CBSE Class 12 Chemistry Aldehydes Ketones and Carboxylic Acids Important Questions Set 04

Very Short Answer Questions (1 Mark)

Question 1. An aromatic organic compound ‘A’ with molecular formula C₈H₈O gives positive DNP and iodoform tests. It neither reduces Tollens’ reagent nor does it decolourise bromine water. Write the structure of ‘A’. [CBSE Comptt. Delhi/OD 2018]
Answer:
C₆H₅COCH₃
Detailed Answer:
‘A’ gives positive DNP test. Therefore, it is an aldehyde or a ketone. Since it does not reduce Tollens’ reagent, ‘A’ must be a ketone. ‘A’ responds to iodoform test. Therefore, it should be a methyl ketone. The molecular formula of ‘A’ indicates high degree of unsaturation, yet it does not decolourise bromine water. This indicates the presence of unsaturation due to an aromatic ring. The molecular formula of ‘A’ indicates that it should be phenyl methyl ketone (acetophenone).

Teacher's Note:
1. Use each clue in turn: DNP = C=O, no Tollens = ketone, iodoform = CH₃CO– group, no Br₂ decolourisation = aromatic ring.

Question 2. (CH₃)₃C–CHO does not undergo aldol condensation. Comment. [CBSE SQP 2017]
Answer:
No α-H is present.

Teacher's Note:
1. The α-carbon here is the quaternary carbon of the tert-butyl group, so no enolate can form; this aldehyde gives Cannizzaro reaction instead.

Question 3. Out of CH₃CH₂COCH₂CH₃ and CH₃CH₂CH₂COCH₃, which gives iodoform test? [CBSE OD 2014]
Answer:
CH₃CH₂CH₂COCH₃ will give iodoform test as it has a terminal ketomethyl group.

Teacher's Note:
1. Only compounds with a CH₃CO– group (or CH₃CH(OH)– that can be oxidised to it) give the iodoform test.

Question 4. Give the chemical test to distinguish between the following pairs of compounds.
(i) Propanal and propanone.
(ii) Benzaldehyde and Benzoic acid. [CBSE Delhi 2012]
Answer:
(i) Propanal gives silver mirror on reaction with Tollens’ reagent while propanone does not give.
(ii) Benzoic acid evolves CO₂ gas with NaHCO₃ while benzaldehyde does not evolve CO₂.

Teacher's Note:
1. Always name the reagent and the observation for both compounds – just naming the test loses marks.

Teacher's Note:
1. Memory aid for C₂ to C₆ dicarboxylic acids: ‘Oh My Such Good Apple’ – oxalic, malonic, succinic, glutaric, adipic.

Question 6. Write IUPAC name of:
CH₃–CH(OH)–CH₂–COOH [CBSE Delhi 2014]
Answer:
3-Hydroxybutanoic acid / 3-hydroxybutan-1-oic acid.

Teacher's Note:
1. The –COOH carbon is C-1, so the OH on the third carbon from it gets locant 3.

Question 7. Write the IUPAC name of:
CH₃–C≡C–CH₂–COOH [CBSE Delhi 2011]
Answer:
Pent-3-ynoic acid.

Teacher's Note:
1. Count from the COOH carbon: the triple bond starts at C-3, giving the ending ‘-3-ynoic acid’.

Question 8. Complete the following reaction:
CH₃–COOH → [Br₂/P] → [CBSE Delhi 2013]
Answer:
CH₃–COOH → [Br₂/P] → CH₂(Br)–COOH

Teacher's Note:
1. This is the Hell–Volhard–Zelinsky reaction: halogen enters only at the α-carbon, and the acid must have an α-H.

Question 9. Illustrate the decarboxylation reaction giving a suitable example: [CBSE Delhi 2012]
Answer:
Decarboxylation refers to the reaction in which carboxylic acid loose carbon dioxide to form hydrocarbons when their sodium salts are heated with sodalime, e.g.,
CH₃COONa → [Sodalime (NaOH + CaO), Δ] → CH₄ + Na₂CO₃

Teacher's Note:
1. The product always has one carbon fewer than the acid salt; CaO keeps NaOH dry and stops it attacking the glass.

Question 10. How many of the following compounds undergo aldol condensation? (Answer is an integer from 0 to 9.)
Methanal, 2-methylpentanal, benzaldehyde, benzophenone, cyclohexanone, 1-phenylpropanone, phenylacetaldehyde, butan-1-ol, 2,2-dimethylbutanal
Answer:
Answer: 4
2-Methylpentanal, cyclohexanone, 1-phenylpropanone and phenylacetaldehyde have α-hydrogens and undergo aldol condensation.

Teacher's Note:
1. Check two things: the compound must be an aldehyde or ketone, and it must have an α-hydrogen – butan-1-ol fails the first test, 2,2-dimethylbutanal the second.

Question 11. Arrange the following compounds in increasing order of their acid strengths:
(CH₃)₂CHCOOH, CH₃CH₂CH(Br)COOH, CH₃CH(Br)CH₂COOH
Answer:
(CH₃)₂CHCOOH < CH₃CH(Br)CH₂COOH < CH₃CH₂CH(Br)COOH

Teacher's Note:
1. The −I effect of Br weakens with distance, so an α-bromo acid is stronger than a β-bromo acid.

Short Answer Questions (2 Marks)

Question 12. Complete the following:
(i) C₆H₅COOH –[SO₃/H₂SO₄]→ ?
(ii) C₆H₅COOH –[conc. HNO₃, conc. H₂SO₄]→ ?
Answer:
(i) m-HO₃S–C₆H₄–COOH (3-sulphobenzoic acid)
(ii) m-O₂N–C₆H₄–COOH (3-nitrobenzoic acid)

Teacher's Note:
1. –COOH is a deactivating, meta-directing group, so ring substitution of benzoic acid gives the meta product.

Question 13. An organic compound X has molecular formula C₅H₁₀O. It does not reduce Fehling's solution but forms a bisulphite compound. It also gives positive iodoform test. What are possible structures of X? Explain your reasoning relating structure.
Answer:
Since X does not reduce Fehling's solution but forms a bisulphite compound, it is a ketone. Ketone giving +ve iodoform test is a methyl ketone.
CH₃–CO–CH₂–CH₂–CH₃ (pentan-2-one) and CH₃–CO–CH(CH₃)–CH₃ (3-methylbutan-2-one) are possible structures of the compound.

Teacher's Note:
1. Iodoform test is positive for CH₃CO– compounds (and CH₃CH(OH)– alcohols), so the C=O must be at C-2.

Question 14. Give the chemical test to distinguish between:
(i) CH₃CHO and CH₃–CO–CH₃
(ii) CH₃CHO and C₆H₅CHO
Answer:
(i) CH₃CHO gives Tollens' reagent test (silver mirror); acetone does not.
CH₃CHO + 2AgNO₃ + 2NH₄OH → CH₃COOH + 2Ag (silver mirror) + 2NH₄NO₃ + H₂O
(ii) CH₃CHO gives brick red ppt. with Fehling's solution; C₆H₅CHO does not.
CH₃CHO + 2CuSO₄ + 5NaOH → CH₃COONa + Cu₂O (brick red ppt.) + 2Na₂SO₄ + 3H₂O

Teacher's Note:
1. In a ‘distinguish’ answer always state what each compound does: one gives the result, the other does not.

Question 15. Which acid of each pair shown here would you expect to be stronger?
(i) CH₃CO₂H or CH₂FCO₂H
(ii) CH₂FCO₂H or CH₂ClCO₂H
(iii) CH₂FCH₂CH₂CO₂H or CH₃CHFCH₂CO₂H
(iv) p-F₃C–C₆H₄–COOH or p-H₃C–C₆H₄–COOH
Answer:
(i) FCH₂COOH
(ii) CH₂FCOOH
(iii) CH₃–CH(F)–CH₂COOH
(iv) p-F₃C–C₆H₄–COOH

Teacher's Note:
1. The closer and the more electronegative the substituent, the stronger its −I effect and the stronger the acid; –CF₃ withdraws electrons while –CH₃ releases them.

Question 16. Carboxylic acids do not give reactions of aldehydes and ketones. Why?
Answer:
It is due to resonance: R–C(=O)–OH ↔ R–C(–O⁻)=O⁺H. Position of >C=O group keeps changing.

Teacher's Note:
1. Resonance with the lone pair of –OH lowers the electrophilic character of the carboxyl carbon, so it does not behave as a true carbonyl group.

Question 17. Write IUPAC names of the following compounds (i) and (ii):

Teacher's Note:
1. In (ii) the ring carbon bearing –COOH is C-1; number towards the nearer substituent and write prefixes alphabetically (bromo, chloro, iodo).

Question 18. Account for the following:
(i) Oxidation of toluene to C₆H₅CHO with CrO₃ is carried out in presence of acetic anhydride.
(ii) Melting point of an acid with even number of carbon atoms is higher than those of its neighbours with odd number of carbon atoms.
Answer:
(i) Acetic anhydride converts the aldehyde formed into benzylidene diacetate, C₆H₅CH(OCOCH₃)₂. It is used to prevent oxidation to benzoic acid.
(ii) Acids with even number of carbon atoms fit better (pack more closely) into the crystal lattice.

Teacher's Note:
1. The benzylidene diacetate is hydrolysed with aqueous acid afterwards to release benzaldehyde.

Question 19. Distinguish between:
(i) C₂H₅OH and CH₃CHO
(ii) C₆H₅COCH₃ and C₆H₅CH₂CHO
Answer:
(i) C₂H₅OH evolves H₂ gas with Na, CH₃CHO does not.
(ii) Acetophenone will give yellow ppt. of iodoform with NaOI while C₆H₅CH₂CHO will not.

Teacher's Note:
1. Tollens' test also works for (i) and (ii): only the aldehyde gives a silver mirror.

Question 20. Complete the following reactions by identifying A, B and C:
(i) A + H₂(g) –[Pd/BaSO₄]→ (CH₃)₂CH–CHO
(ii) (CH₃)₃C–CO–CH₃ + NaOI → B + C
Answer:
(i) ‘A’ is (CH₃)₂CH–COCl (2-methylpropanoyl chloride).
(ii) ‘B’ is CHI₃ and ‘C’ is (CH₃)₃C–COONa.

Teacher's Note:
1. In the haloform reaction the CH₃ of CH₃CO– becomes CHX₃, and the rest becomes the salt of an acid with one carbon less.

Question 21. Benzaldehyde gives a positive test with Tollens' reagent but not with Fehling's and Benedict's solutions. Why?
Answer:
It is due to stronger oxidizing nature of Tollens' reagent as compared to Fehling's and Benedict's solutions, which cannot oxidise benzaldehyde to benzoic acid. In general, all these three can oxidise aliphatic aldehydes.

Teacher's Note:
1. This is the basis of the standard test to tell aliphatic aldehydes from aromatic aldehydes.

Question 22. Aldehydes usually do not form stable hydrates but chloral normally exists as chloral hydrate. Give reason.
Answer:
In case of aldehydes, the reaction is reversible: RCHO + H₂O ⇄ RCH(OH)₂
In case of CCl₃CHO, the Cl atoms increase the +ve charge on carbonyl carbon. Therefore, weak nucleophiles like water readily add to the carbonyl group:
Cl₃C–CHO + H₂O → Cl₃C–CH(OH)₂ (chloral hydrate)
The hydrate is further stabilised by intramolecular hydrogen bonding between the –OH groups and Cl atoms.

Teacher's Note:
1. Electron-withdrawing groups on the α-carbon make the C=O carbon more electrophilic and shift the hydration equilibrium towards the hydrate.

Question 23. Give possible explanation for the following:
(i) Cyclohexanone forms cyanohydrin in good yield but 2,2,6-trimethylcyclohexanone does not.
(ii) There are two –NH₂ groups in semicarbazide. However, only one is involved in formation of semicarbazone.
Answer:
(i) Cyclohexanone + HCN/H⁺ → cyclohexanone cyanohydrin (1-hydroxycyclohexane-1-carbonitrile). In 2,2,6-trimethylcyclohexanone there is steric hindrance for the attack of CN⁻ at C=O due to the three methyl groups at α-positions, so no reaction (no good yield) occurs; cyclohexanone has no such hindrance.
(ii) The –NH₂ group attached to C=O is involved in resonance: H₂N–CO–NH–NH₂ ↔ H₂N⁺=C(O⁻)–NH–NH₂ ↔ H₂N–C(O⁻)=N⁺H–NH₂. As a result the electron density on this NH₂ group decreases and hence it does not act as nucleophile; the free –NH₂ of –NHNH₂ reacts.

Teacher's Note:
1. Semicarbazide behaves like an amide on one side and a hydrazine on the other – only the hydrazine end is nucleophilic.

Question 24. Aldehydes are easily oxidisable yet propanal can conveniently be prepared by the oxidation of propanol by acidic K₂Cr₂O₇. Explain.
Answer:
Propanal (b.p. 322 K) has a boiling point lower than that of propanol (370 K) and below 373 K, so it can be removed by distillation as soon as it is formed. Therefore propanal can easily be prepared from propanol by distillation from the alcohol-acid dichromate solution, before it is oxidised further.

Teacher's Note:
1. The trick is distillation during the reaction: the aldehyde leaves the oxidising mixture before it can become the acid.

Question 25. Do the following conversions in not more than two steps:
(i) Benzoic acid to benzaldehyde (ii) Propanone to propene
Answer:
(i) C₆H₅COOH –[SOCl₂]→ C₆H₅COCl –[H₂, Pd–BaSO₄]→ C₆H₅CHO
(ii) CH₃COCH₃ –[NaBH₄]→ CH₃CH(OH)CH₃ –[conc. H₂SO₄, Δ]→ CH₃CH=CH₂

Teacher's Note:
1. Acid → aldehyde cannot be done directly with ordinary reducing agents; going through the acid chloride (Rosenmund) is the standard route.

Question 26. Convert the following:
(i) Ethyl benzene to benzoic acid (ii) Ethanal to but-2-enal
Answer:
(i) C₆H₅C₂H₅ –[K₂Cr₂O₇/H⁺]→ C₆H₅COOH
(ii) 2CH₃CHO –[dil. NaOH]→ CH₃CH(OH)CH₂CHO –[H₃O⁺/Δ]→ CH₃CH=CHCHO

Teacher's Note:
1. Heating the aldol (3-hydroxybutanal) removes water because the product C=C is conjugated with C=O.

Question 27. Explain why o-hydroxybenzaldehyde is a liquid at room temperature while p-hydroxybenzaldehyde is a high melting solid.
Answer:
Due to intramolecular H-bonding (chelation) between the –OH hydrogen and the C=O oxygen of the adjacent –CHO group, o-hydroxybenzaldehyde exists as discrete molecules while due to intermolecular H-bonding (HO–C₆H₄–CH=O····HO–C₆H₄–CH=O····), p-hydroxybenzaldehyde exists as associated molecules.
To break these intermolecular H-bonds, a large amount of energy is needed. Consequently, p-hydroxybenzaldehyde has a much higher m.p. and b.p. than that of o-hydroxybenzaldehyde. As a result, o-hydroxybenzaldehyde is a liquid at room temperature while p-hydroxybenzaldehyde is a high melting solid.

Teacher's Note:
1. Intramolecular H-bonding lowers m.p./b.p.; intermolecular H-bonding raises them – the same reasoning explains o- vs p-nitrophenol.

Question 28. What is the function of Roschelle salt in Fehling’s solution?
Answer:
In alkaline medium, Cu²⁺ ions get precipitated as Cu(OH)₂. To keep Cu²⁺ ions in solution in alkaline medium, Rochelle salt is added. The insoluble Cu(OH)₂ first formed goes into solution due to the formation of a soluble complex between Cu²⁺ ions and tartrate ion.
Cu²⁺ + 2 tartrate ions [from Rochelle salt, sodium potassium tartrate, KOOC–CH(OH)–CH(OH)–COONa] → soluble copper(II) tartrate complex, in which Cu²⁺ is bonded to the deprotonated –OH oxygens of two tartrate ions (+ 4H⁺ + 2Na⁺ + 2K⁺)

Teacher's Note:
1. Fehling’s A is aqueous CuSO₄; Fehling’s B is alkaline Rochelle salt – they are mixed just before use.

Short Answer Questions (3 Marks)

Question 29. An alcohol A (C₄H₁₀O) on oxidation with acidified potassium dichromate gives acid B (C₄H₈O₂). Compound A when dehydrated with conc. H₂SO₄ at 443 K gives compound C. Treatment of C with aqueous H₂SO₄ gives compound D (C₄H₁₀O) which is an isomer of A. Compound D is resistant to oxidation but compound A can be easily oxidised. Identify A, B, C and D. Name the type of isomerism exhibited by A and D.
Answer:
A : CH₃–CH(CH₃)–CH₂–OH
B : CH₃–CH(CH₃)–COOH
C : CH₃–C(CH₃)=CH₂
D : CH₃–C(CH₃)(OH)–CH₃
A and D are positional isomers.

Teacher's Note:
1. “Resistant to oxidation” points to a 3° alcohol; hydration of 2-methylpropene follows Markovnikov’s rule and gives exactly that.

Question 30. Complete the following reactions:
(i) Cyclopentanone + H₂N–OH →
(ii) Cyclohexene \( \xrightarrow[\Delta]{\text{KMnO}_{4}\text{, H}_{2}\text{SO}_{4}} \)
(iii) Phthalic acid (benzene-1,2-dicarboxylic acid) + NH₃ \( \xrightarrow{\text{Strong heating}} \)
Answer:
(i) Cyclopentanone oxime, (CH₂)₄C=N–OH
(ii) HOOC–CH₂–CH₂–CH₂–CH₂–COOH
(iii) Phthalimide, C₆H₄(CO)₂NH (the two C=O groups on adjacent ring carbons are joined through NH in a five-membered ring)

Teacher's Note:
1. Hot acidic KMnO₄ cleaves the ring C=C completely, so cyclohexene opens to the six-carbon dicarboxylic acid (adipic acid).

Question 31. (i) Account for the following:
(a) Cl–CH₂COOH is a stronger acid than CH₃COOH.
(b) Carboxylic acids do not give reactions of carbonyl group.
(ii) Write the chemical equation to illustrate the following name reaction: Rosenmund reduction.
Answer:
(i) (a) Cl-CH₂COOH has lower pK_a value than acetic acid. Also, Cl group is an electron withdrawing, creating less electron density on oxygen of carboxylic acid making the release of proton easier than acetate ion. Hence, Cl–CH₂COOH is a stronger acid than CH₃COOH.
(b) The carbonyl group in –COOH is inert and does not show nucleophilic addition reaction like carbonyl compound due to resonance stabilisation of carboxylate ion:
R–C(=O)–O⁻ ↔ R–C(–O⁻)=O
(ii) Rosenmund reaction: R–CO–Cl (Acyl chloride) + H₂ \( \xrightarrow[\Delta]{\text{Pd/BaSO}_{4}} \) R–CO–H (Aldehyde) + HCl
CH₃–CO–Cl (Acetyl chloride) + H₂ \( \xrightarrow[\Delta]{\text{Pd/BaSO}_{4}} \) CH₃–CO–H (Acetaldehyde) + HCl

Teacher's Note:
1. The lone pair on the –OH oxygen is delocalised onto the C=O group, so the carbonyl carbon of an acid is much less electrophilic than in aldehydes and ketones.

Question 32. An alkene ‘A’ molecular formula (C₅H₁₀) on ozonolysis gives a mixture of two compounds ‘B’ and ‘C’. Compound ‘B’ gives positive Fehling’s test and also reacts with iodine and NaOH solution. Compound ‘C’ does not give Fehling’s test but forms iodoforms. Identify the compounds ‘A’, ‘B’ and ‘C’ giving suitable explanation and write the reactions of ozonolysis and iodoform formation from either ‘B’ or ‘C’. [NCERT Exemplar] [HOTS]
Answer:
Compound ‘B’ gives both Fehling’s test and iodoform test and therefore it must be an aldehyde with –COCH₃ group. Moreover, compound ‘C’ doesn’t give Fehling’s test but gives positive iodoform test and therefore it must be a ketone with –COCH₃ group. Thus, the compound A, B and C are as follows:
CH₃–CH=C(CH₃)–CH₃ (A: 2-Methylbut-2-ene) \( \xrightarrow[(ii)\,Zn/H_2O]{(i)\,O_3} \) CH₃–CHO (B: Acetaldehyde) + O=C(CH₃)–CH₃ (C: Acetone)
CH₃–CHO + 3I₂ + 4NaOH \( \xrightarrow{\text{Iodoform reaction}} \) HCOONa + CHI₃↓ + 3NaI + 3H₂O
Other isomers of ‘A’ will not give products corresponding to the given test.

Teacher's Note:
1. Ethanal is the only aldehyde that gives the iodoform test – spotting this immediately fixes B.

Question 33. Write down functional isomers of a carbonyl compound with molecular formula C₃H₆O. Which isomer will react faster with HCN and why? Explain the mechanism of the reaction also. Will the reaction lead to the completion with the conversion of whole reactant into product at reaction conditions? If a strong acid is added to the reaction mixture what will be the effect on concentration of the product and why? [NCERT Exemplar] [HOTS]
Answer:
CH₃CH₂CHO (I)    CH₃COCH₃ (II)
(a) Compound I will react faster with HCN due to less steric hinderance and electronic reasons than II.
Mechanism: Nucleophilic addition reaction:
Step 1 (slow): CN⁻ attacks the carbonyl carbon (C^δ+=O^δ−) to form the tetrahedral alkoxide intermediate: CH₃CH₂CHO + CN⁻ ⇄ CH₃CH₂CH(O⁻)CN
Step 2 (fast): The intermediate takes up H⁺: CH₃CH₂CH(O⁻)CN + H⁺ ⇄ CH₃CH₂CH(OH)CN
(b) No, it is a reversible reaction. Hence, equilibrium is established.
(c) Addition of acid inhibits the reaction because the formation of CN⁻ ions is prevented.

Teacher's Note:
1. HCN is a weak acid; a base is added to generate the nucleophile CN⁻, and a strong acid suppresses its ionisation.

Question 34. An unknown aldehyde ‘A’ on reacting with alkali gives a β-hydroxy aldehyde, which loses water to form an unsaturated aldehyde, But-2-enal. Another aldehyde ‘B’ undergoes disproportionation reaction in the presence of conc. alkali to form products C and D. C is an arylalcohol with the formula, C₇H₈O.
(i) Identify A and B.
(ii) Write the sequence of reactions involved.
(iii) Name the product, when ‘B’ reacts with zinc amalgam and hydrochloric acid. [HOTS]
Answer:
(i) A is CH₃CHO (ethanal) and B is C₆H₅CHO (benzaldehyde).
(ii) 2CH₃CHO [A] \( \xrightarrow{NaOH} \) CH₃–CH(OH)–CH₂CHO \( \xrightarrow{-H_2O} \) CH₃–CH=CH–CHO (But-2-enal)
2C₆H₅CHO [B] \( \xrightarrow{\text{Conc. alkali}} \) C₆H₅CH₂OH [C] + C₆H₅COO⁻ [D]
(iii) Toluene.

Teacher's Note:
1. Clemmensen reduction converts the –CHO group of benzaldehyde into –CH₃.

Question 35. (i) Account for the following:
(a) Cl–CH₂COOH is a stronger acid than CH₃COOH.
(b) Carboxylic acids do not give reactions of carbonyl group.
(ii) Out of CH₃CH₂–CO–CH₂–CH₃ and CH₃CH₂–CH₂–CO–CH₃, which gives iodoform test? [CBSE (AI) 2014]
Answer:
(i) (a) Because of –I effect of Cl atom in ClCH₂COOH and +I effect of CH₃ group in CH₃COOH the electron density in the O–H bond in ClCH₂COOH is much lower than CH₃COOH. As a result O–H bond in ClCH₂COOH is much weaker than in CH₃COOH therefore loses a proton more easily than CH₃COOH. Hence ClCH₂COOH acid is stronger acid than CH₃COOH.
(b) Carboxylic acids are resonance hybrid of the following structures:
R–C(=O)–ḤH (I)ndash;OH (I) ↔ R–C(–O⁻)=O⁺–H (II)
Similarly, a carbonyl group of aldehydes and ketones may regarded as resonance hybrid of following structures.
>C=O (III) ↔ >C⁺–O⁻ (IV)
Because of contribution of structure (IV), the carbonyl carbon in aldehydes and ketones is electrophilic. On the other hand, electrophilic character of carboxyl carbon is reduced due to contribution of structure (II). As carbonyl carbon of carboxyl group is less electropositive than carbonyl carbon in aldehydes and ketones, therefore, carboxylic acids do not give nucleophilic addition reactions of aldehydes and ketones.
(ii) CH₃–CH₂–CH₂–COCH₃

Teacher's Note:
1. Only compounds with a CH₃CO– group (or CH₃CH(OH)–) give the iodoform test; pentan-3-one has ethyl groups on both sides.

Question 36. Arrange the following compounds in increasing order of their property as indicated:
(i) Acetaldehyde, Acetone, Methyl tert-butyl ketone (reactivity towards HCN)
(ii) Benzoic acid, 3,4-Dinitrobenzoic acid, 4-Methoxybenzoic acid (acid strength)
(iii) CH₃CH₂CH(Br)COOH, CH₃CH(Br)CH₂COOH, (CH₃)₂CHCOOH (acid strength) [CBSE (AI) 2012]
Answer:
(i) Methyl tert-butyl ketone < Acetone < Acetaldehyde
(ii) 4-Methoxybenzoic acid < Benzoic acid < 3,4-Dinitrobenzoic acid
(iii) (CH₃)₂CHCOOH < CH₃CH(Br)CH₂COOH < CH₃CH₂CH(Br)COOH

Teacher's Note:
1. A para-methoxy group donates electrons by resonance and makes benzoic acid weaker; nitro groups do the opposite.

Long Answer Questions (5 Marks)

Question 37. (i) Describe the following reactions:
(a) Acetylation   (b) Aldol condensation
(ii) Write the main product in the following equations:
(a) CH₃–CO–CH₃ \( \xrightarrow{\text{LiAlH}_{4}} \) ?
(b) C₆H₅CHO \( \xrightarrow[\text{273-283 K}]{\text{HNO}_{3}\text{/H}_{2}\text{SO}_{4}} \) ?
(c) CH₃–COOH \( \xrightarrow{\text{PCl}_{5}} \) ? [CBSE Comptt. Delhi 2015]
Answer:
(i) (a) The acyl groups are introduced at ortho- and para- positions by reaction of chlorobenzene with acyl halide in the presence of anhydrous aluminium chloride (a Lewis acid) as catalyst.
C₆H₅Cl + CH₃–CO–Cl \( \xrightarrow{\text{Anhyd. AlCl}_{3}} \) 2-chloroacetophenone (o-ClC₆H₄COCH₃) + 4-chloroacetophenone (p-ClC₆H₄COCH₃)
(b) Aldehydes and ketones having at least one α-hydrogen undergo a reaction in the presence of dilute alkali as catalyst to form β-hydroxy aldehydes (aldol) or β-hydroxy ketones (ketol), respectively.
2CH₃–CHO \( \xrightarrow{\text{dil. NaOH}} \) CH₃–CH(OH)–CH₂–CHO \( \xrightarrow{\Delta} \) CH₃–CH=CH–CHO
(Award full marks if correct equation is given)
(ii) (a) CH₃–CH(OH)–CH₃
(b) m-Nitrobenzaldehyde (3-NO₂C₆H₄CHO)
(c) CH₃COCl

Teacher's Note:
1. –CHO is a deactivating, meta-directing group, so nitration of benzaldehyde gives the meta product; the low temperature avoids oxidation of –CHO.

Question 38. (i) Draw the structures of the following:
(a) p-Methylbenzaldehyde,
(b) 4-Methylpent-3-en-2-one.
(ii) Give chemical tests to distinguish between the following pairs of compounds:
(a) Benzoic acid and Ethyl benzoate,
(b) Benzaldehyde and Acetophenone
(c) Phenol and Benzoic acid.
Answer:
(i) (a) p-CH₃–C₆H₄–CHO (CHO and CH₃ at opposite positions, 1 and 4, of the benzene ring)
(b) (CH₃)₂C=CHCOCH₃
(ii) (a) On adding sodium bicarbonate to benzoic acid, brisk effervescence of CO₂ is evolved.
C₆H₅COOH (Benzoic acid) + NaHCO₃ → C₆H₅COONa (Sodium benzoate) + CO₂↑ + H₂O
Whereas ethyl benzoate does not.
(b) Acetophenone having at least one –CH₃ group on heating with alkaline solution of iodine forms yellow coloured precipitate of iodoform.
C₆H₅COCH₃ + 3NaOI → C₆H₅COONa + CHI₃↓ (Iodoform, yellow ppt.) + 2NaOH
Whereas benzaldehyde does not.
(c) Benzoic acid reacts with NaHCO₃ giving CO₂ gas with effervescence whereas phenol does not.
C₆H₅COOH (Benzoic acid) + NaHCO₃ → C₆H₅COONa (Sodium benzoate) + CO₂↑ + H₂O

Teacher's Note:
1. Write the observation for both compounds of the pair; stating only the positive result is a common way of losing marks.

Question 39. (i) Draw the structures of the following derivatives:
(a) Propanone oxime,
(b) Semicarbazone of CH₃CHO.
(ii) How will you convert ethanal into the following compounds? Give the chemical equations involved.
(a) CH₃–CH₃
(b) CH₃–CH(OH)–CH₂–CHO
(c) CH₃CH₂OH [CBSE Comptt. OD 2015]
Answer:
(i) (a) (CH₃)₂C=N–OH
(b) CH₃(H)C=N–NH–CO–NH₂
(ii) (a) CH₃CHO \( \xrightarrow[\text{conc HCl}]{\text{Zn-Hg}} \) CH₃–CH₃
(b) 2CH₃–CHO \( \xrightarrow{\text{dil. NaOH}} \) CH₃–CH(OH)–CH₂–CHO
(c) CH₃CHO \( \xrightarrow{\text{LiAlH}_{4}} \) CH₃CH₂OH

Teacher's Note:
1. In oximes and semicarbazones, the C=O oxygen is replaced by =N–; the carbon skeleton of the carbonyl compound is unchanged.

Question 40. (i) Give a plausible explanation for each one of the following:
(a) Although phenoxide ion has more number of resonating structures than carboxylate ion, carboxylic acid is a stronger acid than phenol.
(b) There are two –NH₂ groups in semicarbazide. However, only one is involved in the formation of semicarbazones.
(ii) Carry out the following conversions in not more than two steps:
(a) Phenyl magnesium bromide to benzoic acid.
(b) Acetaldehyde to But-2-enal.
(c) Benzene to m-Nitroacetophenone.
Answer:
(i)
(a) Phenoxide ion has non-equivalent resonance structures in which the negative charge is at the lesser electronegative carbon atom whereas in case of carboxylate ion both the resonating structures are equivalent. The negative charge is delocalised over two electronegative oxygen atoms in carboxylate ion whereas in phenoxide ion, the negative charge less effectively delocalises over one oxygen atom and less electronegative carbon atoms. So, the carboxylate ion is more resonance stabilised than phenoxide ion. Thus, the release of proton from carboxylic acid is much easier than from phenol. Hence, carboxylic acid is a stronger acid than phenol.
(b) Semicarbazide has two –NH₂ groups. One of them, which is directly attached to >C=O is involved in resonance. Thus, electron density on this group decreases and it does not act as a nucleophile. In contrast, the lone pair of electrons on the other –NH₂ group is available for nucleophilic attack.
(ii) (a) PhMgBr + O=C=O → PhCOOMgBr \( \xrightarrow{\text{H}_{2}\text{O}} \) PhCOOH
(b) 2CH₃CHO \( \xrightarrow{\text{OH}^{-}} \) CH₃CH(OH)–CH₂CHO \( \xrightarrow{\text{Heat}} \) CH₃CH=CH–CHO
(c) C₆H₆ \( \xrightarrow{\text{(CH}_{3}\text{CO)}_{2}\text{O/Anhy AlCl}_{3}} \) PhCOCH₃ \( \xrightarrow{\text{conc. H}_{2}\text{SO}_{4}\text{ + conc. HNO}_{3}} \) m–NO₂C₆H₄COCH₃ (or CH₃COCl/AlCl₃ in the first step)

Teacher's Note:
1. In semicarbazide, H₂N–NH–CO–NH₂, the reacting NH₂ is the one on the hydrazine end, away from the C=O group.

Question 41. (i) Give a simple chemical test to distinguish between the pair of organic compounds: Ethanal and Propanal.
(ii) Name and complete the following chemical reaction:
RCH₂COOH \( \xrightarrow{\text{(i) X}_{2}\text{/red P (ii) H}_{2}\text{O}} \)
(iii) Draw the structures of the following derivatives:
(a) The 2,4-Dinitrophenylhydrazone of benzaldehyde,
(b) Acetaldehyde dimethyl acetal
(c) Cyclopropanone oxime.
Answer:
(i) Ethanal and propanal can be distinguished by Iodoform test.
Ethanal gives a yellow precipitate of iodoform with an alkaline solution of NaOH. Propanal does not gives this test.
CH₃CHO + 3NaOI \( \xrightarrow{\text{NaOH/I}_{2}} \) CHI₃ + HCOONa + 2NaOH
(ii) RCH₂COOH \( \xrightarrow{\text{(i) X}_{2}\text{/red P (ii) H}_{2}\text{O}} \) RCH(X)COOH (α-Halo carboxylic acid)
The name of the reaction is Hell-Volhard-Zelinsky reaction.
(iii) (a) C₆H₅CH=NNH–C₆H₃(NO₂)₂ (NO₂ groups at positions 2 and 4 of the ring attached to NH)
(b) CH₃–CH(OCH₃)₂
(c) A three-membered ring in which one ring carbon is double bonded to N–OH: (CH₂)₂C=NOH

Teacher's Note:
1. Ethanal is the only aldehyde that gives the iodoform test, because it is the only one with a CH₃CO– group.