Aldehydes, Ketones and Carboxylic Acids — NCERT Solutions

Benzophenone is a ketone, but it has no alpha-hydrogen atom. Since aldol condensation needs at least one alpha-hydrogen, and Cannizzaro reaction is shown by aldehydes without alpha-hydrogen, benzophenone undergoes neither of these reactions.

So the correct choice is neither.

Cyclohexanone has alpha-hydrogen atoms, so it undergoes aldol condensation.

Therefore, the correct reaction type is aldol condensation.

1-Phenylpropanone is a ketone with at least one alpha-hydrogen atom, so it undergoes aldol condensation.

So the correct reaction type is aldol condensation.

Phenylacetaldehyde contains alpha-hydrogen atoms, so it undergoes aldol condensation.

Therefore, the correct reaction type is aldol condensation.

Butan-1-ol is an alcohol, not an aldehyde or ketone. So it does not undergo aldol condensation or Cannizzaro reaction.

Hence, the correct choice is neither.

2,2-Dimethylbutanal is an aldehyde with no alpha-hydrogen atom because the carbon next to the –CHO group is fully substituted. Therefore, it does not undergo aldol condensation, but it does undergo Cannizzaro reaction.

So the correct reaction type is Cannizzaro reaction.

Use the following conversions from the chapter:

1. Ethanal to butane-1,3-diol: carry out aldol condensation of ethanal to give 3-hydroxybutanal, then reduce the aldehyde and alcohol group appropriately to obtain the diol.

2. Ethanal to but-2-enal: ethanal undergoes aldol condensation in dilute alkali to form 3-hydroxybutanal, which on dehydration gives but-2-enal.

3. Ethanal to but-2-enoic acid: first convert ethanal to but-2-enal by aldol condensation, then oxidise the aldehyde group to the acid.

The chapter specifically shows that ethanal gives aldol products under dilute alkali and that aldehydes are readily oxidised to carboxylic acids.

Ethanal gives butane-1,3-diol by aldol condensation followed by reduction:

1. Aldol addition of ethanal gives 3-hydroxybutanal:

$$2\,\mathrm{CH_3CHO} \xrightarrow{\text{dil. NaOH}} \mathrm{CH_3CH(OH)CH_2CHO}$$

2. Reduction of the aldehyde group gives the diol:

$$\mathrm{CH_3CH(OH)CH_2CHO} \xrightarrow{\mathrm{NaBH_4}} \mathrm{CH_3CH(OH)CH_2CH_2OH}$$

So the product is butane-1,3-diol.

Ethanal undergoes aldol condensation to form 3-hydroxybutanal, which on dehydration gives but-2-enal (crotonaldehyde).

Ethanal is first converted to but-2-enal by aldol condensation, and then the aldehyde is oxidised to the acid:

$$2\,\mathrm{CH_3CHO} \xrightarrow{\text{dil. NaOH}} \mathrm{CH_3CH(OH)CH_2CHO}$$
$$\mathrm{CH_3CH(OH)CH_2CHO} \xrightarrow{\Delta} \mathrm{CH_3CH=CHCHO}$$
$$\mathrm{CH_3CH=CHCHO} \xrightarrow{[O]} \mathrm{CH_3CH=CHCOOH}$$

So the product is but-2-enoic acid.

There are four possible aldol condensation products because either aldehyde can act as the nucleophile or the electrophile.

Let the two aldehydes be propanal and butanal.

1. Propanal acting as nucleophile, propanal acting as electrophile

Self-aldol product:
3-Hydroxy-2-methylpentanal

2. Butanal acting as nucleophile, butanal acting as electrophile

Self-aldol product:
3-Hydroxyoctanal

3. Propanal acting as nucleophile, butanal acting as electrophile

Cross-aldol product:
A hydroxy aldehyde containing the combined carbon skeleton of both compounds.

4. Butanal acting as nucleophile, propanal acting as electrophile

Another cross-aldol product:
A different hydroxy aldehyde due to the opposite pairing.

Because both aldehydes contain $alpha$-hydrogen, cross aldol condensation gives a mixture of four products. The chapter states this directly and illustrates it with ethanal and propanal.

The compound forms a 2,4-DNP derivative, so it contains a carbonyl group. It reduces Tollens’ reagent, so it is an aldehyde. It undergoes Cannizzaro reaction, so it has no α-hydrogen. On vigorous oxidation it gives 1,2-benzenedicarboxylic acid, which means the aldehyde groups are on adjacent positions of the benzene ring. Therefore, the compound is 1,2-benzenedicarbaldehyde, also called phthalaldehyde.

The compound A is an ester because it is hydrolysed with dilute sulphuric acid to give a carboxylic acid B and an alcohol C.

Since oxidation of C gives B, the alcohol must be the corresponding primary alcohol. Also, C dehydrates to but-1-ene, so C = butan-1-ol.

Oxidation of butan-1-ol gives butanoic acid, so B = butanoic acid.

Therefore, A is butyl butanoate.

Equations

$$\mathrm{CH_3CH_2CH_2COOCH_2CH_2CH_2CH_3 + H_2O \,\xrightarrow{dil.\ H_2SO_4}\, CH_3CH_2CH_2COOH + CH_3CH_2CH_2CH_2OH}$$

$$\mathrm{CH_3CH_2CH_2CH_2OH + [O] \xrightarrow{CrO_3} CH_3CH_2CH_2COOH}$$

$$\mathrm{CH_3CH_2CH_2CH_2OH \xrightarrow{dehydration} CH_2=CHCH_2CH_3 + H_2O}$$

For reactivity towards HCN, nucleophilic addition is easier when the carbonyl carbon is less hindered and more electrophilic.

• Ketones are less reactive than aldehydes because they have two alkyl groups, which increase steric hindrance and decrease electrophilicity.
• Among ketones, di-tert-butyl ketone is least reactive because bulky tert-butyl groups hinder attack strongly.
• Methyl tert-butyl ketone is more reactive than di-tert-butyl ketone but less reactive than propanone.
• Among aldehydes, ethanal is more reactive than propanal because it has a smaller alkyl group.

So the increasing order is:
$$\text{Di-tert-butyl ketone} < \text{Methyl tert-butyl ketone} < \text{Propanone} < \text{Acetaldehyde}$$
This corresponds to the given names: Butanone < Propanone < Propanal < Ethanal is not correct for these exact compounds, so the chemically correct order for the list is the one above.

Acidity of carboxylic acids increases with electron-withdrawing groups and decreases with electron-donating alkyl groups.

• Butanoic acid is least acidic among these.
• 2-Methylpropanoic acid has an extra alkyl group, so it is less acidic than butanoic acid.
• In the bromo-substituted acids, Br increases acidity by the –I effect.
• The α-bromo acid is more acidic than the β-bromo acid because the –I effect is stronger when Br is closer to the COOH group.

Therefore, increasing acid strength:
$$\mathrm{CH_3CH_2CH_2COOH < (CH_3)_2CHCOOH < CH_3CH(Br)CH_2COOH < CH_3CH_2CH(Br)COOH}$$

Electron-withdrawing groups increase the acidity of carboxylic acids, while electron-donating groups decrease acidity.

• 4-Methoxybenzoic acid has an electron-donating –OCH3 group, so it is less acidic than benzoic acid.
• Benzoic acid is the reference.
• 4-Nitrobenzoic acid is more acidic because –NO2 is strongly electron withdrawing.
• 3,4-Dinitrobenzoic acid is strongest because it has two –NO2 groups.

Thus, increasing acidity:
$$\text{4-Methoxybenzoic acid} < \text{Benzoic acid} < \text{4-Nitrobenzoic acid} < \text{3,4-Dinitrobenzoic acid}$$

Use Tollens' test or Fehling's test.

• Propanal is an aldehyde, so it is oxidised by Tollens' reagent to give a silver mirror.
• Propanone is a ketone, so it does not respond.

So, on warming with Tollens' reagent:

• Propanal → positive test
• Propanone → negative test

Use the iodoform test.

• Acetophenone is a methyl ketone ($\mathrm{C_6H_5COCH_3}$), so it gives a yellow precipitate of iodoform.
• Benzophenone ($\mathrm{C_6H_5COC_6H_5}$) has no $\mathrm{CH_3CO-}$ group, so it does not give the test.

Therefore:

• Acetophenone → positive
• Benzophenone → negative

A simple test is with sodium bicarbonate.

• Benzoic acid is a carboxylic acid, so it reacts with NaHCO3 to release CO2 and causes effervescence.
• Phenol is much weaker and does not react with sodium bicarbonate.

So:

• Benzoic acid → brisk effervescence
• Phenol → no reaction

Use sodium bicarbonate.

• Benzoic acid reacts with NaHCO3 to produce CO2, so it gives effervescence.
• Ethyl benzoate is an ester, so it does not react with sodium bicarbonate.

Therefore:

• Benzoic acid → effervescence
• Ethyl benzoate → no effervescence

Pentan-2-one is a methyl ketone because it has the group $\mathrm{CH_3CO-}$. So it gives the iodoform test with $\mathrm{I_2/NaOH}$ and forms a yellow precipitate of iodoform. Pentan-3-one does not have a $\mathrm{CH_3CO-}$ group, so it does not give this test.

Acetophenone is a methyl ketone ($\mathrm{C_6H_5COCH_3}$), so it gives the iodoform test with $\mathrm{I_2/NaOH}$ and forms a yellow precipitate of iodoform. Benzaldehyde does not have the $\mathrm{CH_3CO-}$ group, so it does not give this test.

Ethanal gives the iodoform test because it contains the $\mathrm{CH_3CO-}$ unit after oxidation, so it gives a yellow precipitate of iodoform with $\mathrm{I_2/NaOH}$. Propanal does not have this unit, so it does not respond to the test.

From benzene to the required compounds, use these standard sequences:

1. Methyl benzoate: first convert benzene to toluene by Friedel–Crafts alkylation, then oxidise the side chain to benzoic acid, and finally esterify.

• Benzene $$toluene$$ benzoic acid $$ methyl benzoate

2. m-Nitrobenzoic acid: first form benzoic acid from benzene, then nitration. The $-$COOH group is meta-directing, so nitration gives the meta product.

• Benzene $$toluene$$ benzoic acid $$m-nitrobenzoic acid 3. **p-Nitrobenzoic acid**: first prepare **toluene**, then nitration to get mainly **p-nitrotoluene**, and finally oxidise the side chain to the acid. - Benzene$$ toluene $$p-nitrotoluene$$ p-nitrobenzoic acid

4. Phenylacetic acid: convert benzene to benzyl chloride by chloromethylation, then treat with KCN to get benzyl cyanide, and hydrolyse to the acid.

• Benzene $$C6H5CH2Cl$$ C6H5CH2CN $$C6H5CH2COOH 5. **p-Nitrobenzaldehyde**: first prepare **toluene**, nitrate to get mainly **p-nitrotoluene**, then oxidise the methyl group carefully to the aldehyde using a controlled side-chain oxidation reagent. - Benzene$$ toluene $$p-nitrotoluene$$ p-nitrobenzaldehyde

These are the routes given from the chapter’s synthesis patterns and named reactions.

From benzene, first prepare toluene by Friedel-Crafts alkylation, then oxidise the side chain to benzoic acid, and finally esterify with methanol.

$$\mathrm{C_6H_6 \xrightarrow[ ]{CH_3Cl/AlCl_3} C_6H_5CH_3 \xrightarrow[ ]{KMnO_4} C_6H_5COOH \xrightarrow[ ]{CH_3OH/H^+} C_6H_5COOCH_3}$$

So the compound is prepared by oxidation followed by esterification.

Prepare toluene from benzene, then oxidise the side chain to benzoic acid. The carboxyl group is meta-directing, so nitration of benzoic acid gives mainly the meta product.

$$\mathrm{C_6H_6 \xrightarrow[ ]{CH_3Cl/AlCl_3} C_6H_5CH_3 \xrightarrow[ ]{KMnO_4} C_6H_5COOH \xrightarrow[ ]{conc.\ HNO_3/H_2SO_4} m\!\text{-nitrobenzoic acid}}$$

First make toluene from benzene, nitrate it to get p-nitrotoluene, and then oxidise the methyl group to p-nitrobenzoic acid.

$$\mathrm{C_6H_6 \xrightarrow[ ]{CH_3Cl/AlCl_3} C_6H_5CH_3 \xrightarrow[ ]{HNO_3/H_2SO_4} p\!\text{-nitrotoluene} \xrightarrow[ ]{KMnO_4} p\!\text{-nitrobenzoic\ acid}}$$

This follows the chapter’s idea that side-chain oxidation converts alkyl benzenes into the corresponding benzoic acids.

Convert benzene to toluene first, then to benzyl chloride, then to benzyl cyanide with KCN, and hydrolyse the nitrile to the acid.

A valid route is:
$$\mathrm{C_6H_6 \xrightarrow[ ]{CH_3Cl/AlCl_3} C_6H_5CH_3 \xrightarrow[ ]{Cl_2/h\nu} C_6H_5CH_2Cl \xrightarrow[ ]{KCN} C_6H_5CH_2CN \xrightarrow[ ]{H^+/H_2O} C_6H_5CH_2COOH}$$

So the product is phenylacetic acid.

Prepare toluene from benzene, then use p-nitrobenzaldehyde route by directing substitution on the ring and controlled oxidation at the side chain. A suitable route is to nitrate toluene first to obtain p-nitrotoluene, then oxidise the methyl group to the aldehyde using chromyl chloride or CrO3 in acetic anhydride.

$$\mathrm{C_6H_6 \xrightarrow[ ]{CH_3Cl/AlCl_3} C_6H_5CH_3 \xrightarrow[ ]{HNO_3/H_2SO_4} p\!\text{-nitrotoluene} \xrightarrow[ ]{CrO_2Cl_2/H_2O} p\!\text{-nitrobenzaldehyde}}$$

This gives p-nitrobenzaldehyde.

Use these conversions in not more than two steps:

1. Propanone to propene: reduce the carbonyl group to methylene by Clemmensen reduction or Wolff-Kishner reduction, giving propane, then dehydrogenate to propene. In chapter-style conversions, the essential idea is carbonyl reduction followed by elimination to alkene.

2. Benzoic acid to benzaldehyde: convert the acid to acid chloride, then carry out Rosenmund reduction.

3. Ethanol to 3-hydroxybutanal: oxidise ethanol to ethanal, then subject ethanal to aldol reaction with dilute alkali.

4. Benzene to m-nitroacetophenone: first do Friedel-Crafts acylation to form acetophenone; then nitration. The $-$COCH3 group is meta-directing, so nitro group enters meta position.

5. Benzaldehyde to benzophenone: convert benzaldehyde to benzoic acid on oxidation, then to benzoyl chloride, and finally react with benzene in Friedel-Crafts acylation, or use the standard aromatic ketone preparation route.

6. Bromobenzene to 1-phenylethanol: prepare phenylmagnesium bromide from bromobenzene, then add ethanal, followed by hydrolysis.

7. Benzaldehyde to 3-phenylpropan-1-ol: react benzaldehyde with formaldehyde equivalent/Grignard route to extend the chain by two carbons, then reduce appropriately. Standard textbook route uses the carbonyl chemistry of aldehydes to build the required alcohol.

8. Benzaldehyde to $$-hydroxyphenylacetic acid: form the cyanohydrin from benzaldehyde using HCN, then hydrolyse the nitrile group to carboxylic acid.

9. Benzoic acid to m-nitrobenzyl alcohol: first nitrate benzoic acid to m-nitrobenzoic acid; then reduce the carboxyl group to primary alcohol with LiAlH4 or diborane.

These are the standard transformation ideas from the chapter: oxidation, reduction, acylation, nitration, Grignard addition, and hydrolysis.

A suitable two-step conversion is:

1. Reduce propanone to propan-2-ol.
2. Dehydrate propan-2-ol to propene.

$$\mathrm{CH_3COCH_3 \xrightarrow[ ]{NaBH_4} CH_3CHOHCH_3 \xrightarrow[\Delta]{conc.\ H_2SO_4} CH_3CH=CH_2}$$

So the required conversion is through propan-2-ol as the intermediate.

Convert benzoic acid first to benzoyl chloride using $\mathrm{SOCl_2}$ or $\mathrm{PCl_5}$, then carry out Rosenmund reduction to stop at the aldehyde.

$$\mathrm{C_6H_5COOH \xrightarrow[ ]{SOCl_2} C_6H_5COCl \xrightarrow[ ]{H_2/Pd-BaSO_4} C_6H_5CHO}$$

So benzoic acid is converted to benzaldehyde by acid chloride formation followed by Rosenmund reduction.

From the chapter’s examples and reactions, ethanal undergoes aldol condensation with dilute base to give 3-hydroxybutanal:

$$2\,\mathrm{CH_3CHO} \xrightarrow{\text{dil. NaOH}} \mathrm{CH_3CH(OH)CH_2CHO}$$

So the required conversion is done in one step by dilute alkali (aldol reaction).

To get m-nitroacetophenone from benzene, first do Friedel-Crafts acylation to form acetophenone, then nitrate. The carbonyl group is meta-directing, so nitration gives the meta product.

Step 1:
$$\mathrm{C_6H_6} \xrightarrow{\mathrm{CH_3COCl}/\mathrm{AlCl_3}} \mathrm{C_6H_5COCH_3}$$

Step 2:
$$\mathrm{C_6H_5COCH_3} \xrightarrow{\mathrm{HNO_3/H_2SO_4}} m\text{-nitroacetophenone}$$

So the required reagent sequence is CH3COCl/AlCl3, then HNO3/H2SO4.

From the chapter, benzaldehyde to benzophenone can be done by converting benzaldehyde to a benzylic halide/alcohol derivative and then using Friedel-Crafts acylation with benzene, but the standard textbook route for this type of conversion is the introduction of a benzoyl group via benzene + benzoyl chloride / AlCl3. Since the question asks for a conversion from benzaldehyde in not more than two steps, the intended chapter-level answer is not explicitly listed as a named sequence in the text. The direct, standard school-level identification is that benzophenone is prepared by Friedel-Crafts acylation of benzene with benzoyl chloride; therefore the needed answer is that the starting compound must first be converted to benzoyl chloride/benzene equivalent. If asked as a reagent sequence, the textbook content does not give a unique direct two-step route from benzaldehyde to benzophenone.

To convert bromobenzene to 1-phenylethanol in the chapter’s style, first make the Grignard reagent and then add ethanal, followed by hydrolysis:

$$\mathrm{C_6H_5Br} \xrightarrow{\mathrm{Mg/dry\ ether}} \mathrm{C_6H_5MgBr}$$

$$\mathrm{C_6H_5MgBr} + \mathrm{CH_3CHO} \xrightarrow{} \mathrm{C_6H_5CH(OMgBr)CH_3} \xrightarrow{\mathrm{H_3O^+}} \mathrm{C_6H_5CH(OH)CH_3}$$

So the reagents are Mg/dry ether, then ethanal and hydrolysis.

The chapter gives the textbook route to 3-phenylpropan-1-ol from benzaldehyde through side-chain extension by one carbon using a Grignard-type addition followed by reduction, but the standard school-level conversion is:

$$\mathrm{C_6H_5CHO} \xrightarrow{\mathrm{CH_2N_2}} \text{(not in chapter)}$$

This specific transformation is not directly listed in the chapter source. So, based only on the chapter content, there is no unique stated reagent sequence for this exact conversion.

The conversion of benzaldehyde to α-hydroxyphenylacetic acid is not explicitly given in the chapter’s worked examples or summary. The chapter does show that aldehydes can undergo reactions at the carbonyl group, but this particular synthesis is not directly listed. So, from the given chapter source, there is no unique named route to state here.

The chapter explains that benzoic acid can be reduced to a primary alcohol with LiAlH4 or diborane. To make m-nitrobenzyl alcohol, first introduce the nitro group on benzoic acid, then reduce the carboxyl group:

1. Nitration gives m-nitrobenzoic acid because –COOH is meta-directing.
2. Reduce –COOH to –CH2OH with LiAlH4 (or diborane).

So the sequence is:
$$\mathrm{C_6H_5COOH} \xrightarrow{\mathrm{HNO_3/H_2SO_4}} m\text{-nitrobenzoic acid} \xrightarrow{\mathrm{LiAlH_4}} m\text{-nitrobenzyl alcohol}$$

This is the chapter-consistent route.

Acetylation is the introduction of an acetyl group ($\mathrm{CH_3CO-}$) into a molecule.

Example:
$$\mathrm{C_6H_5OH + CH_3COCl \rightarrow C_6H_5OCOCH_3 + HCl}$$
Here phenol gets acetylated to give phenyl acetate.

Cannizzaro reaction is the self oxidation and reduction (disproportionation) of an aldehyde that does not have an α-hydrogen atom when heated with concentrated alkali.

Example:
$$2\,\mathrm{C_6H_5CHO} + \mathrm{KOH} \rightarrow \mathrm{C_6H_5CH_2OH} + \mathrm{C_6H_5COOK}$$
One molecule is reduced to alcohol and the other is oxidised to carboxylate salt.

Cross aldol condensation is aldol condensation between two different aldehydes and/or ketones.

If both reactants have α-hydrogen atoms, a mixture of four products is formed. For example, a mixture of ethanal and propanal gives four aldol products.

Decarboxylation is the process in which a carboxylic acid, usually in the form of its sodium salt, loses carbon dioxide on heating with sodalime ($\mathrm{NaOH}$ and $\mathrm{CaO}$ in the ratio 3:1), forming a hydrocarbon with one carbon atom less.

$$\mathrm{RCOONa} \xrightarrow[\text{Heat}]{\mathrm{NaOH\ +\ CaO}} \mathrm{RH} + \mathrm{Na_2CO_3}$$

Example:
$$\mathrm{CH_3COONa} \xrightarrow[\text{Heat}]{\mathrm{NaOH\ +\ CaO}} \mathrm{CH_4} + \mathrm{Na_2CO_3}$$

The missing syntheses are the standard textbook reactions:

• Carbonyl compounds can be converted into 2,4-DNP derivatives, oximes, hydrazones, semicarbazones, acetals, and ketals by the corresponding reagents.
• Aldehydes on oxidation give acids; ketones usually resist mild oxidation.
• Carboxylic acids form acid chlorides with $\mathrm{SOCl_2}$ / $\mathrm{PCl_5}$, esters with alcohols in acid, and amides with ammonia on heating.

These synthesis completion problems are answered by choosing the reagent or product that fits the named reaction from the chapter.

Cyclohexanone forms cyanohydrin in good yield because its carbonyl carbon is sufficiently accessible. In 2,2,6-trimethylcyclohexanone, three methyl groups create strong steric hindrance around the carbonyl group, so the nucleophile cannot attack the carbonyl carbon easily. Therefore, cyanohydrin formation is poor or does not occur appreciably.

Although semicarbazide has two $-\mathrm{NH_2}$ groups, only the terminal amino group is sufficiently nucleophilic to attack the carbonyl compound and form the semicarbazone. The other nitrogen is involved in the amide-like $\mathrm{C(=O)NH_2}$ part, where its lone pair is delocalised by resonance and is therefore much less available for reaction.

In esterification, the reaction is reversible:

$$\mathrm{RCOOH + R'OH \rightleftharpoons RCOOR' + H_2O}$$

To drive the equilibrium to the product side, the water formed should be removed. If the ester itself is volatile, removing it also helps shift the equilibrium forward. According to Le Chatelier’s principle, removing a product causes more ester to form, increasing the yield.

• Molecular mass $=86$ and composition gives:

Carbon: $\frac{69.77}{12}\approx 5.81$ mol, Hydrogen: $\frac{11.63}{1}=11.63$ mol, Oxygen: $\frac{18.60}{16}=1.16$ mol.
Dividing by $1.16$ gives ratio about $5:10:1$.

• So the molecular formula is $\mathrm{C_5H_{10}O}$.
• It does not reduce Tollens’ reagent, so it is not an aldehyde.
• It gives a sodium hydrogensulphite addition compound, so it is a ketone.
• It gives a positive iodoform test, so it must be a methyl ketone ($\mathrm{CH_3CO-}$ group).
• The ketone with formula $\mathrm{C_5H_{10}O}$ and a methyl ketone group is pentan-2-one:

$\mathrm{CH_3COCH_2CH_2CH_3}$

• On vigorous oxidation, a methyl ketone can cleave to give smaller carboxylic acids; pentan-2-one gives ethanoic acid and propanoic acid, which matches the data.

So the possible structure is pentan-2-one.

• Carboxylic acids are more acidic than phenol because their conjugate base, the carboxylate ion, is much more stable than the phenoxide ion.
• In the carboxylate ion, the negative charge is equally delocalised over two equivalent oxygen atoms through resonance.
• In phenoxide ion, resonance structures are not equivalent and the negative charge is delocalised partly onto less electronegative carbon atoms, so the stabilisation is less effective.
• Also, in carboxylate ion the negative charge remains mainly on oxygen atoms, which are more electronegative and better able to hold negative charge.
• Hence, the carboxylate ion is more stabilised than phenoxide ion, so carboxylic acids are stronger acids than phenol.