CBSE Class 12 Chemistry Aldehydes Ketones Carboxylic Acids Questions Answers

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ALDEHYDES, KETONES AND CARBOXYLIC ACIDS

SECTION-A (One mark Questions)

1. Name one distinguishing test between aldehydes and ketones?

Answer: Aldehydes and ketones can be distinguished by Tollens' test. Aldehydes give a silver mirror on reacting with Tollens' reagent, whereas ketones do not react.

Teacher's Note:
a) Tollens' reagent is ammoniacal silver nitrate, and the silver mirror forms only with the aldehyde group.
b) Ketones lack the easily oxidised C-H of the aldehyde group, so they give no mirror.

2. Give reason why Formaldehyde does not undergo aldol condensation?

Answer: Formaldehyde does not have any α-hydrogen, and therefore it cannot show aldol condensation.

Teacher's Note:
a) Aldol condensation needs an α-hydrogen to form the enolate ion.
b) Formaldehyde, HCHO, has only hydrogen atoms attached to its carbonyl carbon.

3. Carboxylic acids have higher boiling points than alcohols of same no. of carbon atoms?

Answer: Carboxylic acids have more extensive association of molecules through intermolecular hydrogen bonding than alcohols.

Teacher's Note:
a) Carboxylic acids form hydrogen-bonded dimers, which effectively double the molecular size.
b) More hydrogen bonding means more heat is needed to separate the molecules, so the boiling point is higher.

4. Write IUPAC name of CH₃COCH₂COCH₃.

Answer: Pentane-2,4-dione.

Teacher's Note:
a) The chain has five carbons, and the two ketone groups are at positions 2 and 4.

5. What product is obtained when Ethyl benzene is oxidized with alkaline KMnO₄?

Answer: Benzoic acid is formed.

Teacher's Note:
a) Any alkyl side chain on a benzene ring is oxidised to a carboxylic acid group, whatever its length.

6. Give chemical test to distinguish between acetaldehyde and benzaldehyde.

Answer: Acetaldehyde gives the iodoform test, whereas benzaldehyde does not.

Teacher's Note:
a) Only compounds with a CH₃CO- group give a positive iodoform test.

7. Write one chemical to distinguish between Formic acid and Acetic acid.

Answer: Formic acid gives a silver mirror when treated with Tollens' reagent, whereas acetic acid does not.

Teacher's Note:
a) Formic acid has an aldehyde-like CHO group, which reduces Tollens' reagent.

8. Give two important uses of formalin.

Answer: (i) Used as a preservative. (ii) Used for the preparation of Bakelite.

Teacher's Note:
a) Formalin is used to preserve biological specimens.
b) Its reaction with phenol gives Bakelite, a phenol-formaldehyde resin.

9. How is formalin and trioxane related to methanal?

Answer: Formalin is a 40% aqueous solution of methanal, whereas trioxane is a trimer of methanal.

Teacher's Note:
a) The printed key says "methanol" here; formalin is an aqueous solution of methanal (formaldehyde), not methanol.
b) Trioxane is a cyclic trimer of three methanal units.

10. Complete the following reaction and give the name of the major product.
HCHO + CH₃MgX → ?

Answer: HCHO + CH₃MgX → CH₃CH₂OMgX, which on hydrolysis gives CH₃CH₂OH. The major product is ethanol.

Teacher's Note:
a) The printed key gives CH₃CH₂OH with Mg(OH)X directly, which does not show the magnesium alkoxide intermediate; the hydrolysis step is needed.
b) Grignard addition to methanal gives a primary alcohol with one more carbon.

11. Draw the structural formula of Hex-2-en-4-yn-oic acid.

Answer: CH₃-C≡C-CH=CH-COOH

Teacher's Note:
a) Number from the COOH carbon as C-1; the double bond is at C-2 and the triple bond at C-4.

12. Arrange the following in the increasing order of acidic character.
HCOOH, ClCH₂COOH, CF₃COOH, Cl₃CCOOH

Answer: HCOOH < ClCH₂COOH < Cl₃CCOOH < CF₃COOH

Teacher's Note:
a) The printed key gives the order of decreasing acidity, which does not match the "increasing" order asked; the answer above is the requested one.
b) Electron-withdrawing halogens stabilise the carboxylate ion, and fluorine has the strongest effect.

13. Complete the reaction:-
RCONH₂ + 4NaOH + Br₂ → ?

Answer: RCONH₂ + 4NaOH + Br₂ → RNH₂ + 2NaBr + Na₂CO₃ + 2H₂O

Teacher's Note:
a) This is the Hofmann bromamide reaction, which removes one carbon from the amide.

14. Give one chemical test to distinguish between Phenol and benzoic acid.

Answer: On treatment with neutral FeCl₃ solution, phenol gives a violet colour, whereas benzoic acid does not.

Teacher's Note:
a) The violet colour is due to a complex formed by phenol with iron(III).

15. Most of the aromatic acids are solids while acetic acids and others of this series are liquids. Why?

Answer: Aromatic acids have higher molecular weights, and therefore stronger van der Waals forces of attraction than aliphatic acids, so they are solids.

Teacher's Note:
a) Larger molecules have larger surface area and stronger van der Waals forces.

SECTION-B (2 Mark Questions.)

1. Would you expect benzaldehyde to be more or less reactive in nucleophilic addition reaction than Propanal? Explain your answer.

Answer: Benzaldehyde is less reactive than propanal. The carbon atom of the carbonyl group of benzaldehyde is less electrophilic than that of propanal, because the polarity of the carbonyl group is reduced by resonance with the benzene ring.

Teacher's Note:
a) Resonance with the ring delocalises the positive character of the carbonyl carbon.
b) A less electrophilic carbon is attacked more slowly by a nucleophile.

2. Describe the Trans esterification reaction giving an example.

Answer: When an ester reacts with an alcohol to form another ester and another alcohol, the process is called trans-esterification. CH₃COOC₂H₅ + CH₃OH → CH₃COOCH₃ + C₂H₅OH

Teacher's Note:
a) The printed key writes CH₃COOC₂H₅ as the product, which is the starting ester; the correct new ester is methyl acetate, CH₃COOCH₃.
b) The reaction is an equilibrium, and an excess of the alcohol drives it forward.

3. Explain Hell- Volhard -Zelinsky reaction with an example.

Answer: Carboxylic acids having an α-hydrogen atom are halogenated at the α-position on treatment with chlorine or bromine in the presence of a small amount of red phosphorus, to give α-halo carboxylic acids. RCH₂COOH → (i) X₂/P (red); (ii) H₂O/H⁺ → RCH(X)COOH

Teacher's Note:
a) Red phosphorus converts the acid to an acyl halide, which enolises more readily.
b) The α-halo acid is obtained after hydrolysis with water.

4. Give simple chemical tests to distinguish between:-
(i) Pentan-2-one and Pentan-3-one
(ii) Ethanal and propanal

Answer: (i) Pentan-2-one gives the iodoform test on treatment with I₂/NaOH, whereas pentan-3-one does not. (ii) Ethanal gives the iodoform test, whereas propanal does not.

Teacher's Note:
a) Only methyl ketones and acetaldehyde carry the CH₃CO- group needed for iodoform.
b) Pentan-3-one has two ethyl groups and no CH₃CO- unit.

5. Although Phenoxide ion has more number of resonating structures than Carboxylate ion, Carboxylic acids are more acidic than Phenol. Why?

Answer: In the carboxylate ion, the negative charge is delocalised over two oxygen atoms, whereas in the phenoxide ion it is delocalised over one oxygen atom and the ring carbons. The carboxylate ion is therefore more stable, so carboxylic acids are more acidic than phenol.

Teacher's Note:
a) Acidity depends on the stability of the conjugate base, not on the number of resonance structures.
b) Equal sharing of the charge between two oxygens is the key point.

6. Why is there a large difference in the boiling points of butanal and butan-1-ol?

Answer: Butan-1-ol has a higher boiling point because of intermolecular hydrogen bonding between its molecules, which butanal cannot form.

Teacher's Note:
a) The O-H group of the alcohol forms strong hydrogen bonds.
b) Butanal has only a polar C=O group, so its molecules are held by weaker dipole-dipole attraction.

7. Name the electrophile produced in the reaction of benzene with benzoyl chloride in the presence of anhydrous AlCl₃. Name the reaction also.

Answer: The electrophile is the benzoyl cation (benzoylium ion), C₆H₅CO⁺. The reaction is Friedel-Crafts acylation.

Teacher's Note:
a) AlCl₃ removes chloride from benzoyl chloride to generate the acylium ion.
b) The acylium ion attacks the benzene ring to give benzophenone.

8. Arrange the following in decreasing order of their acidic strength and give reason for your answer.
CH₃CH₂OH, CH₃COOH, ClCH₂COOH, FCH₂COOH, C₆H₅CH₂COOH

Answer: FCH₂COOH > ClCH₂COOH > C₆H₅CH₂COOH > CH₃COOH > CH₃CH₂OH. The acidity is due to the positive and negative inductive effects of the groups attached to the acid. The electron-withdrawing F and Cl atoms stabilise the carboxylate ion, while the alkyl group of ethanol donates electrons and makes it the weakest acid.

Teacher's Note:
a) Fluorine has a stronger -I effect than chlorine, so fluoroacetic acid is the strongest acid in this list.
b) Ethanol has no carboxyl group, so its conjugate base is not stabilised by resonance.

9. Write the names associated with the following reactions:-
(i) CH₃COCl + H₂ → CH₃CHO (in presence of Pd-C/BaSO₄)
(ii) CH₃CN → CH₃CHO (i) SnCl₂/HCl (ii) H₂O/H⁺

Answer: (i) Rosenmund reduction. (ii) Stephen reaction.

Teacher's Note:
a) Rosenmund reduction uses a poisoned palladium catalyst so that the reduction stops at the aldehyde.
b) In the Stephen reaction, SnCl₂/HCl reduces the nitrile to an imine salt, which is hydrolysed to the aldehyde.

10. Complete the following reaction sequence:
CH₃COCH₃ → (i) CH₃MgBr, (ii) H₂O → (A) → Na metal, Ether → (B) → CH₃-Br → (C)

Answer: (A) = (CH₃)₃C-OH (2-methylpropan-2-ol); (B) = (CH₃)₃C-O^-Na⁺ (sodium tert-butoxide); (C) = (CH₃)₃C-O-CH₃ (tert-butyl methyl ether).

Teacher's Note:
a) Grignard addition to acetone gives a tertiary alcohol after hydrolysis.
b) Sodium reacts with the alcohol to give the alkoxide, which is then methylated by CH₃Br (Williamson synthesis).

SECTION- C (THREE MARKS QUESTIONS)

1. What happens when:- (i) an aqueous solution of Sodium acetate is electrolysed (ii) Calcium acetate is dry distilled (iii) Sodium benzoate is heated with Sodalime

Answer:
(i) 2CH₃COONa + 2H₂O → C₂H₆ + 2CO₂ + 2NaOH + H₂ (ethane is formed).
(ii) (CH₃COO)₂Ca → CH₃COCH₃ + CaCO₃ (acetone is formed).
(iii) C₆H₅COONa + NaOH → (CaO, heat) C₆H₆ + Na₂CO₃ (benzene is formed).

Teacher's Note:
a) Electrolysis of sodium acetate (Kolbe electrolysis) gives ethane at the anode, with CO₂ released.
b) Soda lime decarboxylates the benzoate to benzene.

2. Write IUPAC names of the following Compounds:-
(i) CH₃CO(CH₂)₄CH₃ (ii) Ph-CH=CH-CHO (iii) [Figure: cyclopentane ring with a CHO group attached]

Answer: (i) Heptan-2-one. (ii) 3-Phenylprop-2-en-1-al. (iii) Cyclopentanecarbaldehyde.

Teacher's Note:
a) The printed key writes "Heptane-2-one"; the correct name is heptan-2-one, because the chain has seven carbons and one ketone group.
b) In (iii), the -CHO group is attached to the ring, so the name ends in carbaldehyde.

3. Complete the following equations:-
(i) CH₃CONH₂ → (P₂O₅, heat) ?
(ii) 2CH₃CHO → (dil. NaOH) ?
(iii) C₆H₅COOH → (HNO₃/H₂SO₄) ?

Answer:
(i) CH₃CONH₂ → CH₃CN + H₂O
(ii) 2CH₃CHO → CH₃CH(OH)CH₂CHO (3-hydroxybutanal)
(iii) C₆H₅COOH → m-O₂N-C₆H₄-COOH (m-nitrobenzoic acid)

Teacher's Note:
a) P₂O₅ removes water from the amide to give the nitrile.
b) Dilute alkali causes aldol addition, giving a β-hydroxy aldehyde, not dehydration, at room temperature.
c) The -COOH group is meta-directing, so nitration gives the meta product.

4. Explain the following:-
(i) Gatterman-Koch reaction
(ii) Clemmensen reduction
(iii) Wolff-Kishner Reduction

Answer:
(i) When benzene or its derivative is treated with carbon monoxide and hydrogen chloride in the presence of anhydrous aluminium chloride or cuprous chloride, benzaldehyde or a substituted benzaldehyde is formed. This is the Gatterman-Koch reaction.
(ii) The carbonyl group of aldehydes and ketones is reduced to a CH₂ group on treatment with zinc amalgam and concentrated hydrochloric acid. This is Clemmensen reduction.
(iii) The carbonyl group is converted to a CH₂ group by treatment with hydrazine, followed by heating with sodium or potassium hydroxide in a high-boiling solvent such as ethylene glycol. This is Wolff-Kishner reduction.

Teacher's Note:
a) Clemmensen reduction works in acid and Wolff-Kishner reduction works in base, so they suit different substrates.
b) The printed key labels both reductions as (ii); the Wolff-Kishner reduction is (iii).

5. Predict the products of the following reactions
(i) cyclopentanone + HO-NH₂ (H⁺)
(ii) cyclohexanone + NH₂-NH-(2,4-dinitrophenyl) (2,4-dinitrophenylhydrazine)
(iii) R-CH=CH-CHO + NH₂-CO-NH-NH₂ (H⁺)

Answer:
(i) Cyclopentanone oxime.
(ii) Cyclohexanone 2,4-dinitrophenylhydrazone (an orange precipitate).
(iii) R-CH=CH-CH=N-NH-CO-NH₂ (a semicarbazone).

Teacher's Note:
a) Hydroxylamine and the other amine derivatives add to C=O and lose one molecule of water.
b) The 2,4-DNP derivative is the one used to identify aldehydes and ketones.

6. An alkene 'A' (Mol. formula C₅H₁₀) on ozonolysis gives a mixture of two compounds 'B' and 'C'. Compound 'B' gives positive Fehling's test and also forms iodoform on treatment with I₂ and NaOH. Compound 'C' does not give Fehling's test but forms iodoform. Identify the compounds A, B and C. Write the reaction for ozonolysis and formation of iodoform from B and C

Answer:
1. A is 2-methylbut-2-ene, CH₃-CH=C(CH₃)-CH₃.
2. Ozonolysis: CH₃-CH=C(CH₃)-CH₃ → (i) O₃, (ii) Zn/H₂O → CH₃CHO (B) + CH₃COCH₃ (C).
3. Iodoform from B: CH₃CHO + 3I₂ + 4NaOH → CHI₃ + HCOONa + 3NaI + 3H₂O.
4. Iodoform from C: CH₃COCH₃ + 3I₂ + 4NaOH → CHI₃ + CH₃COONa + 3NaI + 3H₂O.

Teacher's Note:
a) B is acetaldehyde, which reduces Fehling's solution, while C is acetone, which does not.
b) Both B and C contain CH₃CO-, so both give the iodoform test.

SECTION- D (FIVE MARKS QUESTIONS)

1. 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. (A) forms a 2,4-DNP derivative, so it is an aldehyde or a ketone. It does not reduce Tollens' or Fehling's reagent, so it is a ketone.
2. (A) gives the iodoform test, so it is a methyl ketone, CH₃CO-.
3. The formula C₈H₈O shows a high degree of unsaturation, but (A) does not decolourise bromine water or Baeyer's reagent. The unsaturation must therefore be due to an aromatic ring.
4. Oxidation of a methyl ketone with chromic acid gives a carboxylic acid (B) of formula C₇H₆O₂, which is benzoic acid.
5. Hence (A) is phenyl methyl ketone (acetophenone), C₆H₅COCH₃, and (B) is benzoic acid, C₆H₅COOH.
6. With I₂/NaOH, (A) gives iodoform (yellow precipitate) and sodium benzoate: C₆H₅COCH₃ + 3I₂ + 4NaOH → C₆H₅COONa + CHI₃ + 3NaI + 3H₂O.

Teacher's Note:
a) The iodoform test confirms the CH₃CO- group, and the negative Tollens' and Fehling's tests rule out an aldehyde.
b) Lack of decolourisation of bromine water rules out an alkene or alkyne, so the unsaturation is an aromatic ring.

2. Write chemical reactions to affect the following transformations:
(i) Butan-1-ol to butanoic acid
(ii) Benzyl alcohol to phenyl ethanoic 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

Answer:
(i) CH₃CH₂CH₂CH₂OH → (CrO₃/H₂SO₄, Jones reagent) CH₃CH₂CH₂COOH
(ii) C₆H₅CH₂OH → (HBr) C₆H₅CH₂Br → (KCN) C₆H₅CH₂CN → (H₃O⁺, heat) C₆H₅CH₂COOH
(iii) 3-O₂N-C₆H₄Br → (Mg, dry ether) 3-O₂N-C₆H₄MgBr → (CO₂, dry ice) → (H₃O⁺) 3-O₂N-C₆H₄COOH
(iv) 4-CH₃-C₆H₄-COCH₃ → (KMnO₄/KOH, heat, then H₃O⁺) benzene-1,4-dicarboxylic acid
(v) Cyclohexene → (KMnO₄/H⁺, heat) hexane-1,6-dioic acid (adipic acid)

Teacher's Note:
a) Jones reagent oxidises the primary alcohol all the way to the acid, without stopping at the aldehyde.
b) The benzyl bromide route adds one carbon through the cyanide, which gives phenylethanoic acid.
c) Hot alkaline permanganate oxidises both the acetyl group and the ring methyl, so the product has two COOH groups.

3. 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 hydrogen sulphite and give positive iodoform test. On vigorous oxidation it gives ethanoic and propanoic acid. Write the possible structure of the compound.

Answer:
1. Atomic ratio: C = 69.77/12 = 5.81; H = 11.63/1 = 11.63; O = 18.60/16 = 1.16.
2. Dividing by 1.16 gives C : H : O = 5 : 10 : 1, so the empirical formula is C₅H₁₀O with mass 86.
3. n = 86/86 = 1, so the molecular formula is C₅H₁₀O.
4. It does not reduce Tollens' reagent, so it is a ketone. It forms a bisulphite addition compound and gives the iodoform test, so it is a methyl ketone.
5. Vigorous oxidation gives ethanoic acid and propanoic acid, so the compound is pentan-2-one, CH₃COCH₂CH₂CH₃.

Teacher's Note:
a) Oxidation of pentan-2-one breaks the C-C bond next to the carbonyl, giving CH₃COOH and CH₃CH₂COOH.
b) Pentan-3-one would give propanoic acid only, and it would not give the iodoform test.

4. An organic compound (A) molecular formula C₈H₁₆O₂ was hydrolysed with dil. H₂SO₄ to give a carboxylic acid (B) and 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. Hydrolysis: CH₃CH₂CH₂COOCH₂CH₂CH₂CH₃ + H₂O → (dil. H₂SO₄) CH₃CH₂CH₂COOH (B) + CH₃CH₂CH₂CH₂OH (C).
2. Oxidation of (C): CH₃CH₂CH₂CH₂OH → (CrO₃) CH₃CH₂CH₂COOH (B).
3. Dehydration of (C): CH₃CH₂CH₂CH₂OH → (conc. H₂SO₄, heat) CH₃CH₂CH=CH₂ (but-1-ene).

Teacher's Note:
a) The ester is butyl butanoate, which hydrolyses back to its acid and alcohol.
b) Concentrated acid at high temperature removes water from the alcohol to give the alkene.

5. What is meant by the following terms:- (a) Cyanohydrin (b) Semicarbazone (c) Hemiacetal (d) Ketal (e) 2,4-DNP derivative

Answer:
(a) Cyanohydrin: a compound in which -CN and -OH groups are attached to the same carbon atom, for example CH₃CH(OH)CN.
(b) Semicarbazone: the product formed when an aldehyde or ketone reacts with semicarbazide, for example CH₃CH=N-NHCONH₂.
(c) Hemiacetal: the product formed when an aldehyde reacts with one mole of alcohol in the presence of dry HCl gas, for example CH₃CH(OH)OCH₃.
(d) Ketal: the product formed when a ketone reacts with two moles of alcohol, or with ethylene glycol, in the presence of dry HCl gas.
(e) 2,4-DNP derivative: the orange precipitate formed when an aldehyde or ketone reacts with 2,4-dinitrophenylhydrazine.

Teacher's Note:
a) Cyanohydrin formation is a nucleophilic addition of CN^- to the carbonyl carbon.
b) A hemiacetal is unstable, and it becomes an acetal or ketal with more alcohol.

AMINES

Q.1 Give the IUPAC name of H₂N-CH₂-CH₂-CH=CH₂.

Answer: But-3-en-1-amine.

Teacher's Note:
a) Numbering starts from the end bearing the amino group, so the carbon bearing NH₂ is C-1.
b) The printed key gives 4-amino-but-1-ene, which is not the correct IUPAC form.

Q.2 Write structure of methyl amine?

Answer: CH₃-NH₂

Teacher's Note:
a) Methylamine is the simplest primary aliphatic amine.
b) It is a gas at room temperature with a fishy odour.

Q.3 Write the structure of methyl isocyanides?

Answer: CH₃-NC

Teacher's Note:
a) In an isocyanide, the carbon is bonded directly to nitrogen.
b) Methyl isocyanide is a structural isomer of acetonitrile.

Q.4 Name the tests for Primary amine.

Answer: Carbylamine test.

Teacher's Note:
a) Primary amines give a foul-smelling isocyanide on warming with CHCl₃ and alcoholic KOH.
b) Secondary and tertiary amines do not give this test.

Q.5 Primary amines have higher b.p than tertiary amines.

Answer: Due to intermolecular hydrogen bonding in primary amines.

Teacher's Note:
a) Primary amines have N-H bonds, which allow hydrogen bonding between molecules.
b) Tertiary amines have no N-H, so they boil at lower temperatures.

Q.6 Why is alkyl amine more basic than ammonia?

Answer: Due to the +I effect of the alkyl group, which increases electron density on nitrogen.

Teacher's Note:
a) Alkyl groups push electrons towards nitrogen, so the lone pair binds H⁺ more strongly.

Q.7 why do amine react as nucleophile.

Answer: Due to the lone pair of electrons on nitrogen.

Teacher's Note:
a) The lone pair donates to electron-deficient centres.
b) This is the same reason amines act as bases.

Q.8 Why are aqueous solution of amine basic in nature?

Answer: Because of the high electron density on nitrogen, the amine gains H⁺ from water, releasing OH^- ions.

Teacher's Note:
a) Amines accept protons from water, which leaves OH^- in solution.

Q.9 Name one test to distinguish between ethyl cyanide and ethyl isocyanides.

Answer: Ethyl cyanide on hydrolysis with acids gives propionic acid, whereas ethyl isocyanide with dilute HCl gives ethylamine and formic acid.

Teacher's Note:
a) The two isomers give different products on acid hydrolysis, which is the basis of the test.
b) Ethylamine is a basic gas that can be identified by its smell.

Q.10 Identify A and B
C₆H₅NH₂ → (NaNO₂/HCl) A → (CuBr) B

Answer: A is benzenediazonium chloride, C₆H₅N₂⁺Cl^-. B is bromobenzene, C₆H₅Br.

Teacher's Note:
a) NaNO₂ and HCl at low temperature convert the amine to the diazonium salt.
b) CuBr in the Sandmeyer reaction replaces the diazonium group with bromine.

Q.11 Name the reaction in which amide directly converted into amines.

Answer: Hofmann's bromamide reaction.

Teacher's Note:
a) The reaction uses Br₂ and NaOH on a primary amide.
b) The product amine has one carbon less than the starting amide.

Q.12 Complete the following:
RNH₂ + CHCl₃ + 3KOH → ?

Answer: RNH₂ + CHCl₃ + 3KOH → RNC + 3KCl + 3H₂O (an alkyl isocyanide is formed).

Teacher's Note:
a) This is the carbylamine reaction, and the isocyanide has a foul smell.
b) Three moles of KOH are needed to remove the three chlorines from CHCl₃.

Q.13 Complete the following:
RCONH₂ + Br₂ + 4 NaOH → ?

Answer: RCONH₂ + Br₂ + 4NaOH → RNH₂ + 2NaBr + Na₂CO₃ + 2H₂O

Teacher's Note:
a) This is the Hofmann bromamide reaction, which gives a primary amine.
b) The carbonyl carbon is lost as sodium carbonate.

Q.14 Write the formula of hinsberg's reagent.

Answer: Benzenesulphonyl chloride, C₆H₅SO₂Cl.

Teacher's Note:
a) The printed key gives benzoyl chloride, which is not Hinsberg's reagent; the correct reagent is benzenesulphonyl chloride.
b) It is used to distinguish primary, secondary and tertiary amines.

Q.15 What is meant by diazotization?

Answer: Conversion of primary aromatic amines into diazonium salts.

Teacher's Note:
a) Diazotisation uses NaNO₂ and HCl at 0-5 °C.
b) The diazonium salt is unstable, so it is used at once.

2 marks questions

Q.1 In an increasing order of basic strength:
C₆H₅NH₂, C₆H₅N(CH₃)₂, (C₂H₅)₂NH and CH₃NH₂

Answer: C₆H₅NH₂ < C₆H₅N(CH₃)₂ < CH₃NH₂ < (C₂H₅)₂NH

Teacher's Note:
a) Aromatic amines are weaker bases, because the lone pair is delocalised into the ring.
b) Diethylamine has two electron-releasing ethyl groups, so it is the strongest base in this list.

Q.2 In a decreasing order of basic strength:
Aniline, p-nitro aniline and p-toluidine

Answer: p-Toluidine > Aniline > p-Nitroaniline

Teacher's Note:
a) The methyl group donates electrons to the ring, which makes p-toluidine the most basic.
b) The nitro group withdraws electrons strongly, so p-nitroaniline is the least basic.

Q.3 In an increasing order of pK_b values:
C₂H₅NH₂, C₆H₅NHCH₃, (C₂H₅)₂NH and C₆H₅NH₂

Answer: (C₂H₅)₂NH < C₂H₅NH₂ < C₆H₅NHCH₃ < C₆H₅NH₂

Teacher's Note:
a) A stronger base has a smaller pK_b, so the order is the reverse of basicity.
b) Aniline has the largest pK_b because its lone pair is delocalised in the ring.

Q.4 Write a chemical reaction in which the iodide ion replaces the diazonium group in a diazonium salt.

Answer: C₆H₅N₂⁺Cl^- + KI → C₆H₅I + KCl + N₂

Teacher's Note:
a) Iodide displaces N₂ without a copper catalyst.
b) The nitrogen gas released drives the reaction forward.

Q.5 Why is an alkyl amine more basic than ammonia?

Answer: An alkyl amine is more basic than ammonia because of the +I (inductive) effect of the alkyl group, which releases electrons and increases the electron density on nitrogen.

Teacher's Note:
a) The alkyl group pushes electrons towards nitrogen.
b) Greater electron density makes the lone pair more available to accept H⁺.

3 marks questions

Q.1 Describe the Hofmann's bromamide reaction

Answer: Hofmann's bromamide reaction involves the reaction of bromine with an acid amide in the presence of an alkali. It results in the formation of a primary amine with one carbon less than the parent compound. The alkyl group migrates from the carbonyl carbon, with the elimination of CO₂.
RCONH₂ + Br₂ + 4KOH → RNH₂ + 2KBr + K₂CO₃ + 2H₂O
For example, CH₃CONH₂ + Br₂ + 4KOH → CH₃NH₂ + 2KBr + K₂CO₃ + 2H₂O

Teacher's Note:
a) The amide is first converted to an N-bromoamide, which rearranges.
b) Acetamide gives methylamine.

Q.2 Describe the Gatterman reaction

Answer: Gatterman reaction is a modification of Sandmeyer's reaction in which benzenediazonium chloride is treated with copper powder and halogen acid to form aryl halides. C₆H₅N₂⁺Cl^- → (Cu/HCl) C₆H₅Cl + N₂

Teacher's Note:
a) Copper powder with the halogen acid replaces the diazonium group.
b) Nitrogen gas is evolved, which is the driving force.

Q.3 Describe the coupling reaction

Answer: Coupling reaction is the reaction of diazonium salts with phenols and aromatic amines to form azo compounds of the general formula Ar-N=N-Ar. The coupling of phenol takes place in a mildly alkaline medium at 273-278 K.

Teacher's Note:
a) The diazonium ion is a weak electrophile, so it needs an activated ring such as phenol.
b) The alkaline medium forms the phenoxide ion, which couples more readily.

Q.4 pK_b for aniline is more than that for methylamine.

Answer: In aniline, the lone pair on nitrogen is delocalised over the benzene ring, so the electron density on nitrogen decreases. In methylamine, the +I effect of CH₃ increases the electron density on nitrogen. Aniline is therefore a weaker base than methylamine, and its pK_b is higher.

Teacher's Note:
a) Resonance with the ring is the main reason aniline is a weak base.
b) A higher pK_b means a weaker base.

Q.5 Methylamine Soln in water reacts with ferric chloride Soln to give a precipitate of ferric hydroxide.

Answer: Being more basic than water, methylamine accepts a proton from water, liberating OH^- ions.
CH₃NH₂ + H₂O → CH₃NH₃⁺ + OH^-
These OH^- ions combine with Fe³⁺ ions from FeCl₃ to form a brown precipitate of hydrated ferric oxide, Fe(OH)₃.

Teacher's Note:
a) The OH^- ions produced by methylamine cause the precipitate.
b) The brown precipitate is hydrated ferric oxide, formed from the hydroxide.

5 marks questions

Q.1 An aromatic compound 'A' on treatment with aqueous ammonia and heating forms compound 'B' which on heating with Br₂ and KOH forms a compound 'C' of molecular formula C₆H₇N. Write the structures and IUPAC names of compounds A, B and C.

Answer:
1. A is benzoic acid, C₆H₅COOH.
2. Heating with aqueous ammonia gives ammonium benzoate, which on heating forms B, benzamide, C₆H₅CONH₂.
3. Hofmann bromamide reaction of benzamide with Br₂/KOH removes one carbon, giving C, aniline, C₆H₅NH₂ (benzenamine).

Teacher's Note:
a) Ammonium salts dehydrate to amides on heating.
b) Aniline has the formula C₆H₇N, which confirms the identity of C.

Q.2 Complete the following reactions:
(i) C₆H₅NH₂ + CHCl₃ + alc. KOH →
(ii) C₆H₅N₂Cl + H₃PO₂ + H₂O →
(iii) C₆H₅NH₂ + H₂SO₄ (conc.) →
(iv) C₆H₅N₂Cl + C₂H₅OH →
(v) C₆H₅NO₂ + Fe/HCl →

Answer:
(i) C₆H₅NH₂ + CHCl₃ + 3KOH → C₆H₅NC + 3KCl + 3H₂O (phenyl isocyanide).
(ii) C₆H₅N₂Cl + H₃PO₂ + H₂O → C₆H₆ + N₂ + H₃PO₃ + HCl.
(iii) C₆H₅NH₂ + H₂SO₄ (conc.) → p-H₂N-C₆H₄-SO₃H (sulphanilic acid) + H₂O.
(iv) C₆H₅N₂Cl + C₂H₅OH → C₆H₆ + N₂ + CH₃CHO + HCl.
(v) C₆H₅NO₂ + 6[H] (Fe/HCl) → C₆H₅NH₂ + 2H₂O.

Teacher's Note:
a) Reaction (i) is the carbylamine test, which gives a foul-smelling isocyanide.
b) In (iii) the sulphonic acid group enters the para position, giving sulphanilic acid.
c) In (iv), ethanol reduces the diazonium salt, replacing the group with H.

Ans. Aromatic acids have higher molecular weights. Therefore more Vander Waal’s force of attraction as compared to aliphatic acids and hence they are solids.