CUET UG Chemistry Booster Test - 2 Chemical Reactions and Uses -

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CUET UG Chemistry Booster Test - 2 Chemical Reactions and Uses

⏱️ 30 Minutes

❓ 20 Questions

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

CUET UG Chemistry Booster Test - 2 Chemical Reactions and Uses

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QUESTION 1 OF 20

Statements regarding alcohols as nucleophiles:
1. They react as nucleophiles due to the presence of unshared electron pairs on oxygen.
2. The O–H bond is broken during this reaction.
3. The C–O bond remains intact when alcohol acts as a nucleophile.
4. Alcohols can act as Brønsted bases.

Answer Review

1 Statements regarding alcohols as nucleophiles:
1. They react as nucleophiles due to the presence of unshared electron pairs on oxygen.
2. The O–H bond is broken during this reaction.
3. The C–O bond remains intact when alcohol acts as a nucleophile.
4. Alcohols can act as Brønsted bases.

A (1), (3), (4)

B (1), (2)

C (2), (3)

D (2), (4)

💡 Short Explanation

�� Alcohols act as nucleophiles through the lone pair on oxygen. �� The C–O bond remains intact during nucleophilic attack. �� Alcohols can also behave as Brønsted bases.

✅ Correct Answer Explanation

Statement 1 is correct because the oxygen atom possesses lone pairs that can attack an electrophile. Statement 2 is incorrect because cleavage of the O–H bond occurs when alcohol behaves as a Brønsted acid, not when it acts as a nucleophile. Statement 3 is correct because during nucleophilic reactions the oxygen atom donates its lone pair while the C–O bond remains intact. Statement 4 is correct because alcohols can accept a proton through the lone pair present on oxygen and therefore behave as Brønsted bases. Hence, statements 1, 3 and 4 are correct, so Option A is the correct answer.

❓ Why Other Options Are Incorrect

�� Option B → Includes statement 2, which is incorrect.
�� Option C → Includes statement 2 and omits statement 1.
�� Option D → Includes statement 2, which is incorrect.

🎯 Strategy

Used

Elimination

Application:

Verify each statement using the NCERT concept of nucleophilic behaviour of alcohols.

Final Logic:

Lone Pair Present + C–O Bond Intact + Brønsted Base Property = Correct Statements.

🧠 Memory Trick

Lone Pair Attacks, Bond Stays

📚 Class XII – NCERT Chemistry Part II – Chapter: Alcohols, Phenols and Ethers – Topic: Reactions of Alcohols as Nucleophiles

2 Which species acts as the electrophile when the C–O bond of an alcohol is broken?

A Alkoxide ion

B Protonated alcohol

C Phenoxide ion

D Carboxylate ion

💡 Short Explanation

�� Protonation converts –OH into a good leaving group. �� Protonated alcohol becomes electrophilic. �� Nucleophiles attack the carbon atom.

✅ Correct Answer Explanation

Before C–O bond cleavage, the alcohol is protonated by an acid to form a protonated alcohol. This converts the poor leaving group (–OH) into water, making the carbon atom more susceptible to nucleophilic attack. Thus, the protonated alcohol acts as the electrophilic species. Therefore, option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Alkoxide ion is nucleophilic, not electrophilic.
�� Option C → Phenoxide ion is also nucleophilic.
�� Option D → Carboxylate ions are nucleophiles rather than electrophiles.

🎯 Strategy

Used

Elimination

Application:

Identify the activated species formed before substitution.

Final Logic:

Protonation Creates the Electrophile.

🧠 Memory Trick

H⁺ First, Attack Next

📚 Class XII – NCERT Chemistry Part II – Chapter: Alcohols, Phenols and Ethers – Topic: Reactions Involving Cleavage of C–O Bond

3 Identify the reaction type when phenol reacts with aqueous sodium hydroxide to form sodium phenoxide.

A Electrophilic aromatic substitution

B Acid-base neutralization

C Oxidation

D Reduction

💡 Short Explanation

�� Phenol behaves as a weak acid. �� Sodium hydroxide acts as a base. �� Sodium phenoxide and water are formed.

✅ Correct Answer Explanation

Phenol reacts with aqueous sodium hydroxide by donating a proton to the hydroxide ion. This acid-base reaction produces sodium phenoxide and water. No oxidation, reduction or substitution occurs. Therefore, option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → No substitution on the aromatic ring occurs.
�� Option C → Oxidation does not take place.
�� Option D → Reduction does not occur.

🎯 Strategy

Used

Elimination

Application:

Identify the nature of the reaction based on products formed.

Final Logic:

Phenol + NaOH → Sodium Phenoxide + Water.

🧠 Memory Trick

Phenol + Base = Phenoxide

📚 Class XII – NCERT Chemistry Part II – Chapter: Alcohols, Phenols and Ethers – Topic: Acidic Nature of Phenol

4 Arrange the following in increasing order of acid strength based on pKa values:
1. Ethanol
2. p-Cresol
3. Phenol
4. m-Nitrophenol

A (3), (2), (1), (4)

B (4), (3), (2), (1)

C (1), (4), (3), (2)

D (1), (2), (3), (4)

💡 Short Explanation

�� Alcohols are weaker acids than phenols. �� Methyl groups decrease acidity. �� Nitro groups increase acidity.

✅ Correct Answer Explanation

Ethanol is the weakest acid because it lacks resonance stabilization of its conjugate base. p-Cresol is slightly less acidic than phenol because the methyl group exerts an electron-releasing (+I) effect. Phenol is more acidic due to resonance stabilization of the phenoxide ion. m-Nitrophenol is the strongest acid because the nitro group withdraws electrons through the −I effect, stabilizing the phenoxide ion. Therefore, the increasing order of acid strength is: Ethanol < p-Cresol < Phenol < m-Nitrophenol. Hence, option D is correct.

❓ Why Other Options Are Incorrect

�� Option A → Places ethanol above phenol.
�� Option B → Gives the reverse order.
�� Option C → Places m-nitrophenol before phenol incorrectly.

🎯 Strategy

Used

Option Grouping

Application:

Compare the effects of electron-donating and electron-withdrawing groups.

Final Logic:

Alcohol < Alkyl Phenol < Phenol < Nitro Phenol.

🧠 Memory Trick

Nitro Strong, Methyl Weak

📚 Class XII – NCERT Chemistry Part II – Chapter: Alcohols, Phenols and Ethers – Topic: Acidic Nature of Phenols

5 Why do electron-releasing groups (like –CH₃) decrease the acid strength of alcohols?

A They increase electron density on oxygen, decreasing O–H bond polarity.

B They decrease electron density on oxygen, increasing O–H bond polarity.

C They delocalize the negative charge.

D They form intermolecular hydrogen bonds.

💡 Short Explanation

�� Alkyl groups show the +I effect. �� Electron density on oxygen increases. �� Proton release becomes more difficult.

✅ Correct Answer Explanation

Electron-releasing alkyl groups exert a positive inductive (+I) effect, increasing the electron density on the oxygen atom. As a result, the O–H bond becomes less polar, making it more difficult to release a proton. Consequently, the acidity of alcohols decreases. Therefore, option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Electron-releasing groups do not decrease electron density.
�� Option C → Delocalization is not produced by methyl groups.
�� Option D → Hydrogen bonding does not explain the decrease in acidity.

🎯 Strategy

Used

Elimination

Application:

Recall the effect of the +I effect on acidity.

Final Logic:

+I Effect → Lower Acidity.

🧠 Memory Trick

CH₃ Pushes, Acid Falls

📚 Class XII – NCERT Chemistry Part II – Chapter: Alcohols, Phenols and Ethers – Topic: Acidic Nature of Alcohols

6 Match List-I (Compound) with List-II (Approximate pKa Value)

List I	List II
1. o-Nitrophenol	a. 15.9
2. m-Nitrophenol	b. 10.0
3. Phenol	c. 7.2
4. Ethanol	d. 8.3

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

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

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

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

💡 Short Explanation

�� Nitro groups increase acidity. �� Ethanol is much less acidic than phenol. �� Lower pKa indicates stronger acidity.

✅ Correct Answer Explanation

1 → c : o-Nitrophenol has a pKa of approximately 7.2. 2 → d : m-Nitrophenol has a pKa of approximately 8.3. 3 → b : Phenol has a pKa of approximately 10.0. 4 → a : Ethanol has a pKa of approximately 15.9. Therefore, option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → o-Nitrophenol and m-nitrophenol are interchanged.
�� Option C → Phenol and m-nitrophenol are incorrectly matched.
�� Option D → Multiple incorrect pairings are present.

🎯 Strategy

Used

Option Grouping

Application:

Match each compound with its approximate NCERT acidity trend.

Final Logic:

Lower pKa = Stronger Acid.

🧠 Memory Trick

o-Nitro < m-Nitro < Phenol < Ethanol

📚 Class XII – NCERT Chemistry Part II – Chapter: Alcohols, Phenols and Ethers – Topic: Acidic Nature of Phenols

📄 Passage

Alcohols and phenols react with carboxylic acids, acid chlorides, and acid anhydrides to form esters. The reaction with carboxylic acid and acid anhydride is carried out in the presence of a small amount of concentrated sulphuric acid. The reaction is reversible, and therefore, water is removed as soon as it is formed. The reaction with acid chloride is carried out in the presence of a base (pyridine) so as to neutralise HCl which is formed during the reaction. It shifts the equilibrium to the right-hand side. The introduction of acetyl (CH₃CO) group in alcohols or phenols is known as acetylation. Acetylation of salicylic acid produces aspirin, which possesses analgesic, anti-inflammatory and antipyretic properties.

Based on the passage, why is pyridine used during the reaction of alcohols with acid chlorides?

A To act as a catalyst

B To neutralise the HCl formed and shift the equilibrium to the right

C To remove water from the reaction mixture

D To introduce an acetyl group

💡 Short Explanation

�� Pyridine acts as a base. �� It neutralises HCl formed. �� The equilibrium shifts towards ester formation.

✅ Correct Answer Explanation

As stated in the passage, pyridine is added during the reaction of alcohols with acid chlorides to neutralise the HCl produced. Removal of HCl shifts the equilibrium towards the products, increasing the yield of the ester. Therefore, option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Pyridine primarily acts as a base, not as the catalyst.
�� Option C → Water removal is associated with esterification using carboxylic acids.
�� Option D → The acetyl group is supplied by the acid chloride or acid anhydride, not by pyridine.

🎯 Strategy

Used

Contextual/Tonal Matching

Application:

Locate the exact reason stated in the passage.

Final Logic:

Pyridine Removes HCl → Ester Formation Favoured.

🧠 Memory Trick

Pyridine Pulls HCl

📚 Class XII – NCERT Chemistry Part II – Chapter: Alcohols, Phenols and Ethers – Topic: Esterification and Acetylation

📄 Passage

Based on the passage, what is the structure/name of the group introduced during the acetylation of salicylic acid?

A Salicyl group

B Benzoyl group

C Acetyl (CH₃CO) group

D Carboxyl group

💡 Short Explanation

�� Acetylation introduces an acetyl group. �� The acetyl group is CH₃CO–. �� Aspirin is formed from salicylic acid.

✅ Correct Answer Explanation

The passage states that acetylation is the introduction of an acetyl (CH₃CO) group into alcohols or phenols. Acetylation of salicylic acid introduces this group to form aspirin. Therefore, option C is correct.

❓ Why Other Options Are Incorrect

�� Option A → No salicyl group is introduced.
�� Option B → Benzoyl and acetyl groups are different.
�� Option D → The carboxyl group is already present in salicylic acid.

🎯 Strategy

Used

Contextual/Tonal Matching

Application:

Identify the group explicitly mentioned in the passage.

Final Logic:

Acetylation = CH₃CO Introduction.

🧠 Memory Trick

Acetyl = CH₃CO

📚 Class XII – NCERT Chemistry Part II – Chapter: Alcohols, Phenols and Ethers – Topic: Acetylation

9 Arrange the following compounds in decreasing order of their reactivity with HCl (Lucas test):
1. Tertiary alcohol
2. Secondary alcohol
3. Primary alcohol
4. Phenol

A (1), (2), (3), (4)

B (3), (2), (1), (4)

C (2), (1), (3), (4)

D (1), (3), (2), (4)

💡 Short Explanation

�� Tertiary alcohols react immediately. �� Secondary alcohols react more slowly. �� Primary alcohols react very slowly, while phenol does not undergo the Lucas reaction.

✅ Correct Answer Explanation

Statement 1 represents tertiary alcohols, which react immediately with Lucas reagent because they readily form stable tertiary carbocations. Statement 2 represents secondary alcohols, which react after a short time due to the formation of less stable secondary carbocations. Statement 3 represents primary alcohols, which react very slowly or do not react at room temperature because primary carbocations are highly unstable. Statement 4 represents phenol, which does not react with Lucas reagent because the C–O bond has partial double bond character due to resonance and cannot undergo substitution under these conditions. Therefore, the decreasing order of reactivity is: Tertiary alcohol > Secondary alcohol > Primary alcohol > Phenol. Hence, option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Reverses the actual reactivity order.
�� Option C → Places secondary alcohol above tertiary alcohol.
�� Option D → Places primary alcohol above secondary alcohol.

🎯 Strategy

Used

Option Grouping

Application:

Arrange compounds according to their ability to form carbocations and undergo substitution with Lucas reagent.

Final Logic:

3° Alcohol > 2° Alcohol > 1° Alcohol > Phenol.

🧠 Memory Trick

3° Instant → 2° Minutes → 1° Slow → Phenol Never

📚 Class XII – NCERT Chemistry Part II – Chapter: Alcohols, Phenols and Ethers – Topic: Lucas Test and Reactivity of Alcohols

10 Statements about the Lucas test:
1. It uses a reagent containing concentrated HCl and ZnCl₂.
2. Alcohols are miscible in the reagent.
3. Alkyl halides formed are miscible and produce a clear solution.
4. Tertiary alcohols produce turbidity immediately.

A (1), (2), (3), (4)

B (1), (3), (4)

C (1), (2), (4)

D (2), (3)

💡 Short Explanation

�� Lucas reagent contains concentrated HCl and ZnCl₂. �� Alcohols initially dissolve in the reagent. �� Tertiary alcohols react immediately.

✅ Correct Answer Explanation

Statement 1 is correct because Lucas reagent is a mixture of concentrated hydrochloric acid and anhydrous zinc chloride. Statement 2 is correct because alcohols are initially miscible with the reagent. Statement 3 is incorrect because the alkyl halides formed are insoluble in the reaction mixture and produce turbidity rather than a clear solution. Statement 4 is correct because tertiary alcohols react rapidly, producing immediate turbidity. Therefore, option C is correct.

❓ Why Other Options Are Incorrect

�� Option A → Includes statement 3, which is incorrect.
�� Option B → Includes statement 3, which is incorrect.
�� Option D → Includes statement 3 and omits correct statements 1 and 4.

🎯 Strategy

Used

Elimination

Application:

Verify each statement using the NCERT description of the Lucas test.

Final Logic:

Statements 1, 2 and 4 are correct; statement 3 is incorrect.

🧠 Memory Trick

Lucas: Clear First, Cloudy Later

📚 Class XII – NCERT Chemistry Part II – Chapter: Alcohols, Phenols and Ethers – Topic: Lucas Test

11 What is the IUPAC name of the major product formed when 1-methylcyclohexanol undergoes acid-catalysed dehydration?

A 1-Methylcyclohexene

B Methylenecyclohexane

C Cyclohexene

D Cyclohexanol

💡 Short Explanation

�� Dehydration follows Zaitsev's rule. �� The more substituted alkene is the major product. �� 1-Methylcyclohexene is more stable than methylenecyclohexane.

✅ Correct Answer Explanation

1-Methylcyclohexanol is a tertiary alcohol and undergoes acid-catalysed dehydration through an E1 mechanism. The major product follows Zaitsev's rule, giving the more substituted and more stable alkene, 1-methylcyclohexene. Therefore, option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Methylenecyclohexane is the less substituted alkene and is formed as a minor product.
�� Option C → Cyclohexene would require removal of the methyl group, which does not occur.
�� Option D → Cyclohexanol is the reactant, not the dehydration product.

🎯 Strategy

Used

Elimination

Application:

Choose the most stable alkene formed during dehydration.

Final Logic:

More Substituted Alkene → Major Product.

🧠 Memory Trick

Zaitsev = More Substituted Wins

📚 Class XII – NCERT Chemistry Part II – Chapter: Alcohols, Phenols and Ethers – Topic: Dehydration of Alcohols

12 In the acid-catalysed dehydration of ethanol, which step is the slowest and rate-determining?

A Formation of protonated alcohol

B Formation of carbocation

C Elimination of a proton to form ethene

D Addition of water

💡 Short Explanation

�� Protonation occurs rapidly. �� Carbocation formation is the slowest step. �� Elimination produces the alkene.

✅ Correct Answer Explanation

In acid-catalysed dehydration, the alcohol is first protonated. The slowest and rate-determining step is the removal of water to generate the carbocation. The carbocation then rapidly loses a proton to form the alkene. Therefore, option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Protonation is a fast equilibrium step.
�� Option C → Deprotonation is rapid after carbocation formation.
�� Option D → Water is not added during dehydration.

🎯 Strategy

Used

Elimination

Application:

Identify the highest-energy step in the reaction mechanism.

Final Logic:

Carbocation Formation Controls the Rate.

🧠 Memory Trick

Slow Step = Carbocation

📚 Class XII – NCERT Chemistry Part II – Chapter: Alcohols, Phenols and Ethers – Topic: Dehydration of Alcohols

13 The temperature unit at which the vapours of a primary alcohol are passed over heated copper to undergo dehydrogenation is:

A 573 K

B 443 K

C 337 K

D 298 K

💡 Short Explanation

�� Heated copper acts as the catalyst. �� Dehydrogenation occurs at 573 K. �� Primary alcohols produce aldehydes.

✅ Correct Answer Explanation

According to NCERT, vapours of primary and secondary alcohols passed over heated copper at 573 K undergo dehydrogenation. Primary alcohols produce aldehydes, whereas secondary alcohols produce ketones. Therefore, option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Used for dehydration conditions.
�� Option C → Not used for this reaction.
�� Option D → Too low for dehydrogenation.

🎯 Strategy

Used

Elimination

Application:

Recall the standard NCERT reaction conditions.

Final Logic:

Heated Copper = 573 K.

🧠 Memory Trick

Copper = 573 K

📚 Class XII – NCERT Chemistry Part II – Chapter: Alcohols, Phenols and Ethers – Topic: Oxidation of Alcohols

14 Match List-I (Reactant) with List-II (Product obtained on passing vapours over heated copper at 573 K)

List I	List II
1. Primary alcohol	a. Alkene (Dehydration)
2. Secondary alcohol	b. Ketone
3. Tertiary alcohol	c. Aldehyde
4. Heated Cu at 573 K	d. Catalyst used for dehydrogenation/dehydration

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

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

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

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

💡 Short Explanation

�� Primary alcohols form aldehydes. �� Secondary alcohols form ketones. �� Tertiary alcohols undergo dehydration to alkenes.

✅ Correct Answer Explanation

1 → c : Primary alcohols undergo dehydrogenation to form aldehydes. 2 → b : Secondary alcohols undergo dehydrogenation to form ketones. 3 → a : Tertiary alcohols lack α-hydrogen and therefore undergo dehydration to form alkenes. 4 → d : Heated copper at 573 K acts as the catalyst for these reactions. Therefore, option C is correct.

❓ Why Other Options Are Incorrect

�� Option A → Primary alcohol is incorrectly matched with alkene.
�� Option B → Primary and secondary alcohol products are interchanged.
�� Option D → Secondary and tertiary alcohol products are incorrectly matched.

🎯 Strategy

Used

Option Grouping

Application:

Match each class of alcohol with its characteristic reaction over heated copper.

Final Logic:

1° → Aldehyde, 2° → Ketone, 3° → Alkene.

🧠 Memory Trick

1° Aldehyde, 2° Ketone, 3° Alkene

📚 Class XII – NCERT Chemistry Part II – Chapter: Alcohols, Phenols and Ethers – Topic: Oxidation of Alcohols

15 Which isomer of nitrophenol is steam volatile due to intramolecular hydrogen bonding?

A p-Nitrophenol

B m-Nitrophenol

C o-Nitrophenol

D 2,4,6-Trinitrophenol

💡 Short Explanation

�� o-Nitrophenol forms intramolecular hydrogen bonding. �� Molecules remain less associated. �� Hence, it is steam volatile.

✅ Correct Answer Explanation

o-Nitrophenol exhibits intramolecular hydrogen bonding, preventing intermolecular association between molecules. Consequently, it has a lower boiling point and is steam volatile. In contrast, p-nitrophenol forms intermolecular hydrogen bonding and is less volatile. Therefore, option C is correct.

❓ Why Other Options Are Incorrect

�� Option A → p-Nitrophenol forms intermolecular hydrogen bonding.
�� Option B → m-Nitrophenol does not exhibit the characteristic intramolecular hydrogen bonding responsible for steam volatility.
�� Option D → 2,4,6-Trinitrophenol (picric acid) is not steam volatile.

🎯 Strategy

Used

Elimination

Application:

Compare the type of hydrogen bonding shown by nitrophenol isomers.

Final Logic:

Intramolecular Hydrogen Bonding → Steam Volatile.

🧠 Memory Trick

Ortho = One Molecule Bond

📚 Class XII – NCERT Chemistry Part II – Chapter: Alcohols, Phenols and Ethers – Topic: Nitration of Phenol and Hydrogen Bonding

16 When phenol is treated with bromine water, the product formed as a white precipitate is:

A o-Bromophenol

B p-Bromophenol

C 2,4,6-Tribromophenol

D 2,4-Dibromophenol

💡 Short Explanation

�� Phenol strongly activates the benzene ring. �� Bromination occurs at ortho and para positions. �� A white precipitate of 2,4,6-tribromophenol is formed.

✅ Correct Answer Explanation

The hydroxyl group in phenol strongly activates the benzene ring towards electrophilic substitution. On treatment with bromine water, bromination occurs readily at the two ortho and one para positions to form 2,4,6-tribromophenol as a white precipitate. Therefore, option C is correct.

❓ Why Other Options Are Incorrect

�� Option A → o-Bromophenol is formed under controlled bromination in non-polar solvents, not with bromine water.
�� Option B → p-Bromophenol is not the major product with bromine water.
�� Option D → 2,4-Dibromophenol is not the final product under these conditions.

🎯 Strategy

Used

Elimination

Application:

Recall the characteristic reaction of phenol with bromine water.

Final Logic:

Phenol + Br₂(aq) → 2,4,6-Tribromophenol.

🧠 Memory Trick

Br₂ Water = White Tribromo

📚 Class XII – NCERT Chemistry Part II – Chapter: Alcohols, Phenols and Ethers – Topic: Bromination of Phenol

17 Arrange the sequence of intermediates/reactants in Kolbe's reaction:
1. Phenoxide ion
2. Phenol
3. Reaction with CO₂ followed by acidification
4. Ortho hydroxybenzoic acid

A (1), (2), (3), (4)

B (1), (4), (2), (3)

C (2), (1), (3), (4)

D (3), (2), (1), (4)

💡 Short Explanation

�� Phenol first forms phenoxide ion. �� Phenoxide reacts with CO₂. �� Acidification produces salicylic acid.

✅ Correct Answer Explanation

Statement 1 represents the phenoxide ion formed in the first step. Statement 2 represents phenol, which reacts with sodium hydroxide to form sodium phenoxide. Statement 3 represents the reaction of sodium phenoxide with carbon dioxide followed by acidification. Statement 4 represents ortho hydroxybenzoic acid (salicylic acid), the final product. Therefore, the correct sequence is: Phenol → Phenoxide ion → CO₂ reaction followed by acidification → Ortho hydroxybenzoic acid. Hence, option C is correct.

❓ Why Other Options Are Incorrect

�� Option A → Begins with phenoxide ion instead of phenol.
�� Option B → Places the final product before the intermediate.
�� Option D → Incorrect reaction sequence.

🎯 Strategy

Used

Option Grouping

Application:

Arrange the steps according to the NCERT mechanism of Kolbe's reaction.

Final Logic:

Phenol → Phenoxide → CO₂ → Salicylic Acid.

🧠 Memory Trick

Phenol → Phenoxide → CO₂ → COOH

📚 Class XII – NCERT Chemistry Part II – Chapter: Alcohols, Phenols and Ethers – Topic: Kolbe's Reaction

18 Identify the reaction type that introduces a –CHO group at the ortho position of phenol using CHCl₃ and NaOH.

A Nucleophilic addition

B Electrophilic aromatic substitution

C Free radical halogenation

D Elimination

💡 Short Explanation

�� This is the Reimer–Tiemann reaction. �� A formyl group is introduced. �� The substitution occurs mainly at the ortho position.

✅ Correct Answer Explanation

In the Reimer–Tiemann reaction, phenol reacts with chloroform and sodium hydroxide to generate dichlorocarbene, which acts as the electrophile. The electrophile substitutes mainly at the ortho position of the activated aromatic ring. Hence, the reaction is an electrophilic aromatic substitution. Therefore, option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → No nucleophilic addition occurs.
�� Option C → The reaction is not a free radical process.
�� Option D → Elimination is not involved.

🎯 Strategy

Used

Contextual/Tonal Matching

Application:

Identify the reaction mechanism of the named reaction.

Final Logic:

Reimer–Tiemann = Electrophilic Aromatic Substitution.

🧠 Memory Trick

Reimer = Ring CHO

📚 Class XII – NCERT Chemistry Part II – Chapter: Alcohols, Phenols and Ethers – Topic: Reimer–Tiemann Reaction

19 Statements regarding cleavage of alkyl aryl ethers:
1. They are cleaved at the alkyl–oxygen bond.
2. The aryl–oxygen bond is more stable due to partial double bond character.
3. The reaction yields phenol and an alkyl halide.
4. They yield aryl halide and alcohol.

A (1), (2), (3)

B (2), (3), (4)

C (1), (3), (4)

D (1), (2), (4)

💡 Short Explanation

�� Cleavage occurs at the alkyl–oxygen bond. �� The aryl–oxygen bond is resonance-stabilised. �� Products are phenol and an alkyl halide.

✅ Correct Answer Explanation

Statement 1 is correct because cleavage of alkyl aryl ethers occurs at the alkyl–oxygen bond. Statement 2 is correct because the aryl–oxygen bond possesses partial double bond character due to resonance and is difficult to break. Statement 3 is correct because the products are phenol and an alkyl halide. Statement 4 is incorrect because aryl halides are not formed during this reaction. Therefore, option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Includes statement 4, which is incorrect.
�� Option C → Includes statement 4, which is incorrect.
�� Option D → Includes statement 4, which is incorrect.

🎯 Strategy

Used

Elimination

Application:

Verify each statement using the NCERT mechanism of ether cleavage.

Final Logic:

Statements 1, 2 and 3 are correct; statement 4 is incorrect.

🧠 Memory Trick

Alkyl Breaks, Aryl Stays

📚 Class XII – NCERT Chemistry Part II – Chapter: Alcohols, Phenols and Ethers – Topic: Cleavage of Ethers

20 In anisole, the methoxy group directs the incoming electrophile to which positions?

A Meta only

B Ortho and meta

C Ortho and para

D Para only

💡 Short Explanation

�� The methoxy group is electron-donating. �� It activates the aromatic ring. �� Electrophilic substitution occurs mainly at ortho and para positions.

✅ Correct Answer Explanation

The methoxy (–OCH₃) group donates electron density to the benzene ring through resonance (+M effect), increasing electron density at the ortho and para positions. Consequently, electrophilic substitution occurs predominantly at these positions. Therefore, option C is correct.

❓ Why Other Options Are Incorrect

�� Option A → The methoxy group is not meta-directing.
�� Option B → Meta substitution is not favoured.
�� Option D → Both ortho and para substitution occur.

🎯 Strategy

Used

Elimination

Application:

Recall the directing effect of the methoxy group.

Final Logic:

��OCH₃ = Ortho/Para Director.

🧠 Memory Trick

OCH₃ → O & P

📚 Class XII – NCERT Chemistry Part II – Chapter: Alcohols, Phenols and Ethers – Topic: Electrophilic Substitution Reactions of Anisole

CUET UG Chemistry Booster Test – 2 Chemical Reactions and Uses

CUET UG Chemistry Booster Test - 2 Chemical Reactions and Uses

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