This chapter explores the chemistry of aldehydes, ketones, and carboxylic acids, all of which contain the carbonyl group (C=O). This group is highly reactive due to the difference in electronegativity between carbon and oxygen, making the carbon atom electrophilic (electron-deficient) and the oxygen atom nucleophilic (electron-rich).
Aldehydes have the carbonyl group bonded to at least one hydrogen atom, while ketones have it bonded to two carbon atoms. Carboxylic acids have a carboxyl group (-COOH), which is a carbonyl group with a hydroxyl group (-OH) attached.
Key Concepts Covered:
- Nomenclature: Understanding how to name these compounds using IUPAC rules.
- Structure and Properties: Examining the structure of the carbonyl group and how it influences the physical and chemical properties of these compounds (e.g., boiling point, solubility).
- Preparation Methods: Learning various methods to synthesize aldehydes, ketones, and carboxylic acids from different starting materials (e.g., oxidation of alcohols, ozonolysis of alkenes, Friedel-Crafts acylation).
- Reactions: Studying the characteristic reactions of these compounds, which primarily involve nucleophilic addition to the carbonyl group. This includes reactions with alcohols (acetal formation), ammonia derivatives (imine and Schiff base formation), Grignard reagents, and reducing agents.
- Oxidation and Reduction: Understanding the oxidation reactions of aldehydes and ketones, and the reduction reactions of aldehydes, ketones, and carboxylic acids.
- Acidity of Carboxylic Acids: Exploring the acidic nature of carboxylic acids and the factors that influence their acidity.
- Important Reactions: Delving into specific named reactions like the Aldol condensation, Cannizzaro reaction, and Hell-Volhard-Zelinsky reaction.
Why this chapter is important:
- Ubiquity: Aldehydes, ketones, and carboxylic acids are widely found in nature and have numerous applications in industry and everyday life.
- Reactivity: The carbonyl group’s reactivity makes these compounds versatile intermediates in organic synthesis.
- Biological Significance: Many biologically important molecules, such as carbohydrates, hormones, and enzymes, contain these functional groups.
Exercise
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.
Ans :
Let’s define these terms and provide an example reaction for each:
(i) Cyanohydrin:
- Definition: A cyanohydrin is an organic compound formed by the addition of hydrogen cyanide (HCN) to an aldehyde or ketone. The product contains both a cyano (-CN) and a hydroxyl (-OH) group on the same carbon atom.
Example:
CH3CHO (Acetaldehyde) + HCN → CH3CH(OH)CN (Acetaldehyde cyanohydrin)
(ii) Acetal:
- Definition: An acetal is a functional group where two alkoxy groups (-OR) are bonded to the same carbon atom, which is also bonded to a hydrogen atom. Acetals are formed by the reaction of an aldehyde with two equivalents of an alcohol in the presence of an acid catalyst.
Example:
CH3CHO (Acetaldehyde) + 2 CH3CH2OH (Ethanol) –H+ catalyst–> CH3CH(OCH2CH3)2 (Diethyl acetal) + H2O
(iii) Semicarbazone:
- Definition: A semicarbazone is a derivative of an aldehyde or ketone formed by reaction with semicarbazide (H2N-NH-CO-NH2). It contains the >C=N-NH-CO-NH2 group.
Example:
CH3CHO (Acetaldehyde) + H2N-NH-CO-NH2 (Semicarbazide) → CH3CH=N-NH-CO-NH2 (Acetaldehyde semicarbazone) + H2O
(iv) Aldol:
- Definition: An aldol is a β-hydroxy aldehyde or β-hydroxy ketone, formed by the addition of one aldehyde or ketone molecule to another in the presence of a base catalyst.
Example:
2 CH3CHO (Acetaldehyde) –Base catalyst–> CH3CH(OH)CH2CHO (Aldol)
(v) Hemiacetal:
- Definition: A hemiacetal is a compound formed by the addition of one molecule of alcohol to an aldehyde. It contains an alkoxy group (-OR) and a hydroxyl group (-OH) on the same carbon atom. It is an intermediate in the formation of an acetal.
Example:
CH3CHO (Acetaldehyde) + CH3CH2OH (Ethanol) ⇌ CH3CH(OH)OCH2CH3 (Hemiacetal)
(vi) Oxime:
- Definition: An oxime is a compound formed by the condensation reaction between an aldehyde or ketone and hydroxylamine (NH2OH). It contains the >C=N-OH group.
Example:
(CH3)2C=O (Acetone) + NH2OH (Hydroxylamine) → (CH3)2C=N-OH (Acetone oxime) + H2O
(vii) Ketal:
- Definition: A ketal is similar to an acetal but is formed from a ketone instead of an aldehyde. Two alkoxy groups are bonded to the carbon of the carbonyl group.
Example:
(CH3)2C=O (Acetone) + 2 CH3CH2OH (Ethanol) –H+ catalyst–> (CH3)2C(OCH2CH3)2 (Diethyl ketal) + H2O
(viii) Imine:
- Definition: An imine is a compound containing a carbon-nitrogen double bond (>C=N-) where the nitrogen is attached to an alkyl or aryl group, but not to a hydrogen atom. They are formed by the reaction of aldehydes or ketones with primary amines (R-NH2).
Example:
CH3CHO (Acetaldehyde) + CH3NH2 (Methylamine) → CH3CH=NCH3 (N-Methylimine) + H2O
(ix) 2,4-DNP derivative:
- Definition: This refers to the product formed when an aldehyde or ketone reacts with 2,4-dinitrophenylhydrazine (2,4-DNPH). The product is a 2,4-dinitrophenylhydrazone. These derivatives are often used to identify aldehydes and ketones due to their characteristic colors and melting points.
Example:
CH3CHO (Acetaldehyde) + 2,4-DNPH → Acetaldehyde 2,4-dinitrophenylhydrazone (a yellow/orange solid) + H2O
(x) Schiff’s base:
- Definition: A Schiff base is a type of imine, specifically the condensation product of an aldehyde or ketone with a primary amine (R-NH2).
Example: (Same as the imine example)
CH3CHO (Acetaldehyde) + CH3NH2 (Methylamine) → CH3CH=NCH3 (a Schiff base) + H2O
2. Name the following compounds according to IUPAC system of nomenclature:
(i) CH3CH (CH3)—CH2 CH2—CHO
(ii) CH3CH2COCH(C2H5)CH2CH2Cl
(iii) CH3CH=CHCHO
(iv) CH3COCH2COCH3
(v) CH3CH(CH3)CH2C(CH3)2COCH3
(vi) (CH3)3CCH2COOH.
(vii) OHCC6H4CHO-p
Ans :
(i) CH3CH(CH3)—CH2CH2—CHO
- IUPAC Name: 4-Methylpentanal
(ii) CH3CH2COCH(C2H5)CH2CH2Cl
- IUPAC Name: 5-Chloro-3-ethyl-2-pentanone
(iii) CH3CH=CHCHO
- IUPAC Name: 2-Butenal (or Crotonaldehyde, which is a common name sometimes used)
(iv) CH3COCH2COCH3
- IUPAC Name: 2,4-Pentanedione
(v) CH3CH(CH3)CH2C(CH3)2COCH3
- IUPAC Name: 3,3,5-Trimethyl-2-hexanone
(vi) (CH3)3CCH2COOH
- IUPAC Name: 3,3-Dimethylbutanoic acid
(vii) OHCC6H4CHO-p (Where “p” indicates para substitution)
- IUPAC Name: 1,4-Benzenedicarbaldehyde (or p-Phthalaldehyde)
3. Draw the structures of the following compounds.
(i) 3-Methylbutanal
(ii) p-Methylbenzaldehyde
(iii) 4-Chloropentan-2-one
(iv) p, p’-Dihydroxybenzophenone
(v) p-Nitropropiophenone
(vi) 4-Methylpent-3-en-2-one.
(vii) 3-Bromo-4-phenylpentanoic acid
(viii) Hex-2-en-4-ynoic acid
Ans :
4. Write the IUPAC names of the following ketones and aldehydes. Wherever possible, give also common names.
(i) CH3CO(CH2)4CH3
(ii) CH3CH2CH BrCH2CH(CH3)CHO
(iii) CH3(CH2)5CHO
(iv) Ph—CH=CH—CHO
Ans :
5. Draw structures of the following derivatives:
(i) The 2,4-dinitrophenylhydrazone of benzaldehyde
(ii) Cydopropanone oxime
(iii) Acetaldehydedimethylacetal
(iv) The semicarbazone of cyclobutanone
(v) The ethylene ketal of hexan-3-one
(vi) The methyl hemiacetal of formaldehyde
Ans :
6. Predict the product when cyclohexanecarbaldehyde reacts with following reagents :
(i) C6H5MgBr followed by H30+
(ii) Tollen’s reagent
(iii) Semicarbazide in the weakly acidic medium
(iv) Excess of ethanol in the presence of acid
(v) Zinc amalgam and Cyclohexanecarbaldehyde Semicarbazide
Ans :
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-l-ol 1
(ix) 2,2-Dimethylbutanal
Ans :
Let’s analyze which compounds undergo aldol condensation, Cannizzaro reaction, or neither, and write the structures of the expected products.
Aldol Condensation: Requires α-hydrogens (hydrogen atoms on the carbon adjacent to the carbonyl group).
Cannizzaro Reaction: Requires no α-hydrogens and uses concentrated base. One molecule is oxidized to a carboxylic acid salt, and another is reduced to an alcohol.
Neither: Compounds that lack α-hydrogens and aren’t aldehydes (for Cannizzaro).
Here’s the breakdown:
(i) Methanal (HCHO):
- Reaction: Cannizzaro reaction
- Product: Methanol (CH3OH) + Potassium formate (HCOOK)
(ii) 2-Methylpentanal:
- Reaction: Aldol condensation
- Product: A mixture of β-hydroxy aldehydes. The major product would depend on reaction conditions and which enol is formed. One major product would be 2-methyl-2-(1-hydroxy-2-methylbutyl)pentanal.
(iii) Benzaldehyde:
- Reaction: Cannizzaro reaction
- Product: Benzyl alcohol (C6H5CH2OH) + Potassium benzoate (C6H5COOK)
(iv) Benzophenone:
- Reaction: Neither. No α-hydrogens, not a reactive aldehyde for Cannizzaro, not easily enolized for Aldol.
(v) Cyclohexanone:
- Reaction: Aldol condensation
- Product: The aldol condensation product of cyclohexanone would be a β-hydroxy ketone where two cyclohexanone rings are joined via a new carbon-carbon bond alpha to the original carbonyls.
(vi) 1-Phenylpropanone:
- Reaction: Aldol condensation
- Product: A mixture of β-hydroxy ketones is possible.
(vii) Phenylacetaldehyde:
- Reaction: Aldol Condensation
- Product: A mixture of β-hydroxy aldehydes is possible.
(viii) Butan-1-ol:
- ** It’s an alcohol, not an aldehyde or ketone.
(ix) 2,2-Dimethylbutanal:
- Reaction: Cannizzaro reaction. No α-hydrogens
- Product: 2,2-Dimethylbutanol + potassium 2,2-dimethylbutanoate
8. How will you convert ethanal into the following compounds?
(i) Butane-1,3-diol
(ii) But-2-enal
(iii) But-2-enoic acid
Ans :
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 :
10. An organic compound with the molecular formula C9H10O forms 2,4-DNP derivative, reduces Tollen’s reagent, and undergoes Cannizzaro reaction. On vigorous oxidation, it gives 1,2-benzenedicarboxylic acid. Identify the compound.
Ans :
Molecular Formula C9H10O: This suggests an aromatic compound (due to the relatively high number of carbons compared to hydrogens) with one oxygen atom.
Forms 2,4-DNP derivative: This indicates the presence of a carbonyl group (C=O), as aldehydes and ketones react with 2,4-dinitrophenylhydrazine to form these derivatives.
Reduces Tollen’s reagent: This means the compound is an aldehyde. Tollen’s reagent is a test for aldehydes (but not ketones).
Undergoes Cannizzaro reaction: This confirms that the compound is an aldehyde and that it lacks α-hydrogens (hydrogens on the carbon atom adjacent to the aldehyde group). This is because aldehydes without α-hydrogens undergo Cannizzaro reaction.
Vigorous oxidation gives 1,2-benzenedicarboxylic acid (phthalic acid): This is the most crucial clue. It tells us that the aldehyde group and another substituent are attached to a benzene ring, and that vigorous oxidation converts these two groups into carboxylic acid groups. This pattern points to an ortho-substituted benzene derivative.
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). (Q on dehydration gives but-l-ene. Write equations for the reactions involved.
Ans :
(A) (C8H16O2) + H2O –H+–> (B) + (C): Compound (A) is hydrolyzed to give a carboxylic acid (B) and an alcohol (C). The molecular formula suggests an ester.
(C) –oxidation–> (B): Alcohol (C) is oxidized to carboxylic acid (B). This indicates that (C) is a primary alcohol or a secondary alcohol. If it were a tertiary alcohol, it wouldn’t be easily oxidized to a carboxylic acid.
(C) –dehydration–> but-1-ene: Alcohol (C) on dehydration gives but-1-ene. This tells us that (C) is butan-1-ol (CH3CH2CH2CH2OH).
Therefore:
(A): Butyl butanoate (CH3CH2CH2COOCH2CH2CH2CH3)
(B): Butanoic acid (CH3CH2CH2COOH)
(C): Butan-1-ol (CH3CH2CH2CH2OH)
12. Arrange the following in increasing order of the property indicated :
(i) Acetaldehyde, Acetone, Di tert. butyl ketone, Methyl tert. butyl ketone (reactivity towards HCN). (C.B.S.E. Sample Paper 2011, 2015, C.B.S.E. Delhi 2012)
(ii) CH3CH2CH(Br)COOH, CH3CH(Br)CH2COOH, (CH3)2CHCOOH, CH3CH2CH2COOH (acid strength) (C.B.S.E. Delhi2008)
(iii) Benzoic acid, 4-Nitrobenzoic acid, 3, 5-Dinitrobenzoic acid, 4-Methoxybenzoic acid (acid strength)
Ans :
i) Reactivity towards HCN (increasing order):
The reactivity towards HCN depends on the steric hindrance around the carbonyl group and the electron density on the carbonyl carbon. More hindered ketones are less reactive. Electron-donating groups decrease reactivity, while electron-withdrawing groups increase it.
- Di-tert-butyl ketone (most hindered, two bulky tert-butyl groups)
- Methyl tert-butyl ketone (one bulky tert-butyl group)
- Acetone (two methyl groups)
- Acetaldehyde (one methyl group, one hydrogen – least hindered)
Therefore, the order is: Di-tert-butyl ketone < Methyl tert-butyl ketone < Acetone < Acetaldehyde
(ii) Acid Strength (increasing order):
The acid strength is influenced by the stability of the conjugate base (carboxylate anion). Electron-withdrawing groups stabilize the conjugate base, increasing acidity. The proximity of the substituent also plays a role (the “inductive effect” decreases with distance).
- (CH3)2CHCOOH (Isopropyl group is electron-donating via +I effect, destabilizes the conjugate base)
- CH3CH2CH2COOH (no strong electron-withdrawing or donating groups)
- CH3CH(Br)CH2COOH (Br is electron-withdrawing, increases acidity, but the effect is slightly lessened by being one carbon further from the -COOH)
- CH3CH2CH(Br)COOH (Br is electron-withdrawing and closer to the -COOH, so it has the largest acid-strengthening effect)
Therefore, the order is: (CH3)2CHCOOH < CH3CH2CH2COOH < CH3CH(Br)CH2COOH < CH3CH2CH(Br)COOH
(iii) Acid Strength (increasing order):
Substituents on the aromatic ring affect the acidity of benzoic acid. Electron-withdrawing groups increase acidity, and electron-donating groups decrease it. The number and position of these groups also play a role.
- 4-Methoxybenzoic acid (-OCH3 is an electron-donating group via resonance, so it decreases acidity)
- Benzoic acid (no strong substituent effects)
- 4-Nitrobenzoic acid (-NO2 is a strong electron-withdrawing group via resonance, so it increases acidity)
- 3,5-Dinitrobenzoic acid (two -NO2 groups strongly increase acidity)
Therefore, the order is: 4-Methoxybenzoic acid < Benzoic acid < 4-Nitrobenzoic acid < 3,5-Dinitrobenzoic acid
13. Give simple chemical tests to distinguish between the following pairs of compounds.
(i) PropanalandPropanone
(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) EthanalandPropanal
Ans :
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) Phenylaceticacid
(v) p-nitrobenzaldehyde
Ans :
15. How will you bring about the following conversions in not more than two steps?
(i) PropanonetoPropene
(ii) Benzoic acid to Benzaldehyde
(iii) Ethanol to 3-Hydroxybutanal
(iv) Benzene to m-Nitroacetophenone
(v) Benzaldehyde to Benzophenone –
(vi) Bromobenzeneto 1-PhenylethanoL
(vii) Benzaldehyde to 3-Phenylpropan-1-ol.
(viii) Benzaldehyde to α Hydroxyphenylacetk acid
(ix) Benzoic acid to m-Nitrobenzy 1 alcohol
Ans :
16. Describe the following:
(i) Acetylation
(ii) Cannizzaro reaction
(iii) Cross aldol condensation
(iv) Decarboxylation
Ans :
(i) Acetylation:
- Definition: Acetylation is a chemical reaction where an acetyl group (CH3CO-) is introduced into a molecule. This is commonly done by reacting a compound containing a hydroxyl (-OH), amino (-NH2), or thiol (-SH) group with an acetylating agent, such as acetic anhydride ((CH3CO)2O) or acetyl chloride (CH3COCl).
- Mechanism: Acetylation usually proceeds through a nucleophilic acyl substitution mechanism. The lone pair on the oxygen, nitrogen, or sulfur atom attacks the carbonyl carbon of the acetylating agent, leading to the substitution of the leaving group (e.g., carboxylate anion from acetic anhydride or chloride ion from acetyl chloride) with the acetyl group.
Example: The acetylation of an alcohol with acetic anhydride:
ROH + (CH3CO)2O → ROCOCH3 + CH3COOH
- Uses: Acetylation is used to protect sensitive functional groups during chemical synthesis, to modify the properties of molecules (e.g., aspirin is acetylated salicylic acid), and in the analysis of carbohydrates.
(ii) Cannizzaro Reaction:
- Definition: The Cannizzaro reaction is a disproportionation reaction of aldehydes that lack α-hydrogens (no hydrogen atoms on the carbon adjacent to the aldehyde group). It’s induced by a concentrated base (e.g., concentrated NaOH or KOH).
- Mechanism: The Cannizzaro reaction involves the transfer of a hydride ion (H⁻) from one aldehyde molecule to another. One molecule is oxidized to a carboxylate anion, and the other is reduced to a primary alcohol.
Example: The Cannizzaro reaction of benzaldehyde:
2 C6H5CHO + KOH (conc.) → C6H5COOK + C6H5CH2OH
- Key Feature: This reaction demonstrates that aldehydes without α-hydrogens cannot form enolates and thus cannot undergo typical aldol condensation.
(iii) Cross Aldol Condensation:
- Definition: A cross aldol condensation is an aldol condensation reaction between two different carbonyl compounds (aldehydes or ketones), at least one of which has α-hydrogens.
- Mechanism: The mechanism is similar to the regular aldol condensation, involving the formation of enolates or enols, nucleophilic attack on the other carbonyl group, and proton transfer. However, since there are two different carbonyl compounds, a mixture of products is usually obtained.
Example: The cross aldol condensation of acetaldehyde and propanal:
CH3CHO + CH3CH2CHO –Base catalyst–> Mixture of products (four possible β-hydroxy aldehydes/ketones)
(iv) Decarboxylation:
- Definition: Decarboxylation is the removal of a carboxyl group (-COOH) from a molecule in the form of carbon dioxide (CO2). This is commonly observed with carboxylic acids, especially those with a carbonyl group on the carbon atom next to the carboxyl group (β-keto acids) or those with geminal dicarboxylic acids.
- Mechanism: The mechanism of decarboxylation often involves a cyclic transition state. In the case of β-keto acids, the enol form of the carboxylic acid facilitates the loss of CO2.
- Example: The decarboxylation of a β-keto acid:
RCOCH2COOH –Heat–> RCOCH3 + CO2
17. Complete each synthesis by giving missing starting material, reagent or products.
Ans :
18. Give plausible explanation for each of the following:
(i) Cyclohexanone forms cyanohydrin in good yield but 2,2, fctrimethylcyclohexanone 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) Cyclohexanone forms cyanohydrin in good yield, but 2,2,6-trimethylcyclohexanone does not:
- Steric Hindrance: The formation of a cyanohydrin involves the nucleophilic addition of cyanide ion (CN⁻) to the carbonyl carbon of the ketone. In cyclohexanone, there is relatively little steric hindrance around the carbonyl group, allowing the cyanide ion to approach and attack easily. However, 2,2,6-trimethylcyclohexanone has three methyl groups surrounding the carbonyl group. These bulky methyl groups create significant steric hindrance, making it much more difficult for the cyanide ion to approach and attack the carbonyl carbon. This steric hindrance greatly reduces the rate and equilibrium of cyanohydrin formation.
(ii) There are two -NH2 groups in semicarbazide, but only one is involved in the formation of semicarbazones:
- Resonance and Nucleophilicity: Semicarbazide (H2N-NH-CO-NH2) has two amino (-NH2) groups. However, only one of them reacts with aldehydes or ketones to form semicarbazones. This is because one of the -NH2 groups is directly attached to the carbonyl carbon (C=O) of the urea part of the molecule. The lone pair of electrons on this nitrogen is involved in resonance with the carbonyl group:
O
||
H2N-NH-C-NH2
This resonance decreases the electron density on that nitrogen atom, making it less nucleophilic. The other -NH2 group, which is not directly attached to the carbonyl, is not involved in this resonance. It retains its nucleophilicity and is the one that attacks the carbonyl group of the aldehyde or ketone during semicarbazone formation.
(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:
- Reversibility of the Reaction: The esterification reaction (reaction of a carboxylic acid with an alcohol to form an ester) is a reversible reaction. It reaches an equilibrium.
- Le Chatelier’s Principle: According to Le Chatelier’s principle, if a product is removed from a reaction mixture, the equilibrium will shift to favor the forward reaction (product formation). Both water and the ester are products of the esterification reaction. Removing either of them as they are formed will drive the equilibrium to the right, increasing the yield of the ester. Distillation is commonly used to remove the water or the ester (depending on which has the lower boiling point). This is why it is essential to remove the water or ester as soon as it is formed to maximize the conversion of reactants to products.
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 Tottens’ 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.
Ans :
1. Analyze the given information:
- Composition: 69.77% Carbon, 11.63% Hydrogen, and the rest Oxygen.
- Molecular Mass: 86
- Tollen’s Reagent: Does not reduce, indicating it’s not an aldehyde.
- Sodium Bisulfite Test: Forms an addition compound, suggesting a carbonyl group (C=O), likely a ketone.
- Iodoform Test: Positive, indicating the presence of a CH3-C(=O)- group (a methyl ketone).
- Vigorous Oxidation: Yields ethanoic acid (CH3COOH) and propanoic acid (CH3CH2COOH). This is the most crucial clue.
2. Deduce the structure:
The oxidation products tell us a lot. Since we get two different carboxylic acids, the original compound must be a ketone. The presence of ethanoic acid suggests a CH3-C(=O)- group, and propanoic acid suggests a CH3CH2- group attached to the carbonyl carbon.
Combining these pieces, the compound is likely 2-pentanone (CH3-CO-CH2-CH2-CH3).
3. Verify the deductions:
- Molecular Formula:
- 2-Pentanone (C5H10O) has a molar mass of (5*12) + (10*1) + 16 = 86.
- Iodoform Test:
- The CH3-CO- group will give a positive iodoform test.
- Sodium Bisulfite Test:
- Ketones react with sodium bisulfite to form addition compounds.
- Tollen’s Reagent:
- Ketones do not react with Tollen’s reagent.
- Oxidation:
- Oxidation of 2-pentanone will break the molecule at the carbonyl group, yielding ethanoic acid and propanoic acid.
Therefore, the structure of the compound is 2-pentanone.
20. Although phenoxide ion has more number of resonating structures than carboxylate ion, carboxylic acid is a stronger acid than on phenol. Why?
Ans :
The resonance structures of the phenoxide ion (V-VII), where the negative charge resides on the less electronegative carbon, contribute minimally to its overall stability. This is because oxygen is far more electronegative and prefers to hold the negative charge.
While the phenoxide ion does exhibit resonance, the negative charge is not as effectively delocalized as in the carboxylate ion. In the carboxylate ion (structures I and II), the negative charge is shared equally between two oxygen atoms, leading to significant stabilization. In contrast, while the phenoxide ion has resonance structures where the charge is on carbon (III and IV), the primary resonance structures place the negative charge on the oxygen atom. The delocalization of electrons within the benzene ring itself (as in III and IV) provides comparatively little stabilization to the phenoxide ion.
Because the carboxylate ion is much more resonance-stabilized than the phenoxide ion, carboxylic acids readily lose a proton (H⁺) to form the stable carboxylate anion. Phenols, on the other hand, are less inclined to lose a proton because the resulting phenoxide ion is not as stable. Consequently, carboxylic acids are considerably stronger acids than phenols.
