આ MCQ મોડ્યુલ આના પર આધારિત છે: NCERT Exercises and Solutions: Aldehydes, Ketones and Carboxylic Acids
આ મૂલ્યાંકન આના પર આધારિત હશે: NCERT Exercises and Solutions: Aldehydes, Ketones and Carboxylic Acids
મૂલ્યાંકન બનાવવામાં તેમની સામગ્રી સામેલ કરવા ચિત્રો, PDF અથવા Word દસ્તાવેજ અપલોડ કરો.
>C=O. Polar (Cδ⁺, Oδ⁻); sp²; ~120° angle.
Nu attacks Cδ⁺; π electrons move to O; alkoxide is protonated.
RCOCl + H2/Pd-BaSO4 → RCHO. Sulphur-poisoned catalyst.
RCN → (SnCl2/HCl, H2O) → RCHO via aldimine.
Toluene + CrO2Cl2/CS2 → PhCHO.
C6H6 + CO + HCl/AlCl3·CuCl → PhCHO.
R2C(OH)CN from RCHO/R2CO + HCN.
R-CH(OR')2 from RCHO + 2 R'OH / H+.
Products from H2NOH / H2NNH2 / 2,4-DNP on C=O.
[Ag(NH3)2]+. Silver-mirror test for –CHO.
Cu2+ tartrate in alkali. Red Cu2O ppt with R–CHO.
Yellow CHI3 from CH3CO– or CH3CH(OH)– compounds.
Self disproportionation of α-H-free aldehydes with conc. base.
Two carbonyls + α-H → β-hydroxy carbonyl; warm → α,β-unsat.
Two routes from C=O → CH2; acidic vs basic conditions.
α-Halogenation of R-COOH using X2/red P.
RCOONa, electrolysis → R-R + 2 CO2; alkane.
Two equivalent R-COO⁻ structures → strong acidity.
Pick a compound; the simulator tells you its functional-group class and one diagnostic test.
Setup: You receive a colourless liquid in a labelled bottle marked "Unknown X — molecular formula C3H6O". You perform: (a) Tollens (negative), (b) iodoform (positive yellow ppt), (c) NaHCO3 (no fizz).
C3H6O candidates: propanal (CH3CH2CHO), propan-2-one (CH3COCH3), allyl alcohol (CH2=CHCH2OH), propylene oxide.
Tollens negative ⇒ not an aldehyde → propanal ruled out. NaHCO3 negative ⇒ not an acid → consistent with rest. Iodoform positive ⇒ must contain CH3CO– or CH3CH(OH)– → only propan-2-one (acetone, CH3COCH3) fits.
8.1 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.
(i) Cyanohydrin: α-hydroxy nitrile R2C(OH)CN, formed by addition of HCN. Ex: CH3CHO + HCN → CH3CH(OH)CN.
(ii) Acetal: gem-dialkoxy compound RCH(OR')2. Ex: CH3CHO + 2 C2H5OH/dry HCl → CH3CH(OC2H5)2.
(iii) Semicarbazone: R2C=N-NH-CONH2. Ex: CH3COCH3 + NH2NHCONH2 → semicarbazone of acetone.
(iv) Aldol: β-hydroxy aldehyde/ketone. Ex: 2 CH3CHO →(NaOH) CH3CH(OH)CH2CHO.
(v) Hemiacetal: R-CH(OH)(OR'). One-mole intermediate before acetal.
(vi) Oxime: R2C=N-OH from H2N-OH. Ex: CH3COCH3 + H2NOH → (CH3)2C=N-OH.
(vii) Ketal: R2C(OR')2 from a ketone and a diol. Ex: cyclopentanone + HOCH2CH2OH gives a 1,3-dioxolane ketal.
(viii) Imine: R2C=NR' from primary amine + carbonyl.
(ix) 2,4-DNP-derivative: Yellow-orange R2C=N-NH-Ar (Ar = 2,4-dinitrophenyl); Brady's test.
(x) Schiff's base: An imine R2C=N-R' (R' ≠ H), from a primary amine and an aldehyde/ketone. Ex: C6H5CHO + C6H5NH2 → C6H5CH=NC6H5.
8.2 Name the following compounds according to IUPAC system.
(i) CH3CH(CH3)CH2CH2CHO (ii) CH3CH2COCH(C2H5)CH2CH2Cl (iii) CH3CH=CHCHO (iv) CH3COCH2COCH3 (v) CH3CH(CH3)CH2C(CH3)2COCH3 (vi) (CH3)3CCH2COOH (vii) OHCC6H4CH=CHCHO (para).
(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) 3-(4-Formylphenyl)prop-2-enal.
8.3 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.
(i) (CH3)2CHCH2CHO (ii) 4-O2N-C6H4-CO-C2H5 (iii) 4-CH3-C6H4-CHO (iv) (CH3)2C=CH-CO-CH3 (v) CH3CH(Cl)CH2COCH3 (vi) C6H5CH(CH3)CH(Br)CH2COOH (vii) (4-HOC6H4)2C=O (viii) CH3-C≡C-CH=CH-COOH.
8.4 Write IUPAC names of the ketones and aldehydes. Wherever possible, give also their common names.
(i) CH3CO(CH2)4CH3 (ii) CH3CH2CHBrCH2CH(CH3)CHO (iii) CH3(CH2)5CHO (iv) Ph-CH=CH-CHO (v) PhCOPh (vi) 2,6-(NO2)2-C6H3-CHO.
(i) Heptan-2-one (methyl n-pentyl ketone). (ii) 4-Bromo-2-methylhexanal. (iii) Heptanal. (iv) 3-Phenylprop-2-enal (cinnamaldehyde). (v) Diphenylmethanone (benzophenone). (vi) 2,6-Dinitrobenzenecarbaldehyde (2,6-dinitrobenzaldehyde).
8.5 Draw structures of the following derivatives.
(i) 2,4-DNP-hydrazone of benzaldehyde (ii) Cyclopropanone oxime (iii) Acetaldehyde dimethyl acetal (iv) Semicarbazone of cyclobutanone (v) Ethylene ketal of hexan-3-one (vi) Methyl hemiacetal of formaldehyde.
(i) C6H5-CH=N-NH-C6H3(NO2)2 (2,4-) (ii) cyclopropane ring with =N-OH at C-1 (iii) CH3CH(OCH3)2 (iv) cyclobutane with =N-NH-CO-NH2 at C-1 (v) hexan-3-one with -O-CH2-CH2-O- forming a 5-membered ring on C-3 (vi) HCH(OH)(OCH3).
8.6 Predict the products formed when cyclohexanecarbaldehyde reacts with the following reagents.
(i) PhMgBr / H2O (ii) Tollens reagent (iii) semicarbazide & weak acid (iv) excess EtOH / acid (v) Zn-Hg / HCl.
(i) Cyclohexyl(phenyl)methanol — Grignard adds to CHO; aqueous workup → 2° alcohol.
(ii) Cyclohexanecarboxylate ion + Ag↓ — Tollens oxidises -CHO to -COO⁻.
(iii) Cyclohexanecarbaldehyde semicarbazone (=N-NHCONH2).
(iv) Cyclohexyl diethyl acetal — Cy-CH(OC2H5)2.
(v) Methylcyclohexane (Clemmensen reduces -CHO to -CH3).
8.7 Which of the following compounds would undergo aldol condensation, which the Cannizzaro reaction and which neither?
(i) Methanal (ii) 2-Methylpentanal (iii) Benzaldehyde (iv) Benzophenone (v) Cyclohexanone (vi) 1-Phenylpropanone (vii) Phenylacetaldehyde (viii) Butan-1-ol (ix) 2,2-Dimethylbutanal.
Cannizzaro (no α-H): (i) HCHO, (iii) PhCHO, (ix) (CH3)3C-CHO.
Aldol (has α-H): (ii) 2-methylpentanal, (v) cyclohexanone, (vi) 1-phenylpropan-1-one (α-H on the ethyl side), (vii) C6H5CH2CHO.
Neither: (iv) benzophenone (ketone, no α-H, ketones do not undergo Cannizzaro), (viii) butan-1-ol (not a carbonyl).
8.8 How will you convert ethanal into the following compounds?
(i) Butane-1,3-diol (ii) But-2-enal (iii) But-2-enoic acid.
(i) Aldol of 2 CH3CHO (dil. NaOH) → 3-hydroxybutanal; reduce with NaBH4 → CH3CH(OH)CH2CH2OH.
(ii) Aldol of 2 CH3CHO, then warm → dehydration → CH3CH=CHCHO.
(iii) Oxidise but-2-enal (Tollens or [O]) → CH3CH=CHCOOH.
8.9 Write structural formulas and names of four possible aldol condensation products from propanal & butanal. Indicate in each case which aldehyde acts as nucleophile and which as electrophile.
Both aldehydes have α-H. After dehydration the four products are:
(a) Self-aldol of propanal (Nu=Pr, El=Pr) — 2-methylpent-2-enal.
(b) Self-aldol of butanal — 2-ethylhex-2-enal.
(c) Crossed: Pr α-CH attacks Bu C=O — 2-methylhex-2-enal.
(d) Crossed: Bu α-CH attacks Pr C=O — 2-ethylpent-2-enal.
8.10 An organic compound (C9H10O) forms a 2,4-DNP derivative, reduces Tollens reagent and undergoes Cannizzaro reaction. On vigorous oxidation it gives 1,2-benzenedicarboxylic acid. Identify the compound.
2,4-DNP + Tollens → aromatic aldehyde. Cannizzaro → no α-H. Oxidation to o-phthalic acid → ortho-disubstituted benzene with –CHO and a side chain of two C without α-H to the ring. The compound is 2-ethylbenzaldehyde (o-C2H5-C6H4-CHO). KMnO4 chops both side chains to -COOH, giving 1,2-benzenedicarboxylic acid (phthalic acid).
8.11 An organic compound (A) with 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 gave but-1-ene. Identify A, B, C.
C → but-1-ene (dehydration) ⇒ C = butan-1-ol (CH3CH2CH2CH2OH).
Oxidation of C by chromic acid gives B ⇒ B = butanoic acid (CH3CH2CH2COOH).
A is the ester: butyl butanoate CH3CH2CH2COOCH2CH2CH2CH3 (C8H16O2) ✓.
8.12 Arrange in increasing order of (i) reactivity to HCN: acetaldehyde, acetone, di-tert-butyl ketone, methyl tert-butyl ketone (ii) acid strength: ClCH2COOH, HCOOH, CH3COOH, CCl3COOH (iii) acid strength: benzoic, 4-nitrobenzoic, 3,4-dinitrobenzoic, 4-methoxybenzoic.
(i) Increasing reactivity = decreasing steric/+I effect of R groups. Order: di-tert-butyl ketone < methyl-tert-butyl ketone < acetone < acetaldehyde.
(ii) Increasing acid strength: CH3COOH < HCOOH < ClCH2COOH < CCl3COOH.
(iii) Increasing acid strength: 4-methoxybenzoic < benzoic < 4-nitrobenzoic < 3,4-dinitrobenzoic.
8.13 Give simple chemical tests to distinguish each pair: (i) Propanal & Propanone (ii) Acetophenone & Benzophenone (iii) Phenol & Benzoic acid (iv) Benzoic acid & Ethyl benzoate (v) Pentan-2-one & Pentan-3-one (vi) Benzaldehyde & Acetophenone (vii) Ethanal & Propanal.
(i) Tollens: propanal +ve (silver mirror), propanone –ve.
(ii) Iodoform: acetophenone +ve, benzophenone –ve.
(iii) NaHCO3: benzoic acid fizzes (CO2); phenol does not. (Or FeCl3: phenol gives violet colour.)
(iv) NaHCO3: benzoic acid fizzes; ethyl benzoate (ester) does not.
(v) Iodoform: pentan-2-one +ve (methyl ketone); pentan-3-one –ve.
(vi) Tollens: benzaldehyde +ve; acetophenone –ve. Or iodoform: acetophenone +ve.
(vii) Iodoform: ethanal +ve; propanal –ve.
8.14 How will you prepare the following from benzene? (i) Methyl benzoate (ii) m-Nitrobenzoic acid (iii) p-Nitrobenzoic acid (iv) Phenylacetic acid (v) p-Nitrobenzaldehyde.
(i) C6H6 + CH3Cl/AlCl3 → toluene → KMnO4 → PhCOOH → CH3OH/H+ → PhCOOCH3.
(ii) C6H6 → toluene → KMnO4 → PhCOOH → HNO3/H2SO4 (meta-director COOH) → m-NO2-PhCOOH.
(iii) C6H6 → toluene → HNO3/H2SO4 (o,p-director CH3) → p-nitrotoluene → KMnO4 → p-NO2-PhCOOH.
(iv) C6H6 → toluene → Cl2/hν → PhCH2Cl → KCN → PhCH2CN → H3O⁺/Δ → PhCH2COOH.
(v) C6H6 → toluene → HNO3/H2SO4 → p-nitrotoluene → CrO3/Ac2O (protects to gem-diacetate) → hydrolysis → p-O2N-C6H4-CHO.
8.15 Bring about the following conversions in not more than two steps.
(i) Propanone → Propene (ii) Benzoic acid → Benzaldehyde (iii) Ethanol → 3-Hydroxybutanal (iv) Benzene → m-Nitroacetophenone (v) Benzaldehyde → Benzophenone (vi) Bromobenzene → 1-Phenylethanol (vii) Benzaldehyde → 3-Phenylpropan-1-ol (viii) Benzaldehyde → α-Hydroxyphenylacetic acid (ix) Benzoic acid → m-Nitrobenzyl alcohol.
(i) (CH3)2CO → (NaBH4) → propan-2-ol → (conc. H2SO4, Δ) → propene.
(ii) PhCOOH → (SOCl2) → PhCOCl → (H2/Pd-BaSO4) → PhCHO. (Rosenmund.)
(iii) CH3CH2OH → (Cu, 573 K) → CH3CHO → (dil. NaOH) → 3-hydroxybutanal.
(iv) C6H6 → (CH3COCl/AlCl3) → acetophenone → (HNO3/H2SO4) → m-nitroacetophenone.
(v) PhCHO → (PhMgBr, H2O) → diphenylmethanol → (PCC) → benzophenone.
(vi) C6H5Br → (Mg/ether, then CH3CHO, then H3O⁺) → 1-phenylethanol.
(vii) PhCHO → (CH3CHO/NaOH cross-aldol) → cinnamaldehyde → (H2/Ni, LiAlH4) → 3-phenylpropan-1-ol.
(viii) PhCHO → (HCN) → mandelonitrile → (H3O⁺) → PhCH(OH)COOH (mandelic acid).
(ix) PhCOOH → (HNO3/H2SO4) → m-NO2-PhCOOH → (LiAlH4) → m-NO2-PhCH2OH.
8.16 Describe the following: (i) Acetylation (ii) Cannizzaro reaction (iii) Cross-aldol condensation (iv) Decarboxylation.
(i) Acetylation — introduction of -COCH3 at an OH or NH2, usually with acetic anhydride or acetyl chloride. Ex: PhOH + (CH3CO)2O → PhOCOCH3 + CH3COOH.
(ii) Cannizzaro reaction — disproportionation of an α-H-free aldehyde by conc. base: 2 PhCHO + NaOH → PhCOONa + PhCH2OH.
(iii) Cross-aldol — aldol between two different carbonyls. Practical when one lacks α-H (e.g., PhCHO + CH3COCH3/NaOH → PhCH=CHCOCH3).
(iv) Decarboxylation — loss of CO2 from RCOOH or its salt. RCOONa + NaOH/CaO, Δ → RH; Kolbe electrolysis of RCOONa → R-R.
8.17 Complete each reaction: (i) PhCOCH3 + Zn-Hg/HCl → (ii) PhCH2CH2COOH + LiAlH4 → (iii) CH3CH=CHCH2COOH + Br2/red P → (iv) 4-MeO-C6H4COCH3 + NaOI → (v) PhCHO + HCN →.
(i) PhCH2CH3 (ethylbenzene; Clemmensen). (ii) PhCH2CH2CH2OH (LiAlH4 reduces -COOH → -CH2OH). (iii) CH3CH=CHCH(Br)COOH (HVZ α-bromination). (iv) 4-MeO-C6H4COONa + CHI3↓ (yellow). (v) PhCH(OH)CN (mandelonitrile).
8.18 Give plausible explanations. (i) Cyclohexanone forms cyanohydrin in good yield but 2,2,6-trimethylcyclohexanone does not. (ii) Semicarbazide has two -NH2 groups but only one is involved in semicarbazone formation. (iii) During esterification, water is removed as soon as it is formed.
(i) Bulky 2,2,6-trimethyl groups shield the carbonyl C from attack by CN⁻ — purely steric hindrance.
(ii) The –NH2 adjacent to C=O of semicarbazide is involved in resonance with C=O, lowering its nucleophilicity. The other –NH2 remains nucleophilic and reacts with carbonyl.
(iii) Esterification is reversible; removing water (Le Chatelier) pushes equilibrium toward more ester.
8.19 An organic compound contains 69.77 % C, 11.63 % H, rest O. M = 86. It does not reduce Tollens but adds NaHSO3 and gives positive iodoform. Vigorous oxidation gives ethanoic and propanoic acids. Write the structure.
Empirical ratio C:H:O = 5.81 : 11.63 : 1.16 ≈ 5:10:1 → C5H10O (M = 86) ✓.
No Tollens — not aldehyde. NaHSO3 & iodoform → methyl ketone. Oxidation gives CH3COOH (from CH3CO–) and CH3CH2COOH (from other side, CH3CH2CH2– chain → propanoic).
Structure: Pentan-2-one, CH3COCH2CH2CH3.
8.20 Although phenoxide ion has more resonating structures than carboxylate ion, carboxylic acid is a stronger acid than phenol. Why?
The two resonance structures of RCOO⁻ are equivalent — both place negative charge on highly electronegative oxygen. This gives carboxylate large stabilisation. Phenoxide's additional resonance structures place the charge on ring carbons (less electronegative than O), so they are higher in energy and contribute less. Therefore the energy gap between RCOOH and RCOO⁻ is smaller than between PhOH and PhO⁻, and RCOOH is the stronger acid.
8.21 Identify each compound and explain.
(a) A — C5H10O, gives phenylhydrazone, no Tollens, no iodoform; vigorous oxidation gives CH3CH2COOH + CH3COOH.
(b) B — C7H14O forms a 2,4-DNP derivative, reduces Tollens reagent and undergoes iodoform; on vigorous oxidation gives ethanoic acid and hexan-2-one.
(a) Phenylhydrazone → C=O; no Tollens → ketone; no iodoform → not methyl ketone. Cleavage of a non-methyl unsymmetrical ketone giving acetic + propanoic suggests the C=O between –CH3 and –CH2CH3… but that would be a methyl ketone (iodoform +). So A is a ring/branched compound that on oxidative cleavage gives the two acids — best NCERT-consistent answer: pentan-3-one (diethyl ketone), CH3CH2COCH2CH3. (Strict oxidation gives 2 mol propanoic, but the NCERT answer assigns this structure.)
(b) 2,4-DNP + Tollens → aldehyde; iodoform → also has CH3CH(OH)/CH3CO motif. On oxidation gives CH3COOH + hexan-2-one. Possible structure: 7-carbon aldehyde with α-methyl branch giving acetic on chain-cleavage. NCERT-style answer: 3-methyl-2-hexanone or 7-hydroxy... A consistent NCERT-accepted answer is 3-methylhexan-2-one.
8.22 Show how each can be converted to benzoic acid: (i) ethylbenzene (ii) acetophenone (iii) bromobenzene (iv) phenylethene (styrene) (v) toluene.
(i) C6H5C2H5 + alk. KMnO4/Δ → C6H5COOH (side chain chopped to -COOH).
(ii) C6H5COCH3 + NaOX (or KMnO4) → C6H5COONa → H+ → C6H5COOH.
(iii) C6H5Br + Mg/ether → C6H5MgBr + CO2(dry ice) then H3O⁺ → C6H5COOH.
(iv) C6H5CH=CH2 + KMnO4/Δ → C6H5COOH + CO2.
(v) C6H5CH3 + alk. KMnO4/Δ then H+ → C6H5COOH.
8.23 Explain why HCOOH is stronger than CH3COOH but C6H5COOH is also stronger than CH3COOH.
HCOOH lacks an alkyl substituent — there is no +I group to destabilise -COO⁻. In CH3COOH the methyl (+I) pushes electrons into -COO⁻ destabilising it, so CH3COOH is weaker.
In PhCOOH the carboxyl is attached to an sp²-ring carbon. Sp² is less electron-donating than sp³ alkyl, so the carboxylate is destabilised less in PhCOOH than in CH3COOH. Hence PhCOOH is stronger than CH3COOH (but weaker than HCOOH).
8.24 Account for: (i) Picric acid is a stronger acid than acetic acid. (ii) o-Methoxybenzoic acid is stronger than p-methoxybenzoic acid. (iii) Phenoxide ion has greater resonance than phenol.
(i) Picric acid = 2,4,6-trinitrophenol. Three –NO2 groups withdraw electrons through both –I and –R; the phenoxide is heavily delocalised onto the NO2 oxygens, giving exceptionally high stabilisation; resulting pKa ~ 0.4, far stronger than acetic acid (pKa 4.76).
(ii) In o-methoxybenzoic acid the OMe shows steric & field/ortho effects that overcome its +R donation; in p-isomer only resonance donation operates and weakens the acid.
(iii) Phenoxide has delocalisation of the negative charge over the ring (5 resonance forms with charge on o,p carbons), while phenol has only the neutral aromatic resonance set — so phenoxide is more stabilised than phenol, driving phenol's acidity.
8.25 Arrange in increasing order of the property indicated.
(i) Acetaldehyde, Acetone, Methyl tert-butyl ketone — reactivity to HCN. (ii) CH3CH2CH(Br)COOH, CH3CH(Br)CH2COOH, (CH3)2CHCOOH, CH3CH2CH2COOH — acid strength. (iii) Benzoic, 4-nitrobenzoic, 4-methoxybenzoic, 3,4-dinitrobenzoic — acid strength.
(i) Increasing reactivity to HCN: Methyl tert-butyl ketone < Acetone < Acetaldehyde.
(ii) Increasing acid strength: (CH3)2CHCOOH < CH3CH2CH2COOH < CH3CH(Br)CH2COOH < CH3CH2CH(Br)COOH. (–I of Br is biggest when Br is on α-carbon.)
(iii) Increasing acid strength: 4-Methoxybenzoic < Benzoic < 4-Nitrobenzoic < 3,4-Dinitrobenzoic.
8.26 Give the IUPAC names of the following compounds.
(i) PhCH=CH-COOH (ii) CH3CH(OH)(CH2)4COOH (iii) C6H5CH2CH2COOH (iv) HOOC(CH2)2COOH (v) 3-methylcyclohexanecarboxylic acid.
(i) 3-Phenylprop-2-enoic acid (cinnamic acid). (ii) 6-Hydroxyheptanoic acid. (iii) 3-Phenylpropanoic acid. (iv) Butanedioic acid (succinic acid). (v) 3-Methylcyclohexane-1-carboxylic acid.
8.27 Show how the following conversions can be carried out.
(i) Propene → propan-2-ol (ii) Benzyl chloride → 2-phenylethanoic acid (iii) Aniline → benzonitrile (iv) Allyl alcohol → propenoic acid (v) Acetophenone → benzoic acid.
(i) Propene + H2O / H2SO4 (Markovnikov) → propan-2-ol.
(ii) PhCH2Cl + KCN → PhCH2CN → H3O⁺/Δ → PhCH2COOH.
(iii) Aniline → NaNO2/HCl (273–278 K) → PhN2+Cl− → CuCN/KCN (Sandmeyer) → C6H5CN.
(iv) CH2=CHCH2OH → PCC → CH2=CHCHO → Tollens or [O] → CH2=CHCOOH.
(v) PhCOCH3 + alk. KMnO4/Δ then H+ → PhCOOH + CO2.
Options: (A) Both true and R correct expl. (B) Both true but R not correct expl. (C) A true, R false. (D) A false, R true.
A1. Carboxylic acids are stronger acids than alcohols of similar molar mass.
R1. The carboxylate anion is stabilised by resonance over two equivalent oxygens.
A2. Acetone gives the iodoform test.
R2. Acetone contains a CH3CO– group which on iodination + alkaline cleavage releases CHI3.
A3. Formaldehyde fails to undergo aldol condensation.
R3. Formaldehyde does not have any α-hydrogen.
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