CUET UG Chemistry Booster Test - 3 Chemical Reactions and Uses -

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CUET UG Chemistry Booster Test - 3 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 - 3 Chemical Reactions and Uses

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QUESTION 1 OF 20

📄 Passage

Alcohols are versatile compounds. They react both as nucleophiles and electrophiles. The bond between O–H is broken when alcohols react as nucleophiles. The bond between C–O is broken when they react as electrophiles. Protonated alcohols react in this manner. Based on the cleavage of O–H and C–O bonds, the reactions of alcohols and phenols may be divided into two groups. When alcohols act as Brønsted bases, it is due to the presence of unshared electron pairs on oxygen, which makes them proton acceptors. Conversely, their acidic character is due to the polar nature of the O–H bond.

Based on the passage, why do alcohols act as Brønsted bases?

Answer Review

📄 Passage

Based on the passage, why do alcohols act as Brønsted bases?

A Because they donate protons to stronger bases.

B Due to the polar nature of the O–H bond.

C Because the C–O bond is easily broken.

D Due to the presence of unshared electron pairs on oxygen.

💡 Short Explanation

�� Oxygen possesses lone pairs. �� Alcohols can accept protons. �� Hence, they behave as Brønsted bases.

✅ Correct Answer Explanation

As stated in the passage, alcohols act as Brønsted bases because the oxygen atom possesses unshared electron pairs that can accept a proton. Thus, alcohols behave as proton acceptors. Therefore, option D is correct.

❓ Why Other Options Are Incorrect

�� Option A → Brønsted bases accept protons rather than donate them.
�� Option B → The polar O–H bond explains acidic behaviour, not basic behaviour.
�� Option C → C–O bond cleavage is unrelated to Brønsted basicity.

🎯 Strategy

Used

Contextual/Tonal Matching

Application:

Identify the statement directly supported by the passage.

Final Logic:

Lone Pair → Proton Acceptor → Brønsted Base.

🧠 Memory Trick

Lone Pair = Base

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

📄 Passage

Based on the passage, in which form do alcohols primarily react as electrophiles?

A As alkoxide ions

B As protonated alcohols

C As phenoxide ions

D As neutral unprotonated molecules

💡 Short Explanation

�� Protonation converts –OH into a good leaving group. �� Protonated alcohols undergo C–O bond cleavage. �� They behave as electrophiles.

✅ Correct Answer Explanation

The passage clearly states that alcohols react as electrophiles in their protonated form. Protonation converts the hydroxyl group into water, making it a good leaving group and facilitating nucleophilic attack. Therefore, option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Alkoxide ions are nucleophiles, not electrophiles.
�� Option C → Phenoxide ions are also nucleophilic.
�� Option D → Neutral alcohols are much less reactive as electrophiles.

🎯 Strategy

Used

Contextual/Tonal Matching

Application:

Identify the electrophilic species mentioned in the passage.

Final Logic:

Protonated Alcohol = Electrophile.

🧠 Memory Trick

H⁺ Makes Alcohol Reactive

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

3 The reaction of an alkoxide ion with water suggests that:

A Water is a stronger acid than alcohol.

B Alcohol is a stronger acid than water.

C Hydroxide is a stronger base than alkoxide.

D Alkoxide is a weaker proton acceptor.

💡 Short Explanation

�� Alkoxide ions abstract a proton from water. �� Water donates the proton. �� Therefore, water is the stronger acid.

✅ Correct Answer Explanation

The reaction RO⁻ + H₂O → ROH + OH⁻ shows that water donates a proton to the alkoxide ion. Since proton transfer occurs from water to the alkoxide ion, water behaves as the stronger acid, while alkoxide is the stronger base. Therefore, option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Alcohols are weaker acids than water.
�� Option C → Alkoxide ions are stronger bases than hydroxide ions.
�� Option D → Alkoxide ions are strong proton acceptors.

🎯 Strategy

Used

Elimination

Application:

Compare the acid-base strengths using the proton-transfer reaction.

Final Logic:

Water Donates H⁺ → Stronger Acid.

🧠 Memory Trick

Water Gives, Alkoxide Takes

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

4 Match List-I (Substituent on Phenol) with List-II (Effect on Acidic Strength)

List I	List II
1. Electron withdrawing group at ortho/para	a. Decreases acid strength
2. Electron releasing group (alkyl)	b. Serves as a reference (pKa ≈ 10.0)
3. Nitro group	c. Highly enhances acid strength
4. Unsubstituted phenol	d. Enhances acidic strength

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

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

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

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

💡 Short Explanation

�� Electron-withdrawing groups increase acidity. �� Alkyl groups decrease acidity. �� Phenol serves as the reference compound.

✅ Correct Answer Explanation

1 → c : Electron-withdrawing groups at the ortho or para position strongly stabilize the phenoxide ion and highly enhance acidity. 2 → a : Electron-releasing alkyl groups destabilize the phenoxide ion and decrease acidity. 3 → d : The nitro group is an electron-withdrawing group and enhances acidic strength. 4 → b : Unsubstituted phenol has a pKa of about 10.0 and serves as the reference. Therefore, option D is correct.

❓ Why Other Options Are Incorrect

�� Option A → Multiple incorrect pairings are present.
�� Option B → Electron-withdrawing groups are incorrectly matched.
�� Option C → Nitro and alkyl effects are incorrectly matched.

🎯 Strategy

Used

Option Grouping

Application:

Match each substituent with its effect on phenol acidity.

Final Logic:

EWG ↑ Acidity; EDG ↓ Acidity.

🧠 Memory Trick

Nitro Up, Alkyl Down

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

5 Statements concerning the acidity of alcohols and phenols:
1. Phenol is millions of times more acidic than ethanol.
2. In phenoxide ion, the negative charge is delocalized.
3. Alcohols are stronger acids than water.
4. Alkoxides are stronger bases than hydroxide ions.

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

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

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

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

💡 Short Explanation

�� Phenoxide ion is resonance stabilized. �� Alcohols are weaker acids than water. �� Alkoxides are stronger bases than hydroxide ions.

✅ Correct Answer Explanation

Statement 1 is correct because phenol is much more acidic than ethanol due to resonance stabilization of the phenoxide ion. Statement 2 is correct because the negative charge in the phenoxide ion is delocalized over the aromatic ring. Statement 3 is incorrect because alcohols are weaker acids than water. Statement 4 is correct because alkoxide ions are stronger bases than hydroxide ions owing to the weaker acidity of alcohols. Therefore, option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Includes statement 3, which is incorrect.
�� Option C → Includes statement 3 and omits statement 1.
�� Option D → Includes statement 3, which is incorrect.

🎯 Strategy

Used

Elimination

Application:

Verify each statement using NCERT concepts on acidity and resonance.

Final Logic:

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

🧠 Memory Trick

Phenol Stable, Alkoxide Strong Base

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

6 Arrange the following in decreasing order of their acid strength based on pKa and substituent effects:
1. 2,4,6-Trinitrophenol
2. Phenol
3. 3-Nitrophenol
4. 4-Methylphenol

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

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

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

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

💡 Short Explanation

�� Nitro groups increase acidity. �� Methyl groups decrease acidity. �� Picric acid is the strongest phenol.

✅ Correct Answer Explanation

Statement 1 represents 2,4,6-trinitrophenol (picric acid), which is the strongest acid because three nitro groups strongly stabilize the phenoxide ion. Statement 2 represents phenol, which is less acidic than nitrophenols. Statement 3 represents 3-nitrophenol, whose nitro group increases acidity through the −I effect. Statement 4 represents 4-methylphenol, where the methyl group exerts a +I effect, decreasing acidity. Therefore, the decreasing order of acid strength is: 2,4,6-Trinitrophenol > 3-Nitrophenol > Phenol > 4-Methylphenol. Hence, option D is correct.

❓ Why Other Options Are Incorrect

�� Option A → Places phenol above 3-nitrophenol.
�� Option B → Gives the reverse order.
�� Option C → Places 3-nitrophenol above picric acid.

🎯 Strategy

Used

Option Grouping

Application:

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

Final Logic:

More −NO₂ → Stronger Acid; CH₃ → Weaker Acid.

🧠 Memory Trick

Three Nitro > One Nitro > Phenol > Methyl Phenol

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

7 To drive the reversible esterification equilibrium of an alcohol with a carboxylic acid to the right-hand side, what action is necessary?

A HCl gas is added

B Water is removed as soon as it is formed

C An excess of base is used

D It is carried out at low temperature

💡 Short Explanation

�� Esterification is reversible. �� Removing water shifts equilibrium forward. �� Ester yield increases.

✅ Correct Answer Explanation

The esterification of an alcohol with a carboxylic acid is a reversible reaction. According to Le Chatelier's principle, continuous removal of water shifts the equilibrium towards ester formation, thereby increasing the yield. Therefore, option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → HCl is not used to shift the equilibrium.
�� Option C → A base neutralizes the acid catalyst and suppresses esterification.
�� Option D → Low temperature does not effectively drive the equilibrium towards ester formation.

🎯 Strategy

Used

Elimination

Application:

Recall the condition used to favour ester formation.

Final Logic:

Remove Water → More Ester.

🧠 Memory Trick

Remove H₂O, Make Ester

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

8 Identify the reaction type that produces aspirin from salicylic acid.

A Acylation/Acetylation

B Halogenation

C Alkylation

D Dehydration

💡 Short Explanation

�� Salicylic acid undergoes acetylation. �� An acetyl group is introduced. �� Aspirin is produced.

✅ Correct Answer Explanation

Aspirin is prepared by acetylating salicylic acid using an acetylating agent such as acetic anhydride or acetyl chloride. The reaction introduces an acetyl (CH₃CO–) group and is known as acetylation (acylation). Therefore, option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → No halogen atom is introduced.
�� Option C → No alkyl group is introduced.
�� Option D → Water elimination is not the principal reaction.

🎯 Strategy

Used

Contextual/Tonal Matching

Application:

Identify the named reaction used to prepare aspirin.

Final Logic:

Aspirin = Acetylation.

🧠 Memory Trick

Aspirin = Acetyl

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

9 What is the IUPAC name of the product formed when tert-butyl alcohol reacts with HCl in the Lucas test?

A 2-Chloro-2-methylpropane

B 1-Chlorobutane

C 2-Chlorobutane

D 1-Chloro-2-methylpropane

💡 Short Explanation

�� tert-Butyl alcohol undergoes substitution. �� The –OH group is replaced by chlorine. �� A tertiary alkyl chloride is formed.

✅ Correct Answer Explanation

tert-Butyl alcohol reacts rapidly with concentrated HCl in the presence of ZnCl₂ (Lucas reagent). The hydroxyl group is replaced by chlorine through an SN1 mechanism, producing 2-chloro-2-methylpropane. Therefore, option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → This is the product from butan-1-ol.
�� Option C → This is obtained from butan-2-ol.
�� Option D → This is not the product of tert-butyl alcohol.

🎯 Strategy

Used

Elimination

Application:

Identify the substitution product of tert-butyl alcohol.

Final Logic:

tert-Butyl Alcohol → tert-Butyl Chloride.

🧠 Memory Trick

tert-Butyl → tert-Butyl Chloride

📚 Class XII – NCERT Chemistry Part II – Chapter: Alcohols, Phenols and Ethers – Topic: Reactions of Alcohols with Hydrogen Halides

10 Arrange the ease of cleavage of the C–O bond by hydrogen halides for the following alcohols in decreasing order:
1. 2-Methylpropan-2-ol
2. Propan-2-ol
3. Methanol
4. Propan-1-ol

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

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

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

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

💡 Short Explanation

�� Tertiary alcohols react fastest. �� Secondary alcohols react more slowly. �� Primary alcohols react slower than secondary alcohols.

✅ Correct Answer Explanation

Statement 1 represents 2-methylpropan-2-ol, a tertiary alcohol that undergoes C–O bond cleavage most readily through the SN1 mechanism. Statement 2 represents propan-2-ol, a secondary alcohol with intermediate reactivity. Statement 3 represents methanol, which shows the least tendency for C–O bond cleavage among these alcohols. Statement 4 represents propan-1-ol, a primary alcohol that reacts more slowly. Therefore, the decreasing order of reactivity is: 2-Methylpropan-2-ol > Propan-2-ol > Propan-1-ol > Methanol. Hence, option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Gives the reverse order.
�� Option C → Places secondary alcohol above tertiary alcohol.
�� Option D → Places primary alcohol above secondary alcohol.

🎯 Strategy

Used

Option Grouping

Application:

Arrange alcohols according to carbocation stability and ease of substitution with hydrogen halides.

Final Logic:

3° > 2° > 1° > Methyl Alcohol.

🧠 Memory Trick

3° Fast → 2° Medium → 1° Slow → CH₃OH Slowest

📚 Class XII – NCERT Chemistry Part II – Chapter: Alcohols, Phenols and Ethers – Topic: Reactions of Alcohols with Hydrogen Halides

11 Statements about dehydration of alcohols:
1. Tertiary alcohols are the easiest to dehydrate.
2. The ease of dehydration follows: Primary > Secondary > Tertiary.
3. The rate-determining step forms a carbocation.
4. Protic acids like H₂SO₄ or H₃PO₄ are used.

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

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

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

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

💡 Short Explanation

�� Tertiary alcohols dehydrate most readily. �� Carbocation formation is the slow step. �� Protic acids catalyse dehydration.

✅ Correct Answer Explanation

Statement 1 is correct because tertiary alcohols form the most stable carbocations and therefore undergo dehydration most easily. Statement 2 is incorrect because the correct order is: Tertiary > Secondary > Primary. Statement 3 is correct because carbocation formation is the rate-determining step in acid-catalysed dehydration. Statement 4 is correct because concentrated H₂SO₄ or H₃PO₄ acts as the acid catalyst. Therefore, option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Includes statement 2, which is incorrect.
�� Option C → Includes statement 2, which is incorrect.
�� Option D → Includes statement 2, which is incorrect.

🎯 Strategy

Used

Elimination

Application:

Verify each statement using the NCERT dehydration mechanism.

Final Logic:

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

🧠 Memory Trick

3° First, Carbocation Next

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

12 In the mechanism of dehydration of ethanol, how is the equilibrium driven to the right?

A By adding more acid catalyst

B By removing ethene as it is formed

C By increasing the pressure

D By adding water

💡 Short Explanation

�� Dehydration is reversible. �� Ethene escapes from the reaction mixture. �� Equilibrium shifts towards products.

✅ Correct Answer Explanation

During acid-catalysed dehydration, ethene is continuously removed as it is formed because it is a gas. According to Le Chatelier's principle, removal of the product shifts the equilibrium towards further formation of ethene. Therefore, option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → More catalyst increases the rate but does not shift equilibrium.
�� Option C → Increased pressure opposes gaseous product formation.
�� Option D → Addition of water shifts equilibrium towards alcohol.

🎯 Strategy

Used

Elimination

Application:

Apply Le Chatelier's principle to the reaction.

Final Logic:

Remove Product → More Product Forms.

🧠 Memory Trick

Gas Leaves, Reaction Proceeds

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

13 Arrange the sequence of products formed during the strong oxidation of a primary alcohol:
1. Aldehyde
2. Primary alcohol
3. Carboxylic acid
4. Strong oxidising agent (KMnO₄/K₂Cr₂O₇) completes the oxidation.

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

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

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

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

💡 Short Explanation

�� Primary alcohol first forms an aldehyde. �� Further oxidation gives a carboxylic acid. �� Strong oxidants complete the conversion.

✅ Correct Answer Explanation

Statement 1 represents the primary alcohol. Statement 2 represents the aldehyde formed during the first stage of oxidation. Statement 3 represents the carboxylic acid obtained after further oxidation. Statement 4 is a correct NCERT fact indicating that strong oxidising agents such as KMnO₄ or K₂Cr₂O₇ complete the oxidation to the acid. Therefore, the sequence is: Primary alcohol → Aldehyde → Carboxylic acid. Hence, option C is correct.

❓ Why Other Options Are Incorrect

�� Option A → Begins with the intermediate.
�� Option B → Reverses the oxidation sequence.
�� Option D → Places carboxylic acid before aldehyde.

🎯 Strategy

Used

Option Grouping

Application:

Arrange compounds according to the oxidation pathway.

Final Logic:

Alcohol → Aldehyde → Acid.

🧠 Memory Trick

Alcohol → Aldehyde → Acid

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

14 When tertiary alcohols are subjected to strong oxidising agents (KMnO₄) at elevated temperatures, what happens?

A They form secondary alcohols.

B They form ketones with the same number of carbons.

C Cleavage of C–C bonds takes place forming a mixture of carboxylic acids.

D They are completely unreactive.

💡 Short Explanation

�� Tertiary alcohols lack α-hydrogen. �� Normal oxidation does not occur. �� Strong oxidation cleaves C–C bonds.

✅ Correct Answer Explanation

Tertiary alcohols do not undergo ordinary oxidation because they lack a hydrogen atom on the carbon bearing the hydroxyl group. Under vigorous oxidation with strong oxidising agents at high temperatures, cleavage of carbon-carbon bonds occurs, producing smaller carboxylic acids. Therefore, option C is correct.

❓ Why Other Options Are Incorrect

�� Option A → Oxidation cannot produce secondary alcohols.
�� Option B → Ketones are formed from secondary alcohols.
�� Option D → Under vigorous conditions they undergo C–C bond cleavage.

🎯 Strategy

Used

Elimination

Application:

Recall the oxidation behaviour of tertiary alcohols.

Final Logic:

No α-H → C–C Cleavage.

🧠 Memory Trick

3° Alcohol = Break Carbon Chain

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

15 Match List-I (Reaction Condition) with List-II (Product)

List I	List II
1. Dilute HNO₃ at 298 K	a. Picric acid (poor yield)
2. Concentrated HNO₃	b. Picric acid (high yield industrially)
3. Concentrated H₂SO₄ followed by concentrated HNO₃	c. Mixture of o- and p-nitrophenols
4. Bromine water	d. 2,4,6-Tribromophenol

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

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

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

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

💡 Short Explanation

�� Dilute nitric acid gives mono-nitrophenols. �� Concentrated nitric acid forms picric acid. �� Sulphonation followed by nitration improves picric acid yield.

✅ Correct Answer Explanation

1 → c : Dilute nitric acid at 298 K produces a mixture of ortho- and para-nitrophenols. 2 → a : Direct nitration with concentrated nitric acid gives picric acid in poor yield because of oxidation side reactions. 3 → b : Sulphonation followed by nitration is the industrial method for obtaining picric acid in high yield. 4 → d : Bromine water reacts with phenol to give 2,4,6-tribromophenol as a white precipitate. Therefore, option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Dilute and concentrated nitration products are interchanged.
�� Option C → Multiple incorrect pairings are present.
�� Option D → Concentrated nitric acid and industrial nitration are interchanged.

🎯 Strategy

Used

Option Grouping

Application:

Match each reaction condition with its characteristic product.

Final Logic:

Dilute → Mono Nitro; Conc. → Picric; Sulphonation + Nitration → High Yield; Br₂ Water → Tribromo.

🧠 Memory Trick

Dilute–Mono, Conc.–Picric, Br₂–Tribromo

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

16 The reaction of phenol with bromine in CS₂ at low temperature yields monobromophenols. The temperature at which dilute nitric acid reacts with phenol is:

A 298 K

B 337 K

C 443 K

D 573 K

💡 Short Explanation

�� Dilute nitric acid reacts at room temperature. �� The reaction produces o- and p-nitrophenols. �� The temperature is 298 K.

✅ Correct Answer Explanation

According to NCERT, phenol reacts with dilute nitric acid at 298 K to produce a mixture of ortho- and para-nitrophenols. This mild temperature favours mononitration of the activated aromatic ring. Therefore, option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → 337 K is not the NCERT condition for nitration of phenol.
�� Option C → 443 K is associated with dehydration reactions of alcohols.
�� Option D → 573 K is used in oxidation/dehydrogenation reactions of alcohols.

🎯 Strategy

Used

Elimination

Application:

Recall the standard NCERT temperature for nitration of phenol.

Final Logic:

Dilute HNO₃ + Phenol → 298 K.

🧠 Memory Trick

Nitration = 298 K

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

17 Statements regarding Kolbe's reaction:
1. Phenoxide ion is more reactive than phenol towards electrophilic substitution.
2. Carbon dioxide acts as a weak electrophile.
3. Ortho hydroxybenzoic acid is formed as the main product.
4. Chloroform is used as a reagent.

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

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

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

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

💡 Short Explanation

�� Phenoxide ion is highly activated. �� CO₂ behaves as a weak electrophile. �� Salicylic acid is the major product.

✅ Correct Answer Explanation

Statement 1 is correct because the phenoxide ion has greater electron density than phenol and undergoes electrophilic substitution more readily. Statement 2 is correct because carbon dioxide acts as a weak electrophile in Kolbe's reaction. Statement 3 is correct because the major product obtained after acidification is ortho hydroxybenzoic acid (salicylic acid). Statement 4 is incorrect because chloroform is used in the Reimer–Tiemann reaction, not in Kolbe's reaction. Therefore, option D is correct.

❓ Why Other Options Are Incorrect

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

🎯 Strategy

Used

Elimination

Application:

Verify each statement using the NCERT description of Kolbe's reaction.

Final Logic:

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

🧠 Memory Trick

Kolbe = CO₂, Reimer = CHCl₃

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

18 In the Reimer-Tiemann reaction, what is the intermediate formed before hydrolysis to salicylaldehyde?

A Phenoxide ion

B Substituted benzal chloride

C Benzoquinone

D Ortho hydroxybenzoic acid

💡 Short Explanation

�� Chloroform reacts in alkaline medium. �� A substituted benzal chloride intermediate is formed. �� Hydrolysis gives salicylaldehyde.

✅ Correct Answer Explanation

In the Reimer–Tiemann reaction, phenol reacts with chloroform and sodium hydroxide to produce a substituted benzal chloride intermediate. Hydrolysis of this intermediate yields salicylaldehyde. Therefore, option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Phenoxide ion is the reactant, not the intermediate before hydrolysis.
�� Option C → Benzoquinone is not formed in this reaction.
�� Option D → Ortho-hydroxybenzoic acid is the product of Kolbe's reaction.

🎯 Strategy

Used

Contextual/Tonal Matching

Application:

Identify the characteristic intermediate of the Reimer–Tiemann reaction.

Final Logic:

CHCl₃ → Benzal Chloride → Salicylaldehyde.

🧠 Memory Trick

Reimer = Benzal Chloride

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

19 Arrange the following hydrogen halides in decreasing order of their reactivity towards the cleavage of ethers:
1. HF
2. HBr
3. HCl
4. HI

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

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

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

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

💡 Short Explanation

�� HI is the most reactive. �� HBr is less reactive than HI. �� HCl and HF are much less effective.

✅ Correct Answer Explanation

Statement 1 represents HF, which is the least reactive because of its weak nucleophilicity. Statement 2 represents HBr, which is less reactive than HI but more reactive than HCl. Statement 3 represents HCl, which is comparatively ineffective for ether cleavage. Statement 4 represents HI, which is the most effective hydrogen halide for ether cleavage because iodide is a strong nucleophile and HI is a strong acid. Therefore, the decreasing order is: HI > HBr > HCl > HF. Hence, option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Places HBr above HI.
�� Option C → Gives the reverse trend.
�� Option D → Incorrectly places HBr and HCl.

🎯 Strategy

Used

Option Grouping

Application:

Arrange hydrogen halides according to their effectiveness in ether cleavage.

Final Logic:

HI > HBr > HCl > HF.

🧠 Memory Trick

I > Br > Cl > F

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

20 Identify the reaction type when anisole reacts with an alkyl halide in the presence of anhydrous aluminium chloride.

A Friedel-Crafts alkylation

B Reimer-Tiemann reaction

C Williamson synthesis

D Kolbe's reaction

💡 Short Explanation

�� Anisole activates the benzene ring. �� Alkyl halide provides the electrophile. �� Alkylation occurs on the aromatic ring.

✅ Correct Answer Explanation

Anisole undergoes electrophilic aromatic substitution with alkyl halides in the presence of anhydrous AlCl₃. This reaction is known as Friedel-Crafts alkylation, in which an alkyl group is introduced mainly at the ortho and para positions. Therefore, option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Reimer–Tiemann reaction introduces a –CHO group using CHCl₃ and NaOH.
�� Option C → Williamson synthesis prepares ethers using alkoxides and alkyl halides.
�� Option D → Kolbe's reaction introduces a –COOH group using CO₂.

🎯 Strategy

Used

Contextual/Tonal Matching

Application:

Identify the named reaction from the reagents used.

Final Logic:

Alkyl Halide + AlCl₃ = Friedel-Crafts Alkylation.

🧠 Memory Trick

Alkyl + AlCl₃ = Friedel Alkylation

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

CUET UG Chemistry Booster Test – 3 Chemical Reactions and Uses

CUET UG Chemistry Booster Test - 3 Chemical Reactions and Uses

Test Complete!

Answer Review