CUET UG Chemistry Booster Test -2 Preparation and Physical Properties -

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CUET UG Chemistry Booster Test -2 Preparation and Physical Properties

⏱️ 30 Minutes

❓ 20 Questions

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

CUET UG Chemistry Booster Test -2 Preparation and Physical Properties

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QUESTION 1 OF 20

Clemmensen reduction properties:
Statements:
1. Reduces the carbonyl group to a CH₂ group
2. Uses zinc-amalgam
3. Uses concentrated hydrochloric acid
4. Operates under basic conditions

Answer Review

1 Clemmensen reduction properties:
Statements:
1. Reduces the carbonyl group to a CH₂ group
2. Uses zinc-amalgam
3. Uses concentrated hydrochloric acid
4. Operates under basic conditions

A 1, 2, 3

B 2, 3, 4

C 1, 2, 4

D 1, 3, 4

💡 Short Explanation

�� Clemmensen reduction converts carbonyl groups into CH₂ groups. �� It uses zinc-amalgam (Zn-Hg). �� The reaction is carried out in concentrated HCl.

✅ Correct Answer Explanation

Clemmensen reduction is a reduction reaction in which aldehydes and ketones are converted into hydrocarbons by reducing the carbonyl group (>C=O) to a methylene group (–CH₂–). The reaction employs zinc-amalgam and concentrated hydrochloric acid under strongly acidic conditions.

❓ Why Other Options Are Incorrect

�� Statement 4 is incorrect because Clemmensen reduction occurs in acidic medium, not basic medium.
�� Therefore options B, C and D are incorrect.

🎯 Strategy

Used

Statement-Based MCQ

🧠 Memory Trick

Clemmensen = Zn-Hg + HCl → C=O becomes CH₂.

📚 Aldehydes, Ketones and Carboxylic Acids – Reduction Reactions

2 Identify the reaction type: Treatment of a carbonyl compound with hydrazine followed by heating with sodium or potassium hydroxide in a high boiling solvent like ethylene glycol.

A Clemmensen reduction

B Wolff-Kishner reduction

C Rosenmund reduction

D Stephen reaction

💡 Short Explanation

�� Hydrazine reacts with the carbonyl compound. �� Heating with KOH/NaOH in ethylene glycol follows. �� Carbonyl group is reduced to CH₂.

✅ Correct Answer Explanation

In the Wolff-Kishner reduction, aldehydes and ketones react with hydrazine to form hydrazones. Subsequent heating with sodium or potassium hydroxide in a high-boiling solvent such as ethylene glycol reduces the carbonyl group completely to a methylene group.

❓ Why Other Options Are Incorrect

�� Option A uses Zn-Hg/HCl.
�� Option C converts acyl chlorides to aldehydes.
�� Option D converts nitriles to aldehydes.

🎯 Strategy

Used

Reaction Type Identification

🧠 Memory Trick

Hydrazine + KOH + Heat = Wolff-Kishner.

📚 Aldehydes, Ketones and Carboxylic Acids – Reduction Reactions

3 During Tollens' test, what chemical change happens to the aldehyde molecule itself?

A It is reduced to a primary alcohol

B It is oxidised to the corresponding carboxylate anion

C It remains unreacted while silver precipitates

D It is converted to a cyanohydrin

💡 Short Explanation

�� Tollens' reagent is an oxidizing agent. �� Aldehyde is oxidized. �� Carboxylate ion is formed in alkaline medium.

✅ Correct Answer Explanation

Tollens' reagent, ammoniacal silver nitrate solution, oxidizes aldehydes to their corresponding carboxylate ions. Simultaneously, Ag⁺ ions are reduced to metallic silver, producing the characteristic silver mirror.

❓ Why Other Options Are Incorrect

�� Option A describes reduction, not oxidation.
�� Option C is incorrect because aldehyde reacts.
�� Option D refers to HCN addition.

🎯 Strategy

Used

Concept MCQ

🧠 Memory Trick

Tollens: Aldehyde oxidized, Silver reduced.

📚 Aldehydes, Ketones and Carboxylic Acids – Tollens' Test

4 Arrange the general steps involved in successfully performing Fehling's test on an aldehyde:
(A) Mix equal amounts of Fehling solution A and Fehling solution B
(B) Add the aldehyde to the mixed Fehling's reagent
(C) Heat the reaction mixture
(D) Observe the formation of a reddish-brown precipitate

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

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

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

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

💡 Short Explanation

�� Prepare Fehling's reagent first. �� Add aldehyde sample. �� Heat the mixture. �� Observe the precipitate.

✅ Correct Answer Explanation

Correct sequence: 1. Mix Fehling solution A and B (A) 2. Add aldehyde sample (B) 3. Heat the reaction mixture (C) 4. Observe reddish-brown Cu₂O precipitate (D) Therefore, the correct order is (A), (B), (C), (D).

❓ Why Other Options Are Incorrect

�� Options B, C and D do not follow the proper experimental procedure.

🎯 Strategy

Used

Ordering

🧠 Memory Trick

Mix → Add → Heat → Observe.

📚 Aldehydes, Ketones and Carboxylic Acids – Fehling's Test

5 Match List-I (Reactant) with List-II (Haloform Test Result/Role):

List-I	List-II
1. Acetaldehyde (CH₃CHO)	a. Reagent for the haloform oxidation
2. Acetone (CH₃COCH₃)	b. Gives positive test (due to methyl ketone group)
3. Propanal (CH₃CH₂CHO)	c. Gives positive test (due to being a methyl aldehyde)
4. Sodium hypohalite	d. Gives negative test

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

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

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

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

💡 Short Explanation

�� Acetaldehyde gives positive haloform test. �� Acetone gives positive haloform test. �� Propanal gives negative test. �� Sodium hypohalite acts as reagent.

✅ Correct Answer Explanation

List-I — List-II 1. Acetaldehyde — c. Gives positive test (methyl aldehyde) 2. Acetone — b. Gives positive test (methyl ketone) 3. Propanal — d. Gives negative test 4. Sodium hypohalite — a. Reagent for haloform oxidation Hence the correct matching is: 1-c, 2-b, 3-d, 4-a

❓ Why Other Options Are Incorrect

�� Options B, C and D incorrectly assign positive/negative haloform results and reagent roles.

🎯 Strategy

Used

Match the Following

🧠 Memory Trick

Haloform positive: CH₃CO– and CH₃CHO.

📚 Aldehydes, Ketones and Carboxylic Acids – Haloform Reaction

6 What is the IUPAC name of the carboxylic acid salt produced alongside chloroform when propanone (CH₃COCH₃) reacts with sodium hypochlorite?

A Sodium methanoate

B Sodium ethanoate

C Sodium propanoate

D Sodium butanoate

💡 Short Explanation

�� Propanone is a methyl ketone. �� Methyl ketones undergo the haloform reaction. �� The reaction produces chloroform and sodium ethanoate.

✅ Correct Answer Explanation

Propanone (CH₃COCH₃) contains the methyl ketone group (CH₃CO–). During the haloform reaction with sodium hypochlorite, the methyl group attached to the carbonyl carbon is converted into chloroform (CHCl₃), while the remaining portion becomes the sodium salt of ethanoic acid. Reaction: CH₃COCH₃ + 3Cl₂ + 4NaOH → CHCl₃ + CH₃COONa + 3NaCl + 3H₂O Thus, the carboxylate salt formed is sodium ethanoate.

❓ Why Other Options Are Incorrect

�� Option A: Sodium methanoate is not formed.
�� Option C: Sodium propanoate would require a different carbon skeleton.
�� Option D: Sodium butanoate contains too many carbon atoms.

🎯 Strategy

Used

Naming

🧠 Memory Trick

Methyl ketone + Haloform reaction → Haloform + Carboxylate salt.

📚 Aldehydes, Ketones and Carboxylic Acids – Haloform Reaction

7 The exact number of alpha-hydrogen atoms present in a single molecule of ethanal (CH₃CHO) is:

A 1

B 2

C 3

D 4

💡 Short Explanation

�� Alpha-carbon is adjacent to the carbonyl carbon. �� In ethanal, the α-carbon is the CH₃ group. �� CH₃ contains three hydrogen atoms.

✅ Correct Answer Explanation

Ethanal has the structure: CH₃–CHO The carbon adjacent to the carbonyl carbon is called the α-carbon. In ethanal, this α-carbon is the methyl carbon (CH₃), which contains three hydrogen atoms. Therefore, ethanal possesses 3 α-hydrogen atoms.

❓ Why Other Options Are Incorrect

�� Options A and B underestimate the number of α-hydrogens.
�� Option D exceeds the actual count.

🎯 Strategy

Used

Unit-Based MCQ

🧠 Memory Trick

Ethanal = CH₃CHO → α-carbon = CH₃ → 3 α-H.

📚 Aldehydes, Ketones and Carboxylic Acids – Alpha Hydrogen

8 The conjugate base formed after the removal of an alpha-hydrogen from a ketone is stabilised by:

A The positive charge on the carbonyl carbon

B Delocalisation of the negative charge through resonance

C The steric bulk of the adjacent alkyl groups

D The lack of electronegative atoms in the structure

💡 Short Explanation

�� Removal of α-hydrogen forms an enolate ion. �� Negative charge is delocalized between carbon and oxygen. �� Resonance stabilizes the conjugate base.

✅ Correct Answer Explanation

When an α-hydrogen is removed from a ketone, an enolate ion is formed. The negative charge is not localized on one atom; instead, it is delocalized through resonance between the α-carbon and the oxygen atom of the carbonyl group. This resonance stabilization makes α-hydrogens acidic and stabilizes the conjugate base.

❓ Why Other Options Are Incorrect

�� Option A does not explain stabilization.
�� Option C relates to steric effects, not resonance stabilization.
�� Option D is incorrect because oxygen is highly electronegative.

🎯 Strategy

Used

Concept MCQ

🧠 Memory Trick

α-H removed → Enolate formed → Resonance stabilizes.

📚 Aldehydes, Ketones and Carboxylic Acids – Acidity of Alpha Hydrogen

9 The catalyst typically used in the formation of an aldol from aldehydes and ketones is:

A Concentrated sulphuric acid

B Dilute alkali

C Anhydrous aluminium chloride

D Zinc dust

💡 Short Explanation

�� Aldol reaction proceeds through enolate formation. �� A base removes the α-hydrogen. �� Dilute alkali commonly catalyzes the reaction.

✅ Correct Answer Explanation

Aldol condensation is generally carried out in the presence of dilute alkali such as NaOH or KOH. The base abstracts an α-hydrogen to form an enolate ion, which then attacks another carbonyl molecule to produce the aldol product.

❓ Why Other Options Are Incorrect

�� Option A is not the standard catalyst for aldol formation.
�� Option C is used in Friedel-Crafts reactions.
�� Option D is not involved in aldol condensation.

🎯 Strategy

Used

Concept MCQ

🧠 Memory Trick

Aldol = α-H removal = Dilute alkali.

📚 Aldehydes, Ketones and Carboxylic Acids – Aldol Condensation

10 Arrange the following straight-chain aliphatic aldehydes in decreasing order of the number of carbon atoms in their respective aldol addition products (before dehydration):
(A) Pentanal
(B) Butanal
(C) Propanal
(D) Ethanal

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

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

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

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

💡 Short Explanation

�� Self-aldol doubles the carbon skeleton. �� Larger aldehydes produce larger aldol products. �� Carbon count decreases from pentanal to ethanal.

✅ Correct Answer Explanation

Aldehyde — Carbon Atoms — Aldol Product Carbon Atoms Pentanal — 5 — 10 Butanal — 4 — 8 Propanal — 3 — 6 Ethanal — 2 — 4 Thus, decreasing order of carbon atoms in the aldol products is: Pentanal > Butanal > Propanal > Ethanal Therefore: (A), (B), (C), (D)

❓ Why Other Options Are Incorrect

�� Options B, C and D do not follow the correct decreasing carbon count sequence.

🎯 Strategy

Used

Ordering

🧠 Memory Trick

Self-aldol approximately doubles the number of carbon atoms.

📚 Aldehydes, Ketones and Carboxylic Acids – Aldol Condensation

11 Can ketones be successfully used as one of the reactive components in a cross aldol reaction?

A Yes, they can be used as one component

B No, cross aldol is strictly limited to aldehydes

C Yes, but only in a strictly acidic medium

D No, because all ketones lack alpha hydrogens

💡 Short Explanation

�� Cross aldol can involve aldehydes and/or ketones. �� Ketones with α-hydrogen can participate. �� It is not limited to aldehydes only.

✅ Correct Answer Explanation

Cross aldol condensation occurs between two different aldehydes, two different ketones, or an aldehyde and a ketone, provided at least one compound can form an enolate ion. Therefore, ketones can be used as one component in a cross aldol reaction.

❓ Why Other Options Are Incorrect

�� Option B: Cross aldol is not limited to aldehydes.
�� Option C: It is commonly carried out under basic conditions.
�� Option D: Many ketones do have α-hydrogens.

🎯 Strategy

Used

Concept MCQ

🧠 Memory Trick

Cross aldol = different carbonyl compounds, including ketones.

📚 Aldehydes, Ketones and Carboxylic Acids – Cross Aldol Condensation

12 Cross aldol condensation between ethanal and propanal:
Statements:
1. Produces a mixture of four distinct products
2. Occurs because both reactants contain alpha-hydrogen atoms
3. Involves self-condensation products as well as cross-condensation products
4. Yields only a single pure alkene

A 1, 2, 3

B 2, 3, 4

C 1, 3, 4

D 1, 2, 4

💡 Short Explanation

�� Ethanal and propanal both have α-hydrogens. �� Both can form enolates. �� A mixture of self and cross aldol products forms.

✅ Correct Answer Explanation

When ethanal and propanal undergo cross aldol condensation, both compounds have α-hydrogen atoms. Therefore, each can form an enolate and react with itself or with the other aldehyde. This gives a mixture of four possible products, including self-condensation and cross-condensation products.

❓ Why Other Options Are Incorrect

�� Statement 4 is incorrect because the reaction does not yield only one pure alkene; it gives a mixture of products.

🎯 Strategy

Used

Statement-Based MCQ

🧠 Memory Trick

Both have α-H → multiple aldol products.

📚 Aldehydes, Ketones and Carboxylic Acids – Cross Aldol Condensation

13 Provide the IUPAC name of the primary alcohol produced via the Cannizzaro reaction of benzaldehyde.

A Phenol

B Phenylmethanol

C Benzyl alcohol

D Benzoic acid

💡 Short Explanation

�� Benzaldehyde undergoes Cannizzaro reaction. �� One molecule is reduced. �� Reduced product is benzyl alcohol, IUPAC name phenylmethanol.

✅ Correct Answer Explanation

In the Cannizzaro reaction of benzaldehyde, one molecule is oxidised to benzoate ion and another molecule is reduced to benzyl alcohol. The IUPAC name of benzyl alcohol is phenylmethanol.

❓ Why Other Options Are Incorrect

�� Option A: Phenol contains –OH directly attached to benzene ring.
�� Option C: Common name, not IUPAC name.
�� Option D: Oxidised product, not reduced alcohol product.

🎯 Strategy

Used

Naming

🧠 Memory Trick

Benzyl alcohol = Phenylmethanol.

📚 Aldehydes, Ketones and Carboxylic Acids – Cannizzaro Reaction

14 Identify the reaction type: Heating formaldehyde with concentrated alkali results in the formation of methanol and sodium formate.

A Disproportionation reaction

B Condensation reaction

C Esterification reaction

D Electrophilic addition

💡 Short Explanation

�� Formaldehyde lacks α-hydrogen. �� One molecule is oxidised. �� Another molecule is reduced.

✅ Correct Answer Explanation

Formaldehyde undergoes Cannizzaro reaction in concentrated alkali. During this process, one formaldehyde molecule is oxidised to formate ion, while another is reduced to methanol. Since oxidation and reduction occur simultaneously, it is a disproportionation reaction.

❓ Why Other Options Are Incorrect

�� Option B: No condensation product is formed.
�� Option C: No ester is formed.
�� Option D: Not an electrophilic addition reaction.

🎯 Strategy

Used

Reaction Type Identification

🧠 Memory Trick

Cannizzaro = Disproportionation.

📚 Aldehydes, Ketones and Carboxylic Acids – Cannizzaro Reaction

15 Match List-I (Chemical Component/Concept) with List-II (Reaction Behaviour/Effect):

List-I	List-II
1. Carbonyl group in benzaldehyde	a. Deactivating and meta-directing
2. Carboxyl group in benzoic acid	b. Does not undergo Friedel-Crafts reaction
3. Electrophilic substitution of benzaldehyde	c. Yields meta-substituted products
4. Reactivity of aromatic aldehydes	d. Less reactive than aliphatic aldehydes toward nucleophiles

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

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

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

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

💡 Short Explanation

�� Carbonyl group is deactivating and meta-directing. �� Benzoic acid does not undergo Friedel-Crafts reaction. �� Benzaldehyde gives meta products. �� Aromatic aldehydes are less reactive than aliphatic aldehydes.

✅ Correct Answer Explanation

List-I — Correct Match Carbonyl group in benzaldehyde — Deactivating and meta-directing Carboxyl group in benzoic acid — Does not undergo Friedel-Crafts reaction Electrophilic substitution of benzaldehyde — Yields meta-substituted products Reactivity of aromatic aldehydes — Less reactive than aliphatic aldehydes toward nucleophiles

❓ Why Other Options Are Incorrect

�� Options B, C and D incorrectly assign directing effects and reaction behaviour.

🎯 Strategy

Used

Match the Following

🧠 Memory Trick

Aromatic C=O = deactivating + meta-directing.

📚 Aldehydes, Ketones and Carboxylic Acids – Aromatic Reactions

16 Why does the carbonyl group act as a meta-directing group in aromatic aldehydes and ketones?

A It donates electron density via resonance

B It withdraws electron density via resonance, deactivating the ortho and para positions

C It sterically hinders the meta position entirely

D It forms a unique hydrogen bond with the incoming electrophile

💡 Short Explanation

�� Carbonyl groups show a −M (resonance withdrawing) effect. �� Electron density decreases at ortho and para positions. �� Electrophilic substitution therefore occurs preferentially at the meta position.

✅ Correct Answer Explanation

The carbonyl group (>C=O) withdraws electron density from the benzene ring through resonance and inductive effects. This decreases electron density particularly at the ortho and para positions, making them less favorable for electrophilic attack. Consequently, electrophilic substitution occurs predominantly at the meta position.

❓ Why Other Options Are Incorrect

�� Option A: Carbonyl groups withdraw, not donate, electron density.
�� Option C: Meta direction is not due to steric hindrance.
�� Option D: Hydrogen bonding is not responsible for directing effects.

🎯 Strategy

Used

Concept MCQ

🧠 Memory Trick

C=O pulls electrons away → Ortho/Para deactivated → Meta favored.

📚 Aldehydes, Ketones and Carboxylic Acids – Electrophilic Substitution Reactions

17 Formaldehyde is an essential starting material to prepare Bakelite. What structural type of resin is Bakelite?

A Phenol-formaldehyde resin

B Urea-formaldehyde resin

C Melamine-formaldehyde resin

D Polyvinyl resin

💡 Short Explanation

�� Bakelite is formed from phenol and formaldehyde. �� It is a thermosetting polymer. �� It is known as phenol-formaldehyde resin.

✅ Correct Answer Explanation

Bakelite is produced by the condensation polymerization of phenol with formaldehyde. The resulting polymer is called a phenol-formaldehyde resin and is widely used as a thermosetting plastic.

❓ Why Other Options Are Incorrect

�� Option B: Forms urea-formaldehyde resins.
�� Option C: Forms melamine-formaldehyde resins.
�� Option D: Unrelated polymer class.

🎯 Strategy

Used

Concept MCQ

🧠 Memory Trick

Bakelite = Phenol + Formaldehyde.

📚 Aldehydes, Ketones and Carboxylic Acids – Uses of Formaldehyde

18 Which of the following is NOT primarily manufactured starting from acetaldehyde?

A Acetic acid

B Ethyl acetate

C Bakelite

D Vinyl acetate

💡 Short Explanation

�� Acetaldehyde is a precursor of acetic acid, ethyl acetate and vinyl acetate. �� Bakelite is prepared from phenol and formaldehyde. �� Therefore Bakelite is not manufactured from acetaldehyde.

✅ Correct Answer Explanation

Acetaldehyde is used industrially in the manufacture of acetic acid, ethyl acetate, vinyl acetate, polymers and drugs. Bakelite, however, is produced from phenol and formaldehyde, not acetaldehyde.

❓ Why Other Options Are Incorrect

�� Option A: Prepared using acetaldehyde as a starting material.
�� Option B: Prepared from acetaldehyde.
�� Option D: Prepared from acetaldehyde.

🎯 Strategy

Used

Concept MCQ

🧠 Memory Trick

Acetaldehyde → Acetic acid family products; Bakelite → Formaldehyde + Phenol.

📚 Aldehydes, Ketones and Carboxylic Acids – Industrial Uses

19 Which aldehyde is specifically mentioned in the text as being used in the dye industry?

A Formaldehyde

B Acetaldehyde

C Benzaldehyde

D Acetone

💡 Short Explanation

�� The passage directly mentions benzaldehyde. �� It is used in perfumery and dye industries. �� Hence benzaldehyde is the correct answer.

✅ Correct Answer Explanation

According to the passage, benzaldehyde is used in perfumery and in dye industries. Therefore, the aldehyde specifically associated with the dye industry is benzaldehyde.

❓ Why Other Options Are Incorrect

�� Option A: Used mainly as formalin and polymer precursor.
�� Option B: Used for acetic acid and related products.
�� Option D: Acetone is a ketone, not an aldehyde.

🎯 Strategy

Used

Passage-Based MCQ

🧠 Memory Trick

Benzaldehyde = Perfumes + Dyes.

📚 Aldehydes, Ketones and Carboxylic Acids – Applications

20 Which of the following is stated as a common industrial solvent?

A Ethyl acetate

B Ethyl methyl ketone

C Vinyl acetate

D Formalin

💡 Short Explanation

�� The passage names acetone and ethyl methyl ketone as industrial solvents. �� Ethyl methyl ketone is widely used as a solvent. �� Hence it is the correct answer.

✅ Correct Answer Explanation

The passage explicitly states that acetone and ethyl methyl ketone are common industrial solvents. Therefore, ethyl methyl ketone is the correct answer.

❓ Why Other Options Are Incorrect

�� Option A: Mainly an ester and chemical intermediate.
�� Option C: Used in polymer production.
�� Option D: Formalin is an aqueous formaldehyde solution used as a preservative.

🎯 Strategy

Used

Passage-Based MCQ

🧠 Memory Trick

Industrial solvents mentioned together: Acetone + Ethyl Methyl Ketone.

📚 Aldehydes, Ketones and Carboxylic Acids – Industrial Applications

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