CUET UG Chemistry Booster Test - 3 Structure and Properties -

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CUET UG Chemistry Booster Test - 3 Structure and Properties

⏱️ 30 Minutes

❓ 20 Questions

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

CUET UG Chemistry Booster Test - 3 Structure and Properties

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QUESTION 1 OF 20

In an alcohol molecule, the carbon-oxygen sigma bond involves sp³ hybridization of carbon. Considering the geometry, how does the sp³ character restrict the structural flexibility compared to an sp² bonded oxygen in phenol?

Answer Review

1 In an alcohol molecule, the carbon-oxygen sigma bond involves sp³ hybridization of carbon. Considering the geometry, how does the sp³ character restrict the structural flexibility compared to an sp² bonded oxygen in phenol?

A The sp³ carbon provides a rigid planar geometry.

B The sp³ carbon lacks an unhybridized p-orbital, preventing conjugation with the oxygen lone pair.

C The sp³ carbon allows conjugation, stabilizing the molecule.

D The sp³ carbon strictly mandates a 180° bond angle.

💡 Short Explanation

�� Alcohol carbon is sp³ hybridized. �� sp³ carbon has no unhybridized p-orbital. �� Conjugation with oxygen is absent in alcohols.

✅ Correct Answer Explanation

In alcohols, the carbon bonded to the hydroxyl group is sp³ hybridized. Since all four hybrid orbitals are involved in σ-bond formation, there is no unhybridized p-orbital available for overlap with the lone pair on oxygen. Consequently, resonance or conjugation between oxygen and carbon does not occur. In phenol, however, the oxygen atom is attached to an sp²-hybridized carbon of the benzene ring, allowing overlap of the oxygen lone pair with the aromatic π-system. Therefore, option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → An sp³ carbon has tetrahedral geometry, not planar geometry.
�� Option C → Conjugation occurs in phenol, not in alcohols.
�� Option D → An sp³ carbon has a bond angle of about 109.5°, not 180°.

🎯 Strategy

Used

Elimination

Application:

Identify the orbital arrangement of an sp³ carbon and eliminate options inconsistent with hybridization.

Final Logic:

No unhybridized p-orbital means no conjugation with oxygen.

🧠 Memory Trick

sp³ = No Free p Orbital

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

2 Phenol C-O cleavage properties:
1. Phenols readily undergo nucleophilic substitution to form aryl halides.
2. The C-O bond in phenol is stronger than in aliphatic alcohols.
3. The carbon of the phenyl group is sp² hybridized, giving the C-O bond partial double bond character.
4. HI cleaves anisole to form phenol and methyl iodide.

A (1), (2)

B (1), (4)

C (3), (4)

D (2), (3)

💡 Short Explanation

�� Phenol has a stronger C-O bond. �� Resonance gives partial double bond character. �� HI cleaves anisole at the alkyl-oxygen bond.

✅ Correct Answer Explanation

Statement 1 is incorrect because phenols do not readily undergo nucleophilic substitution to form aryl halides due to the strong aryl C-O bond. Statement 2 is correct because resonance strengthens the C-O bond in phenol. Statement 3 is correct because the phenyl carbon is sp² hybridized, and resonance imparts partial double bond character to the C-O bond. Statement 4 is incorrect because HI cleaves anisole at the methyl-oxygen bond to produce phenol and methyl iodide, not iodobenzene. Therefore, option D is correct.

❓ Why Other Options Are Incorrect

�� Option A → Includes statement 1, which is incorrect.
�� Option B → Includes statements 1 and 4, both of which are incorrect.
�� Option C → Includes statement 4, which is incorrect.

🎯 Strategy

Used

Elimination

Application:

Verify each statement using the reactions and resonance properties of phenols and ethers.

Final Logic:

Only statements 2 and 3 are correct.

🧠 Memory Trick

Phenol Strong C-O, Anisole Gives CH₃I

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

3 The exact bond angle in methanol is noted as 108.9°. This deviation from the idealized 109.5° angle of a perfect sp³ hybridized central atom demonstrates that:
1. Bond pair-bond pair repulsion is stronger than lone pair-lone pair repulsion.
2. Lone pair-lone pair repulsion on the oxygen atom compresses the C-O-H bond angle.
3. The methyl group is highly electronegative.
4. Oxygen is approximately sp³ hybridized in methanol.

A (2), (4)

B (1), (2)

C (2), (3)

D (1), (3)

💡 Short Explanation

�� Oxygen in methanol is approximately sp³ hybridized. �� Lone pairs compress the bond angle. �� The C-O-H angle becomes slightly less than the tetrahedral angle.

✅ Correct Answer Explanation

Statement 1 is incorrect because lone pair-lone pair repulsion is stronger than bond pair-bond pair repulsion. Statement 2 is correct because the two lone pairs on oxygen exert greater repulsion, reducing the C-O-H bond angle from the ideal tetrahedral value. Statement 3 is incorrect because the methyl group is not highly electronegative. Statement 4 is correct because oxygen in methanol is approximately sp³ hybridized with two bond pairs and two lone pairs. Therefore, statements 2 and 4 are correct, making option A the correct answer.

❓ Why Other Options Are Incorrect

�� Option B → Includes statement 1, which is incorrect.
�� Option C → Includes statement 3, which is incorrect.
�� Option D → Includes statements 1 and 3, both of which are incorrect.

🎯 Strategy

Used

Elimination

Application:

Evaluate each statement using VSEPR theory and hybridization concepts.

Final Logic:

Only statements 2 and 4 agree with the structure of methanol.

🧠 Memory Trick

Lone Pairs Compress, sp³ Oxygen

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

4 Match List-I (Molecule) with List-II (Dominant structural feature affecting bond angle)

List I	List II
1. Methanol	a. Lone pair-lone pair repulsion compressing the bond angle below 109.5°
2. Methoxymethane	b. Steric repulsion between two alkyl groups expanding the bond angle above 109.5°
3. Phenol	c. sp² hybridized carbon giving the C–O bond partial double bond character
4. Ethene	d. Pure sp²-sp² bonding framework forming approximately 120° bond angles

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

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

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

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

💡 Short Explanation

�� Methanol has bond-angle compression due to lone pairs. �� Methoxymethane has a wider bond angle because of alkyl-group repulsion. �� Phenol exhibits resonance, while ethene has trigonal planar geometry.

✅ Correct Answer Explanation

1 → a : Methanol has two lone pairs on oxygen, which compress the C–O–H bond angle below the ideal tetrahedral angle. 2 → b : In methoxymethane, repulsion between the two alkyl groups slightly expands the C–O–C bond angle. 3 → c : In phenol, the oxygen lone pair is conjugated with the benzene ring, giving the C–O bond partial double bond character. 4 → d : Ethene consists of sp² hybridized carbon atoms with approximately 120° bond angles. Therefore, option C is correct.

❓ Why Other Options Are Incorrect

�� Option A → Methanol and phenol are incorrectly matched.
�� Option B → Methanol and methoxymethane are incorrectly matched.
�� Option D → Methoxymethane and phenol are incorrectly matched.

🎯 Strategy

Used

Option Grouping

Application:

Match each molecule with its dominant structural feature affecting bond angle.

Final Logic:

Methanol → Lone Pair, Ether → Steric Repulsion, Phenol → Resonance, Ethene → sp² Geometry.

🧠 Memory Trick

Alcohol–Lone Pair, Ether–Bulky Groups, Phenol–Resonance

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

5 Identify reaction type:
Phenol undergoes a reaction with chloroform in the presence of sodium hydroxide to introduce a –CHO group at the ortho position, producing salicylaldehyde. This reaction is known as:

A Kolbe's reaction

B Williamson synthesis

C Reimer-Tiemann reaction

D Friedel-Crafts alkylation

💡 Short Explanation

�� Phenol reacts with CHCl₃ and NaOH. �� A formyl (–CHO) group is introduced. �� Salicylaldehyde is the major product.

✅ Correct Answer Explanation

In the Reimer-Tiemann reaction, phenol reacts with chloroform and aqueous sodium hydroxide. Dichlorocarbene (:CCl₂) generated during the reaction attacks the activated aromatic ring, introducing a formyl (–CHO) group predominantly at the ortho position. On hydrolysis, salicylaldehyde is formed. Therefore, option C is correct.

❓ Why Other Options Are Incorrect

�� Option A → Kolbe's reaction introduces a –COOH group, not a –CHO group.
�� Option B → Williamson synthesis is used for the preparation of ethers.
�� Option D → Friedel-Crafts alkylation introduces alkyl groups, not formyl groups.

🎯 Strategy

Used

Contextual/Tonal Matching

Application:

Identify the named reaction from its reagents and product.

Final Logic:

CHCl₃ + NaOH + Phenol → Reimer-Tiemann Reaction.

🧠 Memory Trick

Reimer = Ring + CHO

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

6 The conjugation effect in phenol involves the unshared electron pair of oxygen interacting with the aromatic ring. What is the direct consequence of this resonance on the electron density of the oxygen atom?

A The electron density on oxygen increases, making it a stronger base.

B The oxygen atom acquires a partial positive charge, facilitating the release of a proton.

C The oxygen atom acquires a full negative charge, increasing bond length.

D The conjugation completely breaks the O–H bond homolytically.

💡 Short Explanation

�� Oxygen donates its lone pair into the benzene ring. �� Resonance decreases electron density on oxygen. �� Phenol becomes more acidic than alcohols.

✅ Correct Answer Explanation

The lone pair of electrons on the oxygen atom overlaps with the π-electron system of the benzene ring through resonance. As a result, electron density is delocalized from oxygen toward the ring, giving the oxygen atom a partial positive charge. This polarization weakens the O–H bond and facilitates the release of a proton, making phenol more acidic than aliphatic alcohols. Therefore, option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Electron density on oxygen decreases rather than increases.
�� Option C → Oxygen does not acquire a full negative charge through resonance.
�� Option D → Resonance does not break the O–H bond homolytically.

🎯 Strategy

Used

Elimination

Application:

Analyze the effect of resonance on electron distribution in phenol.

Final Logic:

Resonance withdraws electron density from oxygen, giving it a partial positive charge.

🧠 Memory Trick

Resonance Pulls from Oxygen

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

7 The industrial preparation of phenol from chlorobenzene (Dow's process) requires extreme conditions. The unit used to describe the pressure for this process is atmospheric pressure. What is the value required?

A 1 atmospheric pressure

B 100 atmospheric pressure

C 320 atmospheric pressure

D 623 atmospheric pressure

💡 Short Explanation

�� Dow's process uses chlorobenzene. �� High temperature and pressure are required. �� The reaction proceeds with aqueous sodium hydroxide.

✅ Correct Answer Explanation

In Dow's process, chlorobenzene is heated with aqueous sodium hydroxide at about 623 K under approximately 320 atmospheric pressure. These extreme conditions are necessary because the C–Cl bond in chlorobenzene is strengthened by resonance and is resistant to nucleophilic substitution. Therefore, option C is correct.

❓ Why Other Options Are Incorrect

�� Option A → The pressure is far too low for Dow's process.
�� Option B → 100 atmospheric pressure is insufficient for the industrial process.
�� Option D → 623 is the reaction temperature (K), not the pressure.

🎯 Strategy

Used

Elimination

Application:

Differentiate between the temperature and pressure values used in Dow's process.

Final Logic:

Dow's Process → 623 K and 320 atmospheric pressure.

🧠 Memory Trick

Dow = 623 K + 320 atm

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

8 Arrange the following strictly in decreasing order of their C-O-C or C-O-H bond angles:
1. Methoxymethane
2. Pure tetrahedral geometry
3. Methanol
4. Oxygen in both methanol and methoxymethane is approximately sp³ hybridized

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

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

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

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

💡 Short Explanation

�� Ether has the largest bond angle. �� Tetrahedral geometry is intermediate. �� Alcohol has the smallest bond angle.

✅ Correct Answer Explanation

Statement 1 represents methoxymethane, whose C–O–C bond angle is slightly greater than the ideal tetrahedral angle because repulsion between the two alkyl groups expands the bond angle. Statement 2 represents the ideal tetrahedral angle of approximately 109.5°. Statement 3 represents methanol, whose C–O–H bond angle is about 108.9° due to lone pair repulsion on oxygen. Statement 4 is correct because oxygen is approximately sp³ hybridized in both methanol and methoxymethane. Therefore, the decreasing order of bond angles is: Methoxymethane > Pure tetrahedral geometry > Methanol. Hence, option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Reverses the correct trend.
�� Option C → Places tetrahedral geometry above methoxymethane.
�� Option D → Places methanol above tetrahedral geometry.

🎯 Strategy

Used

Option Grouping

Application:

Compare the bond-angle values based on lone-pair and steric repulsion.

Final Logic:

Ether angle > 109.5° > Alcohol angle.

🧠 Memory Trick

Ether Opens, Alcohol Closes

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

9 Boiling point and hydrogen bonding statements for substituted phenols:
1. o-Nitrophenol is steam volatile due to intramolecular hydrogen bonding.
2. p-Nitrophenol is less volatile because of intermolecular hydrogen bonding.
3. Intramolecular hydrogen bonding associates molecules together, increasing the boiling point.
4. Ortho and para nitrophenols can be separated by steam distillation.

A (2), (4)

B (3), (4)

C (1), (3)

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

💡 Short Explanation

�� o-Nitrophenol exhibits intramolecular hydrogen bonding. �� p-Nitrophenol exhibits intermolecular hydrogen bonding. �� Steam distillation separates ortho and para isomers.

✅ Correct Answer Explanation

Statement 1 is correct because o-nitrophenol undergoes intramolecular hydrogen bonding, making it steam volatile. Statement 2 is correct because p-nitrophenol forms intermolecular hydrogen bonds, increasing its boiling point and reducing its volatility. Statement 3 is incorrect because intermolecular, not intramolecular, hydrogen bonding associates molecules and raises the boiling point. Statement 4 is correct because o-nitrophenol and p-nitrophenol can be separated by steam distillation owing to their difference in volatility. Therefore, option D is correct.

❓ Why Other Options Are Incorrect

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

🎯 Strategy

Used

Elimination

Application:

Evaluate each statement using the concepts of intra- and intermolecular hydrogen bonding.

Final Logic:

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

🧠 Memory Trick

Ortho = Intra = Steam; Para = Inter = High BP

📚 Class XII – NCERT Chemistry Part II – Chapter: Alcohols, Phenols and Ethers – Topic: Intramolecular and Intermolecular Hydrogen Bonding in Nitrophenols

10 When examining the molecular mass effect on physical properties, moving from methanol to propan-1-ol to pentan-1-ol, which structural factor most significantly increases the boiling point?

A The addition of more hydroxyl groups

B The geometric alteration of the C-O-H bond angle

C The expansion of the hydrophobic surface area increasing van der Waals interactions

D The transition from sp³ to sp² hybridization

💡 Short Explanation

�� All three alcohols contain one hydroxyl group. �� Molecular mass and surface area increase with chain length. �� Stronger van der Waals forces increase boiling point.

✅ Correct Answer Explanation

Methanol, propan-1-ol and pentan-1-ol each contain one hydroxyl group and are capable of intermolecular hydrogen bonding. As the alkyl chain length increases, molecular mass and surface area increase, strengthening van der Waals forces between molecules. This additional intermolecular attraction raises the boiling point from methanol to pentan-1-ol. Therefore, option C is correct.

❓ Why Other Options Are Incorrect

�� Option A → The number of hydroxyl groups remains one in all three alcohols.
�� Option B → The C–O–H bond angle changes very little and does not account for the boiling-point trend.
�� Option D → The carbon bonded to the hydroxyl group remains sp³ hybridized.

🎯 Strategy

Used

Elimination

Application:

Compare the structural changes that occur as the carbon chain length increases.

Final Logic:

Increasing chain length increases van der Waals interactions and boiling point.

🧠 Memory Trick

Longer Chain → Higher BP

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

11 Identify the IUPAC name of the compound which, among isomeric phenols, forms extensive intermolecular hydrogen bonding resulting in a higher boiling point and lower volatility.

A 2-Nitrophenol

B 4-Nitrophenol

C Methoxybenzene

D Ethoxyethane

💡 Short Explanation

�� Para-nitrophenol forms intermolecular hydrogen bonding. �� Intermolecular hydrogen bonding increases boiling point. �� Ortho-nitrophenol exhibits intramolecular hydrogen bonding.

✅ Correct Answer Explanation

4-Nitrophenol (para-nitrophenol) forms extensive intermolecular hydrogen bonding between its molecules. These stronger intermolecular attractions increase the boiling point and decrease volatility. In contrast, 2-nitrophenol predominantly forms intramolecular hydrogen bonding, making it more volatile and steam volatile. Therefore, option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → 2-Nitrophenol exhibits intramolecular hydrogen bonding and is more volatile.
�� Option C → Methoxybenzene is an ether and does not form intermolecular hydrogen bonding.
�� Option D → Ethoxyethane is an ether and lacks an O–H bond for intermolecular hydrogen bonding.

🎯 Strategy

Used

Elimination

Application:

Differentiate between intra- and intermolecular hydrogen bonding.

Final Logic:

Para = Intermolecular H-bonding = Higher boiling point.

🧠 Memory Trick

Para = Partners; Ortho = Own

📚 Class XII – NCERT Chemistry Part II – Chapter: Alcohols, Phenols and Ethers – Topic: Intramolecular and Intermolecular Hydrogen Bonding

12 Compare the phase states and boiling characteristics of alcohols and ethers. Since ethers lack the hydrogen atom attached directly to an electronegative oxygen, they:

A Cannot act as hydrogen bond acceptors in mixtures with water.

B Cannot form hydrogen bonds with other ether molecules, resulting in drastically lower boiling points than isomeric alcohols.

C Form stronger intermolecular van der Waals forces than alkanes of double their mass.

D Readily polymerize at room temperature.

💡 Short Explanation

�� Ethers lack an O–H bond. �� They cannot form intermolecular hydrogen bonds among themselves. �� Their boiling points are lower than those of isomeric alcohols.

✅ Correct Answer Explanation

Hydrogen bonding between molecules requires a hydrogen atom directly bonded to a highly electronegative atom such as oxygen. Ethers contain oxygen but lack an O–H bond, so they cannot form intermolecular hydrogen bonds with other ether molecules. Consequently, ethers have much lower boiling points than isomeric alcohols, although they can act as hydrogen bond acceptors with water. Therefore, option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Ethers can act as hydrogen bond acceptors because oxygen has lone pairs.
�� Option C → There is no such increase in van der Waals forces compared with alkanes of double the molecular mass.
�� Option D → Ethers do not readily polymerize under ordinary conditions.

🎯 Strategy

Used

Elimination

Application:

Identify the structural requirement for intermolecular hydrogen bonding.

Final Logic:

No O–H bond means no intermolecular hydrogen bonding between ether molecules.

🧠 Memory Trick

Ether Accepts, Never Donates

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

13 Water solubility statements:
1. Both ethoxyethane and butan-1-ol are miscible to almost the same extent in water.
2. Pentane is highly soluble in water due to van der Waals forces.
3. The solubility of alcohols in water is due to the formation of hydrogen bonds.
4. The C–O–C bond in ethers repels water completely.

A (1), (3)

B (2), (4)

C (1), (2)

D (3), (4)

💡 Short Explanation

�� Lower ethers and lower alcohols show appreciable water solubility. �� Alcohols form hydrogen bonds with water. �� Hydrocarbons are nearly insoluble in water.

✅ Correct Answer Explanation

Statement 1 is correct because ethoxyethane and butan-1-ol are soluble in water to almost the same extent, as described in NCERT. Statement 2 is incorrect because pentane is a non-polar hydrocarbon and is practically insoluble in water. Statement 3 is correct because alcohol molecules form hydrogen bonds with water through their hydroxyl groups. Statement 4 is incorrect because the oxygen atom in ethers can accept hydrogen bonds from water; therefore, ethers do not completely repel water. Therefore, option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Includes statements 2 and 4, both of which are incorrect.
�� Option C → Includes statement 2, which is incorrect.
�� Option D → Includes statement 4, which is incorrect.

🎯 Strategy

Used

Elimination

Application:

Evaluate each statement using hydrogen bonding and solubility concepts.

Final Logic:

Statements 1 and 3 are correct; statements 2 and 4 are incorrect.

🧠 Memory Trick

Alcohol Bonds, Pentane Avoids

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

14 Arrange the following in decreasing order of their expected solubility in water:
1. Methanol
2. Hexan-1-ol
3. Propan-1-ol
4. Decan-1-ol

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

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

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

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

💡 Short Explanation

�� Water solubility decreases with increasing alkyl chain length. �� Larger alkyl groups are more hydrophobic. �� Lower alcohols are the most soluble.

✅ Correct Answer Explanation

All four compounds contain one hydroxyl group capable of hydrogen bonding with water. However, as the alkyl chain becomes longer, the hydrophobic character increases and reduces solubility. Therefore, the decreasing order of solubility is: Methanol > Propan-1-ol > Hexan-1-ol > Decan-1-ol. Hence, option D is correct.

❓ Why Other Options Are Incorrect

�� Option A → Incorrectly places hexan-1-ol above propan-1-ol.
�� Option B → Completely reverses the actual solubility trend.
�� Option C → Incorrectly places propan-1-ol above methanol and decan-1-ol above hexan-1-ol.

🎯 Strategy

Used

Option Grouping

Application:

Arrange the alcohols according to increasing hydrophobic alkyl chain length.

Final Logic:

Longer carbon chain = Lower water solubility.

🧠 Memory Trick

Long Chain, Less Soluble

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

15 In the reaction of asymmetric ethers with excess hydrogen iodide (HI), the polarity and structural stability of the intermediates dictate the mechanism. If one alkyl group is tertiary, the cleavage predominantly follows which pathway due to carbocation stability?

A SN2, yielding a primary halide

B SN1, yielding a tertiary halide

C E1, yielding an alkene

D Electrophilic addition

💡 Short Explanation

�� Tertiary carbocations are relatively stable. �� Ether cleavage depends on the nature of the alkyl group. �� Tertiary ethers preferentially undergo SN1 cleavage.

✅ Correct Answer Explanation

When an asymmetric ether containing a tertiary alkyl group reacts with excess HI, the ether oxygen is first protonated. Cleavage then occurs by formation of the more stable tertiary carbocation, followed by attack of the iodide ion. This is an SN1 mechanism and produces the corresponding tertiary alkyl iodide. Therefore, option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → SN2 is favored for methyl and primary alkyl groups, not tertiary groups.
�� Option C → Although elimination may compete under certain conditions, the NCERT-described major pathway for tertiary ethers is SN1 cleavage.
�� Option D → No electrophilic addition reaction occurs.

🎯 Strategy

Used

Elimination

Application:

Identify the mechanism based on carbocation stability.

Final Logic:

Tertiary carbocation stability favors the SN1 pathway.

🧠 Memory Trick

Tertiary = Stable = SN1

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

16 Match List-I (Reaction/Property) with List-II (Reason/Mechanism tied to structural polarity)

List I	List II
1. Cleavage of primary alkyl ethers with HI	a. SN1 mechanism due to stable carbocation formation
2. Cleavage of tertiary alkyl ethers with HI	b. sp² hybridized carbon prevents nucleophilic substitution
3. Non-reactivity of phenol C–O bond with I⁻	c. SN2 attack by iodide on the least substituted carbon
4. Action of ether oxygen as a base	d. Protonation of polar ether oxygen forms oxonium ion

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

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

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

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

💡 Short Explanation

�� Primary ethers undergo SN2 cleavage. �� Tertiary ethers undergo SN1 cleavage. �� Ether oxygen is protonated before cleavage.

✅ Correct Answer Explanation

1 → c : Primary alkyl ethers react with HI through an SN2 attack by iodide at the less substituted carbon. 2 → a : Tertiary alkyl ethers undergo cleavage through an SN1 mechanism because tertiary carbocations are relatively stable. 3 → b : The C–O bond in phenol is resistant to nucleophilic substitution because the carbon is sp² hybridized and resonance gives the bond partial double bond character. 4 → d : Ether oxygen behaves as a Lewis base and is protonated by HI to form an oxonium ion before bond cleavage. Therefore, option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Primary and tertiary ether mechanisms are interchanged, and ether protonation is incorrectly matched.
�� Option C → Phenol and tertiary ether mechanisms are incorrectly matched.
�� Option D → Primary ether cleavage and protonation are incorrectly matched.

🎯 Strategy

Used

Option Grouping

Application:

Match each reaction with its corresponding mechanism or structural reason.

Final Logic:

Primary → SN2, Tertiary → SN1, Phenol → sp² Carbon, Ether → Protonation.

🧠 Memory Trick

Primary–SN2, Tertiary–SN1

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

17 When comparing the relative ease of dehydration amongst alcohols (which relies on the formation of a carbocation intermediate), what is the correct relative order of ease of dehydration?

A Primary > Secondary > Tertiary

B Tertiary > Secondary > Primary

C Secondary > Tertiary > Primary

D Primary > Tertiary > Secondary

💡 Short Explanation

�� Dehydration proceeds through carbocation formation. �� More stable carbocations form more readily. �� Tertiary alcohols dehydrate most easily.

✅ Correct Answer Explanation

The acid-catalysed dehydration of alcohols generally proceeds through carbocation formation. Since tertiary carbocations are the most stable, tertiary alcohols undergo dehydration most readily, followed by secondary alcohols. Primary alcohols form unstable primary carbocations and therefore dehydrate least readily. Thus, the order is: Tertiary > Secondary > Primary Hence, option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Completely reverses the carbocation stability trend.
�� Option C → Incorrectly places secondary alcohols above tertiary alcohols.
�� Option D → Incorrectly places primary alcohols first.

🎯 Strategy

Used

Option Grouping

Application:

Arrange the alcohols according to carbocation stability.

Final Logic:

More stable carbocation = Easier dehydration.

🧠 Memory Trick

3° > 2° > 1°

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

18 Identify reaction type:
The reaction of alcohols with active metals like sodium forms sodium alkoxides and releases hydrogen gas. This reaction clearly establishes the cleavage of the O–H bond, proving that alcohols are:

A Brønsted bases only

B Strong electrophiles

C Weakly acidic (Brønsted acids)

D Lewis acids

💡 Short Explanation

�� Alcohols react with active metals. �� The O–H bond is cleaved. �� Hydrogen gas is evolved.

✅ Correct Answer Explanation

Alcohols react with active metals such as sodium to form sodium alkoxides with the evolution of hydrogen gas. 2ROH + 2Na → 2RONa + H₂ This reaction demonstrates that alcohols donate a proton (H⁺) from the hydroxyl group. Therefore, alcohols behave as weak Brønsted acids. Hence, option C is correct.

❓ Why Other Options Are Incorrect

�� Option A → Alcohols can behave as bases, but this reaction demonstrates their acidic character.
�� Option B → Alcohols are not strong electrophiles.
�� Option D → Alcohols are not Lewis acids in this reaction.

🎯 Strategy

Used

Elimination

Application:

Identify the property demonstrated by hydrogen evolution with sodium.

Final Logic:

Release of H₂ indicates proton donation, confirming weak acidity.

🧠 Memory Trick

Na + ROH → H₂ = Acid

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

📄 Passage

The boiling points of alcohols and phenols increase with an increase in the number of carbon atoms. In alcohols, the boiling points decrease with an increase of branching in the carbon chain because of a decrease in van der Waals forces with a decrease in surface area. Furthermore, the –OH group in alcohols and phenols is involved in intermolecular hydrogen bonding. It is interesting to note that boiling points of alcohols and phenols are higher in comparison to other classes of compounds, namely hydrocarbons, ethers, haloalkanes and haloarenes of comparable molecular masses.

Analytically applying the passage, if you compared straight-chain pentan-1-ol with 2,2-dimethylpropan-1-ol (both C₅H₁₂O), how does the structural geometry (branching) dynamically alter the liquid state properties?

A The branched isomer becomes a solid due to perfect stacking.

B The branched isomer assumes a more spherical shape, minimizing the effective surface area for van der Waals interactions and resulting in a lower boiling point.

C The branched isomer forms stronger intermolecular hydrogen bonds, raising the boiling point.

D Branching forces the molecule into an sp² hybridization state.

💡 Short Explanation

�� Branching decreases molecular surface area. �� Weaker van der Waals forces reduce boiling point. �� Hydrogen bonding remains present in both alcohols.

✅ Correct Answer Explanation

The passage states that branching decreases the boiling point of alcohols because it reduces the effective surface area available for intermolecular van der Waals interactions. Although both straight-chain pentan-1-ol and 2,2-dimethylpropan-1-ol possess an –OH group and can form hydrogen bonds, the highly branched isomer is more compact and nearly spherical. This reduces intermolecular contact and weakens van der Waals attractions, resulting in a lower boiling point. Therefore, option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Branching does not convert the alcohol into a solid by perfect stacking.
�� Option C → Branching does not strengthen intermolecular hydrogen bonding.
�� Option D → Branching does not change the hybridization from sp³ to sp².

🎯 Strategy

Used

Contextual/Tonal Matching

Application:

Apply the passage statement that branching decreases boiling point by reducing surface area.

Final Logic:

More branching → Smaller surface area → Weaker van der Waals forces → Lower boiling point.

🧠 Memory Trick

More Branches, Lower BP

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

📄 Passage

Evaluating the passage's comparison between alcohols and structurally similar compounds, why does methoxyethane (an ether) exhibit a fundamentally lower boiling point than its isomer, propan-1-ol, despite having identical molecular mass and surface area characteristics?

A Methoxyethane has a highly branched structure.

B Methoxyethane undergoes spontaneous dehydration.

C Propan-1-ol engages in extensive intermolecular hydrogen bonding networks that methoxyethane is incapable of forming.

D The C–O–C bond in methoxyethane is highly unstable and breaks easily under thermal stress.

💡 Short Explanation

�� Alcohols possess an O–H bond. �� Ethers cannot form intermolecular hydrogen bonds among themselves. �� Hydrogen bonding greatly increases boiling point.

✅ Correct Answer Explanation

Both methoxyethane and propan-1-ol have the same molecular formula and comparable molecular mass. However, propan-1-ol contains an O–H bond that enables extensive intermolecular hydrogen bonding. Methoxyethane lacks a hydrogen atom directly bonded to oxygen and therefore cannot form intermolecular hydrogen bonds with other ether molecules. Consequently, propan-1-ol has a much higher boiling point. Therefore, option C is correct.

❓ Why Other Options Are Incorrect

�� Option A → Methoxyethane is not highly branched, and branching is not the primary reason.
�� Option B → Methoxyethane does not undergo spontaneous dehydration.
�� Option D → The C–O–C bond is stable under ordinary conditions and does not determine the boiling-point difference.

🎯 Strategy

Used

Contextual/Tonal Matching

Application:

Compare intermolecular forces in alcohols and ethers using the information provided in the passage.

Final Logic:

Alcohols form intermolecular hydrogen bonds; ethers do not.

🧠 Memory Trick

OH Builds Bridges, Ether Doesn't

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

CUET UG Chemistry Booster Test – 3 Structure and Properties

CUET UG Chemistry Booster Test - 3 Structure and Properties

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