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Non-Rationalised Science NCERT Notes and Solutions (Class 12th)
Physics Chemistry Biology

Class 12th (Chemistry) Chapters
1. The Solid State 2. Solutions 3. Electrochemistry
4. Chemical Kinetics 5. Surface Chemistry 6. General Principles And Processes Of Isolation Of Elements
7. The P-Block Elements 8. The D-And F-Block Elements 9. Coordination Compounds
10. Haloalkanes And Haloarenes 11. Alcohols, Phenols And Ethers 12. Aldehydes, Ketones And Carboxylic Acids
13. Amines 14. Biomolecules 15. Polymers
16. Chemistry In Everyday Life



Chapter 12 Aldehydes, Ketones And Carboxylic Acids



Nomenclature And Structure Of Carbonyl Group

This unit focuses on organic compounds containing the **carbonyl functional group** (>C=O). This is a very important functional group in organic chemistry.

Compounds where the carbonyl carbon is bonded to a carbon or hydrogen and the oxygen of a hydroxyl group (–OH) are called **carboxylic acids** (–COOH). Derivatives of carboxylic acids include amides (–CONH$_2$), acyl halides (–COX), esters (–COOR), and anhydrides (–COOCO–).

Carbonyl compounds like aldehydes, ketones, and carboxylic acids are widespread in nature, contributing to fragrances and flavours (e.g., vanillin, salicylaldehyde, cinnamaldehyde) and playing vital biochemical roles. They are also industrially important as solvents, adhesives, plastics, perfumes, and pharmaceuticals.


Nomenclature

Aldehydes and ketones are named using both common and IUPAC systems.

(a) Common Names:

(b) IUPAC Names:

Examples of aldehyde and ketone nomenclature:

Structure Common name IUPAC name
$\textsf{HCHO}$FormaldehydeMethanal
$\textsf{CH}_3\text{CHO}$AcetaldehydeEthanal
$\textsf{(CH}_3)_2\text{CHCHO}$Isobutyraldehyde2-Methylpropanal
Structure of 3-Methylcyclohexanecarbaldehyde
$\gamma$-Methylcyclohexanecarbaldehyde3-Methylcyclohexanecarbaldehyde
$\textsf{CH}_3\text{CH(OCH}_3)\text{CHO}$$\alpha$-Methoxypropionaldehyde2-Methoxypropanal
$\textsf{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{CHO}$ValeraldehydePentanal
$\textsf{CH}_2\text{=CHCHO}$AcroleinProp-2-enal
Structure of Benzene-1,2-dicarbaldehyde
PhthaldehydeBenzene-1,2-dicarbaldehyde
Structure of 3-Bromobenzaldehyde
m-Bromobenzaldehyde3-Bromobenzaldehyde
$\textsf{CH}_3\text{CO(CH}_2)_2\text{CH}_3$Methyl n-propyl ketonePentan-2-one
$\textsf{(CH}_3)_2\text{CHCOCH(CH}_3)_2$Diisopropyl ketone2,4-Dimethylpentan-3-one
Structure of 2-Methylcyclohexanone
$\alpha$-Methylcyclohexanone2-Methylcyclohexanone
$\textsf{(CH}_3)_2\text{C=CHCOCH}_3$Mesityl oxide4-Methylpent-3-en-2-one

Structure Of The Carbonyl Group

In the carbonyl group (>C=O), the carbon atom is $\textsf{sp}^2$ hybridised. It forms three sigma ($\sigma$) bonds: one with the oxygen atom and two with other atoms (carbon or hydrogen). The remaining valence electron on carbon is in a p-orbital, which overlaps with a p-orbital on the oxygen atom to form a pi ($\pi$) bond.

The oxygen atom in the carbonyl group also has two non-bonding lone pairs of electrons. The carbonyl carbon and the three atoms bonded to it lie in the same plane, with the $\pi$-electron cloud located above and below this plane. The bond angles around the carbonyl carbon are approximately 120°, consistent with a trigonal planar geometry.

Orbital diagram showing sp2 hybridisation and pi bond formation in a carbonyl group.

The carbon-oxygen double bond is polar because oxygen is more electronegative than carbon. This gives the carbonyl carbon a partial positive charge (electrophilic centre, acting as a Lewis acid) and the carbonyl oxygen a partial negative charge (nucleophilic centre, acting as a Lewis base).

The polarity of the carbonyl group can be represented by resonance between a neutral structure and a dipolar structure:

Resonance structures of the carbonyl group.

This polarity significantly influences the chemical reactivity of aldehydes and ketones.



Preparation Of Aldehydes And Ketones

Several methods are used to synthesise aldehydes and ketones.


Preparation Of Aldehydes And Ketones

These methods can yield both aldehydes and ketones depending on the starting material:


Preparation Of Aldehydes

Methods specific to preparing aldehydes:


Preparation Of Ketones

Methods specific to preparing ketones:

Example 12.1. Give names of the reagents to bring about the following transformations:

(i) Hexan-1-ol to hexanal

(ii) Cyclohexanol to cyclohexanone

(iii) p-Fluorotoluene to p-fluorobenzaldehyde

(iv) Ethanenitrile to ethanal

(v) Allyl alcohol to propenal

(vi) But-2-ene to ethanal

Answer:

(i) Oxidation of a primary alcohol to an aldehyde requires a mild oxidising agent like PCC (Pyridinium Chlorochromate). Reagent: PCC.

(ii) Oxidation of a secondary alcohol to a ketone. Reagents like anhydrous $\textsf{CrO}_3$ or acidified $\textsf{K}_2\text{Cr}_2\text{O}_7$ can be used. Reagent: Anhydrous $\textsf{CrO}_3$.

(iii) Oxidation of a methyl group on an aromatic ring to an aldehyde. Etard reaction using chromyl chloride or oxidation via benzylidene diacetate using $\textsf{CrO}_3$ in acetic anhydride are suitable methods. Reagent: 1. $\textsf{CrO}_2\text{Cl}_2$; 2. $\textsf{H}_3\text{O}^+$ (Etard reaction) OR 1. $\textsf{CrO}_3$ in acetic anhydride; 2. $\textsf{H}_3\text{O}^+$.

(iv) Reduction of a nitrile to an aldehyde. Stephen reaction or DIBAL-H are suitable. Reagent: 1. $\textsf{SnCl}_2$/$\textsf{HCl}$; 2. $\textsf{H}_2\text{O}$ (Stephen reaction) OR 1. DIBAL-H; 2. $\textsf{H}_2\text{O}$.

(v) Oxidation of a primary allylic alcohol to an aldehyde. PCC can be used as it does not oxidise the double bond. Reagent: PCC.

(vi) Ozonolysis of an alkene followed by reductive workup yields carbonyl compounds. But-2-ene ($\textsf{CH}_3\text{CH=CHCH}_3$) will give two molecules of acetaldehyde ($\textsf{CH}_3\text{CHO}$). Reagent: 1. $\textsf{O}_3$; 2. $\textsf{H}_2\text{O-Zn}$ dust.



Physical Properties

Physical characteristics of aldehydes and ketones:

Example comparison of boiling points:

Compound Formula Molecular Mass Boiling Point (K)
n-Butane$\textsf{CH}_3\text{CH}_2\text{CH}_2\text{CH}_3$58273
Methoxyethane$\textsf{CH}_3\text{OCH}_2\text{CH}_3$60281
Propanal$\textsf{CH}_3\text{CH}_2\text{CHO}$58322
Acetone$\textsf{CH}_3\text{COCH}_3$58329
Propan-1-ol$\textsf{CH}_3\text{CH}_2\text{CH}_2\text{OH}$60370

Example 12.2. Arrange the following compounds in the increasing order of their boiling points:

$\textsf{CH}_3\text{CH}_2\text{CH}_2\text{CHO}$ (Butanal), $\textsf{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{OH}$ (Butan-1-ol), $\textsf{H}_5\text{C}_2\text{-O-C}_2\text{H}_5$ (Ethoxyethane), $\textsf{CH}_3\text{CH}_2\text{CH}_2\text{CH}_3$ (n-Butane)

Answer:

Approximate molecular masses are similar (72-74). Intermolecular forces determine boiling points: hydrogen bonding > dipole-dipole > van der Waals.

  • n-Butane (alkane): Only van der Waals forces (weakest).
  • Ethoxyethane (ether): Dipole-dipole and van der Waals forces.
  • Butanal (aldehyde): Dipole-dipole and van der Waals forces. The carbonyl dipole is stronger than the ether dipole.
  • Butan-1-ol (alcohol): Strong intermolecular hydrogen bonding, plus van der Waals forces (strongest).

Increasing order of boiling points: $\textsf{CH}_3\text{CH}_2\text{CH}_2\text{CH}_3$ (n-Butane) < $\textsf{H}_5\text{C}_2\text{-O-C}_2\text{H}_5$ (Ethoxyethane) < $\textsf{CH}_3\text{CH}_2\text{CH}_2\text{CHO}$ (Butanal) < $\textsf{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{OH}$ (Butan-1-ol).



Chemical Reactions

Aldehydes and ketones share many chemical reactions due to the presence of the carbonyl group.


Nucleophilic Addition Reactions

Aldehydes and ketones characteristic reactions are **nucleophilic addition reactions** across the polarised carbon-oxygen double bond. The nucleophile attacks the electrophilic carbonyl carbon.

(i) Mechanism: A nucleophile ($\textsf{Nu}^-$) attacks the $\textsf{sp}^2$ hybridised carbonyl carbon from a direction perpendicular to the plane of the carbonyl group. This attack changes the hybridisation of the carbon from $\textsf{sp}^2$ to $\textsf{sp}^3$, forming a tetrahedral alkoxide intermediate. This intermediate then quickly accepts a proton ($\textsf{H}^+$) from the reaction medium (or acid catalyst) to give the neutral addition product.

Mechanism of nucleophilic addition to a carbonyl group.

(ii) Reactivity: Aldehydes are generally **more reactive** than ketones towards nucleophilic addition reactions. This difference is due to both steric and electronic factors.

Example 12.3. Would you expect benzaldehyde to be more reactive or less reactive in nucleophilic addition reactions than propanal? Explain your answer.

Answer:

Benzaldehyde ($\textsf{C}_6\text{H}_5\text{CHO}$) has a phenyl group attached to the carbonyl carbon, while propanal ($\textsf{CH}_3\text{CH}_2\text{CHO}$) has an ethyl group.

The phenyl group in benzaldehyde is also attached to the carbonyl carbon. The carbonyl group in benzaldehyde is directly attached to the benzene ring. The $\pi$ electrons of the benzene ring can conjugate with the $\pi$ electrons of the carbonyl group. This resonance interaction reduces the partial positive charge on the carbonyl carbon, making it less electrophilic and less susceptible to nucleophilic attack compared to the carbonyl carbon in propanal, where the ethyl group is electron-donating but does not exhibit resonance with the carbonyl.

Resonance structures of benzaldehyde carbonyl.

Therefore, benzaldehyde is **less reactive** than propanal in nucleophilic addition reactions due to the resonance stabilisation of the carbonyl group by the phenyl ring.


(iii) Important Nucleophilic Addition & Addition-Elimination Reactions:


Reduction


Oxidation

Oxidation reactions are used to distinguish aldehydes from ketones.

Mild oxidising agents used for differentiation:

Example 12.4. An organic compound (A) with molecular formula $\textsf{C}_8\text{H}_8\text{O}$ forms 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 $\textsf{C}_7\text{H}_6\text{O}_2$. Identify the compounds (A) and (B) and explain the reactions involved.

Answer:

Compound (A) ($\textsf{C}_8\text{H}_8\text{O}$) forms a 2,4-DNP precipitate, indicating it contains a carbonyl group (aldehyde or ketone).

It does not reduce Tollens' or Fehling's reagent, ruling out aldehyde functionality (except possibly aromatic aldehydes for Fehling's, but further tests clarify). Thus, (A) is likely a **ketone**.

(A) gives a yellow precipitate with $\textsf{I}_2$/$\textsf{NaOH}$ (positive iodoform test), meaning it is a methyl ketone ($\textsf{CH}_3\text{CO–}$ group).

The molecular formula $\textsf{C}_8\text{H}_8\text{O}$ suggests high unsaturation (Degree of Unsaturation = $8 - 8/2 + 1 = 5$). It doesn't decolourise bromine water or Baeyer's reagent, indicating the unsaturation is likely due to an aromatic ring, not C=C or C$\equiv$C bonds.

Compound (B) ($\textsf{C}_7\text{H}_6\text{O}_2$) is a carboxylic acid produced by oxidation of ketone (A). Benzoic acid has the formula $\textsf{C}_7\text{H}_6\text{O}_2$. If (B) is benzoic acid, then (A) must be a methyl ketone attached to a phenyl ring.

With a molecular formula of $\textsf{C}_8\text{H}_8\text{O}$, the only possible aromatic methyl ketone is phenyl methyl ketone, also known as **acetophenone** ($\textsf{C}_6\text{H}_5\text{COCH}_3$).

Let's check if acetophenone fits all the given properties:

  • Molecular formula: $\textsf{C}_6\text{H}_5\text{COCH}_3 = \textsf{C}_8\text{H}_8\text{O}$ (Matches).
  • 2,4-DNP derivative: Yes, ketones form 2,4-DNP derivatives.
  • Yellow precipitate with $\textsf{I}_2$/$\textsf{NaOH}$: Yes, acetophenone is a methyl ketone.
  • Reduces Tollens'/'Fehling's: No, it's a ketone.
  • Decolourises bromine water/Baeyer's: No, the unsaturation is in the stable aromatic ring.
  • Oxidation to carboxylic acid ($\textsf{C}_7\text{H}_6\text{O}_2$): Vigorous oxidation of acetophenone with chromic acid cleaves the C-C bond adjacent to the carbonyl (between the carbonyl carbon and the phenyl ring), oxidising the phenyl group to benzoic acid ($\textsf{C}_6\text{H}_5\text{COOH}$, $\textsf{C}_7\text{H}_6\text{O}_2$).

All properties match. Therefore, Compound (A) is Acetophenone, and Compound (B) is Benzoic acid.

Reactions involved:

Iodoform test on (A): $\textsf{C}_6\text{H}_5\text{COCH}_3 + 3\textsf{NaOI} \rightarrow \textsf{C}_6\text{H}_5\text{COONa} + \textsf{CHI}_3 (\text{yellow ppt}) + 2\textsf{NaOH}$

Iodoform test on acetophenone.

Oxidation of (A) to (B): $\textsf{C}_6\text{H}_5\text{COCH}_3 \xrightarrow{\textsf{Chromic acid}} \textsf{C}_6\text{H}_5\text{COOH}$

Oxidation of acetophenone to benzoic acid.


Reactions Due To $\alpha$-hydrogen

The hydrogen atoms on the carbon atom adjacent to the carbonyl group ($\alpha$-carbon) are called $\alpha$-hydrogens. These hydrogens are acidic due to the electron-withdrawing nature of the carbonyl group and the resonance stabilisation of the carbanion (enolate ion) formed after the removal of an $\alpha$-hydrogen.

Acidity of alpha-hydrogens and resonance stabilisation of enolate ion.

This acidity leads to characteristic reactions:


Other Reactions



Uses Of Aldehydes And Ketones

Aldehydes and ketones are widely used in various industries:



Carboxylic Acids

Carboxylic acids are organic compounds containing the **carboxyl functional group** (–COOH). This group is a combination of a carbonyl ($>\textsf{C=O}$) and a hydroxyl (–OH) group.

Carboxylic acids can be aliphatic ($\textsf{RCOOH}$) or aromatic ($\textsf{ArCOOH}$). Higher aliphatic carboxylic acids with 12 to 18 carbon atoms are known as **fatty acids** and occur naturally as esters in fats and oils.

Carboxylic acids are essential starting materials for synthesising many other functional groups and compounds like anhydrides, esters, acid chlorides, and amides.


Nomenclature

Carboxylic acids are named using both common and IUPAC systems.

Examples of carboxylic acid nomenclature:

Structure Common name IUPAC name
$\textsf{HCOOH}$Formic acidMethanoic acid
$\textsf{CH}_3\text{COOH}$Acetic acidEthanoic acid
$\textsf{CH}_3\text{CH}_2\text{COOH}$Propionic acidPropanoic acid
$\textsf{CH}_3\text{CH}_2\text{CH}_2\text{COOH}$Butyric acidButanoic acid
$\textsf{(CH}_3)_2\text{CHCOOH}$Isobutyric acid2-Methylpropanoic acid
$\textsf{HOOC-COOH}$Oxalic acidEthanedioic acid
$\textsf{HOOC -CH}_2\text{-COOH}$Malonic acidPropanedioic acid
$\textsf{HOOC -(CH}_2)_2\text{-COOH}$Succinic acidButanedioic acid
$\textsf{HOOC -(CH}_2)_3\text{-COOH}$Glutaric acidPentanedioic acid
$\textsf{HOOC -(CH}_2)_4\text{-COOH}$Adipic acidHexanedioic acid
$\textsf{HOOC -CH}_2\text{-CH(COOH)-CH}_2\text{-COOH}$Tricarballylic acid or carballylic acidPropane-1,2,3-tricarboxylic acid
$\textsf{C}_6\text{H}_5\text{COOH}$Benzoic acidBenzoic acid (Benzenecarboxylic acid)
$\textsf{C}_6\text{H}_5\text{CH}_2\text{COOH}$Phenylacetic acid2-Phenylethanoic acid
Structure of Phthalic acid (Benzene-1,2-dicarboxylic acid).
Phthalic acidBenzene-1,2-dicarboxylic acid

Structure Of Carboxyl Group

In the carboxyl group (–COOH), the carbon atom is $\textsf{sp}^2$ hybridised. The bonds to the carboxyl carbon lie in a plane, separated by approximately 120° angles.

The carboxyl group exhibits resonance due to the interaction between the lone pair on the hydroxyl oxygen and the $\pi$ system of the carbonyl group:

Resonance structures of the carboxyl group.

These resonance structures show that the carboxyl carbon is less electrophilic compared to the carbonyl carbon in aldehydes and ketones because the positive charge character is delocalised away from the carbon onto the hydroxyl oxygen.



Methods Of Preparation Of Carboxylic Acids

Various methods are available for synthesising carboxylic acids.


From Primary Alcohols And Aldehydes

Primary alcohols can be oxidised to carboxylic acids using strong oxidising agents like potassium permanganate ($\textsf{KMnO}_4$) (in neutral, acidic, or alkaline conditions), potassium dichromate ($\textsf{K}_2\text{Cr}_2\text{O}_7$) in acidic medium, or chromium trioxide ($\textsf{CrO}_3$) in acidic medium (Jones reagent).

$\textsf{RCH}_2\text{OH} \xrightarrow{\textsf{KMnO}_4 \text{ or } \textsf{K}_2\text{Cr}_2\text{O}_7/\textsf{H}^+ \text{ or Jones reagent}} \textsf{RCOOH}$

Aldehydes are readily oxidised to carboxylic acids even by milder oxidising agents like Tollens' or Fehling's reagents (as discussed in the Aldehydes and Ketones section).


From Alkylbenzenes

Aromatic carboxylic acids can be prepared by the vigorous oxidation of alkyl benzenes. Strong oxidising agents like chromic acid or acidic/alkaline potassium permanganate oxidise the entire alkyl side chain attached to the benzene ring to a carboxyl group, regardless of the chain length (provided it has at least one benzylic hydrogen). Tertiary alkyl groups are not oxidised by this method.

Oxidation of alkylbenzene to benzoic acid.


From Nitriles And Amides

Nitriles ($\textsf{RCN}$) can be hydrolysed in the presence of acid ($\textsf{H}^+$) or base ($\textsf{OH}^-$) catalysts. Acid hydrolysis first yields an amide, which is then further hydrolysed to a carboxylic acid. Basic hydrolysis yields the carboxylate salt, which needs acidification to get the carboxylic acid. Mild conditions can stop the reaction at the amide stage.

Acid hydrolysis: $\textsf{R–C}\equiv\textsf{N} \xrightarrow{\textsf{H}^+/\textsf{H}_2\text{O}} \textsf{RCONH}_2 \xrightarrow{\textsf{H}^+/\textsf{H}_2\text{O}} \textsf{RCOOH}$

Basic hydrolysis: $\textsf{R–C}\equiv\textsf{N} \xrightarrow{\textsf{OH}^-/\textsf{H}_2\text{O}} \textsf{RCOO}^- \xrightarrow{\textsf{H}^+} \textsf{RCOOH}$


From Grignard Reagents

Grignard reagents ($\textsf{RMgX}$) react with solid carbon dioxide (dry ice) to form magnesium carboxylate salts. Subsequent acidification with a mineral acid liberates the carboxylic acid.

Reaction of Grignard reagent with CO2 followed by acidification to form carboxylic acid.

Since Grignard reagents can be prepared from alkyl halides, this method (along with hydrolysis of nitriles) is useful for converting an alkyl halide into a carboxylic acid with one additional carbon atom (ascending the carbon series).


From Acyl Halides And Anhydrides

Acyl halides ($\textsf{RCOCl}$) undergo hydrolysis to give carboxylic acids. This reaction is faster in the presence of aqueous base, yielding the carboxylate ion which is then acidified.

$\textsf{RCOCl} + \textsf{H}_2\text{O} \rightarrow \textsf{RCOOH} + \textsf{HCl}$

$\textsf{RCOCl} + \textsf{NaOH} \rightarrow \textsf{RCOONa} + \textsf{HCl}$ then $\textsf{RCOONa} \xrightarrow{\textsf{H}^+} \textsf{RCOOH}$

Acid anhydrides also hydrolyse with water to give the corresponding carboxylic acid(s).

$\textsf{(RCO)}_2\text{O} + \textsf{H}_2\text{O} \rightarrow 2\textsf{RCOOH}$


From Esters

Esters ($\textsf{RCOOR'}$) can be hydrolysed to carboxylic acids. Acidic hydrolysis (using acid catalyst) directly yields the carboxylic acid and alcohol. Basic hydrolysis (saponification) yields the carboxylate salt and alcohol; subsequent acidification of the salt provides the carboxylic acid.

Acidic hydrolysis: $\textsf{RCOOR'} \xrightarrow{\textsf{H}^+/\textsf{H}_2\text{O}} \textsf{RCOOH} + \textsf{R'OH}$

Basic hydrolysis: $\textsf{RCOOR'} \xrightarrow{\textsf{OH}^-/\textsf{H}_2\text{O}} \textsf{RCOO}^- + \textsf{R'OH}$ then $\textsf{RCOO}^- \xrightarrow{\textsf{H}^+} \textsf{RCOOH}$

Example 12.5. Write chemical reactions to affect the following transformations:

(i) Butan-1-ol to butanoic acid

(ii) Benzyl alcohol to phenylethanoic 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

(vi) Butanal to butanoic acid.

Answer:

(i) Oxidation of primary alcohol to carboxylic acid. Reagent: Strong oxidising agent like alkaline $\textsf{KMnO}_4$ followed by acidification.

$\textsf{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{OH} \xrightarrow{\textsf{Alkaline KMnO}_4, \text{ heat}} \textsf{CH}_3\text{CH}_2\text{CH}_2\text{COO}^- \xrightarrow{\textsf{H}^+} \textsf{CH}_3\text{CH}_2\text{CH}_2\text{COOH}$

(ii) Convert benzyl alcohol ($\textsf{C}_6\text{H}_5\text{CH}_2\text{OH}$) to phenylethanoic acid ($\textsf{C}_6\text{H}_5\text{CH}_2\text{COOH}$). This requires adding one carbon atom. Convert alcohol to alkyl halide, then to nitrile, then hydrolyse. Or convert alcohol to aldehyde, react with Grignard reagent (adds carbon), then oxidise (this would add 2 carbons if aldehyde from benzyl alcohol reacts with CH3MgX). Simpler route: alcohol to alkyl halide, then nitrile, then hydrolysis.

$\textsf{C}_6\text{H}_5\text{CH}_2\text{OH} \xrightarrow{\textsf{SOCl}_2} \textsf{C}_6\text{H}_5\text{CH}_2\text{Cl} \xrightarrow{\textsf{KCN}} \textsf{C}_6\text{H}_5\text{CH}_2\text{CN} \xrightarrow{\textsf{H}^+/\textsf{H}_2\text{O}} \textsf{C}_6\text{H}_5\text{CH}_2\text{COOH}$

(iii) Convert 3-nitrobromobenzene to 3-nitrobenzoic acid. This involves replacing bromine with a carboxyl group, which usually requires adding a carbon. Convert aryl halide to Grignard reagent, then react with $\textsf{CO}_2$, then acidify.

Conversion of 3-nitrobromobenzene to 3-nitrobenzoic acid.

(iv) Convert 4-methylacetophenone to benzene-1,4-dicarboxylic acid (terephthalic acid). This requires oxidising both the methyl group and the acetyl methyl group to carboxyl groups. Vigorous oxidation conditions are needed, e.g., alkaline $\textsf{KMnO}_4$ followed by acidification.

Conversion of 4-methylacetophenone to benzene-1,4-dicarboxylic acid.

(v) Convert cyclohexene to hexane-1,6-dioic acid (adipic acid). This requires oxidative cleavage of the double bond and oxidation of the terminal carbons to carboxylic acids. Strong oxidising agents like alkaline $\textsf{KMnO}_4$ or acidified $\textsf{K}_2\text{Cr}_2\text{O}_7$ are suitable for cleaving the double bond and oxidising fragments to carboxylic acids.

Conversion of cyclohexene to hexane-1,6-dioic acid.

(vi) Convert butanal to butanoic acid. This is oxidation of an aldehyde to a carboxylic acid, which can be done with common oxidising agents like $\textsf{KMnO}_4$ or $\textsf{K}_2\text{Cr}_2\text{O}_7/\textsf{H}^+$. Even mild agents work, but usually strong ones are mentioned for complete oxidation to acid.

$\textsf{CH}_3\text{CH}_2\text{CH}_2\text{CHO} \xrightarrow{\textsf{KMnO}_4/\textsf{OH}^- \text{ or } \textsf{K}_2\text{Cr}_2\text{O}_7/\textsf{H}^+} \textsf{CH}_3\text{CH}_2\text{CH}_2\text{COOH}$



Physical Properties

Physical characteristics of carboxylic acids:



Chemical Reactions

Carboxylic acids undergo reactions involving the cleavage of O–H bonds (due to acidity), C–OH bonds (nucleophilic substitution at acyl carbon), and reactions of the entire carboxyl group or the hydrocarbon part.


Reactions Involving Cleavage Of O–H Bond

The O–H bond in carboxylic acids is polar, allowing proton donation (acidity).

Example 12.8. Which acid of each pair shown here would you expect to be stronger?

(i) $\textsf{CH}_3\text{CO}_2\text{H}$ or $\textsf{CH}_2\text{FCO}_2\text{H}$

(ii) $\textsf{CH}_2\text{FCO}_2\text{H}$ or $\textsf{CH}_2\text{ClCO}_2\text{H}$

(iii) $\textsf{CH}_2\text{FCH}_2\text{CH}_2\text{CO}_2\text{H}$ or $\textsf{CH}_3\text{CHFCH}_2\text{CO}_2\text{H}$

(iv)

Structures for Example 12.8 (iv).

Answer:

Acidity is increased by electron-withdrawing groups (EWG) and decreased by electron-donating groups (EDG). The effect of inductive withdrawal decreases with distance from the carboxyl group. Electronegativity order: F > Cl.

(i) $\textsf{CH}_3\text{COOH}$ vs $\textsf{CH}_2\text{FCOOH}$. Fluorine is an EWG, methyl is weakly EDG. Fluorine stabilises the carboxylate anion more than methyl. Therefore, $\textsf{CH}_2\text{FCO}_2\text{H}$ is stronger.

Stronger acid: $\textsf{CH}_2\text{FCO}_2\text{H}$.

(ii) $\textsf{CH}_2\text{FCO}_2\text{H}$ vs $\textsf{CH}_2\text{ClCO}_2\text{H}$. Both F and Cl are EWGs. Fluorine is more electronegative than chlorine, so it withdraws electrons more strongly, better stabilising the carboxylate anion. Therefore, fluoroacetic acid is stronger than chloroacetic acid.

Stronger acid: $\textsf{CH}_2\text{FCO}_2\text{H}$.

(iii) $\textsf{CH}_2\text{FCH}_2\text{CH}_2\text{CO}_2\text{H}$ vs $\textsf{CH}_3\text{CHFCH}_2\text{CO}_2\text{H}$. Both have a fluorine atom EWG. In the first compound, F is on the $\gamma$-carbon (three carbons away from COOH). In the second compound, F is on the $\beta$-carbon (two carbons away from COOH). The inductive effect of an EWG decreases rapidly with distance. Therefore, the fluorine on the $\beta$-carbon has a stronger acid-strengthening effect than the fluorine on the $\gamma$-carbon.

Stronger acid: $\textsf{CH}_3\text{CHFCH}_2\text{CO}_2\text{H}$.

(iv)

Structures for Example 12.8 (iv).
The first compound is benzoic acid. The second is 4-nitrobenzoic acid. The nitro group ($\textsf{NO}_2$) is a strong EWG, both by induction and resonance. Located at the *para* position, it significantly stabilises the carboxylate anion of benzoic acid. Benzoic acid itself is more acidic than aliphatic acids due to the inductive withdrawal by the $\textsf{sp}^2$ carbons of the ring and resonance stabilisation of the carboxylate into the ring (minor). Adding a strong EWG like $\textsf{NO}_2$ makes it much stronger.

Stronger acid: 4-Nitrobenzoic acid.


Reactions Involving Cleavage Of C–OH Bond

These reactions involve nucleophilic substitution at the acyl carbon, where the hydroxyl group is replaced by a nucleophile. Carboxylic acids react as a source of acyl groups ($\textsf{RCO–}$).


Reactions Involving –COOH Group

These reactions involve the transformation of the entire carboxyl group.


Substitution Reactions In The Hydrocarbon Part



Uses Of Carboxylic Acids

Carboxylic acids have numerous applications:



Intext Questions



Question 12.1. 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:

Question 12.2. Write the structures of products of the following reactions;

(i)

Hydration of ethyne. Ethyne (HC≡CH) reacts with water in the presence of mercuric sulphate (HgSO4) and sulphuric acid (H2SO4).

(ii) $(C_6H_5CH_2)_2Cd + 2CH_3COCl \rightarrow$

(iii)

Hydration of propyne. Propyne (CH3-C≡CH) reacts with water in the presence of mercuric ions (Hg2+) and sulphuric acid (H2SO4).

(iv)

Etard reaction on p-nitrotoluene. The compound has a benzene ring with a methyl group and a nitro group at para positions. It reacts with 1. Chromyl chloride (CrO2Cl2) followed by 2. hydrolysis (H3O+).

Answer:

Question 12.3. Arrange the following compounds in increasing order of their boiling points.

$CH_3CHO, CH_3CH_2OH, CH_3OCH_3, CH_3CH_2CH_3$

Answer:

Question 12.4. 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.

Hint: Consider steric effect and electronic effect.

Answer:

Question 12.5. Predict the products of the following reactions:

(i)

A reaction scheme shows cyclohexanone reacting with semicarbazide (H2N-NH-CO-NH2) in the presence of an acid catalyst (H+).

(ii)

A reaction scheme shows a benzene ring with two aldehyde groups at positions 1 and 4, reacting with an excess of ethylene glycol (HO-CH2-CH2-OH) in the presence of an acid catalyst.

(iii)

A reaction scheme shows a benzene ring attached to a carbon which is double bonded to an oxygen and single bonded to a CH3 group. This compound (acetophenone) reacts with ethylamine (CH3CH2NH2).

(iv)

A reaction scheme shows R-C≡N reacting with reagents labelled as 1. (i) DIBAL-H (ii) H2O, and 2. N2H4, KOH / ethylene glycol, heat.

Answer:

Question 12.7. Show how each of the following compounds can be converted to benzoic acid.

(i) Ethylbenzene

(ii) Acetophenone

(iii) Bromobenzene

(iv) Phenylethene (Styrene)

Answer:

Question 12.8. Which acid of each pair shown here would you expect to be stronger?

(i) $CH_3CO_2H$ or $CH_2FCO_2H$

(ii) $CH_2FCO_2H$ or $CH_2ClCO_2H$

(iii) $CH_2FCH_2CH_2CO_2H$ or $CH_3CHFCH_2CO_2H$

(iv)

Two substituted benzoic acid structures are compared. The first has a trifluoromethyl group (CF3) attached to the benzene ring at the para position. The second has a methyl group (CH3) attached to the benzene ring at the para position.

Answer:



Exercises



Question 12.1. 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

Answer:

Question 12.2. Name the following compounds according to IUPAC system of nomenclature:

(i) $CH_3CH(CH_3)CH_2CH_2CHO$

(ii) $CH_3CH_2COCH(C_2H_5)CH_2CH_2Cl$

(iii) $CH_3CH=CHCHO$

(iv) $CH_3COCH_2COCH_3$

(v) $CH_3CH(CH_3)CH_2C(CH_3)_2COCH_3$

(vi) $(CH_3)_3CCH_2COOH$

(vii) $OHCC_6H_4CHO-p$

Answer:

Question 12.3. 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:

Question 12.4. Write the IUPAC names of the following ketones and aldehydes. Wherever possible, give also common names.

(i) $CH_3CO(CH_2)_4CH_3$

(ii) $CH_3CH_2CHBrCH_2CH(CH_3)CHO$

(iii) $CH_3(CH_2)_5CHO$

(iv) $Ph-CH=CH-CHO$

(v)

The chemical structure of cyclohexanecarbaldehyde. A cyclohexane ring is attached to a CHO group.

(vi) $PhCOPh$

Answer:

Question 12.5. Draw structures of the following derivatives.

(i) The 2,4-dinitrophenylhydrazone of benzaldehyde

(ii) Cyclopropanone oxime

(iii) Acetaldehydedimethylacetal

(iv) The semicarbazone of cyclobutanone

(v) The ethylene ketal of hexan-3-one

(vi) The methyl hemiacetal of formaldehyde

Answer:

Question 12.6. Predict the products formed when cyclohexanecarbaldehyde reacts with following reagents.

(i) $PhMgBr$ and then $H_3O^+$

(ii) Tollens’ reagent

(iii) Semicarbazide and weak acid

(iv) Excess ethanol and acid

(v) Zinc amalgam and dilute hydrochloric acid

Answer:

Question 12.7. 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:

Question 12.8. How will you convert ethanal into the following compounds?

(i) Butane-1,3-diol

(ii) But-2-enal

(iii) But-2-enoic acid

Answer:

Question 12.9. 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:

Question 12.10. An organic compound with the molecular formula $C_9H_{10}O$ 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:

Question 12.11. An organic compound (A) (molecular formula $C_8H_{16}O_2$) 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:

Question 12.12. 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) $CH_3CH_2CH(Br)COOH$, $CH_3CH(Br)CH_2COOH$, $(CH_3)_2CHCOOH$, $CH_3CH_2CH_2COOH$ (acid strength)

(iii) Benzoic acid, 4-Nitrobenzoic acid, 3,4-Dinitrobenzoic acid, 4-Methoxybenzoic acid (acid strength)

Answer:

Question 12.13. 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:

Question 12.14. 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:

Question 12.15. 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 $\alpha$-Hydroxyphenylacetic acid

(ix) Benzoic acid to m-Nitrobenzyl alcohol

Answer:

Question 12.16. Describe the following:

(i) Acetylation

(ii) Cannizzaro reaction

(iii) Cross aldol condensation

(iv) Decarboxylation

Answer:

Question 12.17. Complete each synthesis by giving missing starting material, reagent or products

(i)

A two-step reaction. A benzene ring with an attached O-CO-CH3 group and a CHO group at the para position is reacted first with aqueous NaOH, and then with H+.

(ii)

A reaction starting with cyclohexanone. It reacts with an unspecified reagent to form two products: cyanohydrin of cyclohexanone and 1-phenylcyclohexanol.

(iii)

A reaction of p-methylbenzaldehyde with an unspecified reagent to produce p-methylbenzoic acid.

(iv)

A reaction of a benzene ring with an attached O-C6H5 group reacting with an unspecified reagent to produce p-hydroxyacetophenone.

(v)

A reaction of an unspecified starting material with 1. O3, 2. Zn-H2O to produce two molecules of ethanal (CH3CHO).

(vi)

A reaction of 2-pentanone (CH3-CO-CH2CH2CH3) with an unspecified reagent to produce pent-2-ene.

(vii)

A reaction of an unspecified starting material to produce butanal (CH3CH2CH2CHO).

(viii)

A reaction of but-2-enal (CH3CH=CHCHO) with an unspecified reagent to produce but-2-en-1-ol.

(ix)

A reaction of a benzene ring with a COOH group reacting with an unspecified reagent to produce a benzene ring with a COCl group.

(x)

A reaction of an unspecified starting material with LiAlH4 to produce 2-phenylethanol.

(xi)

A reaction of butanoic acid (CH3CH2CH2COOH) with an unspecified reagent to produce butanoyl chloride.

Answer:

Question 12.18. 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 –$NH_2$ 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:

Question 12.19. 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:

Question 12.20. Although phenoxide ion has more number of resonating structures than carboxylate ion, carboxylic acid is a stronger acid than phenol. Why?

Answer: