CUET UG Chemistry Booster Test - 3 Properties & Reactions (Aldehydes and Ketones) -

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CUET UG Chemistry Booster Test - 3 Properties & Reactions (Aldehydes and Ketones)

⏱️ 30 Minutes

❓ 20 Questions

📌 Answers are locked once submitted — results and explanations appear at the end.

CUET UG Chemistry Booster Test - 3 Properties & Reactions (Aldehydes and Ketones)

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QUESTION 1 OF 20

Evaluate the statements comparing the physical state and association of carbonyls:
1. The dipole-dipole interactions in aldehydes are stronger than London forces in alkanes.
2. Carbonyls exist mostly as gases because they lack hydrogen bonding.
3. Methanal is a gas, but ethanal is a volatile liquid at room temperature.
4. Aldehydes generally exist as crystalline solids regardless of mass.

Answer Review

1 Evaluate the statements comparing the physical state and association of carbonyls:
1. The dipole-dipole interactions in aldehydes are stronger than London forces in alkanes.
2. Carbonyls exist mostly as gases because they lack hydrogen bonding.
3. Methanal is a gas, but ethanal is a volatile liquid at room temperature.
4. Aldehydes generally exist as crystalline solids regardless of mass.

A 1, 2

B 1, 3

C 3, 4

D 2, 4

💡 Short Explanation

�� Aldehydes show dipole-dipole interactions. �� Methanal is a gas. �� Ethanal is a volatile liquid.

✅ Correct Answer Explanation

Aldehydes and ketones possess polar carbonyl groups, so they show dipole-dipole interactions stronger than the London forces in comparable alkanes. Methanal is gaseous, while ethanal is a volatile liquid at room temperature.

❓ Why Other Options Are Incorrect

�� Statement 2 is incorrect because all carbonyl compounds are not mostly gases.
�� Statement 4 is incorrect because physical state depends on molecular mass.

🎯 Strategy

Used

Statement Analysis

🧠 Memory Trick

Methanal gas, ethanal volatile liquid.

📚 Aldehydes, Ketones and Carboxylic Acids – Physical State and Boiling Points

2 The standard unit of molecular mass used to compare the boiling points of organic compounds like propanal and n-butane is implicitly:

A g L⁻¹

B g mol⁻¹

C kg m⁻³

D dimensionless ratio

💡 Short Explanation

�� Molecular mass is expressed per mole. �� Standard unit is g mol⁻¹. �� Used for comparing compounds.

✅ Correct Answer Explanation

Molecular mass in organic chemistry is commonly expressed in grams per mole, written as g mol⁻¹. Therefore, examples like 58 and 60 refer to molecular masses in g mol⁻¹.

❓ Why Other Options Are Incorrect

�� Option A: Unit of concentration/density-like quantity.
�� Option C: Unit of density.
�� Option D: Molecular mass has units.

🎯 Strategy

Used

Unit-Based MCQ

🧠 Memory Trick

Molecular mass = g mol⁻¹.

📚 Aldehydes, Ketones and Carboxylic Acids – Physical Properties

3 Why is the molecular association in aldehydes and ketones considered "weak" when comparing their boiling points to those of alcohols?

A Aldehydes and ketones decompose at lower temperatures.

B The dipole-dipole interactions in carbonyls require less energy to overcome than the intermolecular hydrogen bonds in alcohols.

C The carbonyl group is non-polar.

D Alcohols have a significantly higher molar mass compared to structurally similar carbonyls.

💡 Short Explanation

�� Carbonyls have dipole-dipole forces. �� Alcohols have hydrogen bonding. �� Hydrogen bonding is stronger.

✅ Correct Answer Explanation

Aldehydes and ketones show dipole-dipole interactions due to the polar carbonyl group. However, they do not form strong intermolecular hydrogen bonds with themselves like alcohols. Therefore, their molecular association is weaker than that of alcohols.

❓ Why Other Options Are Incorrect

�� Option A: Decomposition is not the reason.
�� Option C: Carbonyl group is polar.
�� Option D: Comparison is made for similar molecular masses.

🎯 Strategy

Used

Concept MCQ

🧠 Memory Trick

Alcohol H-bonding > Carbonyl dipole forces.

📚 Aldehydes, Ketones and Carboxylic Acids – Boiling Points

4 Arrange the following compounds of comparable molecular mass in strictly decreasing order of their boiling points:
(A) Propan-1-ol
(B) Acetone
(C) Methoxyethane
(D) n-Butane

A (A), (B), (C), (D)

B (D), (C), (B), (A)

C (A), (C), (B), (D)

D (B), (A), (C), (D)

💡 Short Explanation

�� Alcohol has highest boiling point. �� Ketone has strong dipole-dipole interactions. �� Ether has weaker polarity. �� Alkane has weakest forces.

✅ Correct Answer Explanation

Boiling point depends on intermolecular forces. Propan-1-ol forms intermolecular hydrogen bonds, acetone has dipole-dipole interactions, methoxyethane has weaker dipole interactions, and n-butane has only London dispersion forces. Decreasing order: Propan-1-ol > Acetone > Methoxyethane > n-Butane

❓ Why Other Options Are Incorrect

�� Options B, C and D do not follow the correct intermolecular force order.

🎯 Strategy

Used

Ordering

🧠 Memory Trick

Alcohol > Ketone > Ether > Alkane.

📚 Aldehydes, Ketones and Carboxylic Acids – Boiling Point Comparison

5 Regarding the solubility behaviour of carbonyl compounds, which statements are strictly accurate?
1. The solubility of aldehydes and ketones in water decreases rapidly on increasing the length of the alkyl chain.
2. Lower members form hydrogen bonds with water.
3. Higher aldehydes are completely insoluble in organic solvents like benzene.
4. The hydrophilic character overrides the hydrophobic chain in higher members.

A 1, 2

B 3, 4

C 1, 3

D 2, 4

💡 Short Explanation

�� Lower members form H-bonds with water. �� Longer alkyl chain reduces water solubility. �� Higher members dissolve in organic solvents.

✅ Correct Answer Explanation

Lower aldehydes and ketones dissolve in water because their carbonyl oxygen forms hydrogen bonds with water. As the alkyl chain length increases, the hydrophobic part becomes larger, so water solubility decreases rapidly.

❓ Why Other Options Are Incorrect

�� Statement 3 is incorrect because aldehydes and ketones are generally soluble in organic solvents.
�� Statement 4 is incorrect because hydrophobic character dominates in higher members.

🎯 Strategy

Used

Statement Analysis

🧠 Memory Trick

Longer chain = lower water solubility.

📚 Aldehydes, Ketones and Carboxylic Acids – Solubility

6 All aldehydes and ketones are fairly soluble in organic solvents like benzene, ether, and chloroform primarily because:

A Organic solvents readily form hydrogen bonds with them.

B The hydrophobic hydrocarbon chain interacts well with these non-polar/weakly polar solvents.

C They undergo ionization in organic solvents.

D The nucleophilic addition reaction occurs with the solvent.

💡 Short Explanation

�� Organic solvents dissolve organic compounds well. �� Hydrocarbon part interacts with organic solvents. �� No ionization is required.

✅ Correct Answer Explanation

Aldehydes and ketones contain hydrocarbon parts that interact favourably with organic solvents such as benzene, ether and chloroform. Therefore, they are fairly soluble in such solvents.

❓ Why Other Options Are Incorrect

�� Option A: Benzene and chloroform are not strong hydrogen-bonding solvents.
�� Option C: Aldehydes and ketones do not generally ionize in organic solvents.
�� Option D: Solubility does not require nucleophilic addition.

🎯 Strategy

Used

Concept MCQ

🧠 Memory Trick

Organic compound dissolves well in organic solvent.

📚 Aldehydes, Ketones and Carboxylic Acids – Organic Solubility

7 Identify the specific aldehyde whose lower members are characterized by an extreme, sharp pungent nature but whose higher derivatives (like cinnamaldehyde) are highly fragrant:

A Acetone

B Methanal (and lower aliphatic aldehydes)

C Benzophenone

D Ethylene glycol

💡 Short Explanation

�� Lower aldehydes have sharp pungent odours. �� Higher aldehydes often possess pleasant fragrances. �� Cinnamaldehyde is a fragrant aldehyde.

✅ Correct Answer Explanation

Lower aliphatic aldehydes such as methanal and ethanal possess strong, pungent odours. As the molecular size increases, the odour becomes less pungent and more pleasant. Examples include naturally occurring aldehydes such as cinnamaldehyde and vanillin, which are widely used in perfumes and flavouring agents.

❓ Why Other Options Are Incorrect

�� Option A: Acetone is a ketone, not an aldehyde.
�� Option C: Benzophenone is a ketone.
�� Option D: Ethylene glycol is an alcohol.

🎯 Strategy

Used

Naming / Concept MCQ

🧠 Memory Trick

Small aldehydes = pungent smell; large aldehydes = pleasant smell.

📚 Aldehydes, Ketones and Carboxylic Acids – Physical Properties (Odour)

8 Match List-I (Naturally occurring source) with List-II (Fragrant Aldehyde):

List-I	List-II
1. Vanilla beans	a. Cinnamaldehyde
2. Meadow sweet	b. Vanillin
3. Cinnamon	c. Salicylaldehyde

A 1-b, 2-c, 3-a

B 1-a, 2-b, 3-c

C 1-c, 2-a, 3-b

D 1-b, 2-a, 3-c

💡 Short Explanation

�� Vanillin is obtained from vanilla beans. �� Salicylaldehyde is obtained from meadow sweet. �� Cinnamaldehyde is obtained from cinnamon.

✅ Correct Answer Explanation

Naturally occurring aldehydes contribute characteristic fragrances: Source — Fragrant Compound Vanilla beans — Vanillin Meadow sweet — Salicylaldehyde Cinnamon — Cinnamaldehyde Thus, the correct matching is: 1-b, 2-c, 3-a

❓ Why Other Options Are Incorrect

�� Options B, C and D incorrectly match one or more natural sources with their aldehydes.

🎯 Strategy

Used

Match the Following

🧠 Memory Trick

Cinnamon → Cinnamaldehyde

📚 Aldehydes, Ketones and Carboxylic Acids – Importance and Occurrence

9 In the final step of the basic nucleophilic addition mechanism, the electrically neutral product is achieved when the alkoxide intermediate:

A Expels the nucleophile

B Captures a proton from the reaction medium

C Forms a carbon-carbon double bond

D Undergoes decarboxylation

💡 Short Explanation

�� Nucleophile first attacks carbonyl carbon. �� Alkoxide ion intermediate is formed. �� Protonation gives the neutral product.

✅ Correct Answer Explanation

During nucleophilic addition, the nucleophile attacks the electrophilic carbonyl carbon to form a tetrahedral alkoxide intermediate. This intermediate then accepts a proton (H⁺) from the reaction medium, producing the final electrically neutral addition product.

❓ Why Other Options Are Incorrect

�� Option A: Nucleophile remains attached in the product.
�� Option C: No double bond formation occurs in the final step.
�� Option D: Decarboxylation is unrelated to this mechanism.

🎯 Strategy

Used

Concept MCQ

🧠 Memory Trick

Attack → Alkoxide → Protonation → Product.

📚 Aldehydes, Ketones and Carboxylic Acids – Nucleophilic Addition Reactions

10 Identify the reaction classification that describes the overall nucleophilic addition-elimination process observed when ammonia derivatives react with aldehydes:

A Acid-base neutralization

B Nucleophilic addition followed by rapid dehydration

C Free radical polymerization

D Electrophilic substitution

💡 Short Explanation

�� Ammonia derivative first adds to the carbonyl group. �� A tetrahedral intermediate is formed. �� Water is eliminated to produce a C=N derivative.

✅ Correct Answer Explanation

Ammonia derivatives such as hydroxylamine, hydrazine and semicarbazide react with aldehydes through nucleophilic addition to the carbonyl carbon. The intermediate subsequently loses water (dehydration), producing compounds containing the >C=N– group such as oximes, hydrazones and semicarbazones.

❓ Why Other Options Are Incorrect

�� Option A: Not a simple acid-base reaction.
�� Option C: No polymerization occurs.
�� Option D: Carbonyl compounds undergo nucleophilic reactions, not electrophilic substitution.

🎯 Strategy

Used

Reaction Type Identification

🧠 Memory Trick

Addition first, dehydration next → C=N product.

📚 Aldehydes, Ketones and Carboxylic Acids – Reactions with Ammonia Derivatives

11 Arrange the following compounds in decreasing order of electrophilic character of their carbonyl carbon:
(A) Ethanal
(B) Propanal
(C) Propanone
(D) Butanone

A (A), (B), (C), (D)

B (B), (A), (D), (C)

C (D), (C), (B), (A)

D (C), (D), (A), (B)

💡 Short Explanation

�� Aldehydes are more reactive than ketones. �� Smaller alkyl groups increase electrophilicity. �� More alkyl groups reduce carbonyl reactivity.

✅ Correct Answer Explanation

In nucleophilic addition, reactivity depends on electrophilicity and steric hindrance. Aldehydes are more reactive than ketones because they have only one alkyl group and one hydrogen attached to the carbonyl carbon. Ketones have two alkyl groups, which reduce electrophilicity by +I effect and increase steric hindrance. Order: Ethanal > Propanal > Propanone > Butanone

❓ Why Other Options Are Incorrect

�� Options B, C and D do not follow the correct aldehyde > ketone and smaller > larger trend.

🎯 Strategy

Used

Ordering

🧠 Memory Trick

Aldehyde > Ketone; smaller group > larger group.

📚 Aldehydes, Ketones and Carboxylic Acids – Reactivity in Nucleophilic Addition

12 Which statements accurately explain the reactivity trends in carbonyls?
1. Two alkyl groups in ketones reduce the electrophilicity of the carbonyl carbon more effectively than one group in aldehydes.
2. The presence of two relatively large substituents in ketones creates steric hindrance.
3. Benzaldehyde is less reactive than propanal due to resonance stabilization of the carbonyl polarity.
4. Aromatic ketones are generally more reactive than aliphatic aldehydes.

A 1, 2, 3

B 1, 2

C 3, 4

D 2, 3, 4

💡 Short Explanation

�� Ketones are less electrophilic than aldehydes. �� Ketones are more sterically hindered. �� Benzaldehyde is resonance-stabilized.

✅ Correct Answer Explanation

Ketones contain two alkyl groups, which donate electron density and reduce the electrophilic character of the carbonyl carbon. They also create steric hindrance for nucleophilic attack. Benzaldehyde is less reactive than propanal because its carbonyl group is conjugated with the benzene ring, reducing carbonyl polarity.

❓ Why Other Options Are Incorrect

�� Statement 4 is incorrect because aromatic ketones are generally less reactive than aliphatic aldehydes.

🎯 Strategy

Used

Statement Analysis

🧠 Memory Trick

More alkyl/resonance = less carbonyl reactivity.

📚 Aldehydes, Ketones and Carboxylic Acids – Carbonyl Reactivity

13 Based on the passage, why is a base catalyst essential for the efficient reaction of HCN with carbonyl compounds?

A Base neutralizes the intermediate alkoxide.

B Base generates the stronger nucleophile, cyanide ion (CN⁻).

C Pure HCN attacks too rapidly, causing side reactions.

D Base acts as an electrophile to activate the carbonyl oxygen.

💡 Short Explanation

�� Pure HCN reacts slowly. �� Base produces CN⁻ ion. �� CN⁻ is a stronger nucleophile.

✅ Correct Answer Explanation

The passage states that the reaction of aldehydes and ketones with pure HCN is slow. In the presence of base, cyanide ion (CN⁻) is generated. Since CN⁻ is a stronger nucleophile, it readily attacks the carbonyl carbon to form cyanohydrin.

❓ Why Other Options Are Incorrect

�� Option A: Base does not mainly neutralize alkoxide here.
�� Option C: Pure HCN reacts slowly, not too rapidly.
�� Option D: Base does not act as an electrophile.

🎯 Strategy

Used

Passage-Based MCQ

🧠 Memory Trick

Base makes CN⁻; CN⁻ attacks C=O.

📚 Aldehydes, Ketones and Carboxylic Acids – Cyanohydrin Formation

14 Based on the passage, the equilibrium of sodium hydrogensulphite addition lies to the left for most ketones primarily because of:

A Strong electrophilicity of ketones

B Electronic resonance in ketones

C Steric reasons hindering the addition

D The insolubility of the product

💡 Short Explanation

�� Ketones contain two alkyl groups. �� These groups hinder nucleophilic addition. �� Hence equilibrium lies to the left.

✅ Correct Answer Explanation

The passage states that sodium hydrogensulphite addition lies largely to the left for most ketones due to steric reasons. The two alkyl groups around the carbonyl carbon hinder addition product formation.

❓ Why Other Options Are Incorrect

�� Option A: Ketones are generally less electrophilic.
�� Option B: Passage specifically mentions steric reasons.
�� Option D: Hydrogensulphite addition compound is water soluble.

🎯 Strategy

Used

Passage-Based MCQ

🧠 Memory Trick

Ketone crowding pushes equilibrium left.

📚 Aldehydes, Ketones and Carboxylic Acids – Sodium Hydrogensulphite Addition

15 During the formation of an acetal, what is the specific role of the dry hydrogen chloride gas?

A It protonates the oxygen of the carbonyl compound to increase the electrophilicity of the carbonyl carbon.

B It acts as a solvent to dissolve the alcohol.

C It reduces the carbonyl group to an alkane.

D It deprotonates the alcohol to form an alkoxide ion.

💡 Short Explanation

�� Dry HCl acts as acid catalyst. �� It protonates carbonyl oxygen. �� Carbonyl carbon becomes more electrophilic.

✅ Correct Answer Explanation

In acetal formation, dry HCl protonates the carbonyl oxygen. This increases the positive character of the carbonyl carbon, making it more susceptible to attack by alcohol molecules.

❓ Why Other Options Are Incorrect

�� Option B: Dry HCl is not just a solvent.
�� Option C: It does not reduce carbonyl group.
�� Option D: It protonates, not deprotonates.

🎯 Strategy

Used

Concept MCQ

🧠 Memory Trick

Dry HCl activates C=O by protonating oxygen.

📚 Aldehydes, Ketones and Carboxylic Acids – Acetal Formation

16 When an aldehyde's hemiacetal further reacts with a second molecule of alcohol, the resulting gem-dialkoxy compound is specifically named an:

A Imine

B Acetal

C Ethylene glycol ketal

D Oxime

💡 Short Explanation

�� Hemiacetal reacts with another alcohol molecule. �� Water is eliminated. �� A gem-dialkoxy compound is formed.

✅ Correct Answer Explanation

When an aldehyde reacts with one molecule of alcohol, a hemiacetal is formed. On further reaction with a second molecule of alcohol in acidic medium, the hemiacetal converts into an acetal. Acetals contain two alkoxy groups attached to the same carbon atom.

❓ Why Other Options Are Incorrect

�� Option A: Imine is formed with ammonia or amine derivatives.
�� Option C: Ketal is generally formed from ketones.
�� Option D: Oxime is formed with hydroxylamine.

🎯 Strategy

Used

Naming

🧠 Memory Trick

One alcohol = Hemiacetal; Two alcohols = Acetal.

📚 Aldehydes, Ketones and Carboxylic Acids – Acetal Formation

17 Match List-I (Reagent H₂N-Z) with List-II (Name of the corresponding carbonyl derivative):

List-I	List-II
1. Amine (Z = R)	a. Hydrazone
2. Hydrazine (Z = NH₂)	b. Semicarbazone
3. Semicarbazide (Z = NHCONH₂)	c. Substituted imine (Schiff's base)
4. Hydroxylamine (Z = OH)	d. Oxime

A 1-c, 2-a, 3-b, 4-d

B 1-a, 2-b, 3-c, 4-d

C 1-c, 2-d, 3-b, 4-a

D 1-b, 2-c, 3-a, 4-d

💡 Short Explanation

�� Amine gives Schiff's base. �� Hydrazine gives hydrazone. �� Semicarbazide gives semicarbazone. �� Hydroxylamine gives oxime.

✅ Correct Answer Explanation

Reagent — Carbonyl Derivative Amine — Substituted imine / Schiff's base Hydrazine — Hydrazone Semicarbazide — Semicarbazone Hydroxylamine — Oxime Therefore, the correct matching is: 1-c, 2-a, 3-b, 4-d

❓ Why Other Options Are Incorrect

�� Options B, C and D contain incorrect reagent-product pairings.

🎯 Strategy

Used

Match the Following

🧠 Memory Trick

Hydroxylamine = Oxime; Hydrazine = Hydrazone.

📚 Aldehydes, Ketones and Carboxylic Acids – Ammonia Derivatives

18 Identify the reaction mechanism type responsible for driving the equilibrium forward when an ammonia derivative adds to a carbonyl group:

A Catalytic hydrogenation

B Nucleophilic addition followed by rapid dehydration (elimination)

C Electrophilic aromatic substitution

D Vigorous oxidation

💡 Short Explanation

�� Ammonia derivative first attacks carbonyl carbon. �� Addition intermediate is formed. �� Dehydration gives the final C=N product.

✅ Correct Answer Explanation

Ammonia derivatives react with aldehydes and ketones by nucleophilic addition to the carbonyl group. The initially formed intermediate rapidly loses water, forming a compound with a >C=N-Z linkage. This dehydration step drives the equilibrium forward.

❓ Why Other Options Are Incorrect

�� Option A: Hydrogenation is a reduction reaction.
�� Option C: Not an aromatic substitution reaction.
�� Option D: Oxidation is not involved.

🎯 Strategy

Used

Reaction Type Identification

🧠 Memory Trick

Carbonyl + H₂N-Z = Addition, then dehydration.

📚 Aldehydes, Ketones and Carboxylic Acids – Reactions with Ammonia Derivatives

19 Match List-I (Reduction Reagent) with List-II (Functional Group formed):

List-I	List-II
1. Lithium aluminium hydride (LiAlH₄)	a. Alcohol
2. Hydrazine followed by KOH in high boiling solvent	b. CH₂ group (Hydrocarbon)
3. Sodium borohydride (NaBH₄)	c. Carboxylic acid
4. HCN	d. Cyanohydrin

A 1-b, 2-a, 3-c, 4-d

B 1-a, 2-b, 3-c, 4-d

C 1-c, 2-d, 3-a, 4-b

D 1-a, 2-c, 3-d, 4-b

💡 Short Explanation

�� LiAlH₄ reduces carbonyls to alcohols. �� Wolff-Kishner reduction gives hydrocarbons. �� HCN gives cyanohydrin.

✅ Correct Answer Explanation

Lithium aluminium hydride reduces aldehydes and ketones to alcohols. Hydrazine followed by KOH in a high boiling solvent represents Wolff-Kishner reduction, which converts the carbonyl group into a CH₂ group. HCN adds to carbonyl compounds to form cyanohydrins.

❓ Why Other Options Are Incorrect

�� Options A, C and D contain incorrect matching.

🎯 Strategy

Used

Match the Following

🧠 Memory Trick

LiAlH₄ = Alcohol; Wolff-Kishner = CH₂.

📚 Aldehydes, Ketones and Carboxylic Acids – Reduction Reactions

20 Identify the reaction type associated with the conversion of a ketone into a mixture of carboxylic acids having a lesser number of carbon atoms than the parent ketone:

A Mild reduction

B Vigorous oxidation involving carbon-carbon bond cleavage

C Base-catalysed aldol condensation

D Nucleophilic addition

💡 Short Explanation

�� Ketones resist mild oxidation. �� Vigorous oxidation cleaves C-C bonds. �� Smaller carboxylic acids are formed.

✅ Correct Answer Explanation

Ketones are not easily oxidised under mild conditions. However, under vigorous oxidation, carbon-carbon bonds adjacent to the carbonyl group may break, producing a mixture of carboxylic acids with fewer carbon atoms than the original ketone.

❓ Why Other Options Are Incorrect

�� Option A: Mild reduction forms alcohols, not acids.
�� Option C: Aldol condensation forms larger β-hydroxy carbonyl compounds or unsaturated products.
�� Option D: Nucleophilic addition does not cleave the carbon chain into acids.

🎯 Strategy

Used

Reaction Type Identification

🧠 Memory Trick

Ketone + strong oxidation = C-C cleavage + acids.

📚 Aldehydes, Ketones and Carboxylic Acids – Oxidation of Ketones

CUET UG Chemistry Booster Test – 3 Properties & Reactions (Aldehydes and Ketones)

CUET UG Chemistry Booster Test - 3 Properties & Reactions (Aldehydes and Ketones)

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