not used for ID
H2N–OHR2C=N–OH (oxime)Used for identification, m.p.
H2N–NH2R2C=N–NH2 (hydrazone)Wolff-Kishner intermediate
C6H5NH–NH2R2C=N–NHC6H5 (phenylhydrazone)Sugars; ID
2,4-(NO2)2C6H3–NHNH22,4-DNP-hydrazone (yellow-red)Brady's test for C=O
NH2–CO–NHNH2R2C=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).
R2C=O + NH2NH2 → R2C=N-NH2  KOH/glycol, Δ→  R2CH2 + N2↑

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"

R-CHO + 2 [Ag(NH3)2]⁺ + 3 OH⁻ → R-COO⁻ + 2 Ag↓ + 4 NH3 + 2 H2O

A bright silver mirror coats the test tube. Tollens reagent distinguishes aldehydes from ketones.

(b) Fehling's & Benedict's test — "red precipitate"

R-CHO + 2 Cu²⁺ + 5 OH⁻ → R-COO⁻ + Cu2O↓ (red) + 3 H2O

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.

Tollens — Ag mirror + R-CHO Fehling (blue Cu²⁺) heat → red Cu₂O ppt
Fig. 8.5: Tollens' silver-mirror test (left) and Fehling's blue-to-red colour change (right).

(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.

2 CH3CHO  dil. NaOH→  CH3CH(OH)CH2CHO  Δ→  CH3CH=CHCHO + H2O
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.

2 HCHO + conc. KOH → HCOOK + CH3OH
2 C6H5CHO + conc. NaOH → C6H5COONa + C6H5CH2OH

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.

Activity 8.2 — Predict the Aldol Product

Setup: A reaction flask contains ethanal and propanal in dilute NaOH at 0 °C. Both have α-hydrogens.

Predict: How many aldol products are possible (before dehydration)? Which is the "crossed" product?

Four products are possible — the two self-aldols and two crossed aldols:

  1. Self of ethanal: CH3CH(OH)CH2CHO (3-hydroxybutanal)
  2. Self of propanal: CH3CH2CH(OH)CH(CH3)CHO (3-hydroxy-2-methylpentanal)
  3. Crossed (ethanal α → propanal C=O): CH3CH2CH(OH)CH2CHO
  4. 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.

Choose a compound.
Worked Example 8.3 — Cannizzaro vs Aldol L4

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).

Worked Example 8.4 — Reduction Choice L5

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

Intext 8.2 — Distinguishing tests

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

Q1. Why does acetone have a lower boiling point than ethanol despite having a larger molar mass? L4
Ethanol forms intermolecular hydrogen bonds (O–H···O) while acetone's C=O cannot donate H — only dipole-dipole forces between acetone molecules. The stronger H-bond network in ethanol elevates its b.p. above acetone's.
Q2. Which reagent gives a positive iodoform test? L1
  • (a) Pentan-3-one
  • (b) Benzophenone
  • (c) Acetophenone
  • (d) Hexan-2-ol — pick the one that does NOT give it
(a) Pentan-3-one and (b) Benzophenone do NOT give iodoform. (c) Acetophenone is a methyl ketone (CH3COAr) and DOES give iodoform; (d) Hexan-2-ol has a CH3CH(OH)- group and DOES give iodoform.
Q3. Predict the major product when 2,2-dimethylpropanal is treated with conc. NaOH. L3
It lacks α-H, so Cannizzaro reaction occurs: (CH3)3CCOONa + (CH3)3CCH2OH (sodium 2,2-dimethylpropanoate + 2,2-dimethylpropan-1-ol).
Q4. Why does benzaldehyde fail to give Fehling's test? L4
Fehling's reagent is a milder oxidant (Cu²⁺) than Tollens (Ag⁺) and the resonance-stabilised aromatic aldehyde C–H is harder to oxidise. Aromatic aldehydes give Tollens (positive) but Fehling (negative).
Q5. Design a route from acetone to 4-methylpent-3-en-2-one (mesityl oxide). L6
Self aldol of acetone (Ba(OH)2) gives 4-hydroxy-4-methylpentan-2-one (diacetone alcohol); warming or acid catalysis eliminates water to give 4-methylpent-3-en-2-one (mesityl oxide).

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.

Answer: (A) — both true and R correctly explains A.

A2. Benzaldehyde gives Cannizzaro reaction with conc. NaOH.

R2. Benzaldehyde does not possess an α-hydrogen.

Answer: (A) — both true and R correctly explains A.

A3. Wolff-Kishner reduction is preferred over Clemmensen for nitro-substituted ketones.

R3. Acidic conditions of Clemmensen would protonate / reduce the nitro group.

Answer: (A) — both true and R correctly explains A.

Frequently Asked Questions - Properties Reactions

What is the main concept covered in Properties Reactions?
In NCERT Class 12 Chemistry Chapter 8 (Aldehydes, Ketones and Carboxylic Acids), "Properties Reactions" covers the core chemistry principles and reactions students need for board exam success. The MyAiSchool lesson explains the topic with definitions, structural diagrams, reaction mechanisms, worked examples, and interactive simulations. Key reactions, IUPAC names, and chemical reasoning are highlighted throughout aligned with CBSE 2025-26 syllabus.
How is Properties Reactions useful in real-life or applied chemistry?
Real-life applications of "Properties Reactions" from NCERT Class 12 Chemistry Chapter 8 include drug design, polymer industry, food chemistry, electrochemical cells, fuel cells, dyes/pigments, agrochemicals, and biochemistry. The MyAiSchool lesson links every concept to a tangible industrial or biological example so students see chemistry as a problem-solving framework for the molecular world.
What are the key reactions students should memorize for Properties Reactions?
Key reactions in "Properties Reactions" (NCERT Class 12 Chemistry Chapter 8 Aldehydes, Ketones and Carboxylic Acids) are tabulated in the MyAiSchool reaction map. Students should memorize each reaction with its reagent, conditions, mechanism class (SN1/SN2/E1/E2/electrophilic addition/etc), product, and stereochemistry. The Summary section provides a quick-reference reaction chart for last-minute revision.
How does this part connect to other parts of Chapter 8?
NCERT Class 12 Chemistry Chapter 8 (Aldehydes, Ketones and Carboxylic Acids) is structured so each part builds chemical understanding sequentially. "Properties Reactions" connects to neighbouring parts via shared functional groups, reaction mechanisms, and structural concepts. The MyAiSchool lesson cross-references related concepts with internal links so students can navigate the whole chapter as one connected story rather than disconnected fragments.
What types of CBSE board questions come from Properties Reactions?
CBSE board questions from "Properties Reactions" typically include: (1) 1-mark MCQs on definitions and IUPAC naming, (2) 2-mark short-answer reactions/products, (3) 3-mark mechanism questions, (4) 5-mark long-answer combining mechanism + product + stereochemistry + application. The MyAiSchool lesson tags each Competency-Based Question (CBQ) with Bloom level (L1-L6) so students know how to study for each weight.
How can students use the interactive simulation effectively?
The interactive simulation in the "Properties Reactions" lesson allows students to explore reaction outcomes, predict products, or compare reaction conditions, with live visual feedback. To use it effectively: (1) try every option/configuration, (2) compare with the analytical reasoning, (3) check IUPAC names and structural correctness, (4) test edge cases from worked examples. The simulation reinforces conceptual intuition that pure mechanism memorisation cannot provide.
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Chemistry Class 12 Part II – NCERT (2025-26)
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