NCERT Solution Chemistry Class 12 Chapter 12 Aldehydes, Ketones and Carboxylic Acids
NCERT Solution Chemistry Class 12 Chapter 12 Aldehydes, Ketones and Carboxylic Acids all questions and answers. Chemistry Class 12 12th Chapter Aldehydes, Ketones and Carboxylic Acids exercise solution and experts answer. As one of online learning platforms, we (netex.) are excited to offer the NCERT Solution Chemistry Class 12 Chapter 12. This solution is designed to help students who are looking to brush up on their Chemistry concepts on Chapter 12 Aldehydes, Ketones and Carboxylic Acids.
Questions And Answers
(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
(vii) Imine
(ix) 2, 4-DNP-derivative
(x) Schiff’s base
Ans-
(i) Cyanohydrin – carbonyl group is reacts with HCN in presence of weakly acidic medium produce cyanohydrin. oxalic group and cyano group are attached to the same carbon atom.
pH 9−10
CH3CHO+HCN →−−−−−→ CH3CH(OH)CN
(ii) Acetal : . 2 equivalents of monohydric alcohol added to 1 equivalent of aldehyde in presence of dry HCl gas to get acetal with water as byproduct. Two alkoxy groups attached to the terminal C atom.
Dry HCl
CH3CHO+2CH3OH →−−−−−→ CH3CH(OCH3)2 + H2O
(iii) Semi carbazone : Semi carbazone is ammonia derivative of aldehyde or ketone produced when carbonyl compounds treated with semi carbazide in weakly acidic medium.
pH 3.5
H3CCHO+H2NNHCONH2→−−−→ CH3CH = NNHCONH2
−H2O
(iv) Aldol : Aldol is Beta hydroxy aldehyde or ketone formed by condensation of 2 molecules of aldehydes or ketones in which contains alpha hydrogen atoms in presence of dilute aqueous base.
OH−
2CH3CHO →−−→ CH3CHOHCH2CHO
(v) Hemiacetal :
Hemiacetal are Gem alkoxy alcohols produced by addition of one molecule of monohydric alcohol such as methyl alcohol to aldehyde in presence of dry HCl gas.
CH3CHO+CH3OH⇌CH3CH(OH)OCH3
(vi) Oxime : Oximes are obtained when aldehydes or ketones react with hydroxyl amine in weakly acidic medium
CH3CHO+H2NOH→CH3CH+NOH
Vii) Ketal : Ketals are Gem dilkoxy alkanes obtained when the reaction occures in between ketone and ethylene glycol in presence of dry Two alkoxy groups are attached to the same C atom.
CH3COCH3+HOCH2CH2OH →−−−−→ b (CH3)2C(OCH2−)2
(vii) Imine : Imine contains -CH=N- group produced by the reaction between aldehydes or ketones with ammonia derivatives such as hydrazine, hydroxylamine, phenyl hydrazine, semicarbazide etc.
CH3COCH3+PhNH−NH2→(CH3)2C=NNHPh
(ix) 2,4-DNP-derivative : 2,4-DNP-derivatives obtained by the dehydration reaction between aldehydes or ketones and 2,4-dinitrophenylhydrazone in weakly acidic medium.
pH 3.5
CH3COCH3+H2NNH−C6H4(NO2)2 →−−−→ (CH3)2C+NNH−C6H4(NO2)2
−H20
(x) Schiffs base : Schiffs bases are Azomethines that are obtained by the reaction in between aldehydes or ketones and primary amines , either aliphatic or aromatic.
H+
PhCHO+PHNH2 →→ PhCH=NHPh+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
Ans- IUPAC name of given compounds according to IUPAC system of nomenclature is as follows.
(i) 4-Methylpentanal
(ii) 6-chloro-4-ethylhexan-3-one
(iii) but-2-enal
(iv) pentane-2,4-dione
(v) 3,35-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
Ans-
(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
(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) CHO
(vi) PhCOPh
Ans- IUPAC names and common names of given compounds of ketons and aldehydes are as follows –
Sr. No | Compound | IUPAC Name | Common Name |
i. | CH3CO(CH2)4CH3 | Heptan -2-one | Methyl n-pentyl
ketone |
ii. |
CH3CH2CHBrCH2CH(CH3)CHO | 4 – bromo-2- methyl
hexanal |
γ−bromo−α−methyl caproaldehyde |
iii. | CH3(CH2)5CHO | Heptanal | n-heptyl aldehyde |
iv. | Ph-CH=CH-CHO | 3-phenyl
prop-2-en-1-al |
β-phenyl acrolein |
v. | CHO | Cyclopentane
Carbaldehyde |
|
vi. | PhCOPh | Diphenyl
methanone |
Benzophenone |
(5) Draw structures of the following derivatives.
(i) The 2,4-dinitrophenylhydrazone of benzaldehyde
(ii) Cyclopropanone oxime
(iii) Acetaldehydedimethylacetal
(iv) The semi carbazone of cyclobutanone
(v) The ethylene ketal of hexan-3-one
(vi) The methyl hemiacetal of formaldehyde
Ans- (i) 2,4-dinitrophenylhydrazone of benzaldehyde
(ii) Cyclopropanone oxime
(iii) Acetaldehydedimethylacetal
(iv) The semicarbazone of cyclobutanone
(v) ethylene ketal of hexan-3-one
(vi) The methyl hemiacetal of formaldehyde
(6) Predict the products formed when cyclohexane carbaldehyde 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
Ans-
(i) cyclohexane carbaldehyde reacts with PhMgBr and then H3O+ gives cyclohexyl phenylcarbinol
(iii) cyclohexane carbaldehyde reacts with Semicarbazide and weak acid gives clohexanecarbaldehyde.
(iv) cyclohexanecarbaldehyde reacts with excess ethanol and acid gives cyclohexanecarbaldehyde diethyl acetal.
(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
Ans- Compounds (iv)benzophenone , (viii)phenylacetaldehyde shows neither aldol condensation reaction nor cannizaro reaction.
Compounds (ii) 2-Methylpentanal,(v)Cyclohexanone, (vi)1-Phenylpropanone,
(vii) Phenylacetaldehyde undergo aldol condensation because those compounds have alpha hydrogen atom.
Aldol Condensation of Compounds –
Compounds (i)Methanal, (iii) Benzaldehyde, (ix) 2,2-Dimethylbutanal undergo Cannizzaro reaction because they do not have alpha H atoms.
Cannizaro reaction of Compounds are as follows-
(8) How will you convert ethanal into the following compounds?
- Butane-1,3-diol (ii) But-2-enal (iii) But-2-enoic acid
Ans- (i) when ethanal reacts with dil.NaOH produces 3-hydroxybutanal. When it reduced with NaBH4 gives butane-1, 3-diol.
(ii) when ethanal reacts with dil.NaOH produces 3-hydroxybutanal. Dehydration process occurs after heating which gives but-2-enal.
(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.
Ans- Case I) – When aldol condensation undergoes in two molecules of propanal , one molecule acts as electrophile and another acts as nucleophile .it gives 3-hydroxy- 2-methylpentanal.
dil.NaOH
CH3-CH2-CHO + CH3-CH2-CHO →−−−−−→CH3-CH2-CH(OH)-CH(CH3)-CHO
Electrophile nucleophile 3-hydroxy-2-methylpentanal Propanal
Case II) – When aldol condensation undergoes in two molecules of butanal , one molecule acts as electrophile and another acts as nucleophile .it gives 2-ethyl-3- hydroxyhexanal.
Case III) When aldol condensation undergoes in molecules of propanal and butanal, propanal molecule acts as electrophile and butanal acts as nucleophile .it gives 2- ethyl-3-hydroxyhexanal.
Case IV) When aldol condensation undergoes in molecules of propanal and butanal, propanal molecule acts as electrophile and butanal acts as nucleophile .it gives 2- ethyl-3-hydroxyhexanal.
(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.
Ans- The given molecular formula for an organic compound is C9H10O .When 2,4- DNP reacts with aldehyde or ketone it forms 2,4-DNP derivative. if organic compound forms 2,4-DNP derivative it means it is either ketone or aldehyde . also it reduces Tollens reagent , i.e. [Ag(NH3)2]+ complex . therefore it confirms that the compound is an aldehyde since only aldehydes can reduce tollens reagents. if compound undergoes Cannizzaro reaction. It means there is absence of an alpha H atom. If aldehyde undergoes oxidation it gives 1,2-benzenedicarboxylic acid. Therefore we can conclude that im the compound, the -CHO group is directly attached to benzene ring. And hence the compound is definitely ortho-di-substituted benzene. From all the conclusions it is proved that the compound is 2- ethylbenzaldehyde.
Reactions are as follows –
(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.
Ans- The given molecular formula of organic compound A C8H16O2. when ester undergoes hydrolysis in presence of acids such as H2SO4 it forms a mixture of acid and alcohol. Therefore , organic compound (A) is propyl pentanoate and it forms carboxylic acid and alcohol. If oxidation of alcohol (C) gives acid (B) it means the number of carbon atoms in (B) and (C) are same. propyl pentanoate contains eight C atoms. Hence on hydrolysis of propyl pentanoate in presence of dil.H2SO4 gives butanoic acid as product
(B) and butyl alcohol . if dehydration of alcohol butyl alcohol gives but-1-ene. Then butyl alcohol is definitely straight chain alcohol therefore product (C) is butan-1-ol.
(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)
Ans- (i) The increasing order of the reactivity towards HCN –
Di-tert-butyl ketone < Methyl tert-butyl ketone < Acetone < Acetaldehyde
When HCN reacts with a compound it attacks with Nucleophile as CN– ion. when +I effect increase the reactivity towards nucleophiles decreases. Also steric hindrance increases the reactivity decreases.
(ii) the increasing order of acid strength –
(CH3)2CHCOOH < CH3CH2CH2COOH < CH3CH(Br)CH2COOH < CH3CH2CH(Br)COOH.
When positive inductive effect +I decreases and negative inductive effect -I effect increases , then acidic strength of carboxylic acid is increases . (CH3)2CHCOOH has more +I effect than CH3CH2CH2COOH. Therefore , (CH3)2CHCOOH is weaker acid than CH3CH2CH2COOH. when distance increases, -I effect decreases. therefore, the acid strength of CH3CH(Br)CH2COOH is less than the acid strength of CH3CH2CH(Br)COOH.
(iii) The increasing order of acid strength-
4−methoxybenzoicacid < benzoicacid < 4−nitrobenzoicacid < 3,4−dinitrobenzoic acid Acid strength decreases when electron donating groups increases. Therefore , benzoic
acid has higher acid strength than 4−methoxybenzoicacid . Acid strength increases with more electron withdrawing group. Therefore , the acid strength of 4-nitrobenzoic acid is higher than the acid strength of benzoicacid. The acid strength of 4-nitrobenzoic acid is lower than the acid strength of 3,4−dinitrobenzoic acid because in 3,4−dinitrobenzoic acid there two nitro groups are present which are electron withdrawing group while only one nitro group present in 4-nitrobenzoic acid.
(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
Ans- Simple chemical tests to distinguish between the given pairs of compounds is as follows –
- Propanal and propanone :
(1) Fehlings Solution Test –
Aldehydes reduces Fehlings solution and forms brown-red ppt. As propanal is aldehyde so it gives fehlings solution test. But propanone does not give fehling test.
(2) Tollens Reagent Test –
Aldehydes reduces Tollens reagent, and gives silver mirror test. While ketons does not reduce Tollens reagent. Hence silver mirror test can distinguish between propanal and propanone.
(ii) Acetophenone and Benzophenone :
(1) Iodoform Test –
Acetophenone reacts with sodium hypoiodate gives sodium benzoate with yellow ppt of iodoform . yellow ppt of idoform while Benzophenone does not form yellow ppt of iodoform.
(iii) Phenol and Benzoic acid :
(1) Ferric Chloride Test –
When Phenol react with ferric chloride forms Fe-phenol complex having violet colour. While benzoic acid gives light brownish yellow ppt of ferric benzoate , when it reacts with ferric chloride .
(iv) Benzoic acid and Ethyl benzoate :
(1) Sodium Bicarbonate test –
When benzoic acid reacts with sodium bicarbonate it gives sodium benzoate and produce effervescence of CO2 gas. Whereas Ethyl benzoate does not produce CO2 gas.
(2) Iodoform Test-
When ethyl benzoate reacts NaOH, it gives ethyl alcohol. Ethyl alcohol heated with iodine produce yellow ppt of iodoform. benzoic acid does not give this test.
(v) Pentan-2-one and Pentan-3-one :
Pentan-2-one reacts with sodium hypoiodate gives sodium butanoate and forms yellow ppt of iodoform. whereas Pentan-3-one does not forms yellow ppt of iodoform.
Pentan-2-one forms white ppt with NaHSO3 but pentan-3-one does not give sodium bisulphate test .
(vi) Benzaldehyde and Acetophenone :
(1) Tollens Reagent Test –
Aldehydes reduces Tollens reagent , and gives silver mirror test. While ketons does not reduce Tollens reagent. Hence silver mirror test can distinguish between Benzaldehyde and Acetophenone .
(2) Iodoform Test –
Acetophenone reacts with sodium hypoiodate gives sodium benzoate with yellow ppt of iodoform . yellow ppt of idoform. But benzaldehyde does not shows this test.
(vii) Ethanal and Propanal :
(1) Iodoform Test –
When ethanal reacts with sodium hypoiodate gives sodium methanoate and yellow ppt of iodoform . whereas propanal does not give iodoform test.
(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
(iv) p-Nitrobenzoic acid
(iv) Phenylacetic acid
(v) p-Nitrobenzaldehyde.
Ans- (i) Friedel crafts alkylation of benzene gives toluene. It treated with KMnO4 to form benzoic acid. Benzoic acid oxidised with CH3OH, yields Methyl benzoate.
(ii) Benzene to m-nitrobenzoic acid-
Friedel Craft acylation of benzene forms acetophenone and then by nitration produce nitroacetophenone. by oxidation of the nitroacetophenone produce m- nitrobenzoic acid.
(iii) Benzene to p-Nitrobenzoic acid-
Friedel Craft alkylation of benzene forms toluene. Toluene undergoes nitration forms p-nitrotoluene. Minor products can be filtered and undergoes oxidation to gives p-nitrobenzoic acid.
(iv) Benzene to phenyl acetic acid –
Friedel Craft alkylation of benzene forms toluene. Chlorination of Toluene gives benzyl chloride . by addition of KCN it gives benzyl cyanide. It undergoes hydrolysis in acidic medium to give phenyl acetic acid.
(v) Benzene to p – nitrobenzaldehyde-
benzene undergoes nitration to form nitrobenzene. By using Gateman aldehyde synthesis nitrobenzene convert into p-nitrobenzaldehyde.
(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) Benazaldehyde to α-Hydroxyphenylacetic acid
(ix) Benzoic acid to m- Nitrobenzyl alcohol
Ans- (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) Benazaldehyde to α-Hydroxyphenylacetic acid
(ix) Benzoic acid to m- Nitrobenzyl alcohol
(16) Describe the following:
(i) Acetylation
(ii) Cannizzaro reaction
(iii) Cross aldol condensation
(iv) Decarboxylation Ans-
(i) Acetylation-
The substitution of an acetyl functional group (CH3CO) into an organic compound is termed as acetylation. acetylation undergoes in the presence of a base such as dimethylaniline, pyridine etc. In, acetylation the substitution of an acetyl group carried out by replacement of active hydrogen atom. Acetyl chloride and acetic anhydride are acetylating agents used for acetylation.
For example, acetylation of ethanol produces ethyl acetate. Hydrogen atom replaced by acetyl group . hydrochloric acid is formed as byproduct.
pyridine
CH3CH2OH + CH3COCl →−−−−→ CH3COOC2H5 + HCl
Ethanol Acetyl chloride ethyl acetate
(ii) Cannizzaro reaction:
disproportionation reaction or the self- oxidation reduction of aldehydes whose molecules has no α-hydrogens reacts with concentrated alkalis to give alcohol and carboxylic acid , is termed as the Cannizzaro reaction.
2CH3-CHO + Conc.KOH→−−−−−→ CH3-CH2-OH + CH3COOK
Ethanal ethanol potassium ethanoate
In the reaction, two molecules of ethanal treated with concentrated potassium hydroxide, one is reduced to ethanol and the other is oxidized to potassium ethanoate.
(iii) Cross-aldol condensation:
cross aldol condensation is aldol condensation between two different aldehydes, or two different ketones, or an aldehyde and a ketone, when both the reactants contain α-hydrogens there are 2 products formed as self product and 2 crossed product formed with cross aldol condensation
For example, ethanal and propanal react to give four products.
(iv) Decarboxylation:
The reaction in which carboxylic acids eliminate carboxyl group and liberates carbon dioxide gas to form hydrocarbons is known as decarboxylation reaction.
Sodium acetate treated with soda-lime which is mixture of NaOH and CaO in 3:1 ratio produce methane and sodium carbonate.
(17) Complete each synthesis by giving missing starting material, reagent
or products
Ans- Synthesis of given products are follows -:
(18) Give plausible explanation for each of the following:
(i) Cyclohexanone forms cyanohydrin in good yield but 2,2,6-trimethylcyclo- hexanone 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.
Ans- (i) When HCN reacts with a compound it attacks with nucleophile as CN– ion. But when steric hindrance affects on the attack of nucleophile . cyclohexanone is ketonic group has negligible stearic hindrance. While three methyl groups are present on alpha position in 2,2,6-trimethylcyclohexanone. therefore, the attack of cyanide nucleophile is sterically hindered. Hence, cyclohexanone forms cyanohydrin in good yield but 2,2,6-trimethylcyclohexanone does not.
(ii) Semicarbazide contains two −NH2 groups. The lone pair of electron on nitrogen atom is delocalized through resonance. As a result , lone pair cannot be donated to electrophile. therefore −NH2 group cannot act as a nucleophile and cannot participate in reaction. Therefore, out of two −NH2 groups in semicarbazide, only one is involved in the formation of semicarbazones.
(iii) When carboxylic acid and an alcohol in the presence of an acid catalyst it forms ester and water. it is reversible reaction therefore either ether or water should be removed as soon as it is formed. So we can get desirable product in forward direction.
(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. structure of the compound.
Ans- An organic compound contains Carbon – 69.77%
Hydrogen- 11.63%
Oxygen = 100−(69.77+11.63) = 18.60%
The molecular mass of the compound – 86
It shows that the molecular formula is same as empirical formula. if the compound does not reduce Tollens reagent then definitely it is not an aldehyde because aldehyde can reduce Tollens regent and shows silver mirror test .
It forms an addition compound with sodium hydrogen sulphite therefore it contains carbonyl group. If it give positive iodoform test then it is a methyl ketone. On vigorous oxidation it gives ethanoic and propanoic acid. From all the above conclusions, the organic compound is 2-pentanone.
There are 5 resonating structures of phenoxide ion shown in the above figure . In structure II, III and IV we can observed that the less electronegative carbon atoms carries negative charge. Hence these structures has very less resonance stability. Hence, these structures are negligible and can be eliminated. While structures I and V carries negative charge on the more electronegative oxygen atom. Therefore these structures possess resonance stability. But in these resonating structures i.e. I and V, the negative charge is localized on the same oxygen atom.
There are two resonating structure of carboxylate ion as shown in the above. Now, in resonating structures I’ and II’, the negative charge is delocalized over two oxygen atoms. While in phenoxide ion there is only oxygen atom on which negative charge is localised . hence, the resonating structures of carboxylate ion possesses more resonating stability than those of phenoxide ion. Therefore , carboxylic acid is a stronger acid than phenol.
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