CBSE Class 12 Chemistry Aldehydes Ketones Carboxylic Acids Exam Questions

This Class 12 Chemistry hand-out on Aldehydes, Ketones and Carboxylic Acids is arranged from the exam point of view into reasoning questions, IUPAC naming, mechanisms, name reactions, distinguishing tests, conversions and previous years questions. Every question is given with a complete answer and a Teacher's Note, along with the key mechanisms and name reactions.

ALDEHYDES, KETONES AND CARBOXYLIC ACIDS

This particular chapter is divided in the following parts with the exam pont of view.
1. Reasoning
2. IUPAC Naming
3. Mechanisms
4. Name Reactions
5. Conversions

1. REASONINGS

Q1. Dipole moment of Aldehyde/Ketone is more than Alcohols.

Answer:
1. In an aldehyde or ketone the carbonyl group (C=O) has π-electrons, which are loosely held and shift towards the oxygen atom.
2. This shift is much larger than the shift of the σ electrons in the C-O bond of alcohols, so the carbonyl carbon bears a greater positive charge and oxygen a greater negative charge.
3. Hence the dipole moment of aldehydes and ketones is greater than that of alcohols.

Teacher's Note:
a) The carbonyl bond is polarised as C(δ+)=O(δ-) because oxygen is more electronegative than carbon.
b) Always say that the π electrons, not σ electrons, are the ones that shift; that is the key word.

Q2. Aldehydes are more reactive than ketones towards nucleophlic substitution.

Answer:
1. (a) A ketone has two alkyl groups (+I effect), which reduce the positive charge on the carbonyl carbon and make the C=O bond less polar.
2. (b) The two bulky alkyl groups in a ketone cause steric hindrance to the approaching nucleophile, so ketones are less reactive.

Teacher's Note:
a) The reaction meant here is nucleophilic addition; carbonyl compounds do not undergo nucleophilic substitution at the C=O carbon.
b) Mention both the electronic (+I) and the steric reason to earn full marks.
c) Aldehydes have one alkyl group and one small H atom, so both effects are weaker than in ketones.

Q3. Benzaldehyde gives tollen's test but not with fehling solution

Answer:
1. Benzaldehyde is an aromatic aldehyde. Its C=O group is in conjugation with the benzene ring, and the +R effect of the ring reduces the positive charge on the carbonyl carbon, so the aldehyde is less reactive.
2. Fehling's solution is a weak oxidising agent and cannot oxidise such an aromatic aldehyde.
3. Tollens' reagent is a stronger oxidising agent than Fehling's solution, so it oxidises benzaldehyde and gives the silver mirror.

Teacher's Note:
a) The printed key says that the +R effect makes C=O more polar; the correct reason is that the +R effect makes the carbonyl carbon less electrophilic and the aldehyde less reactive.
b) Only aliphatic aldehydes respond to Fehling's test; aromatic aldehydes do not.
c) Do not write that Tollens' reagent is a weak reagent; here it is the stronger of the two.

Q4. p-fluorobenzoic acid is weaker acid than p-chlorobenzoic acid.

Answer:
1. In p-fluorobenzoic acid the +R effect of fluorine (2p orbital of F overlapping well with the 2p orbital of carbon) outweighs its -I effect.
2. In p-chlorobenzoic acid the 3p orbital of Cl overlaps poorly with the 2p orbital of carbon, so the +R effect is weak and the -I effect dominates.
3. Hence p-chlorobenzoic acid is the stronger acid and p-fluorobenzoic acid is the weaker acid.

Teacher's Note:
a) Size matching of the 2p-2p orbitals decides the strength of the +R effect.
b) Do not apply the usual order of -I effect (F more than Cl) blindly in the para position of benzoic acid.

Q5. Why carboxylic acid do not give the characteristics rection of carbonyl group > C = O.

Answer: In a carboxylic acid the lone pair of the oxygen of the OH group is delocalised towards the carbonyl group, as shown by the resonance structures. This makes the carbonyl carbon much less electrophilic than the carbonyl carbon of aldehydes and ketones, so carboxylic acids do not show the typical nucleophilic addition reactions.

Teacher's Note:
a) Resonance in the -COOH group reduces the positive charge on the carbonyl carbon.
b) Write the resonance structures of the -COOH group in your answer if the question carries more marks.

Q6. Arrange the following in increasing oder of nucleophilic addition Reaction.
(a) Ethnal, Propanal, Propanone, Butanone
(b) Benzalehyde, p-tolualdehyde, p-nitrobenzaldehyde, Acetophenone

Answer:
1. (a) Butanone < Propanone < Propanal < Ethanal (due to +I effect and steric hindrance of the alkyl groups)
2. (b) Acetophenone < p-tolualdehyde < Benzaldehyde < p-nitrobenzaldehyde (the electron withdrawing -NO₂ group increases reactivity and the electron donating -CH₃ group decreases it)

Teacher's Note:
a) The printed key gives the orders in decreasing sequence (Ethanal first), but the question asks for the increasing order, so the sequence is reversed here.
b) Electron withdrawing groups (EWG) raise the positive charge on the carbonyl carbon; electron donating groups (EDG) lower it.
c) Ketones are always less reactive than aldehydes, and acetophenone is the least reactive because of the +R effect of the ring plus the methyl group.

Q7. Arrange in acidic character.
(a) CH₂COOH < FCH₂COOH (-I effect)
(b) FCH₂COOH > ClCH₂COOH (F is more EWG)
(c) FCH₂CH₂CH₂COOH < CH₃CHFCOOH (EWG is nearer)

Answer:
1. (a) CH₃COOH < FCH₂COOH, because the -I effect of fluorine stabilises the carboxylate ion.
2. (b) FCH₂COOH > ClCH₂COOH, because fluorine is more electronegative and a stronger electron withdrawing group than chlorine.
3. (c) FCH₂CH₂CH₂COOH < CH₃CHFCOOH, because the electron withdrawing fluorine is nearer to the -COOH group in the second acid and its -I effect falls with distance.

Teacher's Note:
a) In part (a) the printed formula CH₂COOH should read CH₃COOH, the acetic acid that is being compared.
b) The -I effect decreases rapidly as the distance of the substituent from the -COOH group increases.
c) Remember the rule: more stable carboxylate ion means a stronger acid.

Define :

Cyanohydrin :
>C=O + HCN → (pH 9-10) >C(OH)(CN) → cyanohydrin

Answer: A cyanohydrin is the addition product formed when an aldehyde or ketone reacts with hydrogen cyanide: >C=O + HCN → (pH 9-10) >C(OH)(CN). It has an OH group and a CN group on the same carbon.

Teacher's Note:
a) The reaction is carried out at pH 9-10 so that enough CN^- ion, the real nucleophile, is present.
b) Cyanohydrins are useful because hydrolysis of the CN group gives hydroxy acids.

Semicarbazone :
>C=O + H₂N-NHCONH₂ → (pH 3-5) >C=N-NHCONH₂

Answer: A semicarbazone is the derivative formed when an aldehyde or ketone reacts with semicarbazide: >C=O + H₂N-NHCONH₂ → (pH 3-5) >C=N-NHCONH₂ + H₂O.

Teacher's Note:
a) The reaction needs a weakly acidic medium, pH 3-5.
b) Semicarbazones are crystalline solids, used to identify the carbonyl compound.

Hemiacetal :
>C=O + H-OCH₃ → (Dry HCl) >C(OH)(OCH₃)

Answer: A hemiacetal is the compound with one OH group and one OR group on the same carbon, formed when an aldehyde reacts with one molecule of an alcohol in presence of dry HCl: >C=O + H-OCH₃ → (Dry HCl) >C(OH)(OCH₃).

Teacher's Note:
a) Hemiacetals are unstable; with a second molecule of alcohol they give an acetal.
b) The prefix 'hemi' means half, because only one OR group is present.

Ketal :
figure

Answer: A ketal is the product formed when a ketone reacts with two molecules of a monohydric alcohol, or with one molecule of ethylene glycol, in the presence of dry HCl gas. It has two OR groups on the same carbon: R₂C=O + 2R'OH → (dry HCl) R₂C(OR')₂ + H₂O. With ethylene glycol, a cyclic ketal (five-membered ring) is formed.

Teacher's Note:
a) A ketal is the ketone analogue of an acetal.
b) Ketals are stable to bases and are used to protect the carbonyl group.

2, 4-dNP :
figure

Answer: 2,4-DNP is 2,4-dinitrophenylhydrazine. It reacts with the carbonyl group of aldehydes and ketones to give the 2,4-dinitrophenylhydrazone, an orange, yellow or red crystalline solid: >C=O + H₂N-NH-C₆H₃(NO₂)₂ → >C=N-NH-C₆H₃(NO₂)₂ + H₂O.

Teacher's Note:
a) The coloured precipitate is the standard test for a carbonyl group.
b) It does not distinguish between an aldehyde and a ketone.

Acetal :
figure

Answer: An acetal is the product formed when an aldehyde reacts with two molecules of a monohydric alcohol in the presence of dry HCl gas: R-CHO + 2R'OH → (dry HCl) R-CH(OR')₂ + H₂O. It has two OR groups on the same carbon.

Teacher's Note:
a) The hemiacetal is the intermediate in the formation of an acetal.
b) Acetals are hydrolysed back to the aldehyde by dilute acid.

Aldol :

Answer: An aldol is a β-hydroxy aldehyde (or β-hydroxy ketone, called a ketol) formed when two molecules of an aldehyde or ketone containing α-hydrogen react in the presence of dilute alkali: 2CH₃CHO → (dil. NaOH) CH₃-CH(OH)-CH₂-CHO (3-hydroxybutanal).

Teacher's Note:
a) The name aldol comes from ALDehyde and alcOHOL.
b) On heating, an aldol loses water to give an α,β-unsaturated carbonyl compound.

Schiff base :
>C=O + H₂-NR → (H⁺) >C=N-R

Answer: A Schiff base (substituted imine) is the compound formed by the reaction of an aldehyde or ketone with a primary amine in the presence of an acid: >C=O + H₂N-R → (H⁺) >C=N-R + H₂O.

Teacher's Note:
a) The C=N bond is the characteristic feature of a Schiff base.
b) Only primary amines (R-NH₂) give Schiff bases.

Q1. Arrange the following:
(a) Acetaldehyde, acetone, di-tert-butylketone (Reactivity towards HCN)
(b) CH₃CH₂ CH(Br)COOH, CH₃CH(Br)CH₂COOH, (CH₃)₂CHCOOH, CH₃CH₂COOH (Acid Strength)
(c) Benzoic Acid, 4-nitrobenzoic acid, 3, 4-Dinitrobenzoic acid, 4-methoxybenzoic acid (Acid strength)

Answer:
1. (a) CH₃CHO > (CH₃)₂CO > (CH₃)₃C-CO-C(CH₃)₃ (decreasing reactivity; explained by the +I effect and steric hindrance)
2. (b) CH₃CH₂CH(Br)COOH > CH₃CH(Br)CH₂COOH > CH₃CH₂COOH > (CH₃)₂CHCOOH (decreasing acid strength; the electron donating alkyl groups lower the acidity)
3. (c) 3,4-Dinitrobenzoic acid > 4-nitrobenzoic acid > benzoic acid > 4-methoxybenzoic acid (decreasing acid strength; electron withdrawing groups raise the acidity and electron donating groups lower it)

Teacher's Note:
a) The printed key for (a) lists a methyl tert-butyl ketone that is not in the question; the order above uses only the three compounds asked.
b) In (b) the nearer the Br atom to the -COOH group, the stronger the acid.
c) Each additional EWG on the ring adds to the stabilisation of the benzoate ion.

Q2. Cyclohexanone forms cyanohydrin in good yield but 2. 2, 6 trimethylcyclohexanone does not.

Answer:
1. In 2,2,6-trimethylcyclohexanone the three methyl groups are electron donating (+I effect) groups, which reduce the positive charge on the carbonyl carbon.
2. The three methyl groups next to the carbonyl group also cause steric hindrance to the attack of the CN^- nucleophile.
3. So cyclohexanone gives the cyanohydrin in good yield, but 2,2,6-trimethylcyclohexanone does not.

Teacher's Note:
a) Two reasons earn full marks: the +I effect (electronic) and steric hindrance (spatial).
b) The methyl groups sit on the carbons next to C=O, where they block the path of the nucleophile.

Q3. There are two NH₂ group in semicarbazide. however only one i.e., involving in teh formation of semicarbazone.

Answer: The NH₂ group attached to the C=O group is involved in resonance with the carbonyl group, so its lone pair of electrons is delocalised and it is not available for nucleophilic attack. The NH₂ group attached to the NH group is not in resonance, so its lone pair is available and only this NH₂ group takes part in the formation of the semicarbazone.

Teacher's Note:
a) Resonance: H₂N-C(=O)-NH-NH₂ ↔ ⁺H₂N=C(O^-)-NH-NH₂.
b) A lone pair that is delocalised cannot act as a nucleophile.

Q4. In estrification process, the water or the ester formed should be removed.

Answer: Esterification is a reversible reaction: RCOOH + R'OH ⇌ RCOOR' + H₂O. To shift the equilibrium in the forward direction (Le Chatelier's principle), the water or the ester should be removed as soon as it is formed.

Teacher's Note:
a) Concentrated H₂SO₄ is also used because it acts as a dehydrating agent and a catalyst.
b) Using excess of the alcohol also shifts the equilibrium towards the ester.

Q5. Although phenoxide has more number of resonating structure still it is low acidic then carboxylate ion.

Answer: In the phenoxide ion the negative charge is delocalised over the carbon atoms of the ring, which are less electronegative, and only one of the resonating structures has the charge on oxygen. In the carboxylate ion the negative charge is shared equally between two electronegative oxygen atoms, in two equivalent structures. So the carboxylate ion is more stable than the phenoxide ion, and carboxylic acids are stronger acids than phenols.

Teacher's Note:
a) The number of resonating structures is not the only factor; the electronegativity of the atom carrying the charge matters.
b) Equivalent resonating structures give the carboxylate ion extra stability.

Q6. Why does benzoic acid not undergo friedel crafts Reaction.

Answer: The -COOH group is a strongly electron withdrawing group (-R effect), which deactivates the benzene ring towards electrophilic substitution. In addition, the anhydrous AlCl₃ catalyst gets attached to the -COOH group (the lone pair on oxygen coordinates with AlCl₃), which further deactivates the ring. So benzoic acid does not undergo the Friedel-Crafts reaction.

Teacher's Note:
a) Friedel-Crafts reactions fail on rings that carry strong deactivating groups such as -COOH, -NO₂ and -CHO.
b) The catalyst AlCl₃ is a Lewis acid and reacts with the Lewis base sites of the -COOH group.

2. MECHANISMS

Mechanism of nucleophilic addition reactions:
Step 1 (slow step): The nucleophile (Nu^-) attacks the electrophilic carbonyl carbon of the planar >C(δ+)=O(δ-) group. The carbon changes from sp² to sp³ and a tetrahedral intermediate (with O^-) is formed.
Step 2 (fast step): The negatively charged oxygen of the tetrahedral intermediate takes up H⁺ to give the addition product, >C(Nu)(OH).

[Figure: Planar carbonyl group with substituents shown as wedge and hash bonds and partial charges δ+ on C and δ- on O; slow step arrow to the Tetrahedral intermediate with Nu and O^- on the carbon; fast step arrow with H⁺ to the Addition Product with Nu and OH on the carbon]

Mechanism of esterification of carboxylic acids:
Step 1: The carbonyl oxygen of the carboxylic acid is protonated by H⁺ (R-C(=O)OH + H⁺ ⇌ R-C(=O⁺H)OH).
Step 2: The alcohol (R'-OH) attacks the protonated carbonyl carbon and a tetrahedral intermediate, R-C(OH)₂-O⁺(H)R', is formed.
Step 3: Proton transfer takes place from the oxygen carrying the R' group to one of the OH groups.
Step 4: Loss of a water molecule (-HOH) gives the protonated ester, R-C(=O⁺H)-OR'.
Step 5: Loss of H⁺ (-H⁺) gives the ester, R-C(=O)-OR'.

[Figure: Reaction scheme of esterification - Carboxylic acid ⇌ (H⁺) protonated acid ⇌ (R'-OH) Tetrahedral intermediate ⇌ (Proton transfer) intermediate with OH₂⁺ → (-HOH) Protonated ester → (-H⁺) Ester]

3. NAME REACTIONS

1. Rosenmund reduction:
Benzoyl chloride (C₆H₅COCl) → (H₂, Pd-BaSO₄) Benzaldehyde (C₆H₅CHO)
2. Stephen reaction:
RCN + SnCl₂ + HCl → RCH=NH → (H₃O⁺) RCHO
3. Etard reaction:
Toluene + CrO₂Cl₂ → (CS₂) Chromium complex, C₆H₅CH(OCrOHCl₂)₂ → (H₃O⁺) Benzaldehyde
4. Gatterman-Koch:
Benzene → (CO, HCl; Anhyd. AlCl₃/CuCl) Benzaldehyde
5. Friedel-Crafts acylation reaction:
Benzene + Ar/R-CO-Cl → (Anhyd. AlCl₃) C₆H₅-CO-Ar/R
6. Clemmensen Reduction:
>C=O → (Zn-Hg, HCl) >CH₂ + H₂O
7. Wolff Kishnar Reduction:
>C=O → (NH₂NH₂, -H₂O) >C=NNH₂ → (KOH/ethylene glycol, heat) >CH₂ + N₂
8. Aldol Condensation:
2CH₃-CHO (Ethanal) ⇌ (dil. NaOH) CH₃-CH(OH)-CH₂-CHO (3-Hydroxybutanal, Aldol) → (Δ, -H₂O) CH₃-CH=CH-CHO (But-2-enal, Aldol condensation product)
9. Cannizzaro Reaction:
HCHO + HCHO + conc. KOH → CH₃OH (Methanol) + HCOOK (Potassium formate)
10. Hell-Volhard-Zelinsky reaction:
R-CH₂-COOH → ((i) X₂/Red phosphorus, (ii) H₂O) R-CH(X)-COOH, X = Cl, Br (α-Halocarboxylic acid)

Distinguish Between two compounds:

1. Test for carbonyl group (-C=O):
(a) 2,4 DNP test: Any carbonyl group reacts with 2,4 DNP to give a red, yellow or orange solid.
>C=O + H₂N-NH-C₆H₃(NO₂)₂ → >C=N-NH-C₆H₃(NO₂)₂ (Orange dye)
(b) Addition of sodium hydrogensulphite:
>C=O + NaHSO₃ ⇌ >C(OH)(SO₃Na) (proton transfer; White Crystalline)

2. Test For Aldehyde:
(a) Tollen's Reagent: this test is given by all the aliphatic, aromatic aldehyde and formic acid due to strong oxidising agent.
RCHO + 2[Ag(NH₃)₂]⁺ + 3OH^- → RCOO^- + 2Ag + 2H₂O + 4NH₃
(b) Fehling Reagents: this test is given by all the aliphatic aldehyde due to weak oxidising agent.
RCHO + 2Cu²⁺ + 5OH^- → RCOO^- + Cu₂O + 3H₂O (Red-brown ppt)
(c) Haloform reaction: It is given by ethanol, all the secondary alcohol with OH group at second carbon (CH₃CHOH-) and a carbonyl group with CH₃-C=O.
R-CO-CH₃ → (NaOX) R-COONa + CHX₃ [X = Cl, Br, I]

3. Test for Carboxylic Acid:
They react with NaHCO₃ to give CO₂ gas.
RCOOH + NaHCO₃ → RCOONa + CO₂ + H₂O

Q1. An organic compound (A) with molecular formula C₈H₈O forms an orange-red precipitate with 2,4-DNP reagent and gives yellow precipitate on heating with iodine in the presence of sodium hydroxide. It neither reduces Tollens' or Fehlings' reagent, nor does it decolourise bromine water or Baeyer's reagent. On drastic oxidation with chromic acid, it gives a carboxylic acid (B) having molecular formula C₇H₆O₂. Identify the compounds (A) and (B) and explain the reactions involved.

Answer:
1. Compound (A) is acetophenone, C₆H₅COCH₃, and compound (B) is benzoic acid, C₆H₅COOH.
2. Orange-red precipitate with 2,4-DNP shows a carbonyl group: C₆H₅COCH₃ + H₂N-NH-C₆H₃(NO₂)₂ → C₆H₅C(CH₃)=N-NH-C₆H₃(NO₂)₂ + H₂O
3. Yellow precipitate with I₂/NaOH is the iodoform test, which shows the CH₃-CO- group: C₆H₅COCH₃ + 3I₂ + 4NaOH → C₆H₅COONa + CHI₃ + 3NaI + 3H₂O
4. No reduction of Tollens' or Fehling's reagent shows it is a ketone and not an aldehyde; no decolourisation of bromine water or Baeyer's reagent shows there is no C=C bond (the side chain is not unsaturated).
5. Drastic oxidation: C₆H₅COCH₃ → ([O], chromic acid) C₆H₅COOH (C₇H₆O₂), benzoic acid.

Teacher's Note:
a) C₈H₈O has 5 degrees of unsaturation: 4 for the benzene ring and 1 for the C=O group.
b) A positive iodoform test with a ketone points to a methyl ketone.
c) Check the formula of B: C₇H₆O₂ is benzoic acid, which confirms the ring is attached directly to the carbonyl carbon.
d) State the name of every reaction to show the examiner each clue is used.

Q2. An organic compound with the molecular formula C₉H₁₀O forms 2,4-DNP derivative, reduces Tollens' reagent and undergoes Cannizzaro reaction. On vigorous oxidation, it gives 1,2-benzenedicarboxylic acid. Identify the compound.

Answer:
1. The compound is 2-ethylbenzaldehyde (o-ethylbenzaldehyde), o-C₂H₅-C₆H₄-CHO.
2. It forms a 2,4-DNP derivative and reduces Tollens' reagent, so it is an aldehyde.
3. It undergoes the Cannizzaro reaction, so the CHO group has no α-hydrogen; it is attached directly to the benzene ring.
4. On vigorous oxidation both side groups (-CHO and -C₂H₅) change to -COOH and give 1,2-benzenedicarboxylic acid (phthalic acid), which shows that the two groups are in the ortho position.

Teacher's Note:
a) Formula check: C₆H₄ + C₂H₅ + CHO = C₉H₁₀O.
b) Any alkyl side chain on a benzene ring is oxidised to -COOH, whatever its length.
c) Do not choose a phenyl ketone; a ketone would not reduce Tollens' reagent.

Q3. An organic compound (A) (molecular formula C₈H₁₆O₂) was hydrolysed with dilute sulphuric acid to give a carboxylic acid (B) and an alcohol (C). Oxidation of (C) with chromic acid produced (B). (C) on dehydration gives but-1-ene. Write equations for the reactions involved.

Answer:
1. (A) is butyl butanoate, CH₃CH₂CH₂COOCH₂CH₂CH₂CH₃; (B) is butanoic acid; (C) is butan-1-ol.
2. Hydrolysis: CH₃CH₂CH₂COOC₄H₉ + H₂O → (dil. H₂SO₄) CH₃CH₂CH₂COOH (B) + CH₃CH₂CH₂CH₂OH (C)
3. Oxidation: CH₃CH₂CH₂CH₂OH → (chromic acid) CH₃CH₂CH₂COOH (B)
4. Dehydration: CH₃CH₂CH₂CH₂OH → (conc. H₂SO₄, heat) CH₃CH₂CH=CH₂ (but-1-ene) + H₂O

Teacher's Note:
a) Oxidation of C gives B, so the acid and the alcohol have the same number of carbon atoms (4 each).
b) Dehydration to but-1-ene (not but-2-ene) shows that the OH group is on the end carbon, so C is butan-1-ol.
c) Formula check: C₄H₇O₂ + C₄H₉ = C₈H₁₆O₂.

Q4. Give simple chemical tests to distinguish between the following pairs of compounds.
(i) Propanal and Propanone (ii) Acetophenone and Benzophenone
(iii) Phenol and Benzoic acid (iv) Benzoic acid and Ethyl benzoate
(v) Pentan-2-one and Pentan-3-one (vi) Benzaldehyde and Acetophenone
(vii) Ethanal and Propanal

Answer:
1. (i) Propanal gives a silver mirror with Tollens' reagent (and a red-brown precipitate with Fehling's solution); propanone does not.
2. (ii) Acetophenone (a methyl ketone) gives a yellow precipitate of iodoform with I₂/NaOH (NaOI); benzophenone does not.
3. (iii) Benzoic acid gives brisk effervescence of CO₂ with sodium hydrogencarbonate; phenol does not (phenol gives a violet colour with neutral FeCl₃).
4. (iv) Benzoic acid gives brisk effervescence of CO₂ with sodium hydrogencarbonate; ethyl benzoate does not.
5. (v) Pentan-2-one (a methyl ketone) gives a yellow precipitate of iodoform with I₂/NaOH; pentan-3-one does not.
6. (vi) Benzaldehyde gives a silver mirror with Tollens' reagent; acetophenone does not (acetophenone gives the iodoform test; benzaldehyde does not).
7. (vii) Ethanal gives a yellow precipitate of iodoform with I₂/NaOH; propanal does not.

Teacher's Note:
a) The iodoform test works for compounds with the CH₃-CO- group, or CH₃-CH(OH)- group.
b) Choose one clear test for each pair and state what is seen with each compound.
c) Only the aldehyde in a pair responds to Tollens' reagent; ketones never do.
d) Never use the 2,4-DNP test for such pairs, since both members give a coloured product.

Q5. An organic compound contains 69.77% carbon, 11.63% hydrogen and rest oxygen. The molecular mass of the compound is 86. It does not reduce Tollens' reagent but forms an addition compound with sodium hydrogensulphite and give positive iodoform test. On vigorous oxidation it gives ethanoic and propanoic acid. Write the possible structure of the compound.

Answer:
1. Percentage of oxygen = 100 - (69.77 + 11.63) = 18.60%
2. Number of C atoms = \( \frac{69.77 \times 86}{12 \times 100} \) = 5.0
3. Number of H atoms = \( \frac{11.63 \times 86}{1 \times 100} \) = 10.0
4. Number of O atoms = \( \frac{18.60 \times 86}{16 \times 100} \) = 1.0, so the molecular formula is C₅H₁₀O.
5. It does not reduce Tollens' reagent but adds NaHSO₃, so it is a ketone; the positive iodoform test shows it is a methyl ketone.
6. On vigorous oxidation it gives ethanoic acid (CH₃COOH) and propanoic acid (CH₃CH₂COOH), so the compound is pentan-2-one, CH₃-CO-CH₂CH₂CH₃.
7. Oxidation: CH₃COCH₂CH₂CH₃ → ([O]) CH₃COOH + CH₃CH₂COOH

Teacher's Note:
a) Degrees of unsaturation of C₅H₁₀O is 1, which is accounted for by the C=O group.
b) Pentan-3-one is rejected because it would give propanoic acid only and no iodoform test.
c) Use the percentage and molecular mass to get the number of atoms first, then use the tests.
d) Check: mass of C₅H₁₀O = 60 + 10 + 16 = 86.

4. CONVERSIONS

Q1. Give names of the reagents to bring about the following transformations:
(i) Hexan-1-ol to hexanal
(ii) Cyclohexanol to cyclohexanone
(iii) p-Fluorotoluene to p-fluorobenzaldehyde
(iv) Ethanenitrile to ethanal
(v) Allyl alcohol to propenal
(vi) But-2-ene to ethanal

Answer:
1. (i) Hexan-1-ol to hexanal: C₅H₅NH⁺CrO₃Cl^- (pyridinium chlorochromate, PCC)
2. (ii) Cyclohexanol to cyclohexanone: K₂Cr₂O₇ in acidic medium (or CrO₃)
3. (iii) p-Fluorotoluene to p-fluorobenzaldehyde: CrO₃ in the presence of acetic anhydride, followed by hydrolysis; or 1. CrO₂Cl₂ 2. H₂O (Etard reaction)
4. (iv) Ethanenitrile to ethanal: (Diisobutyl)aluminium hydride (DIBAL-H), followed by hydrolysis; or SnCl₂/HCl then H₃O⁺ (Stephen reaction)
5. (v) Allyl alcohol to propenal: PCC (pyridinium chlorochromate), which does not attack the C=C bond
6. (vi) But-2-ene to ethanal: O₃ then H₂O/Zn dust (ozonolysis)

Teacher's Note:
a) PCC is a mild oxidising agent that stops at the aldehyde stage and does not touch a C=C bond.
b) In (iii) a strong oxidising agent would give p-fluorobenzoic acid, so the aldehyde must be protected as in the CrO₃/acetic anhydride method.
c) Hydrolysis with Zn dust in ozonolysis prevents further oxidation of ethanal.
d) Write the reagent name along with the formula, as the question asks for names of the reagents.

Q2. Write the structures of products of the following reactions.
(i) Benzene + C₂H₅COCl → (Anhyd. AlCl₃, CS₂)
(ii) (C₆H₅CH₂)₂Cd + 2CH₃COCl →
(iii) H₃C-C≡C-H → (Hg²⁺, H₂SO₄)
(iv) p-Nitrotoluene (CH₃ and NO₂ para on the benzene ring) → ((i) CrO₂Cl₂, (ii) H₃O⁺)

Answer:
1. (i) C₆H₆ + C₂H₅COCl → (Anhyd. AlCl₃, CS₂) C₆H₅-CO-C₂H₅ (Propiophenone) + HCl
2. (ii) (C₆H₅CH₂)₂Cd + 2CH₃COCl → 2C₆H₅CH₂-CO-CH₃ (1-Phenylpropan-2-one) + CdCl₂
3. (iii) H₃C-C≡C-H + H₂O → (Hg²⁺, H₂SO₄) [H₃C-C(OH)=CH₂] → (tautomerises) H₃C-CO-CH₃ (Propanone)
4. (iv) p-O₂N-C₆H₄-CH₃ → (CrO₂Cl₂) p-O₂N-C₆H₄-CH(OCrCl₂OH)₂ → (H₃O⁺) p-O₂N-C₆H₄-CHO (p-Nitrobenzaldehyde)

Teacher's Note:
a) The printed key writes CaCl₂ in (ii) and names the product 1-phenylpropane; the by-product is cadmium chloride (CdCl₂) and the product is 1-phenylpropan-2-one.
b) In (iii) the enol form is unstable and changes to the keto form (propanone) by tautomerism; the hydration follows Markovnikov's rule.
c) Reaction (iv) is the Etard reaction, which stops at the aldehyde.
d) An acyl chloride with a dialkylcadmium reagent gives a ketone.

Q3. Predict the products of the following reactions:
(i) Cyclopentanone + HO-NH₂ → (H⁺)
(ii) Cyclohexanone + NH₂-NH-C₆H₃(NO₂)₂ (2,4-dinitrophenylhydrazine) →
(iii) R-CH=CH-CHO + NH₂-CO-NH-NH₂ → (H⁺)
(iv) Acetophenone (C₆H₅COCH₃) + CH₃CH₂NH₂ → (H⁺)

Answer:
1. (i) Cyclopentanone oxime: the ring carbon is joined to =N-OH.
2. (ii) Cyclohexanone 2,4-dinitrophenylhydrazone: the ring carbon is joined to =N-NH-C₆H₃(NO₂)₂.
3. (iii) R-CH=CH-CH=N-NH-CO-NH₂ (semicarbazone of the aldehyde).
4. (iv) C₆H₅-C(CH₃)=N-CH₂CH₃ (a Schiff base, the substituted imine).

Teacher's Note:
a) All four are condensation reactions in which the C=O oxygen is lost as water: >C=O + H₂N-Z → >C=N-Z + H₂O.
b) The C=C bond in (iii) is not affected; the reagent attacks only the carbonyl group.

Q4. Write chemical reactions to affect the following transformations:
(i) Butan-1-ol to butanoic acid
(ii) Benzyl alcohol to phenylethanoic acid
(iii) 3-Nitrobromobenzene to 3-nitrobenzoic acid
(iv) 4-Methylacetophenone to benzene-1,4-dicarboxylic acid
(v) Cyclohexene to hexane-1,6-dioic acid
(vi) Butanal to butanoic acid.

Answer:
1. (i) CH₃CH₂CH₂CH₂OH (Butan-1-ol) → (CrO₃-H₂SO₄, Jones reagent) CH₃CH₂CH₂COOH (Butanoic acid)
2. (ii) C₆H₅CH₂OH → (HBr) C₆H₅CH₂Br → (KCN) C₆H₅CH₂CN → (H₃O⁺, Δ) C₆H₅CH₂COOH (Phenylethanoic acid)
3. (iii) 3-O₂N-C₆H₄-Br → (Mg, ether) 3-O₂N-C₆H₄-MgBr → (CO₂, dry ice) 3-O₂N-C₆H₄-COOMgBr → (H₃O⁺) 3-Nitrobenzoic acid
4. (iv) 4-CH₃-C₆H₄-COCH₃ → (KMnO₄/KOH) Dipotassium benzene-1,4-dicarboxylate (KOOC-C₆H₄-COOK) → (dil. H₂SO₄) Benzene-1,4-dicarboxylic acid (Terephthalic acid)
5. (v) Cyclohexene → (KMnO₄/KOH, heat) Hexane-1,6-dioic acid, HOOC-(CH₂)₄-COOH (Adipic acid)
6. (vi) CH₃CH₂CH₂CHO (Butanal) → (Ammoniacal AgNO₃, Tollens' reagent) CH₃CH₂CH₂COOH (Butanoic acid)

Teacher's Note:
a) The KCN step in (ii) adds one carbon, which is why benzyl alcohol gives phenylethanoic acid.
b) Any alkyl or acyl side chain on a benzene ring is oxidised to -COOH by alkaline KMnO₄, as in (iv).
c) The Grignard route in (iii) is the accepted textbook answer; in practice the nitro group can react with a Grignard reagent, and CuCN followed by hydrolysis of the nitrile is the safer laboratory method.
d) Strong oxidising agents are not needed in (vi) because an aldehyde is oxidised even by mild reagents.

Q5. Show how each of the following compounds can be converted to benzoic acid.
(i) Ethylbenzene (ii) Acetophenone (iii) Bromobenzene (iv) Phenylethene (Styrene)

Answer:
1. (i) C₆H₅CH₂CH₃ → (KMnO₄/KOH, heat) C₆H₅COOK → (H₃O⁺) C₆H₅COOH
2. (ii) C₆H₅COCH₃ → (KMnO₄/KOH, heat) C₆H₅COOK → (H₃O⁺) C₆H₅COOH
3. (iii) C₆H₅Br → (Mg, ether) C₆H₅MgBr → (CO₂, dry ice) C₆H₅COOMgBr → (H₃O⁺) C₆H₅COOH
4. (iv) C₆H₅CH=CH₂ → (KMnO₄/KOH, heat) C₆H₅COOK + HCOOK → (H₃O⁺) C₆H₅COOH

Teacher's Note:
a) The whole side chain, whatever its length, is oxidised to a single -COOH group.
b) In (iii) the Grignard reagent adds to CO₂, which adds one carbon atom.
c) The salt formed in alkaline medium must be acidified to get the free acid.

Q6. Predict the products formed when cyclohexanecarbaldehyde reacts with following reagents.
(i) PhMgBr and then H₃O⁺ (ii) Tollens' reagent
(iii) Semicarbazide and weak acid (iv) Excess ethanol and acid
(v) Zinc amalgam and dilute hydrochloric acid

Answer:
1. (i) Cyclohexyl phenyl carbinol, C₆H₁₁-CH(OH)-C₆H₅ (via the intermediate C₆H₁₁-CH(OMgBr)-Ph)
2. (ii) Cyclohexanecarboxylate ion: C₆H₁₁CHO + 2[Ag(NH₃)₂]⁺ + 3OH^- → C₆H₁₁COO^- + 2Ag↓ + 4NH₃ + 2H₂O (silver mirror)
3. (iii) Cyclohexanecarbaldehyde semicarbazone, C₆H₁₁-CH=N-NH-CO-NH₂
4. (iv) Cyclohexanecarbaldehyde diethyl acetal, C₆H₁₁-CH(OC₂H₅)₂ + H₂O (dry HCl gas)
5. (v) Methylcyclohexane, C₆H₁₁-CH₃ (Clemmensen reduction)

Teacher's Note:
a) The printed key writes 3H₂O in (ii); the balanced equation has 2H₂O (atoms and charge both balance).
b) A Grignard reagent adds to an aldehyde to give a secondary alcohol after hydrolysis.
c) Acetal formation needs excess alcohol and dry HCl, and is a protecting method for aldehydes.
d) Clemmensen reduction converts C=O to CH₂; it is carried out in acid medium.

Q7. Which of the following compounds would undergo aldol condensation, which the Cannizzaro reaction and which neither? Write the structures of the expected products of aldol condensation and Cannizzaro reaction.
(i) Methanal (ii) 2-Methylpentanal (iii) Benzaldehyde (iv) Benzophenone (v) Cyclohexanone (vi) 1-Phenylpropanone (vii) Phenylacetaldehyde (viii) Butan-1-ol (ix) 2,2-Dimethylbutanal

Answer:
1. Aldol condensation: (ii), (v), (vi) and (vii), because each has α-hydrogen.
2. Cannizzaro reaction: (i), (iii) and (ix), because each is an aldehyde with no α-hydrogen.
3. Neither: (iv) and (viii).
4. Aldol product of (ii): 2 CH₃CH₂CH₂CH(CH₃)CHO → CH₃CH₂CH₂CH(CH₃)-CH(OH)-C(CH₃)(CHO)-CH₂CH₂CH₃
5. Aldol product of (v): two cyclohexanone molecules → 2-(1-hydroxycyclohexyl)cyclohexanone, which loses water on heating to give 2-(cyclohexen-1-yl)cyclohexanone.
6. Cannizzaro products: (i) 2HCHO + conc. KOH → CH₃OH + HCOOK; (iii) 2C₆H₅CHO + conc. KOH → C₆H₅CH₂OH + C₆H₅COOK; (ix) 2CH₃CH₂C(CH₃)₂CHO + conc. KOH → CH₃CH₂C(CH₃)₂CH₂OH + CH₃CH₂C(CH₃)₂COOK

Teacher's Note:
a) Aldol condensation needs at least one α-hydrogen; Cannizzaro needs an aldehyde with none.
b) Benzophenone has no α-hydrogen and is not an aldehyde, and butan-1-ol is not a carbonyl compound, so both undergo neither reaction.
c) In the Cannizzaro reaction one molecule is oxidised to the acid salt and the other is reduced to the alcohol.
d) 1-Phenylpropanone (C₆H₅COCH₂CH₃) also has α-hydrogens on its CH₂ group.

Q8. How will you convert ethanal into the following compounds?
(i) Butane-1,3-diol (ii) But-2-enal (iii) But-2-enoic acid

Answer:
1. (i) 2CH₃CHO → (dil. NaOH) CH₃-CH(OH)-CH₂-CHO (3-Hydroxybutanal) → (NaBH₄, reduction) CH₃-CH(OH)-CH₂-CH₂-OH (Butane-1,3-diol)
2. (ii) 2CH₃CHO → (dil. NaOH) CH₃-CH(OH)-CH₂-CHO → (Δ, -H₂O) CH₃-CH=CH-CHO (But-2-enal)
3. (iii) CH₃-CH=CH-CHO → ([Ag(NH₃)₂]⁺ OH^-, Tollens' reagent) CH₃CH=CHCOOH (But-2-enoic acid)

Teacher's Note:
a) All three conversions begin with the aldol condensation of ethanal.
b) Tollens' reagent oxidises the CHO group but leaves the C=C bond untouched, which is why it is used in (iii).

Q9. Write structural formulas and names of four possible aldol condensation products from propanal and butanal. In each case, indicate which aldehyde acts as nucleophile and which as electrophile.

Answer:
1. (i) Two molecules of propanal, one acting as a nucleophile and the other as an electrophile: 2CH₃CH₂CHO → (dil. NaOH) CH₃CH₂-CH(OH)-CH(CH₃)-CHO, 3-Hydroxy-2-methylpentanal.
2. (ii) Two molecules of butanal, one acting as a nucleophile and the other as an electrophile: 2CH₃CH₂CH₂CHO → (dil. NaOH) CH₃CH₂CH₂-CH(OH)-CH(CH₂CH₃)-CHO, 2-Ethyl-3-hydroxyhexanal.
3. (iii) One molecule each of propanal and butanal, in which propanal acts as the nucleophile and butanal acts as the electrophile: CH₃CH₂CH₂CHO + CH₃CH₂CHO → CH₃CH₂CH₂-CH(OH)-CH(CH₃)-CHO, 3-Hydroxy-2-methylhexanal.
4. (iv) One molecule each of propanal and butanal, in which propanal acts as the electrophile and butanal acts as the nucleophile: CH₃CH₂CHO + CH₃CH₂CH₂CHO → CH₃CH₂-CH(OH)-CH(CH₂CH₃)-CHO, 2-Ethyl-3-hydroxypentanal.

Teacher's Note:
a) The nucleophile is the aldehyde that loses its α-hydrogen and attacks through the α-carbon; the electrophile is the aldehyde whose carbonyl carbon is attacked.
b) The printed key gives the names 3-hydroxy-2-methylpentanal for (iii) and 2-ethyl-3-hydroxyhexanal for (iv) with the wrong structures; the correct products and names are as given here (3-hydroxy-2-methylhexanal and 2-ethyl-3-hydroxypentanal).
c) To name an aldol, take the longest chain containing CHO as C1; the OH is on C3 and the substituent from the nucleophile is on C2.
d) Count the carbons of the products as a check: (i) 6, (ii) 8, (iii) 7, (iv) 7.

Q10. Complete each synthesis by giving missing starting material, reagent or products.
(i) Ethylbenzene → (KMnO₄, KOH, heat) ?
(ii) Phthalic acid → (SOCl₂, heat) ?
(iii) C₆H₅CHO + H₂NCONHNH₂ → ?
(iv) Benzene + Benzoyl chloride → (Anhyd. AlCl₃, F.C. acylation) ?
(v) 4-Oxocyclohexanecarbaldehyde → ([Ag(NH₃)₂]⁺) ?
(vi) 2-Formylbenzoic acid → (NaCN/HCl) ?

Answer:
1. (i) Potassium benzoate, C₆H₅COOK
2. (ii) Phthaloyl chloride, 1,2-C₆H₄(COCl)₂
3. (iii) Benzaldehyde semicarbazone, C₆H₅CH=N-NH-CO-NH₂ + H₂O
4. (iv) Benzophenone, C₆H₅-CO-C₆H₅ + HCl
5. (v) 4-Oxocyclohexanecarboxylate anion (the ketone group is not oxidised)
6. (vi) 2-[Cyano(hydroxy)methyl]benzoic acid, the cyanohydrin of the aldehyde group: 2-HOOC-C₆H₄-CH(OH)CN

Teacher's Note:
a) The printed key shows a cyclohexane ring for (vi); the ring of 2-formylbenzoic acid is a benzene ring, so the product is the benzene-ring cyanohydrin named above.
b) Tollens' reagent oxidises only the aldehyde group in (v), never the ring ketone.
c) Thionyl chloride converts both -COOH groups to -COCl, with SO₂ and HCl as gases.
d) Friedel-Crafts acylation with an acyl chloride gives a ketone and stops at one substitution because the ketone deactivates the ring.

PREVIOUS YEARS QUESTIONS

2011

1. Draw the structure of 3-methylbutanal.

Answer: 3-Methylbutanal is (CH₃)₂CH-CH₂-CHO. Its structure is: CH₃-CH(CH₃)-CH₂-CHO, a four-carbon chain with CHO as C1 and a methyl group on C3.

Teacher's Note:
a) Number from the CHO carbon (C1), then place the methyl group on C3.
b) Count the carbons: 4 in the chain plus 1 in the branch = C₅H₁₀O.

2. (a) Give the chemical test to distinguish between
(i) Propanal and propanone, (ii) Benzaldehyde and acetophenone

Answer:
1. (i) Propanal and propanone: Add Tollens' reagent (ammoniacal silver nitrate) and warm. Propanal gives a silver mirror: CH₃CH₂CHO + 2[Ag(NH₃)₂]⁺ + 3OH^- → CH₃CH₂COO^- + 2Ag + 2H₂O + 4NH₃. Propanone gives no silver mirror.
2. (ii) Benzaldehyde and acetophenone: Add iodine and sodium hydroxide solution (iodoform test) and warm. Acetophenone gives a yellow precipitate of iodoform (CHI₃) because of the CH₃-CO- group; benzaldehyde does not. (Benzaldehyde also gives a silver mirror with Tollens' reagent, but acetophenone does not.)

Teacher's Note:
a) Aldehydes respond to Tollens' reagent; ketones do not.
b) The iodoform test is given by methyl ketones, so it separates acetophenone from benzaldehyde.
c) State what you observe in both compounds of the pair.

3. (a) Describe the following giving linked chemical equations :
(i) Cannizzaro reaction (ii) Decarboxylation

Answer:
1. (i) Cannizzaro reaction: Aldehydes that have no α-hydrogen, on treatment with concentrated alkali, undergo self oxidation and reduction (disproportionation) to give a carboxylate salt and an alcohol. 2HCHO + conc. KOH → CH₃OH + HCOOK; similarly 2C₆H₅CHO + conc. NaOH → C₆H₅CH₂OH + C₆H₅COONa
2. (ii) Decarboxylation: Carboxylic acids lose CO₂ to form hydrocarbons when their sodium salts are heated with soda lime (NaOH and CaO): R-COONa + NaOH → (CaO, heat) R-H + Na₂CO₃; for example CH₃COONa + NaOH → (CaO, heat) CH₄ + Na₂CO₃

Teacher's Note:
a) In the Cannizzaro reaction one molecule is oxidised and the other is reduced; both products must be named.
b) In decarboxylation the carbon chain becomes shorter by one carbon.
c) Give an example with every reaction, as the question says linked equations.

(b) Complete the following chemical equations :
(i) Ethylbenzene → (KMnO₄, KOH, heat)
(ii) Phthalic acid → (SOCl₂, heat)
(iii) C₆H₅CONH₂ → (H₃O⁺, heat)

Answer:
1. (i) C₆H₅CH₂CH₃ → (KMnO₄, KOH, heat) C₆H₅COOK (potassium benzoate)
2. (ii) 1,2-C₆H₄(COOH)₂ → (SOCl₂, heat) 1,2-C₆H₄(COCl)₂ (phthaloyl chloride) + SO₂ + HCl
3. (iii) C₆H₅CONH₂ + H₂O → (H₃O⁺, heat) C₆H₅COOH (benzoic acid) + NH₄⁺

Teacher's Note:
a) Alkaline KMnO₄ oxidises the whole alkyl side chain to -COOH, shown as its potassium salt.
b) Hydrolysis of an amide in acid gives the acid and an ammonium salt.

2010

1. Draw the structural formula of 1-phenylpropan-1-one molecule.

Answer: 1-Phenylpropan-1-one is C₆H₅-CO-CH₂-CH₃ (propiophenone): a benzene ring joined to the carbonyl carbon, which is joined to an ethyl group.

Teacher's Note:
a) The 'phenyl' is on C1 of propan-1-one, the same carbon as the C=O group.
b) Formula check: C₉H₁₀O.

2. An organic compound (A) (molecular formula C₈H₁₆O₂) was hydrolysed with dilute sulphuric acid to give a carboxylic acid (B) and an alcohol (C). Oxidation of (C) with chromic acid also produced (B). On dehydration (C) gives but-1-ene. Write the equations for the reactions involved.

Answer:
1. (A) is butyl butanoate, CH₃CH₂CH₂COOCH₂CH₂CH₂CH₃; (B) is butanoic acid; (C) is butan-1-ol.
2. Hydrolysis: CH₃CH₂CH₂COOC₄H₉ + H₂O → (dil. H₂SO₄) CH₃CH₂CH₂COOH + CH₃CH₂CH₂CH₂OH
3. Oxidation: CH₃CH₂CH₂CH₂OH → (chromic acid) CH₃CH₂CH₂COOH
4. Dehydration: CH₃CH₂CH₂CH₂OH → (conc. H₂SO₄, heat) CH₃CH₂CH=CH₂ + H₂O

Teacher's Note:
a) The acid and alcohol have the same number of carbon atoms because oxidation of the alcohol gives the same acid.
b) Dehydration to but-1-ene shows the OH is on the terminal carbon.
c) Check the carbon count: 4 + 4 = 8 carbons in the ester.

2009

1. (a) Illustrate the following name reaction by giving example.
(i) Cannizzaro's reaction (ii) Clemmensen reduction

Answer:
1. (i) Cannizzaro's reaction: Aldehydes with no α-hydrogen undergo disproportionation with concentrated alkali to give an alcohol and a carboxylate salt. 2C₆H₅CHO + conc. NaOH → C₆H₅CH₂OH + C₆H₅COONa
2. (ii) Clemmensen reduction: The carbonyl group of an aldehyde or ketone is reduced to a CH₂ group with zinc amalgam and concentrated hydrochloric acid. CH₃COCH₃ → (Zn-Hg, conc. HCl) CH₃CH₂CH₃ + H₂O

Teacher's Note:
a) The Cannizzaro reaction needs an aldehyde with no α-hydrogen.
b) Clemmensen reduction is done in acid; its alkaline counterpart is the Wolff-Kishner reduction.
c) Always give the reagents above the arrow.

(b) An organic compound contains 69.77% carbon, 11.63% hydrogen and rest oxygen. The molecular mass of the compound is 86. It does not reduce Tollens' reagent but forms an addition compound with sodium hydrogensulphite and give positive iodoform test. On vigorous oxidation it gives ethanoic and propanoic acid. Write the possible structure of the compound.

Answer:
1. Percentage of oxygen = 100 - (69.77 + 11.63) = 18.60%
2. Number of C atoms = \( \frac{69.77 \times 86}{12 \times 100} \) = 5; number of H atoms = \( \frac{11.63 \times 86}{100} \) = 10; number of O atoms = \( \frac{18.60 \times 86}{16 \times 100} \) = 1, so the molecular formula is C₅H₁₀O.
3. No reduction of Tollens' reagent and the addition of NaHSO₃ show that it is a ketone; the positive iodoform test shows it is a methyl ketone.
4. Oxidation gives ethanoic acid and propanoic acid, so the compound is pentan-2-one: CH₃COCH₂CH₂CH₃ → ([O]) CH₃COOH + CH₃CH₂COOH

Teacher's Note:
a) Work out the molecular formula first, then use the tests one by one.
b) Pentan-3-one would give only propanoic acid and no iodoform test, so it is ruled out.
c) Formula mass check: 60 + 10 + 16 = 86.
d) The two acids come from the cleavage of the bond between C2 and C3 of the ketone.

2. (a) How are the following obtained?
(i) Benzoic acid from ethyl benzene (ii) Benzaldehyde from toluene

Answer:
1. (i) Ethylbenzene is oxidised with alkaline potassium permanganate to potassium benzoate, which on acidification gives benzoic acid: C₆H₅CH₂CH₃ → (KMnO₄/KOH, heat) C₆H₅COOK → (H₃O⁺) C₆H₅COOH
2. (ii) Toluene is oxidised with chromyl chloride (Etard reaction): C₆H₅CH₃ + CrO₂Cl₂ → (CS₂) chromium complex, C₆H₅CH(OCrOHCl₂)₂ → (H₃O⁺) C₆H₅CHO. (Toluene can also be given side chain chlorination with Cl₂/hν to benzal chloride and then hydrolysed with water at 373 K.)

Teacher's Note:
a) The strong oxidant KMnO₄ oxidises the whole side chain, but chromyl chloride stops at the aldehyde.
b) Name the Etard reaction in (ii) to earn the mark for the named reaction.

(b) Complete each synthesis by giving the missing material, reagent or products:
(i) C₆H₅COCl → (H₂, Pd-BaSO₄) ............
(ii) Benzene + ................... → (Anhydrous AlCl₃) C₆H₅-CO-CH₃
(iii) Benzene + ................... → (...................) C₆H₅-CO-C₆H₅

Answer:
1. (i) C₆H₅CHO (benzaldehyde), formed by Rosenmund reduction
2. (ii) The missing reagent is CH₃COCl (acetyl chloride): C₆H₆ + CH₃COCl → (Anhydrous AlCl₃) C₆H₅COCH₃ + HCl
3. (iii) The missing reagent is C₆H₅COCl (benzoyl chloride) with anhydrous AlCl₃: C₆H₆ + C₆H₅COCl → (Anhydrous AlCl₃) C₆H₅COC₆H₅ (benzophenone) + HCl

Teacher's Note:
a) (ii) and (iii) are Friedel-Crafts acylation reactions.
b) Pd-BaSO₄ is a poisoned catalyst; it stops reduction at the aldehyde stage.
c) Check every product with the carbon count shown in the question.

2008

1. Arrange the following compounds in an increasing order of their acid strengths:
(CH₃)₂CHCOOH, CH₃CH₂CH(Br)COOH, CH₃CH(Br)CH₂COOH [1 Mark]

Answer: (CH₃)₂CHCOOH < CH₃CH(Br)CH₂COOH < CH₃CH₂CH(Br)COOH

Teacher's Note:
a) The nearer the electron withdrawing Br atom is to the -COOH group, the stronger the acid; the electron donating alkyl groups reduce the acidity.

2. (a) Complete the following reaction statements by giving the missing starting material, reagent or product as required: [5 Marks]
(i) ............... → (O₃, Zn-H₂O) 2 cyclohexanone
(ii) Methylenecyclohexane (cyclohexane ring with =CH₂) → (...............) cyclohexanecarbaldehyde (ring with -CHO)
(iii) Ethylbenzene → (KMnO₄, KOH, heat) ............

Answer:
1. (i) The missing starting material is cyclohexylidenecyclohexane (two cyclohexane rings joined by a C=C bond): C₆H₁₀=C₆H₁₀ → (O₃, Zn-H₂O) 2 C₆H₁₀=O
2. (ii) The missing reagents are (i) B₂H₆, (ii) H₂O₂/OH^- (hydroboration-oxidation) to give cyclohexylmethanol, followed by PCC (pyridinium chlorochromate) to give cyclohexanecarbaldehyde.
3. (iii) The missing product is potassium benzoate, C₆H₅COOK (benzoic acid, C₆H₅COOH, after acidification).

Teacher's Note:
a) Ozonolysis splits C=C into two C=O groups, so two identical ketone molecules come from a symmetrical alkene.
b) In (ii) the alcohol obtained by hydroboration is the primary alcohol; PCC then stops at the aldehyde stage.
c) In (iii) alkaline KMnO₄ oxidises the whole ethyl group to -COOK.
d) The first step gives an anti-Markovnikov product, which is why hydroboration is chosen.

(b) Describe the following reactions:
(i) Cannizzaro reaction (ii) Cross aldol condensation

Answer:
1. (i) Cannizzaro reaction: An aldehyde without α-hydrogen, on heating with concentrated alkali, undergoes disproportionation: one molecule is oxidised to the carboxylate salt and the other is reduced to the alcohol. 2HCHO + conc. KOH → CH₃OH + HCOOK
2. (ii) Cross aldol condensation: Aldol condensation between two different aldehydes (or ketones) having α-hydrogen in presence of dilute alkali gives a mixture of four products. CH₃CHO + CH₃CH₂CHO → (1. NaOH, 2. Δ) CH₃CH=CHCHO (but-2-enal) + CH₃CH₂CH=C(CH₃)CHO (2-methylpent-2-enal) + CH₃CH=C(CH₃)CHO (2-methylbut-2-enal) + CH₃CH₂CH=CHCHO (pent-2-enal)

Teacher's Note:
a) Write one example with each reaction.
b) A cross aldol reaction between two aldehydes with α-hydrogen gives four products: two from self condensation and two from cross condensation.
c) In the Cannizzaro reaction, only aldehydes with no α-hydrogen take part.

3. (a) How would you account for the following: [5 Marks]
(i) Aldehydes are more reactive than ketones towards nucleophiles.
(ii) The boiling points of aldehydes and ketones are lower than of the corresponding acids.
(iii) The aldehydes and ketones undergo a number of addition reactions.

Answer:
1. (i) A ketone has two electron releasing alkyl groups, which reduce the positive charge on the carbonyl carbon and also cause steric hindrance to the incoming nucleophile; an aldehyde has only one alkyl group, so it is more reactive.
2. (ii) Carboxylic acids form strong intermolecular hydrogen bonds, and exist as dimers, whereas aldehydes and ketones are held only by weaker dipole-dipole attractions, so their boiling points are lower.
3. (iii) The carbonyl group is polarised (C(δ+)=O(δ-)) and the carbonyl carbon is electrophilic; the C=O π bond is easily broken by the attack of nucleophiles, so aldehydes and ketones readily undergo nucleophilic addition reactions.

Teacher's Note:
a) Give both the electronic and the steric reason in (i).
b) In (ii) the key word is 'intermolecular hydrogen bonding' in carboxylic acids.
c) Mention that in (iii) the carbon-oxygen double bond is polar, with a partial positive charge on carbon.
d) Allot about one mark for each point and write complete reasons.

(b) Give chemical tests to distinguish between:
(i) Acetaldehyde and benzaldehyde (ii) Propanone and propanol

Answer:
1. (i) Acetaldehyde and benzaldehyde: Warm with iodine and sodium hydroxide (iodoform test). Acetaldehyde gives a yellow precipitate of iodoform; benzaldehyde does not. (Acetaldehyde also reduces Fehling's solution to a red-brown precipitate, whereas benzaldehyde does not.)
2. (ii) Propanone and propanol: Add 2,4-DNP reagent. Propanone gives an orange-yellow precipitate; propanol does not. (Alternatively, propanone gives a yellow precipitate in the iodoform test, as it contains the CH₃-CO- group; propan-1-ol does not.)

Teacher's Note:
a) Pick the test that shows a clear visible change with only one compound of the pair.
b) Fehling's test works for aliphatic aldehydes only, not for aromatic ones.

4. Write the IUPAC name of the compound: CH₃CH(CH₃)CO(CH₃)CHCH₃ [1 Mark]

Answer: 2,4-Dimethylpentan-3-one. The compound is (CH₃)₂CH-CO-CH(CH₃)₂, with five carbon atoms in the main chain, the ketone group on C3 and one methyl group each on C2 and C4.

Teacher's Note:
a) The longest chain has 5 carbon atoms with the C=O in the middle, so the numbering gives the lowest locant 3 to the carbonyl carbon.

2007

5. Give chemical tests to distinguish between the following pairs of compounds : [3 Marks]
(i) Propanal and propanone (ii) Methyl acetate and ethyl acetate
(iii) Benzaldehyde and benzoic acid

Answer:
1. (i) Propanal gives a silver mirror with Tollens' reagent (and a red-brown precipitate with Fehling's solution); propanone does not.
2. (ii) Hydrolyse each ester with dilute NaOH. Methyl acetate gives methanol and ethyl acetate gives ethanol. Warm the product with iodine and NaOH: ethanol gives a yellow precipitate of iodoform, methanol does not.
3. (iii) Add sodium hydrogencarbonate solution. Benzoic acid gives brisk effervescence of CO₂; benzaldehyde does not. (Benzaldehyde gives a silver mirror with Tollens' reagent, benzoic acid does not.)

Teacher's Note:
a) Each pair needs one test and the observation for both compounds.
b) The ester test in (ii) works through the alcohol obtained after hydrolysis.
c) Benzoic acid is a stronger acid than carbonic acid, so it liberates CO₂ from the hydrogencarbonate.

6. Name the following compound according to IUPAC system : CH₃COCH₂COCH₃ [1 Mark]

Answer: Pentane-2,4-dione

Teacher's Note:
a) The chain has five carbons with C=O groups on C2 and C4, so the suffix is -dione.

7. Write one chemical equation for each, to illustrate the following reactions: [3 Marks]
(i) Rosenmund reduction (ii) Cannizzaro reaction (iii) Trans esterification

Answer:
1. (i) Rosenmund reduction: C₆H₅COCl + H₂ → (Pd-BaSO₄) C₆H₅CHO + HCl
2. (ii) Cannizzaro reaction: 2HCHO + conc. KOH → CH₃OH + HCOOK
3. (iii) Trans esterification: R-COOR' + R''OH → (H⁺) R-COOR'' + R'OH (for example CH₃COOC₂H₅ + CH₃OH ⇌ CH₃COOCH₃ + C₂H₅OH)

Teacher's Note:
a) Each equation must show the reagent or catalyst above the arrow.
b) In trans esterification, one alkoxy group of an ester is exchanged with that of an alcohol.
c) In (i) the catalyst is deliberately poisoned with BaSO₄ to avoid reduction of the aldehyde to the alcohol.

2006

7. Describe the following giving suitable examples: [2 Marks]
(i) Cannizzaro reaction (ii) Aldol condensation

Answer:
1. (i) Cannizzaro reaction: Aldehydes with no α-hydrogen, when heated with concentrated alkali, give an alcohol and the salt of a carboxylic acid. 2C₆H₅CHO + conc. NaOH → C₆H₅CH₂OH + C₆H₅COONa
2. (ii) Aldol condensation: Aldehydes and ketones with α-hydrogen, in presence of dilute alkali, give β-hydroxy aldehydes or ketones (aldols), which on heating lose water to give α,β-unsaturated carbonyl compounds. 2CH₃CHO → (dil. NaOH) CH₃CH(OH)CH₂CHO → (Δ, -H₂O) CH₃CH=CHCHO

Teacher's Note:
a) One example for each reaction is enough for 2 marks.
b) Aldol condensation needs α-hydrogen; the Cannizzaro reaction needs its absence.

8. Give a chemical test to distinguish between ethanal and propanal. [1 Mark]

Answer: Iodoform test: Ethanal on warming with iodine and sodium hydroxide gives a yellow precipitate of iodoform (CHI₃), because it has the CH₃-CO- group; propanal does not.

Teacher's Note:
a) Ethanal is the only aldehyde that gives the iodoform test.

9. Write the IUPAC name of CH₃COCH₂COCH₃. [1 Mark]

Answer: Pentane-2,4-dione

Teacher's Note:
a) Two C=O groups on C2 and C4 of a five-carbon chain make it a dione.

10. (a) Write the steps and conditions involved in the following conversions:
(i) Acetophenone to 2-phenyl-2-butanol [1 Mark]
(ii) Propene to acetone [1 Mark]

Answer:
1. (i) Acetophenone is treated with ethylmagnesium bromide (C₂H₅MgBr) in dry ether, followed by hydrolysis with water/dilute acid: C₆H₅COCH₃ + C₂H₅MgBr → (dry ether) C₆H₅C(CH₃)(C₂H₅)OMgBr → (H₂O) C₆H₅C(CH₃)(C₂H₅)OH (2-phenylbutan-2-ol)
2. (ii) Propene is hydrated (H₂O, H⁺, Markovnikov addition) to propan-2-ol, which is oxidised with chromic acid (CrO₃ or K₂Cr₂O₇/H⁺) to acetone: CH₃CH=CH₂ → (H₂O/H⁺) CH₃CH(OH)CH₃ → ([O]) CH₃COCH₃

Teacher's Note:
a) A Grignard reagent adds to a ketone to give a tertiary alcohol after hydrolysis.
b) Hydration of propene follows Markovnikov's rule, so the OH goes to the middle carbon.

(b) Give a chemical test to distinguish between Methyl acetate and Ethyl acetate. [1 Mark]

Answer: Hydrolyse each ester with dilute NaOH and warm the product with iodine and NaOH. Ethyl acetate gives ethanol, which gives a yellow precipitate of iodoform; methyl acetate gives methanol, which does not.

Teacher's Note:
a) Ethanol (CH₃CH(OH)-) gives the iodoform test, but methanol does not.

2005

11. How are formalin and trioxane related to methanal? [1 Mark]

Answer: Formalin is a 40% aqueous solution of methanal (formaldehyde). Trioxane is the cyclic trimer of methanal, formed when methanal polymerises on standing or on heating with traces of acid.

Teacher's Note:
a) Do not call formalin a polymer; it is simply methanal in water.

12. (a) Write the structural formula of 3-phenylprop-2-enal.

Answer: 3-Phenylprop-2-enal (cinnamaldehyde) is C₆H₅-CH=CH-CHO, with the CHO carbon as C1, the double bond between C2 and C3 and the phenyl group on C3.

Teacher's Note:
a) Locate the double bond between C2 and C3 and the phenyl on C3, as the name shows.
b) The formula is C₉H₈O.

(b) Write one chemical equation each to illustrate the following reactions: [3 Marks]
(i) Aldol condensation (ii) Cannizzaro's reaction

Answer:
1. (i) Aldol condensation: 2CH₃CHO → (dil. NaOH) CH₃CH(OH)CH₂CHO → (Δ, -H₂O) CH₃CH=CHCHO
2. (ii) Cannizzaro's reaction: 2HCHO + conc. KOH → CH₃OH + HCOOK

Teacher's Note:
a) Show the reagent above the arrow, as the examiner looks for it.
b) Use an aldehyde with α-hydrogen in (i) and one without α-hydrogen in (ii).
c) Write the aldol first and then the condensation product on heating.

13. (a) Write the structural formula of hex-2-en-4-ynoic acid.

Answer: Hex-2-en-4-ynoic acid is HOOC-CH=CH-C≡C-CH₃: COOH is C1, a double bond between C2 and C3, a triple bond between C4 and C5, and CH₃ is C6.

Teacher's Note:
a) In names with both double and triple bonds, the 'en' is written before the 'yn' and the -oic acid suffix is last.
b) Count six carbons in the chain, including the carboxyl carbon.

(b) To illustrate the following reactions write one chemical equation for each: [3 Marks]
(i) Cross aldol condensation (ii) Hoffmann bromamide reaction

Answer:
1. (i) Cross aldol condensation: CH₃CHO + CH₃CH₂CHO → (dil. NaOH, then Δ) CH₃CH=CHCHO + CH₃CH₂CH=C(CH₃)CHO + CH₃CH=C(CH₃)CHO + CH₃CH₂CH=CHCHO (mixture of four products)
2. (ii) Hoffmann bromamide reaction: R-CONH₂ + Br₂ + 4NaOH → R-NH₂ + Na₂CO₃ + 2NaBr + 2H₂O (for example CH₃CONH₂ gives CH₃NH₂)

Teacher's Note:
a) The Hoffmann bromamide reaction gives a primary amine with one carbon atom less than the amide.
b) A cross aldol reaction between two different aldehydes with α-hydrogen gives four products.
c) Check that the equation for (ii) is balanced: 1 Br₂ and 4 NaOH on the left.