KEY CONCEPTS

1. CLASSIFICATION

Aldehydes — CHO group at end of carbon chain

• HCHO (methanal/formaldehyde)
• CH₃CHO (ethanal/acetaldehyde)

Ketones — C=O between two carbon groups

• CH₃COCH₃ (propanone/acetone)
• CH₃COC₂H₅ (butanone)

Carboxylic Acids — COOH group

• HCOOH (methanoic/formic acid)
• CH₃COOH (ethanoic/acetic acid)

2. IUPAC NOMENCLATURE

Aldehydes: alkanal

• HCHO → Methanal
• CH₃CHO → Ethanal
• CH₃CH₂CHO → Propanal

Ketones: alkan-x-one

• CH₃COCH₃ → Propan-2-one
• CH₃COC₂H₅ → Butan-2-one

Carboxylic acids: alkanoic acid

• HCOOH → Methanoic acid
• CH₃COOH → Ethanoic acid
• CH₃CH₂COOH → Propanoic acid

3. PREPARATION OF ALDEHYDES AND KETONES

From alcohols:
RCH₂OH ==PCC==> RCHO (1° → aldehyde)
R₂CHOH ==K₂Cr₂O₇/H₂SO₄==> R₂C=O (2° → ketone)

Rosenmund reduction:
RCOCl + H₂ ==Pd/BaSO₄==> RCHO + HCl

Stephen's reduction:
RCN ==SnCl₂/HCl, then H₂O==> RCHO

Etard reaction:
C₆H₅CH₃ + CrO₂Cl₂ ---> C₆H₅CHO (benzaldehyde)

Friedel-Crafts acylation (ketone):
C₆H₆ + RCOCl ==anhydrous AlCl₃==> C₆H₅COR + HCl

Ozonolysis:
R-CH=CH-R' ==O₃, then Zn/H₂O==> RCHO + R'CHO

4. PREPARATION OF CARBOXYLIC ACIDS

From alcohols/aldehydes:
RCH₂OH ==KMnO₄==> RCOOH
RCHO ==KMnO₄==> RCOOH

From toluene:
C₆H₅CH₃ ==KMnO₄/H⁺==> C₆H₅COOH

From Grignard + CO₂:
RMgX + CO₂ ==dry ether==> RCOOMgX ==H₃O⁺==> RCOOH

From nitrile:
RCN + H₂O ==H⁺ or OH⁻==> RCOOH + NH₃

5. PHYSICAL PROPERTIES

• Aldehydes/ketones: dipole-dipole interactions (no O-H H-bonding among themselves)
• BP: aldehydes/ketones > alkanes but < alcohols of same MW
• Carboxylic acids: highest BP — form dimers via two O-H---O H-bonds
• RCOOH exists as dimer in non-polar solvents
• Lower members of all three classes are water soluble

6. NUCLEOPHILIC ADDITION — REACTIVITY ORDER

HCHO > CH₃CHO > CH₃COCH₃ > PhCHO > PhCOCH₃ > Ph₂CO

Reasons:

• More alkyl groups → more steric hindrance → less reactive
• Alkyl groups donate electrons → reduce δ+ on C → less reactive
• Aldehydes always more reactive than ketones

7. REACTIONS OF ALDEHYDES AND KETONES

A. Addition Reactions

HCN addition:
RCHO + HCN ---> RCH(OH)CN (cyanohydrin)

NaHSO₃ addition:
RCHO + NaHSO₃ ---> RCH(OH)SO₃Na (white ppt)
(aldehydes + methyl ketones only)

Grignard addition:
RCHO + R'MgX ==then H₂O==> RR'CHOH (2° alcohol)
R₂CO + R'MgX ==then H₂O==> RR'R''COH (3° alcohol)

2,4-DNP test:
RCHO + 2,4-dinitrophenylhydrazine ---> orange/yellow ppt
(confirms C=O group)

Oxime formation:
RCHO + NH₂OH ---> RCH=NOH + H₂O

B. Reduction

To alcohol:
RCHO ==NaBH₄ or LiAlH₄==> RCH₂OH

Clemmensen (acidic medium):
RCHO ==Zn-Hg/conc HCl==> RCH₃ (C=O → CH₂)

Wolff-Kishner (basic medium):
RCHO ==NH₂NH₂, KOH==> RCH₃ (C=O → CH₂)

C. Oxidation

Tollens test:
RCHO + 2[Ag(NH₃)₂]⁺ + 2OH⁻ ---> RCOOH + 2Ag↓ + 4NH₃ + H₂O
(silver mirror — aldehydes only)

Fehling's test:
RCHO + Cu²⁺ (Fehling's) ---> RCOOH + Cu₂O↓ (brick red ppt)
(aliphatic aldehydes only — not benzaldehyde)

D. Aldol Condensation

2CH₃CHO ==dil NaOH==> CH₃CH(OH)CH₂CHO (aldol)
On heating: CH₃CH=CHCHO + H₂O (crotonaldehyde)

E. Cannizzaro Reaction
(only for aldehydes without alpha-H)
2HCHO ==conc NaOH==> CH₃OH + HCOONa
(disproportionation — one oxidised, one reduced)

F. Iodoform Reaction
(CH₃CHO and methyl ketones CH₃COR)
CH₃CHO + I₂ + NaOH ---> CHI₃↓ + HCOONa
(yellow ppt = iodoform)

8. REACTIONS OF CARBOXYLIC ACIDS

A. Acidic nature
RCOOH + NaOH ---> RCOONa + H₂O
RCOOH + NaHCO₃ ---> RCOONa + H₂O + CO₂
(CO₂ evolved — distinguishes RCOOH from phenol)

Acidity order:
CCl₃COOH > HCOOH > CH₃COOH > C₂H₅COOH
(electron-withdrawing groups increase acidity)

B. Derivatives

Acid chloride:
RCOOH + PCl₅ ---> RCOCl + POCl₃ + HCl

Anhydride:
2RCOOH ==P₂O₅==> (RCO)₂O + H₂O

Ester:
RCOOH + R'OH ==conc H₂SO₄==> RCOOR' + H₂O

Amide:
RCOOH + NH₃ ---> RCOONH₄ ==heat==> RCONH₂ + H₂O

C. Reduction
RCOOH ==LiAlH₄==> RCH₂OH
(NaBH₄ does NOT reduce COOH)

D. HVZ Reaction
CH₃COOH + Cl₂ ==red P==> ClCH₂COOH + HCl
(alpha halogenation)

E. Decarboxylation
CH₃COONa + NaOH ==CaO, heat==> CH₄ + Na₂CO₃

9. DISTINCTION TESTS

Tollens:

• Aldehyde → silver mirror
• Ketone → no reaction

Fehling's:

• Aliphatic aldehyde → brick red ppt
• Benzaldehyde + ketones → no reaction

Schiff's:

• Aldehyde → pink/magenta colour
• Ketone → no reaction

Iodoform:

• CH₃CHO, methyl ketones → yellow ppt
• Others → no reaction

10. FORMIC ACID — SPECIAL CASE

HCOOH has both CHO and COOH group.

• Gives Tollens test (silver mirror)
• Gives Fehling's positive
• Gives CO₂ with NaHCO₃
• Decomposes: HCOOH ==conc H₂SO₄==> CO + H₂O

Q&A — 20 QUESTIONS
CBSE + JEE Mains Mix

Q1 (CBSE 1M)
Increasing order of BP:
CH₃CH₃, CH₃CHO, CH₃CH₂OH

Ans: CH₃CH₃ < CH₃CHO < CH₃CH₂OH
Alkane (no H-bond) < aldehyde (dipole-dipole) < alcohol (O-H H-bonding)

Q2 (JEE MCQ)
Which gives Fehling's positive?
(a) Benzaldehyde
(b) Acetone
(c) Acetaldehyde
(d) Acetophenone

Ans: (c) Acetaldehyde
Aliphatic aldehydes only give Fehling's positive.

Q3 (CBSE 2M)
Why are aldehydes more reactive than ketones toward nucleophilic addition?

Ans:

1. Steric effect: ketones have two alkyl groups → more hindrance → nucleophile blocked.
2. Electronic effect: alkyl groups donate electrons → reduce δ+ on C → less attractive to nucleophile.
   Aldehydes have one alkyl group → less hindrance + more δ+ → more reactive.

Q4 (JEE)
Write Cannizzaro reaction of HCHO with conc NaOH.

Ans:
2HCHO ==conc NaOH==> CH₃OH + HCOONa
Condition: only aldehydes without alpha-H (HCHO, C₆H₅CHO)
One molecule oxidised → formate; one reduced → methanol.

Q5 (CBSE 2M)
Write aldol condensation of acetaldehyde.

Ans:
2CH₃CHO ==dil NaOH==> CH₃CH(OH)CH₂CHO (aldol)
On heating:
CH₃CH(OH)CH₂CHO ---> CH₃CH=CHCHO + H₂O (crotonaldehyde)

Q6 (JEE MCQ)
Rosenmund reduction converts:
(a) Nitrile → aldehyde
(b) Acid chloride → aldehyde
(c) Acid chloride → alcohol
(d) Ester → aldehyde

Ans: (b) Acid chloride → aldehyde
RCOCl + H₂ ==Pd/BaSO₄==> RCHO + HCl

Q7 (CBSE 3M)
Distinguish between:
(i) Ethanal and propanal
(ii) Ethanal and propanone
(iii) Benzaldehyde and acetophenone

Ans:
(i) Iodoform test:
CH₃CHO → yellow ppt (CHI₃)
CH₃CH₂CHO → no iodoform

(ii) Fehling's test:
CH₃CHO → brick red ppt
CH₃COCH₃ → no reaction

(iii) Tollens test:
C₆H₅CHO → silver mirror
C₆H₅COCH₃ → no reaction

Q8 (JEE MCQ)
NaBH₄ reduces:
(a) RCOOH
(b) RCHO
(c) Both
(d) Neither

Ans: (b) RCHO (and R₂CO)
NaBH₄ reduces aldehydes and ketones only.
LiAlH₄ needed to reduce COOH.

Q9 (CBSE 2M)
Write HVZ reaction.

Ans:
CH₃COOH + Cl₂ ==red P==> ClCH₂COOH + HCl
(chloroacetic acid — alpha halogenation)
Red P acts as catalyst by forming PCl₃ in situ.

Q10 (JEE Numerical)
How many aldehyde/ketone isomers for C₃H₆O?

Ans: 2 isomers

1. CH₃CH₂CHO → propanal (aldehyde)
2. CH₃COCH₃ → propanone (ketone)

Q11 (CBSE 2M)
How does formic acid differ from other carboxylic acids?

Ans: HCOOH has CHO group in addition to COOH.

• Gives Tollens test → silver mirror
• Gives Fehling's → brick red ppt
• Decomposes: HCOOH ==conc H₂SO₄==> CO + H₂O
  No other carboxylic acid shows these properties.

Q12 (JEE MCQ)
Which gives iodoform test?
(a) CH₃CH₂CHO
(b) CH₃CHO
(c) C₆H₅CHO
(d) HCHO

Ans: (b) CH₃CHO
CH₃CHO + I₂ + NaOH ---> CHI₃↓ + HCOONa
(methyl ketones also give positive)

Q13 (CBSE 3M)
Explain why:
(i) RCOOH has higher BP than RCHO of similar MW
(ii) Carboxylic acids stronger than phenols
(iii) Cl₃CCOOH stronger acid than CH₃COOH

Ans:
(i) RCOOH forms dimers via two O-H---O H-bonds → very high BP.
RCHO has only dipole-dipole interactions → lower BP.

(ii) RCOO⁻ stabilised by resonance + inductive effect of C=O → very stable → stronger acid.
C₆H₅O⁻ stabilised by ring resonance only → weaker acid.

(iii) Three Cl atoms strongly electron-withdrawing → weaken O-H bond → easier H⁺ release → stronger acid.
CH₃ is electron-donating → harder H⁺ release → weaker acid.

Q14 (JEE)
Clemmensen vs Wolff-Kishner reduction.

Ans:
Clemmensen:
CH₃COCH₃ ==Zn-Hg/conc HCl==> CH₃CH₂CH₃
(acidic medium — for acid-stable compounds)

Wolff-Kishner:
CH₃COCH₃ ==NH₂NH₂, KOH==> CH₃CH₂CH₃
(basic medium — for base-stable compounds)

Both convert C=O → CH₂ under opposite pH conditions.

Q15 (CBSE 2M)
Write Stephen's reduction.

Ans:
RCN ==SnCl₂/HCl, then H₂O==> RCHO
CH₃CN ==SnCl₂/HCl==> CH₃CHO (ethanal)
Nitrile → aldehyde (one carbon retained)

Q16 (JEE MCQ)
Correct acidity order:
(a) HCOOH > CH₃COOH > CCl₃COOH
(b) CCl₃COOH > HCOOH > CH₃COOH
(c) CH₃COOH > HCOOH > CCl₃COOH
(d) CCl₃COOH > CH₃COOH > HCOOH

Ans: (b) CCl₃COOH > HCOOH > CH₃COOH
CCl₃ = strong -I effect → most acidic
CH₃ = +I effect → least acidic

Q17 (CBSE 2M)
How is benzoic acid prepared from toluene?

Ans:
C₆H₅CH₃ ==KMnO₄/H⁺, heat==> C₆H₅COOH + H₂O
Acidic KMnO₄ oxidises methyl group to COOH.

Q18 (JEE Assertion-Reason)
A: Acetaldehyde gives Tollens test but acetone does not.
R: Aldehydes can be oxidised easily; ketones resist mild oxidation.

Ans: Both A and R correct. R correctly explains A.
Aldehyde has H on carbonyl carbon → easily oxidised.
Ketone has no such H → resists mild oxidation.

Q19 (CBSE 3M)
Prepare benzoic acid from benzene.

Ans:
Step 1 — Friedel-Crafts alkylation:
C₆H₆ + CH₃Cl ==anhydrous AlCl₃==> C₆H₅CH₃ (toluene)

Step 2 — Oxidation:
C₆H₅CH₃ ==KMnO₄/H⁺==> C₆H₅COOH (benzoic acid)

Q20 (JEE MCQ)
2,4-DNP test confirms presence of:
(a) -OH group
(b) -COOH group
(c) C=O group
(d) -NH₂ group

Ans: (c) C=O group
Both aldehydes and ketones give orange/yellow ppt with 2,4-dinitrophenylhydrazine.