Aldehydes Ketones and Carboxylic Acids — Class 12 Chemistry NCERT Solutions (Free)
Free step-by-step NCERT solutions for Class 12 Chemistry chapter "Aldehydes Ketones and Carboxylic Acids" — 8 important questions with detailed answers for CBSE board exam preparation.
TL;DR: Free step-by-step NCERT solutions for Class 12 Chemistry chapter "Aldehydes Ketones and Carboxylic Acids" — 8 important questions with detailed answer…
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NCERT Solutions for Class 12 Chemistry — Aldehydes Ketones and Carboxylic Acids. Step-by-step answers to all 8 textbook questions from this chapter, written for CBSE board preparation and free to use.
Aldehydes Ketones and Carboxylic Acids — All 8 Questions Solved
Q1. What is the difference between an aldehyde and a ketone? Give structural formulas and one example of each.
Aldehyde: A compound containing a carbonyl group (C=O) with the carbon bonded to at least one hydrogen atom. The carbonyl carbon is at the end of a carbon chain.
General formula: R-CHO or RCHO
Example: CH3CHO (acetaldehyde or ethanal)
Structural formula: H is always bonded to the carbonyl carbon
Ketone: A compound containing a carbonyl group (C=O) where the carbonyl carbon is bonded to two alkyl or aryl groups. The carbonyl carbon is always in the middle of the carbon chain.
General formula: R-CO-R' (where R and R' are alkyl/aryl groups)
Example: CH3COCH3 (acetone or propanone)
Structural formula: No hydrogen is bonded to the carbonyl carbon
Key differences:
1. Position: Aldehydes have the C=O at the end (aldehyde group is -CHO); ketones have C=O in the middle
2. Oxidation level: Aldehydes are more easily oxidized to carboxylic acids; ketones are oxidized only under drastic conditions
3. Reactivity: Aldehydes are generally more reactive than ketones due to less steric hindrance
4. Tests: Aldehydes give positive tests with Tollens' reagent and Fehling's reagent; ketones do not
5. Reduction: Both can be reduced to alcohols, but aldehydes form primary alcohols and ketones form secondary alcohols
6. Nomenclature: Aldehydes use suffix -al; ketones use suffix -one
Q2. Explain the mechanism of nucleophilic addition to carbonyl groups. Use the reaction of acetaldehyde with HCN as an example.
Nucleophilic addition mechanism to carbonyl compounds:
Reaction: CH3CHO (acetaldehyde) + HCN (hydrogen cyanide) → CH3CH(OH)CN (cyanohydrin)
Mechanism (two steps):
Step 1: Nucleophilic attack by CN-
The CN- ion (from HCN dissociation or formation) acts as a nucleophile.
It attacks the electrophilic carbonyl carbon from above or below the C=O plane.
The π electrons of C=O move to oxygen, forming a tetrahedral intermediate with negative charge on oxygen.
Structure: CH3-CH(-CN)-O- (tetrahedral intermediate)
Step 2: Protonation of alkoxide ion
The negatively charged oxygen (alkoxide) is protonated by H+ from HCN or solvent.
Formation of C-OH bond
Final product: CH3CH(OH)CN (cyanohydrin, secondary alcohol)
Full equation: CH3CHO + HCN → CH3CH(OH)CN
Alternatively written: CH3CHO + HCN → CH3CHOHCN
Reactivity comparison (aldehyde vs ketone):
Aldehydes are more reactive because:
1. Less steric hindrance around the carbonyl carbon (only one alkyl group)
2. Stronger C=O dipole in some cases
3. Formed carbocation intermediate is more stable (less substituted carbon leads to easier attack)
Factors affecting reactivity:
1. Electron-withdrawing groups increase reactivity of C=O
2. Electron-donating groups decrease reactivity
3. Steric hindrance decreases reactivity
4. The nucleophile strength is important (CN- is a good nucleophile)
General mechanism applies to all nucleophilic additions:
- Addition of alcohols: acetals/hemiacetals formation
- Addition of ammonia derivatives: Schiff bases, oximes, hydrazones
- Addition of bisulfite: bisulfite adducts
Q3. Describe the Aldol condensation reaction. Write the equation for the aldol condensation of acetaldehyde.
Aldol condensation: Reaction between two molecules of an aldehyde (or ketone) in the presence of a dilute base, resulting in C-C bond formation. The product is a β-hydroxy aldehyde (aldol) which may further dehydrate to form an α,β-unsaturated aldehyde.
Alternative name: Aldol reaction (when the product is an aldol), Aldol condensation (when the product is dehydrated)
Reaction of acetaldehyde:
2 CH3CHO + dilute NaOH → CH3CH(OH)CH2CHO (aldol product, 3-hydroxybutanal)
Conditions: Dilute base (NaOH), room temperature or slight heating
Further dehydration (heating):
CH3CH(OH)CH2CHO + heat → CH3CH=CHCHO (crotonaldehyde or but-2-enal) + H2O
Product is an α,β-unsaturated aldehyde (conjugated system)
Complete equation showing both steps:
2 CH3CHO + dilute NaOH → CH3CH=CHCHO + H2O (overall)
This is often called aldol condensation
Mechanism (under basic conditions):
1. Deprotonation: Base removes α-hydrogen from one acetaldehyde molecule
CH3CHO + OH- → CH2=CHO- (enolate) + H2O
Forms an enolate ion (carbanion resonance with negative charge on O)
2. Nucleophilic addition: Enolate attacks the carbonyl carbon of another acetaldehyde
CH2=CHO- + CH3CHO → CH3CH(OH)CH2CHO-
Forms C-C bond
3. Protonation: The alkoxide is protonated
CH3CH(OH)CH2CHO- + H2O → CH3CH(OH)CH2CHO + OH-
4. Dehydration (E1cb mechanism, when heated): Loss of water to form double bond
CH3CH(OH)CH2CHO → CH3CH=CHCHO + H2O
Applications:
1. Formation of C-C bonds in organic synthesis
2. Production of synthetic fragrances and flavoring compounds
3. Starting material for further transformations
Limitation: Requires substrate with α-hydrogens (so aldehydes, ketones with α-H)
Formaldehyde cannot undergo aldol condensation (no α-H)
Q4. Explain the oxidation of aldehydes to carboxylic acids. Write the equation for oxidation of acetaldehyde with Tollens' reagent.
Oxidation of aldehydes to carboxylic acids:
Aldehydes are easily oxidized to carboxylic acids (one-step oxidation):
R-CHO + [O] → R-COOH
Use of Tollens' reagent:
Reaction: CH3CHO + 2 Ag(NH3)2+ + H2O → CH3COONH4 + 2 Ag↓ + 3 NH3 + H+
Or written more clearly:
CH3CHO + 2 [Ag(NH3)2]+ + 2 OH- → CH3COO- + 2 Ag↓ + 4 NH3 + H2O
Tollens' reagent:
- Silver ammonia complex: [Ag(NH3)2]+
- Prepared by adding ammonia to AgNO3 solution until the white precipitate of Ag2O dissolves
- The product is a clear, colorless solution
What happens (silver mirror test):
1. The aldehyde is oxidized to carboxylic acid (acetaldehyde → acetic acid)
2. The silver ion is reduced to metallic silver: Ag+ + e- → Ag
3. A shiny silver mirror deposits on the flask walls
4. This is called the silver mirror test or Tollens test
Mechanism:
1. The aldehyde adds water (hydration) in ammonia solution
2. The hydrated aldehyde is oxidized by Ag+ ions
3. Ag+ oxidizes the aldehyde C to carboxylic acid level
4. Ag+ is reduced to Ag metal
Other oxidizing agents:
1. Dilute permanganate (KMnO4 in dilute, neutral or slightly basic solution)
2 CH3CHO + 2 KMnO4 → 2 CH3COOK + 2 MnO2↓ + H2O (in neutral solution)
2. Chromic acid (H2CrO4 or K2Cr2O7 in H2SO4, used for ketones and alcohols)
CH3CHO + [O] → CH3COOH (produced as acidic salt)
3. Atmospheric oxygen (slow oxidation, photocatalytic)
2 CH3CHO + O2 → 2 CH3COOH (slow, can occur spontaneously in air)
Why aldehydes are more easily oxidized than ketones:
1. Aldehydes have a C-H bond on the carbonyl carbon (easily oxidized)
2. Ketones have two C-C bonds on the carbonyl carbon (require harsh conditions)
3. The α-hydrogen of aldehydes is more acidic and easily removed
4. The resulting carboxylic acid is more stable than further oxidation products of ketones
Comparative reactivity:
- Primary alcohols → aldehydes → carboxylic acids (easy oxidation)
- Secondary alcohols → ketones (with difficulty, need strong oxidizing agents, often no further oxidation)
- Tertiary alcohols → no oxidation (no H on carbon bearing OH)
Q5. What is a carboxylic acid? Write the IUPAC name of CH3CH2CH2COOH and describe its structure.
Carboxylic acid: An organic compound containing a carboxyl group (-COOH) bonded to a carbon atom.
Structure of -COOH group:
- Contains a carbonyl group (C=O) bonded to a hydroxyl group (-OH)
- The carbon is bonded to the oxygen with a C=O double bond
- The same carbon is bonded to -OH with a single bond
- General formula: R-COOH where R is an alkyl or aryl group
Example: CH3CH2CH2COOH
Structure:
O
||
CH3-CH2-CH2-C-OH
IUPAC name: Butanoic acid (or butyric acid - common name)
Naming steps:
1. Identify the longest carbon chain including the carboxyl carbon: 4 carbons
2. Use suffix -oic acid for the functional group
3. The carboxyl carbon is always C1 (position 1) in the numbering
4. Numbering: CH3(4) - CH2(3) - CH2(2) - COOH(1)
5. No substituents, so name is simply: butanoic acid
Characteristics of carboxylic acids:
1. Acidity: Weakly acidic due to resonance stabilization of conjugate base
R-COOH ⇌ R-COO- + H+
Ka values typically 10^-4 to 10^-5
2. Hydrogen bonding: Can form hydrogen bonds through both the OH (donor) and C=O (acceptor)
This leads to dimeric aggregation in solution
3. Boiling points: Higher than alcohols of similar molar mass due to hydrogen bonding and dimer formation
4. Solubility: Lower alcohols and carboxylic acids are miscible with water
Longer chain carboxylic acids are less soluble due to hydrophobic alkyl chain
5. Reactions: Form esters with alcohols, react with bases, can be reduced, can undergo decarboxylation
Examples of carboxylic acids:
- Formic acid (HCOOH): methanoic acid
- Acetic acid (CH3COOH): ethanoic acid
- Propionic acid (C2H5COOH): propanoic acid
- Butyric acid (C3H7COOH): butanoic acid
- Benzoic acid (C6H5COOH): benzenecarboxylic acid
Q6. Explain the esterification of carboxylic acids. Write the equation for the esterification of acetic acid with ethanol.
Esterification: Reaction between a carboxylic acid and an alcohol to form an ester and water. The reaction is reversible and typically acid-catalyzed.
Reaction equation:
CH3COOH + C2H5OH ⇌ CH3COOC2H5 + H2O
(acetic acid + ethanol → ethyl acetate + water)
Conditions:
1. Acid catalyst (usually concentrated H2SO4 or HCl)
2. Heat (reflux or warming)
3. Removal of water (using drying agents like P2O5 or molecular sieves to shift equilibrium forward)
Alternative equation showing the reversible nature:
R-COOH + R'-OH ⇌ R-COO-R' + H2O
(Acid + Alcohol ⇌ Ester + Water)
Mechanism (Fischer esterification):
Step 1: Protonation of carboxylic acid
CH3COOH + H+ → CH3COO-H2+ (protonated carboxylic acid)
The carbonyl oxygen is protonated, making the carbonyl carbon more electrophilic
Step 2: Nucleophilic addition by alcohol
CH3COO-H2+ + C2H5OH → CH3C(OC2H5)(OH)2+ (tetrahedral intermediate)
The alcohol oxygen attacks the carbonyl carbon
Step 3: Loss of water and formation of oxonium ion
CH3C(OC2H5)(OH)2+ → CH3CO(OC2H5)+ + H2O
Water is eliminated from the tetrahedral intermediate
Step 4: Deprotonation
CH3CO(OC2H5)+ → CH3COOC2H5 + H+ (regenerates catalyst)
The oxonium ion loses a proton to form the ester
Key points:
1. The reaction is reversible (equilibrium)
2. Acid catalyst is regenerated
3. Water must be removed to drive equilibrium to ester formation
4. The OH that leaves comes from the carboxylic acid (proven by O-18 labeling studies)
5. This is a nucleophilic acyl substitution reaction
Factors affecting yield:
1. Concentration of acid and alcohol
2. Temperature (higher temperature favors reaction but increases water evaporation)
3. Removal of water product
4. Nature of alkyl group in alcohol (steric hindrance affects rate)
5. Nature of acid (electron-withdrawing groups increase reactivity)
Reverse reaction (hydrolysis of esters):
Esters can be hydrolyzed back to acid and alcohol under acidic or basic conditions
Acidic hydrolysis: R-COO-R' + H2O ⇌ R-COOH + R'-OH (same mechanism, reversed)
Basic hydrolysis (saponification): R-COO-R' + NaOH → R-COONa + R'-OH (irreversible, not equilibrium)
Q7. Describe the tests for distinguishing between aldehydes and ketones. Which test is most reliable?
Tests to distinguish aldehydes from ketones:
Test 1: Tollens' test (silver mirror test) - MOST RELIABLE
Procedure:
- Add Tollens' reagent (ammoniacal silver nitrate solution)
- Warm gently
Observations:
- Aldehydes: Positive test - bright silver mirror forms on the flask walls
CH3CHO + 2 [Ag(NH3)2]+ + 2 OH- → CH3COO- + 2 Ag↓ + 3 NH3 + H2O
- Ketones: Negative test - no silver mirror, no change
Inter: This is a highly reliable and visually dramatic test
Test 2: Fehling's test
Procedure:
- Add Fehling's reagent (copper complex in basic solution)
- Warm
Observations:
- Aldehydes: Positive test - brick-red precipitate of Cu2O forms
CH3CHO + 2 Cu2+ (complex) + 5 OH- → CH3COO- + Cu2O↓ + 3 H2O
The blue solution turns brick-red
- Ketones: Negative test - no precipitate, solution remains blue
Similarity to Tollens': Both tests are based on oxidation of the aldehyde to carboxylic acid
Test 3: 2,4-DNP test (2,4-dinitrophenylhydrazone test)
Procedure:
- Add 2,4-dinitrophenylhydrazine (2,4-DNP) in methanol/sulfuric acid
Observations:
- Aldehydes: Bright yellow or orange-red precipitate forms
R-CHO + H2N-NH-C6H3(NO2)2 → R-CH=N-NH-C6H3(NO2)2 + H2O
- Ketones: Bright yellow or orange-red precipitate forms
R-CO-R' + H2N-NH-C6H3(NO2)2 → R-C(=N-NH-C6H3(NO2)2)-R' + H2O
Conclusion: Both aldehydes AND ketones give positive tests (not selective)
This test is used to confirm the presence of a carbonyl group, not to distinguish aldehydes from ketones
Test 4: Iodoform test
Procedure:
- Add I2 and NaOH
Observations:
- Aldehydes with CH3 group (acetaldehyde): Positive - yellow precipitate of CHI3
CH3CHO + 3 I2 + 4 NaOH → CHI3↓ + CH3COONa + 3 NaI + 3 H2O
- Ketones with CH3 group (acetone): Positive - yellow precipitate of CHI3
CH3COCH3 + 3 I2 + 4 NaOH → CHI3↓ + CH3COONa + 3 NaI + 3 H2O
- Aldehydes and ketones without CH3 group: Negative
Conclusion: Not selective between aldehydes and ketones (both can give positive)
Only works for compounds with a CH3CO- group
Test 5: Schiff's test
Procedure:
- Add Schiff's reagent (basic fuchsin solution, decolorized)
Observations:
- Aldehydes: Positive - magenta/pink color returns
- Ketones: Negative - no color change
MECHANISM:
Aldehydes react with the amino group of fuchsin to form a magenta colored imine
Ketones generally do not (reaction too slow at room temperature)
RELIABILITY: Good, though some aromatic ketones and bulky ketones may give weak positive tests
Summary table:
Test | Aldehyde | Ketone | Reliability
Tollens | Positive | Negative | BEST
Fehling | Positive | Negative | BEST
Schiff | Positive | Negative | Good
2,4-DNP | Positive | Positive | Confirms C=O, not selective
Iodoform | May be +ve | May be +ve | Only for CH3CO compounds
MOST RELIABLE: Tollens' test or Fehling's test (both give distinctive positive for aldehydes, negative for ketones)
Q8. Explain the decarboxylation of carboxylic acids. Give an example and describe the conditions required.
Decarboxylation of carboxylic acids: Removal of the carboxyl group (-COOH) as carbon dioxide (CO2), resulting in the formation of a hydrocarbon.
General reaction:
R-COOH → R-H + CO2↑
Example: Acetic acid to methane
CH3COOH + heat → CH4 + CO2↑
Conditions for decarboxylation:
1. Strong heating (above 400°C in some cases)
2. Often requires a catalyst
3. Presence of an alkali or mild bases (soda lime mixture)
4. Sometimes reduced pressure (vacuum)
Common decarboxylation methods:
Method 1: Heating with soda lime (MOST COMMON)
Conditions: Soda lime (mixture of NaOH and CaO) + heat (strong heating)
Reaction: R-COOH + NaOH → R-COONa + H2O
R-COONa + CaO → R-H + Na2CO3 + CaO
Product: Hydrocarbon (R-H)
Example: CH3COOH + soda lime → CH4 + Na2CO3 + H2O
Method 2: Kolbe reaction (for carboxylic acids with alkyl groups)
Conditions: Dry soda lime + heat
Reaction: 2 R-COOH → R-R + 2 CO2 + H2O (dimerization of carboxylate)
Actually: Aromatcyclic acids give coupling products
Method 3: Decarboxylation via silver salts or lead salts
Conditions: Silver carboxylate or lead oxide + heat
Reaction: R-COOAg → R-Ag (initially) → R-H + products
Useful for special substrates
Method 4: Thermal decomposition (less common)
Conditions: Very high temperature (400-500°C) with no catalyst
Reaction: R-COOH → R• (radical) + COOH• → products
Factors affecting ease of decarboxylation:
1. Nature of R group:
- Aryl groups (benzene ring): Easier decarboxylation
- Alkyl groups: Less easy
- Electron-donating groups on aromatic ring: Facilitate decarboxylation
- Electron-withdrawing groups: Inhibit decarboxylation
2. Position of carboxyl group:
- α-carboxylic acids: May decarboxylate under milder conditions
- β-dicarboxylic acids: Easier decarboxylation
3. Stability of resulting carbanion:
- If the resulting R- would be stable, decarboxylation is favored
- β-dicarbonyl compounds decarboxylate easily
Examples:
1. Benzoic acid → Benzene
C6H5COOH + soda lime → C6H6 + CO2 (easy, aryl group)
2. Malonic acid → Acetic acid
HOOC-CH2-COOH + heat → CH3COOH + CO2 (β-dicarbonyl acid, easier)
3. Oxaloacetic acid decarboxylation in metabolism:
Important in biological systems (enzymatic decarboxylation)
Biological significance:
- Many enzymatic reactions involve decarboxylation
- Important in metabolic pathways (citric acid cycle)
- Acetyl-CoA formation involves decarboxylation
- Amino acid decarboxylation produces important neurotransmitters
Industrial applications:
- Preparation of alkanes from carboxylic acids
- Synthesis of aromatic compounds
- Degradation pathways in organic chemistry
Aldehydes Ketones and Carboxylic Acids — Practice MCQs with Answers
Attempt these 20 multiple-choice questions after working through the solutions above. Each carries the correct option and the reasoning behind it, so a wrong answer tells you which idea to revisit.
MCQ 1. What is the IUPAC name for CH3CH2COCH3?
- A. Butan-2-one
- B. Propanone
- C. Butanal
- D. Methyl ethyl ketone
Answer: A. Butan-2-one — The longest carbon chain containing the ketone group has four carbons. The ketone group is located at the second carbon atom, hence Butan-2-one.
MCQ 2. Which of the following reagents is used for the Rosenmund reduction?
- A. Pd/BaSO4, S or Quinoline
- B. Zn-Hg/conc. HCl
- C. LiAlH4
- D. NaBH4
Answer: A. Pd/BaSO4, S or Quinoline — Rosenmund reduction involves the catalytic hydrogenation of acyl chlorides over palladium poisoned by barium sulfate, along with sulfur or quinoline, to yield aldehydes.
MCQ 3. Aldehydes and ketones undergo which type of reaction with HCN?
- A. Electrophilic substitution
- B. Nucleophilic addition
- C. Electrophilic addition
- D. Nucleophilic substitution
Answer: B. Nucleophilic addition — The carbon atom of the carbonyl group is electrophilic due to the electron-withdrawing oxygen, making it susceptible to nucleophilic attack, such as by the cyanide ion (CN-).
MCQ 4. Which of the following compounds will not give a positive Aldol condensation test?
- A. Ethanal
- B. Propanal
- C. Benzaldehyde
- D. Butanal
Answer: C. Benzaldehyde — Aldol condensation requires the presence of alpha-hydrogens. Benzaldehyde does not have any alpha-hydrogens directly attached to the carbon adjacent to the carbonyl group, so it cannot undergo aldol condensation.
MCQ 5. Which reagent is used in Tollens' test to distinguish between aldehydes and ketones?
- A. Ammoniacal silver nitrate solution
- B. Fehling's solution
- C. 2,4-Dinitrophenylhydrazine
- D. Benedict's solution
Answer: A. Ammoniacal silver nitrate solution — Tollens' reagent, ammoniacal silver nitrate, oxidizes aldehydes to carboxylic acids while being reduced to metallic silver, forming a silver mirror. Ketones generally do not react with Tollens' reagent.
MCQ 6. Which of the following is the strongest acid?
- A. Ethanoic acid
- B. Propanoic acid
- C. Formic acid
- D. Benzoic acid
Answer: C. Formic acid — Formic acid is generally the strongest among simple aliphatic carboxylic acids due to the absence of electron-donating alkyl groups that destabilize the conjugate base. Benzoic acid is also strong but formic acid is typically slightly stronger.
MCQ 7. What is the product formed when propanone is treated with Zn-Hg/conc. HCl?
- A. Propan-1-ol
- B. Propane
- C. Propan-2-ol
- D. Propene
Answer: B. Propane — Clemmensen reduction reduces carbonyl compounds (aldehydes and ketones) to alkanes using zinc amalgam and concentrated hydrochloric acid. Propanone (CH3COCH3) is thus reduced to propane (CH3CH2CH3).
MCQ 8. Which of the following compounds will undergo Cannizzaro reaction?
- A. Acetone
- B. Acetaldehyde
- C. Benzaldehyde
- D. Propanal
Answer: C. Benzaldehyde — Cannizzaro reaction is given by aldehydes which do not have an alpha-hydrogen atom. Benzaldehyde lacks an alpha-hydrogen, unlike acetaldehyde and propanal.
MCQ 9. The reaction of benzene with CO and HCl in the presence of anhydrous AlCl3/CuCl yields:
- A. Benzophenone
- B. Benzoic acid
- C. Benzaldehyde
- D. Phenol
Answer: C. Benzaldehyde — This is the Gattermann-Koch reaction, which is used for the formylation of aromatic hydrocarbons (like benzene) to yield benzaldehyde.
MCQ 10. Identify the product 'X' in the following reaction: RCN + SnCl2/HCl -> 'X' -> RCHO
- A. Imine
- B. Amide
- C. Nitrile
- D. Carboxylic acid
Answer: A. Imine — In the Stephen reaction, nitriles are reduced to imines using SnCl2 and HCl, which upon hydrolysis yield aldehydes.
MCQ 11. Arrange the following compounds in increasing order of their boiling points: Butanal, Butan-1-ol, Butane, Ethoxyethane.
- A. Butane < Ethoxyethane < Butanal < Butan-1-ol
- B. Butane < Butanal < Ethoxyethane < Butan-1-ol
- C. Ethoxyethane < Butane < Butanal < Butan-1-ol
- D. Butan-1-ol < Butanal < Ethoxyethane < Butane
Answer: A. Butane < Ethoxyethane < Butanal < Butan-1-ol — Boiling points depend on intermolecular forces. Butane (alkane) has weakest London dispersion forces, Ethoxyethane (ether) has dipole-dipole interactions, Butanal (aldehyde) has strong dipole-dipole interactions, and Butan-1-ol (alcohol) has the strongest intermolecular hydrogen bonding.
MCQ 12. The carbonyl carbon in aldehydes and ketones is an electrophilic center. However, the carbonyl group in aromatic aldehydes and ketones is:
- A. Ortho-para directing
- B. Meta-directing
- C. Both ortho-para and meta directing
- D. Non-directing
Answer: B. Meta-directing — The carbonyl group is an electron-withdrawing group, which deactivates the benzene ring towards electrophilic substitution and directs incoming electrophiles to the meta-position.
MCQ 13. Which of the following compounds will not give a positive iodoform test?
- A. Acetone
- B. Acetaldehyde
- C. Ethanol
- D. Propanal
Answer: D. Propanal — The iodoform test is given by compounds containing a methyl ketone group (-COCH3) or a methyl secondary alcohol group (-CH(OH)CH3). Acetone, acetaldehyde, and ethanol all have the structural requirement for the iodoform test, while propanal does not.
MCQ 14. Aldehydes are easily oxidized to carboxylic acids. Which of the following reagents can be used for this oxidation?
- A. K2Cr2O7/H+
- B. KMnO4/H+
- C. Tollens' reagent
- D. All of the above
Answer: D. All of the above — Aldehydes are readily oxidized to carboxylic acids by strong oxidizing agents like K2Cr2O7/H+ and KMnO4/H+, and even by mild oxidizing agents like Tollens' reagent or Fehling's reagent.
MCQ 15. Reaction of a Grignard reagent (RMgX) with carbon dioxide (CO2) followed by hydrolysis yields:
- A. Aldehyde
- B. Ketone
- C. Carboxylic acid
- D. Alcohol
Answer: C. Carboxylic acid — Grignard reagents react with carbon dioxide (dry ice) to form an addition product (an intermediate magnesium carboxylate), which upon hydrolysis with acid gives a carboxylic acid.
MCQ 16. Which of the following reagents would convert a carboxylic acid into an acyl chloride?
- A. PCl3
- B. PCl5
- C. SOCl2
- D. All of the above
Answer: D. All of the above — Carboxylic acids react with phosphorus trichloride (PCl3), phosphorus pentachloride (PCl5), and thionyl chloride (SOCl2) to form acyl chlorides. Thionyl chloride is often preferred due to gaseous by-products.
MCQ 17. What is the main product when sodium ethanoate is heated with soda lime?
- A. Ethane
- B. Methane
- C. Ethene
- D. Ethanol
Answer: B. Methane — When the sodium salt of a carboxylic acid is heated with soda lime (NaOH + CaO), it undergoes decarboxylation, removing CO2 and producing an alkane with one less carbon atom. Sodium ethanoate yields methane.
MCQ 18. When ethanal and propanal react in the presence of dilute NaOH, how many distinct cross-aldol products (excluding self-condensation) are possible?
- A. 2
- B. 3
- C. 4
- D. 6
Answer: A. 2 — In a cross aldol condensation between two different aldehydes (ethanal and propanal), each can form an enolate and each can act as an electrophile, leading to two distinct cross-aldol products.
MCQ 19. Which of the following compounds is typically used as an acylating agent in Friedel-Crafts acylation?
- A. RCOOH
- B. RCOCl
- C. RCHO
- D. RCH2OH
Answer: B. RCOCl — Acyl chlorides (RCOCl) or acid anhydrides are commonly used as acylating agents in Friedel-Crafts acylation in the presence of a Lewis acid like anhydrous AlCl3.
MCQ 20. Arrange the following compounds in decreasing order of their reactivity towards nucleophilic addition reactions: Ethanal, Propanone, Methanal, Benzaldehyde.
- A. Methanal > Ethanal > Benzaldehyde > Propanone
- B. Ethanal > Methanal > Propanone > Benzaldehyde
- C. Propanone > Ethanal > Methanal > Benzaldehyde
- D. Benzaldehyde > Methanal > Ethanal > Propanone
Answer: A. Methanal > Ethanal > Benzaldehyde > Propanone — Reactivity towards nucleophilic addition reactions is highest for aldehydes and lowest for ketones, due to steric hindrance and electron-donating effects of alkyl groups. Aromatic aldehydes are generally less reactive than aliphatic aldehydes but more reactive than ketones due to resonance effects and steric hindrance. Therefore, the order is Methanal > Ethanal > Benzaldehyde > Propanone.
How to Use These Solutions
Attempt each question yourself first and only then compare with the worked answer. Marks in the CBSE board exam are awarded for the METHOD as much as the final result, so reproduce the steps rather than memorising the last line. Where a solution states a law, a formula or a definition, learn that wording — examiners look for it.
Related practice for this chapter: chapter MCQs, previous-year questions, revision notes and sample papers.