Intext Questions
1. Write the structures of the following compounds. (i) α-Methoxypropionaldehyde (ii) 3-Hydroxybutanal (iii) 2-Hydroxycyclopentane carbaldehyde (iv) 4-Oxopentanal (v) Di-sec. butyl ketone (vi) 4-Fluoroacetophenone
Answer
(i) CH₃–CH(OCH₃)–CHO
(ii) CH₃–CH(OH)–CH₂–CHO
(iii) A cyclopentane ring with –CHO on C-1 and –OH on the adjacent C-2
(iv) CH₃–CO–CH₂–CH₂–CHO
(v) CH₃CH₂–CH(CH₃)–CO–CH(CH₃)–CH₂CH₃
(vi) F–C₆H₄–CO–CH₃ (F para to the acetyl group)
2. Write the structures of products of the following reactions: (i) C₆H₆ + C₂H₅COCl —anhyd. AlCl₃, CS₂→ (ii) (C₆H₅CH₂)₂Cd + 2CH₃COCl → (iii) H₃C–C≡C–H —Hg²⁺, H₂SO₄→ (iv) p-nitrotoluene —1. CrO₂Cl₂ 2. H₃O⁺→
Answer
(i) C₆H₅–CO–CH₂CH₃ (propiophenone, 1-phenylpropan-1-one) + HCl
(ii) 2 C₆H₅CH₂–CO–CH₃ (1-phenylpropan-2-one) + CdCl₂
(iii) CH₃–CO–CH₃ (propanone), via the enol CH₃C(OH)=CH₂
(iv) p-O₂N–C₆H₄–CHO (4-nitrobenzaldehyde) – Etard reaction
3. Arrange the following compounds in increasing order of their boiling points. CH₃CHO, CH₃CH₂OH, CH₃OCH₃, CH₃CH₂CH₃
Answer
CH₃CH₂CH₃ < CH₃OCH₃ < CH₃CHO < CH₃CH₂OH. Propane has only weak dispersion forces; the ether is weakly polar; the aldehyde has strong dipole–dipole attraction; ethanol is held by hydrogen bonds.
4. Arrange the following compounds in increasing order of their reactivity in nucleophilic addition reactions. (i) Ethanal, Propanal, Propanone, Butanone. (ii) Benzaldehyde, p-Tolualdehyde, p-Nitrobenzaldehyde, Acetophenone.
Answer
(i) Butanone < Propanone < Propanal < Ethanal. Alkyl groups release electrons (reduce the positive charge on carbonyl carbon) and add steric bulk, so reactivity falls as alkyl groups increase.
(ii) Acetophenone < p-Tolualdehyde < Benzaldehyde < p-Nitrobenzaldehyde. The ketone is least reactive (two groups on C=O); –CH₃ at para position releases electrons and lowers reactivity; –NO₂ withdraws electrons and makes the carbonyl carbon more electrophilic.
5. Predict the products of the following reactions: (i) cyclopentanone + HO–NH₂ —H⁺→ (ii) cyclohexanone + 2,4-dinitrophenylhydrazine (NH₂–NH–C₆H₃(NO₂)₂) → (iii) R–CH=CH–CHO + NH₂–CO–NH–NH₂ —H⁺→ (iv) C₆H₅–CO–CH₃ + CH₃CH₂NH₂ —H⁺→
Answer
(i) (CH₂)₄C=N–OH, cyclopentanone oxime + H₂O
(ii) (CH₂)₅C=N–NH–C₆H₃(NO₂)₂, cyclohexanone 2,4-dinitrophenylhydrazone + H₂O
(iii) R–CH=CH–CH=N–NH–CO–NH₂ (semicarbazone) + H₂O
(iv) C₆H₅–C(CH₃)=N–CH₂CH₃ (an imine / Schiff base) + H₂O
6. Give the IUPAC names of the following compounds: (i) PhCH₂CH₂COOH (ii) (CH₃)₂C=CHCOOH (iii) cyclopentane with –COOH and a –CH₃ on the adjacent carbon (iv) benzoic acid with –NO₂ groups at 2, 4 and 6 positions
Answer
(i) 3-Phenylpropanoic acid
(ii) 3-Methylbut-2-enoic acid
(iii) 2-Methylcyclopentanecarboxylic acid
(iv) 2,4,6-Trinitrobenzoic acid
7. Show how each of the following compounds can be converted to benzoic acid. (i) Ethylbenzene (ii) Acetophenone (iii) Bromobenzene (iv) Phenylethene (Styrene)
Answer
(i) C₆H₅CH₂CH₃ —KMnO₄/KOH, heat
⇒ C₆H₅COOK —H₃O⁺
⇒ C₆H₅COOH (the whole side chain is oxidised to –COOH)
(ii) C₆H₅COCH₃ —KMnO₄/KOH, heat
⇒ C₆H₅COOK —H₃O⁺
⇒ C₆H₅COOH (or by the haloform reaction: C₆H₅COCH₃ + NaOI
⇒ C₆H₅COONa + CHI₃, then H⁺)
(iii) C₆H₅Br + Mg —dry ether
⇒ C₆H₅MgBr; C₆H₅MgBr + CO₂ (dry ice)
⇒ C₆H₅COOMgBr —H₃O⁺
⇒ C₆H₅COOH
(iv) C₆H₅CH=CH₂ —KMnO₄/KOH, heat
⇒ C₆H₅COOK —H₃O⁺
⇒ C₆H₅COOH
8. 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) F₃C–C₆H₄–COOH (p) or H₃C–C₆H₄–COOH (p)
Answer
Electron-withdrawing groups stabilise the carboxylate ion and increase acidity.
(i) CH₂FCO₂H – F withdraws electrons (–I effect).
(ii) CH₂FCO₂H – F is more electronegative than Cl, so its –I effect is stronger.
(iii) CH₃CHFCH₂CO₂H – here F is nearer to –COOH; –I effect weakens with distance.
(iv) F₃C–C₆H₄–COOH – –CF₃ is strongly electron-withdrawing, while –CH₃ is electron-releasing and lowers acidity.
Exercises
1. What is meant by the following terms? Give an example of the reaction in each case. (i) Cyanohydrin (ii) Acetal (iii) Semicarbazone (iv) Aldol (v) Hemiacetal (vi) Oxime (vii) Ketal (viii) Imine (ix) 2,4-DNP-derivative (x) Schiff's base
Answer
(i) Cyanohydrin: a compound with –OH and –CN on the same carbon, formed by adding HCN to an aldehyde or ketone (base catalysed). CH₃CHO + HCN
⇒ CH₃CH(OH)CN
(ii) Acetal: a gem-dialkoxy compound formed from an aldehyde and two molecules of a monohydric alcohol in dry HCl. CH₃CHO + 2C₂H₅OH —dry HCl
⇒ CH₃CH(OC₂H₅)₂ + H₂O
(iii) Semicarbazone: product of an aldehyde/ketone with semicarbazide. (CH₃)₂C=O + H₂N–NH–CO–NH₂
⇒ (CH₃)₂C=N–NH–CO–NH₂ + H₂O
(iv) Aldol: a β-hydroxy aldehyde (or ketone) formed when an aldehyde with α-H reacts with dilute alkali. 2CH₃CHO —dil. NaOH
⇒ CH₃CH(OH)CH₂CHO (3-hydroxybutanal)
(v) Hemiacetal: an alkoxy-alcohol (–OH and –OR on the same carbon), the intermediate when one molecule of alcohol adds to an aldehyde. CH₃CHO + C₂H₅OH ⇌ CH₃CH(OH)OC₂H₅
(vi) Oxime: product with hydroxylamine. CH₃CHO + NH₂OH
⇒ CH₃CH=N–OH (acetaldoxime) + H₂O
(vii) Ketal: cyclic gem-dialkoxy compound from a ketone and ethylene glycol in dry HCl. (CH₃)₂C=O + HOCH₂CH₂OH
⇒ (CH₃)₂C(–OCH₂CH₂O–) (ethylene ketal) + H₂O
(viii) Imine: a compound containing C=N, formed from a carbonyl compound and ammonia or a primary amine. CH₃CHO + NH₃
⇒ CH₃CH=NH + H₂O
(ix) 2,4-DNP derivative: the coloured (yellow/orange) 2,4-dinitrophenylhydrazone formed with 2,4-dinitrophenylhydrazine (Brady's reagent); used to identify carbonyl compounds. (CH₃)₂C=O + H₂N–NH–C₆H₃(NO₂)₂
⇒ (CH₃)₂C=N–NH–C₆H₃(NO₂)₂ + H₂O
(x) Schiff's base: a substituted imine (C=N–R) formed from an aldehyde/ketone and a primary amine. C₆H₅CHO + C₆H₅NH₂
⇒ C₆H₅CH=N–C₆H₅ (benzylideneaniline) + H₂O
2. Name the following compounds according to IUPAC system of nomenclature: (i) CH₃CH(CH₃)CH₂CH₂CHO (ii) CH₃CH₂COCH(C₂H₅)CH₂CH₂Cl (iii) CH₃CH=CHCHO (iv) CH₃COCH₂COCH₃ (v) CH₃CH(CH₃)CH₂C(CH₃)₂COCH₃ (vi) (CH₃)₃CCH₂COOH (vii) OHCC₆H₄CHO-p
Answer
(i) 4-Methylpentanal
(ii) 6-Chloro-4-ethylhexan-3-one
(iii) But-2-enal
(iv) Pentane-2,4-dione
(v) 3,3,5-Trimethylhexan-2-one
(vi) 3,3-Dimethylbutanoic acid
(vii) Benzene-1,4-dicarbaldehyde
3. Draw the structures of the following compounds. (i) 3-Methylbutanal (ii) p-Nitropropiophenone (iii) p-Methylbenzaldehyde (iv) 4-Methylpent-3-en-2-one (v) 4-Chloropentan-2-one (vi) 3-Bromo-4-phenylpentanoic acid (vii) p,p'-Dihydroxybenzophenone (viii) Hex-2-en-4-ynoic acid
Answer
(i) (CH₃)₂CH–CH₂–CHO
(ii) p-O₂N–C₆H₄–CO–CH₂CH₃
(iii) p-CH₃–C₆H₄–CHO
(iv) (CH₃)₂C=CH–CO–CH₃
(v) CH₃–CO–CH₂–CH(Cl)–CH₃
(vi) CH₃–CH(C₆H₅)–CH(Br)–CH₂–COOH
(vii) HO–C₆H₄–CO–C₆H₄–OH (both –OH para to the C=O)
(viii) CH₃–C≡C–CH=CH–COOH
4. Write the IUPAC names of the following ketones and aldehydes. Wherever possible, give also common names. (i) CH₃CO(CH₂)₄CH₃ (ii) CH₃CH₂CHBrCH₂CH(CH₃)CHO (iii) CH₃(CH₂)₅CHO (iv) Ph–CH=CH–CHO (v) cyclopentane–CHO (vi) PhCOPh
Answer
(i) Heptan-2-one (methyl n-pentyl ketone)
(ii) 4-Bromo-2-methylhexanal (γ-bromo-α-methylcaproaldehyde)
(iii) Heptanal (n-heptaldehyde)
(iv) 3-Phenylprop-2-enal (cinnamaldehyde)
(v) Cyclopentanecarbaldehyde
(vi) Diphenylmethanone (benzophenone)
5. Draw structures of the following derivatives. (i) The 2,4-dinitrophenylhydrazone of benzaldehyde (ii) Cyclopropanone oxime (iii) Acetaldehyde dimethylacetal (iv) The semicarbazone of cyclobutanone (v) The ethylene ketal of hexan-3-one (vi) The methyl hemiacetal of formaldehyde
Answer
(i) C₆H₅–CH=N–NH–C₆H₃(NO₂)₂ (NO₂ groups at 2 and 4 of the phenyl ring)
(ii) (CH₂)₂C=N–OH (the C=N carbon is part of a three-membered ring)
(iii) CH₃–CH(OCH₃)₂
(iv) (CH₂)₃C=N–NH–CO–NH₂ (C=N carbon in a four-membered ring)
(v) CH₃CH₂–C(CH₂CH₂CH₃)(–O–CH₂–CH₂–O–), i.e. C-3 of hexane is part of a five-membered 1,3-dioxolane ring
(vi) HO–CH₂–OCH₃
6. 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
(i) C₆H₁₁–CH(OH)–C₆H₅ (cyclohexylphenylmethanol)
(ii) C₆H₁₁–COO⁻ (cyclohexanecarboxylate ion) + silver mirror
(iii) C₆H₁₁–CH=N–NH–CO–NH₂ (semicarbazone)
(iv) C₆H₁₁–CH(OC₂H₅)₂ (diethyl acetal)
(v) C₆H₁₁–CH₃ (methylcyclohexane) – Clemmensen reduction
7. 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
Aldol needs at least one α-H; Cannizzaro occurs with aldehydes having no α-H.
Aldol condensation: (ii), (v), (vi), (vii). Cannizzaro: (i), (iii), (ix). Neither: (iv) (ketone, no α-H) and (viii) (an alcohol, not a carbonyl compound).
(i) 2HCHO + conc. NaOH → CH₃OH + HCOONa
(ii) 2CH₃CH₂CH₂CH(CH₃)CHO —dil. NaOH
⇒ CH₃CH₂CH₂CH(CH₃)–CH(OH)–C(CH₃)(CH₂CH₂CH₃)–CHO (3-hydroxy-2,4-dimethyl-2-propylheptanal)
(iii) 2C₆H₅CHO + conc. NaOH
⇒ C₆H₅CH₂OH + C₆H₅COONa
(v) 2 cyclohexanone —dil. NaOH
⇒ 2-(1-hydroxycyclohexyl)cyclohexanone —heat, –H₂O
⇒ 2-cyclohexylidenecyclohexanone
(vi) 2C₆H₅COCH₂CH₃ —dil. NaOH
⇒ C₆H₅C(OH)(C₂H₅)–CH(CH₃)–COC₆H₅ —heat
⇒ C₆H₅C(C₂H₅)=C(CH₃)–COC₆H₅
(vii) 2C₆H₅CH₂CHO —dil. NaOH
⇒ C₆H₅CH₂CH(OH)CH(C₆H₅)CHO (3-hydroxy-2,4-diphenylbutanal) —heat
⇒ C₆H₅CH₂CH=C(C₆H₅)CHO (2,4-diphenylbut-2-enal)
(ix) 2CH₃CH₂C(CH₃)₂CHO + conc. NaOH
⇒ CH₃CH₂C(CH₃)₂CH₂OH + CH₃CH₂C(CH₃)₂COONa
8. How will you convert ethanal into the following compounds? (i) Butane-1,3-diol (ii) But-2-enal (iii) But-2-enoic acid
Answer
(i) 2CH₃CHO —dil. NaOH
⇒ CH₃CH(OH)CH₂CHO —NaBH₄
⇒ CH₃CH(OH)CH₂CH₂OH
(ii) 2CH₃CHO —dil. NaOH
⇒ CH₃CH(OH)CH₂CHO —heat, –H₂O
⇒ CH₃CH=CHCHO
(iii) CH₃CH=CHCHO (from ii) —Tollens' reagent, then H⁺
⇒ CH₃CH=CHCOOH (mild oxidant does not affect C=C)
9. 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
(a) Propanal (nucleophile) + propanal (electrophile): CH₃CH₂CH(OH)CH(CH₃)CHO – 3-hydroxy-2-methylpentanal
(b) Butanal (nucleophile) + butanal (electrophile): CH₃CH₂CH₂CH(OH)CH(C₂H₅)CHO – 2-ethyl-3-hydroxyhexanal
(c) Propanal (nucleophile, its enolate) + butanal (electrophile): CH₃CH₂CH₂CH(OH)CH(CH₃)CHO – 3-hydroxy-2-methylhexanal
(d) Butanal (nucleophile) + propanal (electrophile): CH₃CH₂CH(OH)CH(C₂H₅)CHO – 2-ethyl-3-hydroxypentanal
On heating, each loses water to give the α,β-unsaturated aldehyde (e.g. 2-methylpent-2-enal from (a)).
10. 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
2,4-DNP derivative
⇒ carbonyl compound; reduces Tollens'
⇒ aldehyde; Cannizzaro
⇒ the –CHO is on a carbon with no α-H (attached directly to the ring); oxidation to phthalic acid
⇒ the ring has two side chains in ortho positions. C₉H₁₀O minus C₆H₄ and CHO leaves C₂H₅. The compound is 2-ethylbenzaldehyde, o-C₂H₅–C₆H₄–CHO.
o-C₂H₅C₆H₄CHO —KMnO₄/KOH, then H⁺
⇒ o-C₆H₄(COOH)₂
11. 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
C gives but-1-ene on dehydration
⇒ C = butan-1-ol; its oxidation gives B = butanoic acid; so A = butyl butanoate (C₈H₁₆O₂).
CH₃CH₂CH₂COOCH₂CH₂CH₂CH₃ + H₂O —dil. H₂SO₄
⇒ CH₃CH₂CH₂COOH (B) + CH₃CH₂CH₂CH₂OH (C)
CH₃CH₂CH₂CH₂OH —H₂CrO₄→ CH₃CH₂CH₂COOH
CH₃CH₂CH₂CH₂OH —conc. H₂SO₄, heat
⇒ CH₃CH₂CH=CH₂ + H₂O
12. Arrange the following compounds in increasing order of their property as indicated: (i) Acetaldehyde, Acetone, Di-tert-butyl ketone, Methyl tert-butyl ketone (reactivity towards HCN) (ii) CH₃CH₂CH(Br)COOH, CH₃CH(Br)CH₂COOH, (CH₃)₂CHCOOH, CH₃CH₂CH₂COOH (acid strength) (iii) Benzoic acid, 4-Nitrobenzoic acid, 3,4-Dinitrobenzoic acid, 4-Methoxybenzoic acid (acid strength)
Answer
(i) Di-tert-butyl ketone < Methyl tert-butyl ketone < Acetone < Acetaldehyde (steric hindrance and +I effect increase towards the left).
(ii) (CH₃)₂CHCOOH < CH₃CH₂CH₂COOH < CH₃CH(Br)CH₂COOH < CH₃CH₂CH(Br)COOH (branching adds +I; Br nearer to –COOH has stronger –I).
(iii) 4-Methoxybenzoic acid < Benzoic acid < 4-Nitrobenzoic acid < 3,4-Dinitrobenzoic acid (–OCH₃ releases electrons; more –NO₂ groups withdraw more).
13. 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
(i) Tollens' test: propanal gives a silver mirror (or red ppt with Fehling's solution); propanone does not. (Propanone gives the iodoform test; propanal does not.)
(ii) Iodoform test (I₂/NaOH): acetophenone (C₆H₅COCH₃) gives a yellow ppt of CHI₃; benzophenone does not.
(iii) NaHCO₃ test: benzoic acid gives brisk effervescence of CO₂; phenol does not. Also neutral FeCl₃ gives violet colour with phenol and a buff ppt with benzoic acid.
(iv) NaHCO₃ test: benzoic acid gives effervescence; ethyl benzoate does not.
(v) Iodoform test: pentan-2-one (CH₃CO– group) gives yellow CHI₃; pentan-3-one does not.
(vi) Tollens' test: benzaldehyde gives a silver mirror; acetophenone does not. Iodoform test: acetophenone positive, benzaldehyde negative.
(vii) Iodoform test: ethanal (CH₃CHO) gives a yellow ppt of CHI₃; propanal does not.
14. How will you prepare the following compounds from benzene? You may use any inorganic reagent and any organic reagent having not more than one carbon atom (i) Methyl benzoate (ii) m-Nitrobenzoic acid (iii) p-Nitrobenzoic acid (iv) Phenylacetic acid (v) p-Nitrobenzaldehyde.
Answer
Start in each case with C₆H₆ + CH₃Cl —anhyd. AlCl₃
⇒ C₆H₅CH₃ (toluene).
(i) C₆H₅CH₃ —KMnO₄/KOH, heat; H₃O⁺
⇒ C₆H₅COOH —CH₃OH, conc. H₂SO₄
⇒ C₆H₅COOCH₃
(ii) C₆H₅CH₃
⇒ C₆H₅COOH (as above) —conc. HNO₃/conc. H₂SO₄
⇒ m-O₂NC₆H₄COOH (–COOH is meta-directing)
(iii) C₆H₅CH₃ —conc. HNO₃/conc. H₂SO₄
⇒ p-O₂NC₆H₄CH₃ (separated from ortho) —KMnO₄/KOH; H₃O⁺
⇒ p-O₂NC₆H₄COOH
(iv) C₆H₅CH₃ —Cl₂, hν
⇒ C₆H₅CH₂Cl —KCN
⇒ C₆H₅CH₂CN —H₃O⁺
⇒ C₆H₅CH₂COOH
(v) C₆H₅CH₃ —HNO₃/H₂SO₄
⇒ p-O₂NC₆H₄CH₃ —CrO₂Cl₂ (CS₂), then H₃O⁺
⇒ p-O₂NC₆H₄CHO (Etard reaction)
15. How will you bring about the following conversions in not more than two steps? (i) Propanone to Propene (ii) Benzoic acid to Benzaldehyde (iii) Ethanol to 3-Hydroxybutanal (iv) Benzene to m-Nitroacetophenone (v) Benzaldehyde to Benzophenone (vi) Bromobenzene to 1-Phenylethanol (vii) Benzaldehyde to 3-Phenylpropan-1-ol (viii) Benzaldehyde to α-Hydroxyphenylacetic acid (ix) Benzoic acid to m-Nitrobenzyl alcohol
Answer
(i) CH₃COCH₃ —NaBH₄
⇒ CH₃CH(OH)CH₃ —conc. H₂SO₄, heat
⇒ CH₃CH=CH₂
(ii) C₆H₅COOH —SOCl₂
⇒ C₆H₅COCl —H₂, Pd/BaSO₄ (Rosenmund)
⇒ C₆H₅CHO
(iii) CH₃CH₂OH —PCC
⇒ CH₃CHO —dil. NaOH
⇒ CH₃CH(OH)CH₂CHO
(iv) C₆H₆ —CH₃COCl, anhyd. AlCl₃
⇒ C₆H₅COCH₃ —conc. HNO₃/H₂SO₄
⇒ m-O₂NC₆H₄COCH₃
(v) C₆H₅CHO —C₆H₅MgBr, then H₃O⁺
⇒ (C₆H₅)₂CHOH —CrO₃ (or PCC)
⇒ (C₆H₅)₂C=O
(vi) C₆H₅Br —Mg, dry ether
⇒ C₆H₅MgBr —CH₃CHO, then H₃O⁺
⇒ C₆H₅CH(OH)CH₃
(vii) C₆H₅CHO + CH₃CHO —dil. NaOH, heat
⇒ C₆H₅CH=CHCHO —H₂, Ni (excess)
⇒ C₆H₅CH₂CH₂CH₂OH
(viii) C₆H₅CHO —HCN
⇒ C₆H₅CH(OH)CN —H₃O⁺
⇒ C₆H₅CH(OH)COOH
(ix) C₆H₅COOH —conc. HNO₃/H₂SO₄
⇒ m-O₂NC₆H₄COOH —B₂H₆, then H₃O⁺
⇒ m-O₂NC₆H₄CH₂OH (diborane reduces –COOH but not –NO₂)
16. Describe the following: (i) Acetylation (ii) Cannizzaro reaction (iii) Cross aldol condensation (iv) Decarboxylation
Answer
(i) Acetylation: introduction of an acetyl group (CH₃CO–) in place of an active hydrogen of –OH, –NH₂ etc., using acetyl chloride or acetic anhydride, usually with a base such as pyridine. C₂H₅OH + CH₃COCl
⇒ CH₃COOC₂H₅ + HCl; C₆H₅NH₂ + (CH₃CO)₂O
⇒ C₆H₅NHCOCH₃ + CH₃COOH.
(ii) Cannizzaro reaction: aldehydes without α-H, heated with concentrated alkali, undergo self oxidation–reduction (disproportionation): one molecule is reduced to the alcohol and another oxidised to the carboxylic acid salt. 2HCHO + conc. KOH
⇒ CH₃OH + HCOOK.
(iii) Cross aldol condensation: aldol condensation between two different aldehydes/ketones. If both have α-H, four products form. E.g. CH₃CHO + CH₃CH₂CHO —NaOH
⇒ but-2-enal, pent-2-enal, 2-methylbut-2-enal and 2-methylpent-2-enal. It is useful when one partner has no α-H: C₆H₅CHO + CH₃COCH₃ —NaOH
⇒ C₆H₅CH=CHCOCH₃ (Claisen–Schmidt).
(iv) Decarboxylation: loss of CO₂ from a carboxylic acid; heating the sodium salt with soda lime (NaOH + CaO) gives an alkane with one carbon less. CH₃COONa + NaOH —CaO, heat
⇒ CH₄ + Na₂CO₃. (Kolbe electrolysis of salts also decarboxylates.)
17. Complete each synthesis by giving missing starting material, reagent or products: (i) C₆H₅CH₂CH₃ —KMnO₄, KOH, heat→ ? (ii) o-C₆H₄(COOH)₂ —SOCl₂, heat→ ? (iii) C₆H₅CHO —H₂NCONHNH₂→ ? (iv) C₆H₆ —?→ C₆H₅COC₆H₅ (v) 4-oxocyclohexane-1-carbaldehyde —[Ag(NH₃)₂]⁺→ ? (vi) o-OHC–C₆H₄–COOH —NaCN/HCl→ ? (vii) C₆H₅CHO + CH₃CH₂CHO —dil. NaOH, heat→ ? (viii) CH₃COCH₂COOC₂H₅ —(i) NaBH₄ (ii) H⁺→ ? (ix) cyclohexanol —CrO₃→ ? (x) methylenecyclohexane —?→ cyclohexanecarbaldehyde (xi) ? —(i) O₃ (ii) Zn–H₂O→ 2 cyclohexanone
Answer
(i) C₆H₅COOK (potassium benzoate; benzoic acid after acidification)
(ii) o-C₆H₄(COCl)₂ (phthaloyl dichloride)
(iii) C₆H₅CH=N–NH–CO–NH₂ (benzaldehyde semicarbazone)
(iv) Reagent: C₆H₅COCl, anhydrous AlCl₃ (Friedel–Crafts acylation)
(v) 4-Oxocyclohexane-1-carboxylate ion (only –CHO is oxidised; the ring ketone is untouched) + silver mirror
(vi) o-HOOC–C₆H₄–CH(OH)CN (cyanohydrin of the aldehyde group)
(vii) C₆H₅CH=C(CH₃)CHO (2-methyl-3-phenylprop-2-enal) as the main cross-aldol product (along with the self-condensation product of propanal, CH₃CH₂CH=C(CH₃)CHO)
(viii) CH₃CH(OH)CH₂COOC₂H₅ (NaBH₄ reduces the ketone, not the ester)
(ix) Cyclohexanone
(x) Reagents: (1) B₂H₆, then H₂O₂/OH⁻
⇒ cyclohexylmethanol (anti-Markovnikov); (2) PCC
⇒ cyclohexanecarbaldehyde
(xi) Starting material: cyclohexylidenecyclohexane, (CH₂)₅C=C(CH₂)₅
18. Give plausible explanation for each of 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 the formation of semicarbazones. (iii) During the preparation of esters from a carboxylic acid and an alcohol in the presence of an acid catalyst, the water or the ester should be removed as soon as it is formed.
Answer
(i) In 2,2,6-trimethylcyclohexanone the three methyl groups next to the C=O crowd the carbonyl carbon and hinder the approach of CN⁻; cyclohexanone has no such hindrance, so it adds HCN readily.
(ii) In H₂N–CO–NH–NH₂ the –NH₂ attached to C=O has its lone pair delocalised onto the carbonyl oxygen by resonance, so it is not nucleophilic. Only the terminal –NH₂ of the –NH–NH₂ end is free to attack the carbonyl compound.
(iii) Esterification is reversible: RCOOH + R'OH ⇌ RCOOR' + H₂O. Removing water or ester as it forms shifts the equilibrium forward (Le Chatelier) and prevents hydrolysis of the ester, increasing the yield.
19. 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
O = 100 – 69.77 – 11.63 = 18.60%.
Moles: C = 69.77/12 = 5.81; H = 11.63/1 = 11.63; O = 18.60/16 = 1.16
⇒ ratio 5 : 10 : 1
⇒ empirical formula C₅H₁₀O (mass 86) = molecular formula.
No Tollens' reaction
⇒ ketone; NaHSO₃ addition and iodoform test
⇒ methyl ketone (CH₃CO–); oxidation to CH₃COOH + CH₃CH₂COOH
⇒ the chain breaks between C-2 and C-3 of a straight chain.
The compound is pentan-2-one, CH₃–CO–CH₂–CH₂–CH₃.
20. Although phenoxide ion has more number of resonating structures than carboxylate ion, carboxylic acid is a stronger acid than phenol. Why?
Answer
In the carboxylate ion, the two resonance structures are equivalent and the negative charge sits on two electronegative oxygen atoms, giving strong stabilisation. In the phenoxide ion, the charge is delocalised over one oxygen and less electronegative ring carbons; the structures are not equivalent and those with charge on carbon contribute little. So the carboxylate ion is much more stable than the phenoxide ion, and carboxylic acids are stronger acids.
Important Extra Questions
What is the Hell–Volhard–Zelinsky reaction?
Answer
Carboxylic acids having an α-H react with Cl₂ or Br₂ in presence of red phosphorus to give α-halo acids: CH₃COOH —Cl₂/red P, then H₂O
⇒ ClCH₂COOH.
Distinguish between Clemmensen and Wolff–Kishner reductions.
Answer
Both convert C=O to CH₂. Clemmensen uses Zn–Hg and conc. HCl (acidic); Wolff–Kishner uses NH₂NH₂ then KOH in ethylene glycol on heating (basic).
Why do carboxylic acids not give the usual reactions of the carbonyl group?
Answer
Resonance with the –OH lone pair reduces the electrophilic character of the carboxyl carbon, so it does not add nucleophiles like HCN, NH₂OH etc.
Name the reagent used in the Rosenmund reduction and state its role.
Answer
H₂ with Pd on BaSO₄ (poisoned catalyst); it reduces acyl chloride to aldehyde and the poisoning prevents further reduction to alcohol: RCOCl + H₂
⇒ RCHO + HCl.
Why is formic acid stronger than acetic acid?
Answer
The CH₃ group in acetic acid releases electrons (+I), destabilising the acetate ion; formate has no such group, so HCOOH (pKa 3.75) is stronger than CH₃COOH (pKa 4.76).