CUET UG Chemistry Booster Test - 2 Preparation Methods -

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CUET UG Chemistry Booster Test - 2 Preparation Methods

⏱️ 30 Minutes

❓ 20 Questions

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

CUET UG Chemistry Booster Test - 2 Preparation Methods

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QUESTION 1 OF 20

Regarding the acid catalysed hydration mechanism of alkenes, which statements are true?
1. Step 1 involves protonation of the alkene by H₃O⁺ to form a carbocation.
2. Step 2 involves nucleophilic attack of water on the carbocation.
3. Step 3 involves deprotonation to form an alcohol.
4. The reaction follows Markovnikov addition in unsymmetrical alkenes.

Answer Review

1 Regarding the acid catalysed hydration mechanism of alkenes, which statements are true?
1. Step 1 involves protonation of the alkene by H₃O⁺ to form a carbocation.
2. Step 2 involves nucleophilic attack of water on the carbocation.
3. Step 3 involves deprotonation to form an alcohol.
4. The reaction follows Markovnikov addition in unsymmetrical alkenes.

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

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

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

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

💡 Short Explanation

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

✅ Correct Answer Explanation

Statement 1 is correct because the first step is protonation of the alkene by hydronium ion, forming the more stable carbocation. Statement 2 is correct because water, acting as a nucleophile, attacks the carbocation to form a protonated alcohol. Statement 3 is correct because deprotonation of the protonated alcohol gives the final alcohol. Statement 4 is a correct statement about acid-catalysed hydration of unsymmetrical alkenes, but it is not part of the reaction mechanism being asked in the question. The question specifically asks for the mechanistic steps; therefore, only statements 1, 2 and 3 satisfy the requirement. Hence, option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Omits the essential nucleophilic attack step.
�� Option C → Omits the initial protonation step.
�� Option D → Includes statement 4, which is not a mechanistic step.

🎯 Strategy

Used

Elimination

Application:

Verify each statement against the sequence of steps in the hydration mechanism.

Final Logic:

Mechanism = Protonation → Water Attack → Deprotonation.

🧠 Memory Trick

P → A → D (Protonation, Attack, Deprotonation)

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

2 During the hydration of an alkene, the nucleophilic attack of water occurs on which intermediate species?

A Carbanion

B Carbocation

C Free radical

D Transition metal complex

💡 Short Explanation

�� Protonation forms a carbocation. �� Water acts as the nucleophile. �� The reaction proceeds through electrophilic addition.

✅ Correct Answer Explanation

During acid-catalysed hydration, the alkene first reacts with H₃O⁺ to form the most stable carbocation. Water then attacks this positively charged intermediate to produce a protonated alcohol, which finally loses a proton to give the alcohol. Therefore, option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → No carbanion intermediate is formed.
�� Option C → The mechanism is ionic, not free-radical.
�� Option D → No transition-metal catalyst is involved.

🎯 Strategy

Used

Elimination

Application:

Identify the intermediate formed immediately before nucleophilic attack.

Final Logic:

Water always attacks the carbocation.

🧠 Memory Trick

Water Attacks C⁺

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

3 What is the expected IUPAC name of the primary alcohol product when propene undergoes hydroboration-oxidation?

A Propan-2-ol

B Propan-1-ol

C Propane-1,2-diol

D 2-Methylpropan-1-ol

💡 Short Explanation

�� Hydroboration follows anti-Markovnikov addition. �� Boron attaches to the less substituted carbon. �� Oxidation replaces boron with –OH.

✅ Correct Answer Explanation

Hydroboration-oxidation proceeds through anti-Markovnikov addition. In propene, boron attaches to the terminal carbon (the less substituted carbon), and subsequent oxidation replaces boron with a hydroxyl group. The product obtained is propan-1-ol. Therefore, option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Propan-2-ol is formed by acid-catalysed hydration, not hydroboration.
�� Option C → No diol is produced in this reaction.
�� Option D → The carbon skeleton of propene does not change.

🎯 Strategy

Used

Contextual/Tonal Matching

Application:

Relate hydroboration with anti-Markovnikov addition.

Final Logic:

Hydroboration → Anti-Markovnikov → Propan-1-ol.

🧠 Memory Trick

Hydroboration = 1° Alcohol

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

4 In the addition of borane to a double bond, boron gets attached to the sp² carbon carrying which relative number of hydrogen atoms?

A Fewer hydrogen atoms

B No hydrogen atoms

C An equal number of hydrogen atoms

D Greater number of hydrogen atoms

💡 Short Explanation

�� Hydroboration is anti-Markovnikov. �� Boron attaches to the less substituted carbon. �� The less substituted carbon carries more hydrogen atoms.

✅ Correct Answer Explanation

During hydroboration, boron adds to the less substituted carbon atom of the double bond because of steric and electronic factors. This carbon generally bears the greater number of hydrogen atoms. Subsequent oxidation replaces boron with the hydroxyl group without changing the orientation. Therefore, option D is correct.

❓ Why Other Options Are Incorrect

�� Option A → Boron does not attach to the carbon having fewer hydrogen atoms.
�� Option B → The carbon receiving boron normally possesses one or more hydrogen atoms.
�� Option C → Addition is regioselective, not equal.

🎯 Strategy

Used

Elimination

Application:

Recall the regioselectivity of hydroboration.

Final Logic:

Boron goes to the carbon with more hydrogen atoms.

🧠 Memory Trick

Boron Loves More H

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

5 Which reagent can be used to reduce an aldehyde to a primary alcohol or a ketone to a secondary alcohol?

A Acidified KMnO₄

B NaBH₄

C Concentrated HNO₃

D Zinc dust

💡 Short Explanation

�� NaBH₄ is a mild reducing agent. �� Aldehydes give primary alcohols. �� Ketones give secondary alcohols.

✅ Correct Answer Explanation

Sodium borohydride (NaBH₄) is a selective reducing agent that converts aldehydes into primary alcohols and ketones into secondary alcohols. It supplies hydride ions to the carbonyl carbon without affecting many other functional groups. Therefore, option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Acidified KMnO₄ is a strong oxidising agent.
�� Option C → Concentrated HNO₃ is also an oxidising agent.
�� Option D → Zinc dust is not used for reducing aldehydes and ketones to alcohols.

🎯 Strategy

Used

Elimination

Application:

Identify the reagent known for selective reduction of carbonyl compounds.

Final Logic:

NaBH₄ reduces aldehydes and ketones to alcohols.

🧠 Memory Trick

NaBH₄ = Carbonyl to Alcohol

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

6 Arrange the following in decreasing order of the degree (3°, 2°, 1°) of alcohol formed in the respective reactions.
1. Ketone
2. Aldehyde
3. Carboxylic acid
4. Ketone + Grignard reagent followed by hydrolysis

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

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

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

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

💡 Short Explanation

�� Grignard addition to ketones gives tertiary alcohols. �� Ketone reduction gives secondary alcohols. �� Aldehydes and carboxylic acids give primary alcohols.

✅ Correct Answer Explanation

Statement 1 represents ketones, which on catalytic hydrogenation are reduced to secondary alcohols. Statement 2 represents aldehydes, which are reduced to primary alcohols. Statement 3 represents carboxylic acids, which on reduction also produce primary alcohols. Statement 4 represents the reaction of a ketone with a Grignard reagent followed by hydrolysis, producing a tertiary alcohol. Therefore, the decreasing order of alcohol degree is: Ketone + Grignard reagent > Ketone > Aldehyde = Carboxylic acid Hence, option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Places secondary alcohol above tertiary alcohol.
�� Option C → Places primary alcohols above secondary alcohol.
�� Option D → Completely reverses the correct trend.

🎯 Strategy

Used

Option Grouping

Application:

Compare the degree of alcohol formed in each reaction.

Final Logic:

3° > 2° > 1°.

🧠 Memory Trick

Grignard → 3°, Ketone → 2°, Aldehyde/Acid → 1°.

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

7 The transformation of RCOOH to RCH₂OH using LiAlH₄ followed by H₂O is an example of what type of reaction?

A Electrophilic substitution

B Oxidation

C Carboxylic acid reduction

D Dehydration

💡 Short Explanation

�� LiAlH₄ is a strong reducing agent. �� Carboxylic acids are converted into primary alcohols. �� Water completes the reduction after hydrolysis.

✅ Correct Answer Explanation

Lithium aluminium hydride (LiAlH₄) reduces carboxylic acids to primary alcohols by supplying hydride ions to the carbonyl carbon. Hydrolysis with water then converts the intermediate into the corresponding primary alcohol. Therefore, option C is correct.

❓ Why Other Options Are Incorrect

�� Option A → No electrophilic substitution occurs.
�� Option B → The reaction is reduction, not oxidation.
�� Option D → No elimination of water takes place.

🎯 Strategy

Used

Elimination

Application:

Identify the reaction from the reagent and product formed.

Final Logic:

LiAlH₄ converts carboxylic acids into primary alcohols.

🧠 Memory Trick

LiAlH₄ = Acid to Alcohol

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

8 Due to the high expense of LiAlH₄, what is the intermediate synthesized commercially before catalytic hydrogenation is used to get the alcohol?

A Acid chloride

B Amide

C Ester

D Anhydride

💡 Short Explanation

�� LiAlH₄ is expensive. �� Carboxylic acids are first converted into esters. �� Esters are then reduced by catalytic hydrogenation.

✅ Correct Answer Explanation

Commercially, direct reduction of carboxylic acids with LiAlH₄ is avoided because the reagent is expensive. Instead, the acid is first converted into an ester, which is subsequently reduced by catalytic hydrogenation to obtain the corresponding alcohol. Therefore, option C is correct.

❓ Why Other Options Are Incorrect

�� Option A → Acid chlorides are not the commercial intermediate mentioned in NCERT.
�� Option B → Amides are not used for this commercial route.
�� Option D → Anhydrides are also not the commercial intermediate.

🎯 Strategy

Used

Elimination

Application:

Differentiate the industrial method from the laboratory method.

Final Logic:

Commercial route: Acid → Ester → Alcohol.

🧠 Memory Trick

Acid → Ester → Alcohol

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

9 During the nucleophilic addition of a Grignard reagent (RMgX) to a carbonyl group, which part acts as the nucleophile?

A The carbonyl carbon

B The carbonyl oxygen

C The alkyl group (R) of the Grignard reagent

D The magnesium atom

💡 Short Explanation

�� Grignard reagents possess a polar C–Mg bond. �� The alkyl group behaves like a carbanion. �� It attacks the electrophilic carbonyl carbon.

✅ Correct Answer Explanation

The carbon-magnesium bond in a Grignard reagent is highly polar because magnesium is more electropositive than carbon. As a result, the alkyl group (R) acquires partial negative character and behaves as a nucleophile. It attacks the electrophilic carbonyl carbon to form an alkoxide intermediate, which on hydrolysis gives an alcohol. Therefore, option C is correct.

❓ Why Other Options Are Incorrect

�� Option A → The carbonyl carbon is the electrophilic centre.
�� Option B → The carbonyl oxygen does not act as the nucleophile.
�� Option D → Magnesium coordinates with oxygen but is not the nucleophile.

🎯 Strategy

Used

Elimination

Application:

Identify the electron-rich species that attacks the carbonyl carbon.

Final Logic:

R⁻ equivalent attacks the carbonyl carbon.

🧠 Memory Trick

Grignard = R⁻ Attack

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

10 Give the IUPAC name of the product formed by the reaction of propanone with methylmagnesium bromide followed by hydrolysis.

A Propan-2-ol

B Butan-2-ol

C 2-Methylpropan-1-ol

D 2-Methylpropan-2-ol

💡 Short Explanation

�� Grignard reagents add an alkyl group to carbonyl compounds. �� Ketones form tertiary alcohols. �� Hydrolysis completes the reaction.

✅ Correct Answer Explanation

Propanone reacts with methylmagnesium bromide through nucleophilic addition. The methyl group adds to the carbonyl carbon, forming a tertiary alkoxide intermediate. Hydrolysis converts this intermediate into 2-methylpropan-2-ol, a tertiary alcohol. Therefore, option D is correct.

❓ Why Other Options Are Incorrect

�� Option A → Propan-2-ol is obtained by reduction of propanone, not by Grignard addition.
�� Option B → Butan-2-ol has an incorrect carbon skeleton.
�� Option C → 2-Methylpropan-1-ol is a primary alcohol and is not formed.

🎯 Strategy

Used

Contextual/Tonal Matching

Application:

Determine the product by adding the methyl group to the ketone followed by hydrolysis.

Final Logic:

Ketone + Grignard → Tertiary Alcohol.

🧠 Memory Trick

Ketone + RMgX = 3° Alcohol

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

11 Regarding the NaOH fusion of chlorobenzene, consider the following statements:
1. Chlorobenzene is fused with NaOH at 623 K.
2. The pressure maintained is 320 atmospheric pressure.
3. The immediate product of fusion is phenol.
4. The immediate product is sodium phenoxide.

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

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

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

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

💡 Short Explanation

�� Dow's process uses high temperature and pressure. �� Sodium phenoxide is formed first. �� Acidification yields phenol.

✅ Correct Answer Explanation

Statement 1 is correct because chlorobenzene is fused with aqueous sodium hydroxide at 623 K. Statement 2 is correct because the reaction is carried out at about 320 atmospheric pressure. Statement 3 is incorrect because phenol is not formed directly after fusion. Statement 4 is correct because the immediate product of the fusion reaction is sodium phenoxide, which is subsequently acidified to produce phenol. Therefore, option B is correct.

❓ Why Other Options Are Incorrect

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

🎯 Strategy

Used

Elimination

Application:

Verify each statement using the sequence of Dow's process.

Final Logic:

Fusion → Sodium Phenoxide → Acidification → Phenol.

🧠 Memory Trick

Fusion First, Phenol Later

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

12 Match List-I (Process Steps of Phenol Preparation from Chlorobenzene) with List-II (Details)

List I	List II
1. Reactant	a. Acidification
2. Condition	b. 623 K, 320 atm
3. Intermediate	c. Chlorobenzene + NaOH
4. Final Step	d. Sodium phenoxide

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

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

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

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

💡 Short Explanation

�� Chlorobenzene reacts with NaOH. �� High temperature and pressure are required. �� Sodium phenoxide is the intermediate.

✅ Correct Answer Explanation

1 → c : The reactants are chlorobenzene and aqueous sodium hydroxide. 2 → b : The reaction is carried out at 623 K and 320 atmospheric pressure. 3 → d : Sodium phenoxide is formed as the intermediate. 4 → a : Acidification of sodium phenoxide produces phenol. Therefore, option C is correct.

❓ Why Other Options Are Incorrect

�� Option A → Reactant and condition are interchanged.
�� Option B → Conditions and intermediate are incorrectly matched.
�� Option D → Reactant and final step are incorrectly matched.

🎯 Strategy

Used

Option Grouping

Application:

Match each stage of Dow's process with its corresponding description.

Final Logic:

Reactant → Condition → Intermediate → Acidification.

🧠 Memory Trick

React → Heat → Phenoxide → Acid

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

13 In the sulphonic acid method for phenol preparation, what is the structure of the intermediate formed upon heating benzene sulphonic acid with molten sodium hydroxide?

A Sodium phenoxide

B Cumene hydroperoxide

C Benzene diazonium chloride

D Benzoquinone

💡 Short Explanation

�� Benzene sulphonic acid undergoes alkali fusion. �� Sodium phenoxide is the intermediate. �� Acidification yields phenol.

✅ Correct Answer Explanation

In the sulphonic acid method, benzene is first converted into benzene sulphonic acid. On heating with molten sodium hydroxide, the sulphonic acid group is replaced by the hydroxyl group, forming sodium phenoxide as the intermediate. Subsequent acidification converts sodium phenoxide into phenol. Therefore, option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Cumene hydroperoxide is an intermediate in the cumene process.
�� Option C → Benzene diazonium chloride is formed from diazotisation of aniline, not from benzene sulphonic acid.
�� Option D → Benzoquinone is not formed in this method.

🎯 Strategy

Used

Elimination

Application:

Identify the intermediate formed during alkali fusion.

Final Logic:

Benzene Sulphonic Acid → Sodium Phenoxide → Phenol.

🧠 Memory Trick

Fusion First, Phenoxide Next

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

14 The temperature at which an aromatic primary amine reacts with nitrous acid to form a diazonium salt is given in which specific unit range?

A 273–278 °C

B 273–278 K

C 0–5 K

D 273–278 °F

💡 Short Explanation

�� Diazotisation is carried out at low temperature. �� The temperature is maintained at 273–278 K. �� Higher temperatures decompose diazonium salts.

✅ Correct Answer Explanation

Aromatic primary amines react with nitrous acid at 273–278 K (0–5°C) to form stable diazonium salts. Maintaining this low temperature prevents decomposition of the diazonium salt and ensures efficient diazotisation. Therefore, option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → 273–278°C is far too high.
�� Option C → 0–5 K is close to absolute zero and is incorrect.
�� Option D → 273–278°F is not the temperature specified in NCERT.

🎯 Strategy

Used

Elimination

Application:

Recall the temperature range used for diazotisation.

Final Logic:

Diazotisation occurs at 273–278 K.

🧠 Memory Trick

Diazo = 273–278 K

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

15 Identify the reaction type occurring when cumene is converted to cumene hydroperoxide in the presence of air.

A Reduction

B Electrophilic substitution

C Cumene oxidation

D Hydrolysis

💡 Short Explanation

�� Cumene reacts with oxygen from air. �� Cumene hydroperoxide is formed. �� This is the first step of the cumene process.

✅ Correct Answer Explanation

In the industrial cumene process, cumene (isopropylbenzene) is oxidised by atmospheric oxygen to form cumene hydroperoxide. This oxidation step is the key initial stage in the commercial manufacture of phenol and acetone. Therefore, option C is correct.

❓ Why Other Options Are Incorrect

�� Option A → Oxygen is added; therefore, the reaction is oxidation, not reduction.
�� Option B → No electrophilic substitution occurs.
�� Option D → Hydrolysis occurs in the subsequent acid treatment, not in this step.

🎯 Strategy

Used

Contextual/Tonal Matching

Application:

Identify the reaction from the reactant and product.

Final Logic:

Cumene + Air → Cumene Hydroperoxide = Oxidation.

🧠 Memory Trick

Air Oxidises Cumene

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

16 What reagent is used to cleave cumene hydroperoxide into phenol and acetone?

A Concentrated NaOH

B Zinc dust

C Hydrogen gas

D Dilute acid

💡 Short Explanation

�� Cumene hydroperoxide is the intermediate. �� Dilute acid cleaves the peroxide. �� Phenol and acetone are produced.

✅ Correct Answer Explanation

In the cumene process, cumene hydroperoxide is treated with dilute acid. The acid catalyses its cleavage to produce phenol as the main product and acetone as the valuable by-product. Therefore, option D is correct.

❓ Why Other Options Are Incorrect

�� Option A → Concentrated NaOH is not used in this step.
�� Option B → Zinc dust is not used in the cumene process.
�� Option C → Hydrogen gas is not involved in peroxide cleavage.

🎯 Strategy

Used

Elimination

Application:

Recall the reagent used for cleavage of cumene hydroperoxide.

Final Logic:

Cumene Hydroperoxide + Dilute Acid → Phenol + Acetone.

🧠 Memory Trick

Peroxide + Acid = Phenol

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

📄 Passage

Alcohols undergo dehydration in the presence of protic acids (H₂SO₄, H₃PO₄). The formation of the reaction product, alkene or ether depends on the reaction conditions. For example, ethanol is dehydrated to ethene in the presence of sulphuric acid at 443 K. At 413 K, ethoxyethane is the main product. The formation of ether is a nucleophilic bimolecular reaction (SN2) involving the attack of an alcohol molecule on a protonated alcohol. The method is suitable for the preparation of ethers having primary alkyl groups only.

For the preparation of ethers by alcohol dehydration, the alkyl group of the alcohol must ideally be:

A Unhindered and primary

B Highly branched and tertiary

C Aromatic

D Secondary

💡 Short Explanation

�� Ether formation proceeds through an SN2 mechanism. �� Primary alcohols favour substitution. �� Branched alcohols favour elimination.

✅ Correct Answer Explanation

The passage states that ether formation by dehydration occurs through an SN2 mechanism in which one alcohol molecule attacks a protonated alcohol molecule. Since SN2 reactions require minimal steric hindrance, primary alcohols are the most suitable substrates. Secondary and tertiary alcohols undergo elimination more readily. Therefore, option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Tertiary alcohols preferentially undergo elimination.
�� Option C → Aromatic alcohols are not suitable for this method.
�� Option D → Secondary alcohols readily undergo competing elimination.

🎯 Strategy

Used

Contextual/Tonal Matching

Application:

Apply the passage stating that ether formation proceeds via an SN2 mechanism.

Final Logic:

SN2 requires an unhindered primary alkyl group.

🧠 Memory Trick

SN2 Loves 1°

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

📄 Passage

If secondary or tertiary alcohols undergo dehydration under these conditions, which pathway competes and which products are mainly formed?

A Substitution competing over elimination; ethers are formed.

B Elimination competes over substitution; alkenes are easily formed.

C Oxidation takes over; ketones are formed.

D Polymerization occurs; plastics are formed.

💡 Short Explanation

�� Secondary and tertiary alcohols undergo elimination readily. �� Alkenes become the major products. �� Ether formation is favoured only for primary alcohols.

✅ Correct Answer Explanation

The passage states that dehydration of primary alcohols at 413 K produces ethers through an SN2 mechanism. However, secondary and tertiary alcohols preferentially undergo elimination because of the greater stability of the resulting carbocations and the steric hindrance that disfavors SN2 substitution. Consequently, alkenes are formed as the major products. Therefore, option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Elimination predominates over substitution.
�� Option C → Oxidation is not involved in this reaction.
�� Option D → Polymerisation does not occur under these conditions.

🎯 Strategy

Used

Contextual/Tonal Matching

Application:

Use the passage to determine the competing pathway for secondary and tertiary alcohols.

Final Logic:

2° and 3° Alcohols → Elimination → Alkene.

🧠 Memory Trick

Higher Alcohol → Higher Elimination

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

19 Match List-I (Reactants in Williamson Synthesis) with List-II (Roles/Nature)

List I	List II
1. Primary alkyl halide	a. Gives best ether yield (SN2)
2. Sodium alkoxide	b. Acts as nucleophile and base
3. Tertiary alkyl halide	c. Yields alkene exclusively
4. Alkoxide ion	d. SN2 attacker

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

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

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

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

💡 Short Explanation

�� Primary alkyl halides favour SN2. �� Alkoxide ions act as nucleophiles. �� Tertiary alkyl halides undergo elimination.

✅ Correct Answer Explanation

1 → a : Primary alkyl halides undergo SN2 reactions efficiently, giving the best ether yield. 2 → b : Sodium alkoxide provides the alkoxide ion, which acts as both a nucleophile and a strong base. 3 → c : Tertiary alkyl halides undergo elimination rather than SN2 substitution, producing alkenes. 4 → d : The alkoxide ion attacks the primary alkyl halide through the SN2 mechanism. Therefore, option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Primary alkyl halide and sodium alkoxide are incorrectly matched.
�� Option C → Tertiary alkyl halide and alkoxide ion are incorrectly matched.
�� Option D → Primary alkyl halide and sodium alkoxide are interchanged.

🎯 Strategy

Used

Option Grouping

Application:

Match each reactant with its role in Williamson ether synthesis.

Final Logic:

Primary → Best SN2; Alkoxide → Nucleophile; Tertiary → Elimination.

🧠 Memory Trick

Primary Wins, Tertiary Eliminates

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

20 Why is the reaction of sodium ethoxide with tert-butyl bromide not appropriate for preparing tert-butyl ethyl ether?

A Sodium ethoxide acts as a strong base leading to elimination.

B Tert-butyl bromide is completely unreactive.

C The SN2 mechanism is too fast.

D It strictly forms an alcohol.

💡 Short Explanation

�� Tertiary alkyl halides are sterically hindered. �� SN2 substitution is not favoured. �� Elimination predominates.

✅ Correct Answer Explanation

Williamson ether synthesis proceeds through an SN2 mechanism, which is highly unfavorable with tertiary alkyl halides because of steric hindrance. Sodium ethoxide behaves as a strong base and abstracts a β-hydrogen from tert-butyl bromide, leading predominantly to elimination and the formation of 2-methylpropene instead of the desired ether. Therefore, option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Tert-butyl bromide is reactive but undergoes elimination.
�� Option C → SN2 is hindered, not accelerated.
�� Option D → An alcohol is not formed in this reaction.

🎯 Strategy

Used

Elimination

Application:

Determine whether substitution or elimination is favoured for a tertiary alkyl halide.

Final Logic:

Tertiary Halide + Strong Base → Elimination.

🧠 Memory Trick

3° Halide = Elimination

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

CUET UG Chemistry Booster Test – 2 Preparation Methods

CUET UG Chemistry Booster Test - 2 Preparation Methods

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