CUET UG Chemistry Booster Test - 3 Preparation Methods -

🚀

CUET UG Chemistry Booster Test - 3 Preparation Methods

⏱️ 30 Minutes

❓ 20 Questions

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

CUET UG Chemistry Booster Test - 3 Preparation Methods

1 / 20

QUESTION 1 OF 20

Arrange the following carbocations formed during acid hydration of alkenes in decreasing order of stability:
1. Primary carbocation
2. Tertiary carbocation
3. Secondary carbocation
4. Methyl carbocation

Answer Review

1 Arrange the following carbocations formed during acid hydration of alkenes in decreasing order of stability:
1. Primary carbocation
2. Tertiary carbocation
3. Secondary carbocation
4. Methyl carbocation

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

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

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

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

💡 Short Explanation

�� Carbocation stability determines the reaction pathway. �� Alkyl groups stabilize carbocations through the +I effect and hyperconjugation. �� Tertiary carbocations are the most stable.

✅ Correct Answer Explanation

Statement 1 represents a primary carbocation, which has low stability due to minimal electron-donating alkyl groups. Statement 2 represents a tertiary carbocation, which is the most stable because three alkyl groups stabilize the positively charged carbon through hyperconjugation and the +I effect. Statement 3 represents a secondary carbocation, which has intermediate stability. Statement 4 represents a methyl carbocation, which is the least stable because it has no alkyl groups to stabilize the positive charge. Therefore, the decreasing order of stability is: Tertiary carbocation > Secondary carbocation > Primary carbocation > Methyl carbocation Hence, option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Incorrectly places the primary carbocation above the tertiary carbocation.
�� Option C → Places the secondary carbocation above the tertiary carbocation and the primary carbocation above the tertiary carbocation.
�� Option D → Incorrectly places the primary carbocation above the secondary carbocation.

🎯 Strategy

Used

Option Grouping

Application:

Arrange the carbocations according to hyperconjugation and the +I effect.

Final Logic:

3° > 2° > 1° > CH₃⁺.

🧠 Memory Trick

3° Wins, CH₃ Loses

📚 Class XII – NCERT Chemistry Part II – Chapter: Alcohols, Phenols and Ethers – Topic: Acid-Catalysed Hydration of Alkenes (Carbocation Stability)

2 Which of the following statements are correct regarding the mechanism of acid-catalysed hydration?
1. The reaction skips the carbocation intermediate.
2. Nucleophilic attack of water occurs on the carbocation formed.
3. The final step is deprotonation to form an alcohol.
4. Protonation of the alkene is the first step of the mechanism.

A (1), (2)

B (3), (4)

C (2), (3)

D (1), (3)

💡 Short Explanation

�� Protonation forms a carbocation. �� Water attacks the carbocation. �� Deprotonation produces the alcohol.

✅ Correct Answer Explanation

Statement 1 is incorrect because acid-catalysed hydration proceeds through a carbocation intermediate. Statement 2 is correct because water attacks the carbocation to form a protonated alcohol. Statement 3 is correct because the final step is deprotonation, yielding the alcohol. Statement 4 is a correct NCERT statement but is not included in the correct option provided. Based on the given options, the correct combination is statements 2 and 3. Therefore, option C is correct.

❓ Why Other Options Are Incorrect

�� Option A → Includes statement 1, which is incorrect.
�� Option B → Omits statement 2, an essential mechanistic step.
�� Option D → Includes statement 1, which is incorrect.

🎯 Strategy

Used

Elimination

Application:

Verify each mechanistic step of acid-catalysed hydration.

Final Logic:

Water attacks the carbocation, followed by deprotonation.

🧠 Memory Trick

Attack → Deprotonate

📚 Class XII – NCERT Chemistry Part II – Chapter: Alcohols, Phenols and Ethers – Topic: Acid-Catalysed Hydration of Alkenes

3 Match List-I (Alkene Reaction) with List-II (Specific Regiochemical Outcome)

List I	List II
1. Acid-catalysed hydration	a. Uses H₂O₂/NaOH
2. Hydroboration-oxidation	b. Boron attaches to the less substituted sp² carbon
3. Diborane addition step	c. Markovnikov addition of water
4. Oxidation of trialkylborane	d. Overall anti-Markovnikov alcohol formation

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

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

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

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

💡 Short Explanation

�� Acid hydration follows Markovnikov addition. �� Hydroboration gives anti-Markovnikov alcohols. �� Oxidation requires H₂O₂/NaOH.

✅ Correct Answer Explanation

1 → c : Acid-catalysed hydration follows Markovnikov addition of water. 2 → d : Hydroboration-oxidation produces anti-Markovnikov alcohols. 3 → b : During hydroboration, boron attaches to the less substituted carbon atom. 4 → a : Trialkylborane is oxidised using H₂O₂/NaOH. Therefore, option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Acid hydration and oxidation conditions are incorrectly matched.
�� Option C → Multiple reactions are incorrectly paired.
�� Option D → Diborane addition and oxidation are incorrectly matched.

🎯 Strategy

Used

Option Grouping

Application:

Associate each reaction with its characteristic regiochemical outcome.

Final Logic:

Markovnikov → Acid Hydration; Anti-Markovnikov → Hydroboration.

🧠 Memory Trick

Acid = Markovnikov; Borane = Anti-Markovnikov

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

4 Why does hydroboration-oxidation ultimately yield an alcohol that appears to violate Markovnikov's rule?

A The boron atom acts as a nucleophile.

B Boron attaches to the carbon with more hydrogens, and is later replaced by an OH group.

C Hydrogen peroxide reverses the bond polarity.

D Carbocation rearrangement strictly prevents it.

💡 Short Explanation

�� Hydroboration is an anti-Markovnikov addition. �� Boron adds to the less substituted carbon. �� Oxidation replaces boron with –OH without changing its position.

✅ Correct Answer Explanation

During hydroboration, boron preferentially attaches to the less substituted carbon atom, which carries the greater number of hydrogen atoms. In the oxidation step using H₂O₂/NaOH, the C–B bond is replaced by a C–OH bond without altering the carbon skeleton. As a result, the hydroxyl group finally appears on the less substituted carbon, giving an anti-Markovnikov alcohol. Therefore, option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → The alkyl group of the borane is not acting as the nucleophile in determining regioselectivity.
�� Option C → Hydrogen peroxide replaces boron with –OH but does not reverse bond polarity.
�� Option D → No carbocation intermediate is formed in hydroboration; therefore, rearrangement does not occur.

🎯 Strategy

Used

Elimination

Application:

Recall the mechanism of hydroboration followed by oxidation.

Final Logic:

Boron attaches first to the carbon with more hydrogens and is then replaced by –OH.

🧠 Memory Trick

Boron First, OH Later

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

5 Between catalytic hydrogenation and lithium aluminium hydride (LiAlH₄), why is the latter generally reserved only for preparing special chemicals?

A LiAlH₄ is highly toxic and gaseous.

B LiAlH₄ cannot reduce carboxylic acids.

C LiAlH₄ is an expensive reagent.

D LiAlH₄ produces poor yields.

💡 Short Explanation

�� LiAlH₄ is a powerful reducing agent. �� It is costly for large-scale use. �� Commercial processes prefer catalytic hydrogenation.

✅ Correct Answer Explanation

Lithium aluminium hydride (LiAlH₄) efficiently reduces aldehydes, ketones and carboxylic acids to alcohols. However, because it is an expensive reagent, it is generally reserved for laboratory synthesis and the preparation of special chemicals. Industrially, catalytic hydrogenation is preferred due to its lower cost. Therefore, option C is correct.

❓ Why Other Options Are Incorrect

�� Option A → LiAlH₄ is not a gaseous reagent.
�� Option B → LiAlH₄ readily reduces carboxylic acids.
�� Option D → LiAlH₄ generally gives excellent yields.

🎯 Strategy

Used

Elimination

Application:

Compare laboratory and industrial reducing agents.

Final Logic:

Effective but expensive → Used mainly for special chemicals.

🧠 Memory Trick

LiAlH₄ = Strong but Costly

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

6 Regarding ketone reduction, consider the following statements:
1. Ketones are reduced to primary alcohols.
2. Ketones can be reduced using hydrogen in the presence of finely divided Pt, Pd or Ni.
3. NaBH₄ and LiAlH₄ can also reduce ketones.
4. Ketones yield secondary alcohols upon reduction.

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

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

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

D (3), (4)

💡 Short Explanation

�� Ketones produce secondary alcohols. �� Catalytic hydrogenation reduces ketones. �� NaBH₄ and LiAlH₄ are effective reducing agents.

✅ Correct Answer Explanation

Statement 1 is incorrect because ketones are reduced to secondary alcohols, not primary alcohols. Statement 2 is correct because catalytic hydrogenation using finely divided Pt, Pd or Ni converts ketones into secondary alcohols. Statement 3 is correct because both sodium borohydride (NaBH₄) and lithium aluminium hydride (LiAlH₄) reduce ketones efficiently. Statement 4 is correct because the product obtained from ketone reduction is always a secondary alcohol. Therefore, option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Includes statement 1, which is incorrect.
�� Option C → Includes statement 1, which is incorrect.
�� Option D → Omits statement 2, which is also correct.

🎯 Strategy

Used

Elimination

Application:

Verify each statement using NCERT reduction reactions of ketones.

Final Logic:

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

🧠 Memory Trick

Ketone → 2° Alcohol

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

7 During the commercial conversion of acids to alcohols, acids are first converted into esters. What is the subsequent step?

A Reduction using hydrogen in the presence of a catalyst.

B Treatment with LiAlH₄

C Oxidation with KMnO₄

D Reaction with Grignard reagent

💡 Short Explanation

�� Direct use of LiAlH₄ is expensive. �� Acids are first converted into esters. �� Catalytic hydrogenation then produces alcohols.

✅ Correct Answer Explanation

Commercially, carboxylic acids are first converted into esters because direct reduction with LiAlH₄ is expensive. The esters are then reduced using hydrogen in the presence of a suitable catalyst to obtain the corresponding alcohols economically. Therefore, option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → LiAlH₄ is generally avoided in commercial production because of its high cost.
�� Option C → Oxidation does not convert esters into alcohols.
�� Option D → Grignard reagents are not used in this commercial process.

🎯 Strategy

Used

Elimination

Application:

Recall the industrial method for reducing carboxylic acids.

Final Logic:

Acid → Ester → Catalytic Hydrogenation → Alcohol.

🧠 Memory Trick

Acid → Ester → H₂ → Alcohol

📚 Class XII – NCERT Chemistry Part II – Chapter: Alcohols, Phenols and Ethers – Topic: Commercial Preparation of Alcohols from Carboxylic Acids

8 The transformation sequence RCOOH → RCOOR' → RCH₂OH + R'OH utilizes which type of reaction in its second step?

A Esterification

B Catalytic hydrogenation (Ester reduction)

C Nucleophilic addition

D Decarboxylation

💡 Short Explanation

�� Carboxylic acids are first converted into esters. �� Esters are then reduced by catalytic hydrogenation. �� Alcohols are obtained in the final step.

✅ Correct Answer Explanation

In the commercial preparation of alcohols from carboxylic acids, the acids are first converted into esters by esterification. The esters are then reduced using hydrogen in the presence of a suitable catalyst. This catalytic hydrogenation converts the ester into the corresponding primary alcohol along with the alcohol derived from the alkoxy group. Therefore, option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Esterification is the first step, not the second.
�� Option C → The second step is reduction, not nucleophilic addition.
�� Option D → No decarboxylation occurs in this sequence.

🎯 Strategy

Used

Contextual/Tonal Matching

Application:

Identify the reaction occurring in the second step of the sequence.

Final Logic:

Acid → Ester → Catalytic Hydrogenation → Alcohol.

🧠 Memory Trick

Ester + H₂ = Alcohol

📚 Class XII – NCERT Chemistry Part II – Chapter: Alcohols, Phenols and Ethers – Topic: Commercial Preparation of Alcohols from Carboxylic Acids

📄 Passage

Alcohols are produced by the reaction of Grignard reagents with aldehydes and ketones. The first step of the reaction is the nucleophilic addition of Grignard reagent to the carbonyl group to form an adduct. Hydrolysis of the adduct yields an alcohol. The reaction produces a primary alcohol with methanal, a secondary alcohol with other aldehydes and a tertiary alcohol with ketones.

Which type of fundamental reaction defines the first step of this preparation method?

A Electrophilic substitution

B Nucleophilic substitution

C Nucleophilic addition

D Radical addition

💡 Short Explanation

�� Grignard reagents act as nucleophiles. �� They attack the carbonyl carbon. �� An alkoxide adduct is formed first.

✅ Correct Answer Explanation

As stated in the passage, the first step of the Grignard reaction is the nucleophilic addition of the alkyl group of the Grignard reagent to the electrophilic carbonyl carbon. This produces an alkoxide (adduct), which on hydrolysis forms the corresponding alcohol. Therefore, option C is correct.

❓ Why Other Options Are Incorrect

�� Option A → No electrophilic substitution occurs.
�� Option B → The reaction is addition, not substitution.
�� Option D → No free-radical intermediate is involved.

🎯 Strategy

Used

Contextual/Tonal Matching

Application:

Identify the reaction type directly stated in the passage.

Final Logic:

Grignard attacks the carbonyl by nucleophilic addition.

🧠 Memory Trick

Grignard = Nucleophilic Addition

📚 Class XII – NCERT Chemistry Part II – Chapter: Alcohols, Phenols and Ethers – Topic: Preparation of Alcohols Using Grignard Reagents

📄 Passage

If the adduct formed from an unbranched aldehyde (other than methanal) undergoes the hydrolysis step, the final product is strictly a:

A Primary alcohol

B Carboxylic acid

C Tertiary alcohol

D Secondary alcohol

💡 Short Explanation

�� Methanal gives primary alcohols. �� Other aldehydes give secondary alcohols. �� Ketones give tertiary alcohols.

✅ Correct Answer Explanation

The passage clearly states that when Grignard reagents react with aldehydes other than methanal, the nucleophilic addition product (alkoxide) undergoes hydrolysis to produce a secondary alcohol. This is because the carbonyl carbon becomes bonded to two carbon groups after the addition reaction. Therefore, option D is correct.

❓ Why Other Options Are Incorrect

�� Option A → Primary alcohols are obtained only from methanal.
�� Option B → Hydrolysis produces an alcohol, not a carboxylic acid.
�� Option C → Tertiary alcohols are produced from ketones.

🎯 Strategy

Used

Contextual/Tonal Matching

Application:

Relate the type of carbonyl compound to the alcohol formed after the Grignard reaction.

Final Logic:

Other Aldehydes → Secondary Alcohol.

🧠 Memory Trick

Methanal = 1°, Aldehyde = 2°, Ketone = 3°

📚 Class XII – NCERT Chemistry Part II – Chapter: Alcohols, Phenols and Ethers – Topic: Preparation of Alcohols Using Grignard Reagents

11 Why are drastic conditions (623 K, 320 atm) required to fuse chlorobenzene with NaOH in the preparation of phenol?

A The C–Cl bond in chlorobenzene has partial double bond character due to resonance.

B Chlorobenzene is highly soluble in NaOH.

C NaOH is a weak nucleophile.

D The reaction is highly exothermic and needs containment.

💡 Short Explanation

�� Resonance strengthens the C–Cl bond. �� The bond acquires partial double bond character. �� High temperature and pressure are needed for substitution.

✅ Correct Answer Explanation

In chlorobenzene, the lone pair of chlorine participates in resonance with the benzene ring, giving the C–Cl bond partial double bond character. This makes the bond shorter and stronger than a normal C–Cl bond, making nucleophilic substitution difficult. Therefore, chlorobenzene must be fused with aqueous NaOH at 623 K and 320 atm to produce sodium phenoxide, which is later acidified to form phenol. Therefore, option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Chlorobenzene is not highly soluble in aqueous NaOH.
�� Option C → Hydroxide ion is a strong nucleophile.
�� Option D → The drastic conditions are required because of bond strength, not because the reaction is highly exothermic.

🎯 Strategy

Used

Elimination

Application:

Identify the structural reason for the low reactivity of chlorobenzene.

Final Logic:

Resonance strengthens the C–Cl bond, requiring harsh reaction conditions.

🧠 Memory Trick

Resonance = Strong C–Cl

📚 Class XII – NCERT Chemistry Part II – Chapter: Alcohols, Phenols and Ethers – Topic: Preparation of Phenol from Chlorobenzene (Dow's Process)

12 In the preparation of phenol from haloarenes, what happens immediately upon acidification of the reaction mixture after fusion?

A Chlorobenzene regenerates.

B Sodium phenoxide is converted to phenol.

C Benzene is produced.

D Anisole is formed.

💡 Short Explanation

�� Fusion with NaOH forms sodium phenoxide. �� Acidification converts sodium phenoxide into phenol. �� This is the final step of Dow's process.

✅ Correct Answer Explanation

During the preparation of phenol from chlorobenzene, fusion with aqueous sodium hydroxide at 623 K and 320 atm produces sodium phenoxide. On acidification with a dilute mineral acid, sodium phenoxide is protonated to give phenol. Therefore, option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Chlorobenzene is not regenerated.
�� Option C → Benzene is not formed during acidification.
�� Option D → Anisole is not produced in this reaction.

🎯 Strategy

Used

Elimination

Application:

Recall the final step of Dow's process.

Final Logic:

Sodium Phenoxide + Acid → Phenol.

🧠 Memory Trick

Phenoxide + H⁺ = Phenol

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

13 When preparing phenol via the sulphonic acid method, the benzene ring is first treated with oleum. What is the active functional group introduced to the benzene ring before NaOH fusion?

A –OH

B –NO₂

C –Cl

D –SO₃H (Sulphonic acid group)

💡 Short Explanation

�� Oleum sulphonates benzene. �� The sulphonic acid group is introduced. �� Alkali fusion later replaces this group with –OH.

✅ Correct Answer Explanation

Oleum (fuming sulphuric acid) sulphonates benzene to form benzene sulphonic acid by introducing the –SO₃H functional group. During subsequent fusion with molten sodium hydroxide, this group is replaced by –OH to produce sodium phenoxide, which upon acidification gives phenol. Therefore, option D is correct.

❓ Why Other Options Are Incorrect

�� Option A → The hydroxyl group is introduced only after alkali fusion.
�� Option B → Nitration introduces the –NO₂ group, not sulphonation.
�� Option C → Chlorination is not carried out in this method.

🎯 Strategy

Used

Elimination

Application:

Identify the functional group introduced during sulphonation.

Final Logic:

Oleum introduces the –SO₃H group.

🧠 Memory Trick

Oleum = SO₃H

📚 Class XII – NCERT Chemistry Part II – Chapter: Alcohols, Phenols and Ethers – Topic: Preparation of Phenol from Benzene Sulphonic Acid

14 What is the name of the intermediate salt that is hydrolysed by warming with water or treating with dilute acids to prepare phenol?

A Benzene diazonium chloride

B Sodium phenoxide

C Cumene hydroperoxide

D Benzenesulphonic acid

💡 Short Explanation

�� Aromatic primary amines form diazonium salts. �� The diazonium salt undergoes hydrolysis. �� Phenol is obtained after evolution of nitrogen.

✅ Correct Answer Explanation

Aromatic primary amines first undergo diazotisation to form benzene diazonium chloride. On warming with water or treatment with dilute acids, the diazonium group is replaced by a hydroxyl group to produce phenol with the evolution of nitrogen gas. Therefore, option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Sodium phenoxide is obtained in the Dow process, not by diazotisation.
�� Option C → Cumene hydroperoxide is the intermediate in the cumene process.
�� Option D → Benzenesulphonic acid is an intermediate in the sulphonic acid method.

🎯 Strategy

Used

Contextual/Tonal Matching

Application:

Identify the intermediate involved in the diazotisation method.

Final Logic:

Diazonium Salt + H₂O → Phenol.

🧠 Memory Trick

Diazonium → Phenol

📚 Class XII – NCERT Chemistry Part II – Chapter: Alcohols, Phenols and Ethers – Topic: Preparation of Phenol from Diazonium Salts

15 Arrange the steps of the industrial preparation of phenol from cumene in the correct sequence:
1. Treatment with dilute acid
2. Oxidation in the presence of air
3. Formation of cumene hydroperoxide
4. Formation of phenol and acetone

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

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

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

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

💡 Short Explanation

�� Cumene is first oxidised. �� Cumene hydroperoxide is formed. �� Acid treatment yields phenol and acetone.

✅ Correct Answer Explanation

Statement 2 is the first step in which cumene undergoes oxidation in the presence of air. Statement 3 follows, producing cumene hydroperoxide. Statement 1 is the next step, where cumene hydroperoxide is treated with dilute acid. Statement 4 is the final outcome, producing phenol and acetone. Therefore, the correct sequence is: Oxidation in air → Cumene hydroperoxide → Treatment with dilute acid → Formation of phenol and acetone. Hence, option C is correct.

❓ Why Other Options Are Incorrect

�� Option A → Begins with acid treatment before oxidation.
�� Option B → Places cumene hydroperoxide formation before oxidation.
�� Option D → Acid treatment cannot occur before cumene hydroperoxide formation.

🎯 Strategy

Used

Option Grouping

Application:

Arrange the industrial steps of the cumene process in chronological order.

Final Logic:

Air Oxidation → Hydroperoxide → Acid Treatment → Phenol + Acetone.

🧠 Memory Trick

Air → Peroxide → Acid → Phenol

📚 Class XII – NCERT Chemistry Part II – Chapter: Alcohols, Phenols and Ethers – Topic: Industrial Preparation of Phenol (Cumene Process)

16 Match List-I (Compounds in the Cumene Process) with List-II (Descriptions)

List I	List II
1. Cumene	a. Main product
2. Cumene hydroperoxide	b. Important large-quantity by-product
3. Acetone	c. Starting hydrocarbon
4. Phenol	d. Intermediate formed by air oxidation

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

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

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

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

💡 Short Explanation

�� Cumene is the starting material. �� Cumene hydroperoxide is the intermediate. �� Phenol is the main product, while acetone is the by-product.

✅ Correct Answer Explanation

1 → c : Cumene (isopropylbenzene) is the starting hydrocarbon. 2 → d : Air oxidation converts cumene into cumene hydroperoxide. 3 → b : Acetone is obtained as an important by-product in large quantities. 4 → a : Phenol is the principal product of the cumene process. Therefore, option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Starting material and intermediate are interchanged.
�� Option C → All major compounds are incorrectly matched.
�� Option D → Intermediate and main product are incorrectly matched.

🎯 Strategy

Used

Option Grouping

Application:

Match each compound with its role in the industrial cumene process.

Final Logic:

Cumene → Hydroperoxide → Phenol + Acetone.

🧠 Memory Trick

Start–Intermediate–By-product–Product

📚 Class XII – NCERT Chemistry Part II – Chapter: Alcohols, Phenols and Ethers – Topic: Industrial Preparation of Phenol (Cumene Process)

17 In the preparation of ethers by acid dehydration of primary alcohols, what kind of substitution characterizes the reaction?

A Unimolecular nucleophilic substitution (SN1)

B Bimolecular nucleophilic substitution (SN2)

C Electrophilic aromatic substitution

D Bimolecular elimination (E2)

💡 Short Explanation

�� Ether formation occurs through substitution. �� Primary alcohols favour SN2. �� One alcohol attacks a protonated alcohol molecule.

✅ Correct Answer Explanation

During acid-catalysed dehydration at 413 K, one alcohol molecule attacks a protonated alcohol molecule through a bimolecular nucleophilic substitution (SN2) mechanism to form an ether. This method is suitable mainly for primary alcohols because steric hindrance is minimal. Therefore, option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → SN1 is not the mechanism for ether formation from primary alcohols.
�� Option C → No aromatic substitution is involved.
�� Option D → Elimination predominates at higher temperatures and with secondary or tertiary alcohols.

🎯 Strategy

Used

Contextual/Tonal Matching

Application:

Identify the mechanism involved in ether formation.

Final Logic:

Primary Alcohol + Protonated Alcohol → SN2 → Ether.

🧠 Memory Trick

Primary Alcohol = SN2 Ether

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

18 At what specific temperature value (in Kelvin) does the acid dehydration of ethanol shift from producing ethoxyethane to producing ethene?

A 298 K

B 351 K

C 413 K

D 443 K

💡 Short Explanation

�� Ether formation occurs at 413 K. �� Higher temperature favours elimination. �� Ethene is obtained at 443 K.

✅ Correct Answer Explanation

When ethanol is heated with concentrated sulphuric acid at 413 K, intermolecular dehydration produces ethoxyethane. Increasing the temperature to 443 K favours intramolecular dehydration, resulting in the formation of ethene. Thus, the shift from ether formation to alkene formation occurs at 443 K. Therefore, option D is correct.

❓ Why Other Options Are Incorrect

�� Option A → Too low for dehydration.
�� Option B → Not the NCERT temperature for either product.
�� Option C → Produces ethoxyethane, not ethene.

🎯 Strategy

Used

Elimination

Application:

Recall the temperatures associated with ethanol dehydration.

Final Logic:

413 K → Ether; 443 K → Ethene.

🧠 Memory Trick

413 = Ether; 443 = Ethene

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

19 Write the appropriate reagents for the Williamson synthesis of tert-butyl ethyl ether to avoid elimination reactions.

A Ethyl bromide and Sodium tert-butoxide

B Tert-butyl bromide and Sodium ethoxide

C Tert-butyl alcohol and Ethyl alcohol

D Ethene and tert-butanol

💡 Short Explanation

�� Primary alkyl halides favour SN2. �� Tertiary alkyl halides undergo elimination. �� Use the bulky group as the alkoxide.

✅ Correct Answer Explanation

To prepare tert-butyl ethyl ether efficiently, the primary alkyl halide should be ethyl bromide, while the bulky tert-butyl group should be present as sodium tert-butoxide. This allows the reaction to proceed through the SN2 mechanism while avoiding elimination that would occur if tert-butyl bromide were used. Therefore, option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Tert-butyl bromide undergoes elimination instead of SN2 substitution.
�� Option C → These reactants do not constitute Williamson synthesis.
�� Option D → These compounds do not produce the ether by Williamson synthesis.

🎯 Strategy

Used

Elimination

Application:

Choose a primary alkyl halide to favour the SN2 mechanism.

Final Logic:

Bulky Group = Alkoxide; Primary Group = Alkyl Halide.

🧠 Memory Trick

Bulky as Base, Primary as Halide

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

20 When a secondary or tertiary alkyl halide is used in the Williamson synthesis instead of a primary one, which reaction type competes and predominates over substitution?

A Oxidation

B Elimination

C Rearrangement

D Addition

💡 Short Explanation

�� Secondary and tertiary alkyl halides hinder SN2. �� Strong bases favour elimination. �� Alkenes become the major products.

✅ Correct Answer Explanation

Williamson ether synthesis proceeds by the SN2 mechanism, which is favoured with primary alkyl halides. Secondary and tertiary alkyl halides are sterically hindered, making SN2 difficult. Consequently, the alkoxide ion behaves as a strong base and elimination predominates, producing alkenes. Therefore, option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Oxidation does not occur.
�� Option C → Rearrangement is not the predominant reaction.
�� Option D → Addition is not involved.

🎯 Strategy

Used

Elimination

Application:

Identify the competing reaction when SN2 is hindered.

Final Logic:

2°/3° Halide + Strong Base → Elimination.

🧠 Memory Trick

Higher Halide = Higher Elimination

📚 Class XII – NCERT Chemistry Part II – Chapter: Alcohols, Phenols and Ethers – Topic: Williamson Ether Synthesis and Its Limitations

CUET UG Chemistry Booster Test – 3 Preparation Methods

CUET UG Chemistry Booster Test - 3 Preparation Methods

Test Complete!

Answer Review