CUET UG Chemistry Booster Test - 2 Structure and Properties -

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CUET UG Chemistry Booster Test - 2 Structure and Properties

⏱️ 30 Minutes

❓ 20 Questions

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

CUET UG Chemistry Booster Test - 2 Structure and Properties

1 / 20

QUESTION 1 OF 20

In the formation of the C-O bond in methanol, the bond is classified as a sigma bond. This type of bonding dictates that the electron density is concentrated:

Answer Review

1 In the formation of the C-O bond in methanol, the bond is classified as a sigma bond. This type of bonding dictates that the electron density is concentrated:

A Above and below the internuclear axis

B Symmetrically along the internuclear axis between the sp³ hybridized carbon and sp³ hybridized oxygen

C Exclusively on the oxygen atom

D Outside the tetrahedral framework

💡 Short Explanation

�� Sigma bonds result from head-on orbital overlap. �� Electron density lies along the internuclear axis. �� Methanol contains a C(sp³)-O(sp³) sigma bond.

✅ Correct Answer Explanation

A sigma bond is formed by the head-on overlap of orbitals. In methanol, the carbon atom is sp³ hybridized and the oxygen atom is also approximately sp³ hybridized. The overlap occurs directly along the line joining the two nuclei, leading to electron density concentrated symmetrically along the internuclear axis. Therefore, option B correctly describes the nature of the C–O sigma bond.

❓ Why Other Options Are Incorrect

�� Option A → Above and below the internuclear axis describes a π bond, not a σ bond.
�� Option C → Electron density is shared between carbon and oxygen, not exclusively on oxygen.
�� Option D → Sigma bond electron density lies within the bonding framework, not outside it.

🎯 Strategy

Used

Elimination

Application:

Identify the defining feature of a sigma bond and eliminate descriptions of π bonding or incorrect electron distribution.

Final Logic:

Sigma bonds always possess electron density along the internuclear axis.

🧠 Memory Trick

σ = Straight Overlap

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

2 Carbon hybridization effects:
1. In aliphatic alcohols, the -OH is attached to an sp³ hybridized carbon.
2. In phenols, the -OH is attached to an sp² hybridized carbon.
3. The sp³ hybridized carbon forms a stronger bond with oxygen than an sp² hybridized carbon does.
4. The bond between oxygen and an sp² hybridized carbon is stronger and has partial double bond character.

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

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

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

D (1), (3)

💡 Short Explanation

�� Alcohols contain O–H attached to sp³ carbon. �� Phenols contain O–H attached to aromatic sp² carbon. �� Resonance gives partial double bond character in phenols.

✅ Correct Answer Explanation

Statement 1 is correct because in aliphatic alcohols the hydroxyl group is attached to an sp³ hybridized carbon atom. Statement 2 is correct because in phenols the hydroxyl group is directly attached to an aromatic ring carbon, which is sp² hybridized. Statement 3 is incorrect because the C–O bond involving an sp² carbon in phenol is stronger than the corresponding bond involving an sp³ carbon in alcohols. Statement 4 is correct because resonance between the oxygen lone pair and the aromatic ring gives the C–O bond partial double bond character, making it shorter and stronger. 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.
�� Option D → Includes statement 3, which is incorrect.

🎯 Strategy

Used

Elimination

Application:

Check each statement against the structural differences between alcohols and phenols.

Final Logic:

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

🧠 Memory Trick

Phenol = Resonance = Stronger C–O

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

3 Arrange the following in decreasing order of their approximate C-O-C or C-O-H bond angles:
1. Pure tetrahedral angle
2. Methoxymethane (ether)
3. Methanol (alcohol)
4. Oxygen

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

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

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

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

💡 Short Explanation

�� Ethers have bond angles slightly greater than tetrahedral. �� Methanol has bond angle slightly less than tetrahedral. �� Oxygen is approximately sp³ hybridized in both molecules.

✅ Correct Answer Explanation

Statement 1 represents the ideal tetrahedral angle of 109°28'. Statement 2 represents methoxymethane (ether), whose C-O-C bond angle is slightly greater than the tetrahedral angle because repulsion between the two methyl groups widens the bond angle. Statement 3 represents methanol, whose C-O-H bond angle is slightly less than the tetrahedral angle 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: (2) Methoxymethane > (1) Tetrahedral angle > (3) Methanol Hence, option C is correct.

❓ Why Other Options Are Incorrect

�� Option A → Places the tetrahedral angle above methoxymethane.
�� Option B → Reverses the actual bond-angle trend.
�� Option D → Places methanol above the tetrahedral angle.

🎯 Strategy

Used

Option Grouping

Application:

Compare the relative bond-angle values associated with the three structural situations.

Final Logic:

Ether angle > Tetrahedral angle > 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

4 The bond angle in alcohols is slightly less than 109°28'. Conceptually, this deviation is best explained by:

A The large atomic radius of the carbon atom

B The attractive forces between the alkyl groups

C The strong repulsion between the unshared electron pairs on the oxygen atom

D The high electronegativity of the hydrogen atom

💡 Short Explanation

�� Oxygen contains two lone pairs. �� Lone pair repulsion affects geometry. �� Bond angle decreases below tetrahedral value.

✅ Correct Answer Explanation

Oxygen in alcohols possesses two lone pairs of electrons. Lone pair–lone pair and lone pair–bond pair repulsions are stronger than bond pair–bond pair repulsions. These repulsions compress the bond angle slightly below the ideal tetrahedral value of 109°28'. Therefore, option C is correct.

❓ Why Other Options Are Incorrect

�� Option A → Carbon size is not responsible for bond-angle compression.
�� Option B → Attractive forces between alkyl groups are not the controlling factor.
�� Option D → Hydrogen electronegativity does not explain the deviation.

🎯 Strategy

Used

Elimination

Application:

Identify the dominant factor controlling molecular geometry around oxygen.

Final Logic:

Lone pair repulsion reduces the bond angle.

🧠 Memory Trick

Lone Pairs Push Harder

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

5 What is the IUPAC name of the phenol derivative commonly known as anisole?

A Ethoxybenzene

B Methoxybenzene

C Phenoxybenzene

D Methyl phenyl alcohol

💡 Short Explanation

�� Anisole is an aromatic ether. �� Contains a methoxy group attached to benzene. �� IUPAC naming uses alkoxybenzene format.

✅ Correct Answer Explanation

Anisole consists of a methoxy group (-OCH₃) attached to a benzene ring. According to IUPAC nomenclature, it is named methoxybenzene. Therefore, option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Ethoxybenzene contains an ethoxy group, not methoxy.
�� Option C → Refers to diphenyl ether type structure.
�� Option D → Not a valid IUPAC name.

🎯 Strategy

Used

Substitution

Application:

Replace the common name with the corresponding IUPAC alkoxybenzene name.

Final Logic:

Anisole = Methoxy + Benzene.

🧠 Memory Trick

Anisole = Methoxybenzene

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

6 Identify reaction type:
The –OH group attached to the benzene ring in phenols directs incoming groups to ortho and para positions because it activates the ring. This type of reaction is an:

A Electrophilic aromatic substitution

B Nucleophilic aromatic substitution

C Elimination reaction

D Free radical addition

💡 Short Explanation

�� Phenol activates the aromatic ring. �� OH group is ortho-para directing. �� Aromatic substitution occurs via electrophiles.

✅ Correct Answer Explanation

The hydroxyl group donates electron density into the benzene ring through resonance, increasing electron density at ortho and para positions. Therefore, electrophiles preferentially attack these positions. Such reactions belong to electrophilic aromatic substitution.

❓ Why Other Options Are Incorrect

�� Option B → Nucleophilic substitution is not characteristic of activated phenol rings.
�� Option C → No elimination process occurs.
�� Option D → Free radical addition is not involved.

🎯 Strategy

Used

Contextual/Tonal Matching

Application:

Identify the reaction based on ring activation and directing effects.

Final Logic:

Activated aromatic rings undergo electrophilic substitution.

🧠 Memory Trick

Phenol Loves Electrophiles

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

7 The temperature unit used to specify the exact conditions for the acidic dehydration of ethanol to ethoxyethane is the Kelvin. At what temperature is ethoxyethane the main product?

A 443 K

B 373 K

C 413 K

D 298 K

💡 Short Explanation

�� Dehydration conditions determine products. �� Lower temperature favors ether formation. �� Higher temperature favors alkene formation.

✅ Correct Answer Explanation

When ethanol is heated with concentrated sulfuric acid at about 413 K, intermolecular dehydration occurs producing ethoxyethane. At 443 K, intramolecular dehydration predominates, yielding ethene. Therefore, option C is correct.

❓ Why Other Options Are Incorrect

�� Option A → 443 K mainly produces ethene.
�� Option B → Insufficient for the specified dehydration condition.
�� Option D → No dehydration occurs under normal conditions.

🎯 Strategy

Used

Elimination

Application:

Recall temperature-specific outcomes of ethanol dehydration.

Final Logic:

413 K favors ether formation; 443 K favors alkene formation.

🧠 Memory Trick

413 → Ether, 443 → Ethene

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

8 In methoxymethane, the bond angle is slightly greater than the standard tetrahedral angle. Why does this structural variation occur despite oxygen having two lone pairs?

A The two lone pairs repel each other less than in alcohols.

B The repulsive interaction between the two bulky methyl (-R) groups overcomes the lone pair repulsion.

C The carbon atoms are sp² hybridized.

D Conjugation with an aromatic ring alters the geometry.

💡 Short Explanation

�� Ether contains two alkyl groups. �� Alkyl-group repulsion increases bond angle. �� Angle becomes slightly greater than tetrahedral.

✅ Correct Answer Explanation

In methoxymethane, the repulsion between the two methyl groups attached to oxygen is significant. This repulsive interaction pushes the groups apart and increases the C–O–C bond angle beyond the ideal tetrahedral value. Therefore, option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Lone pair behavior is not the main reason.
�� Option C → Carbon atoms are sp³ hybridized.
�� Option D → Methoxymethane contains no aromatic ring.

🎯 Strategy

Used

Elimination

Application:

Identify the dominant repulsive interaction affecting geometry.

Final Logic:

Alkyl-group repulsion widens the bond angle.

🧠 Memory Trick

Big Groups Push Apart

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

9 Match List-I (Compound) with List-II (Boiling Point in K)

List I	List II
1. n-Pentane	a. 390 K
2. Ethoxyethane	b. 309.1 K
3. Butan-1-ol	c. 307.6 K
4. Water	d. 373 K

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

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

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

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

💡 Short Explanation

�� Alcohols exhibit intermolecular hydrogen bonding. �� Ethers possess lower boiling points than alcohols. �� Alkanes have the weakest intermolecular forces.

✅ Correct Answer Explanation

1 → b : n-Pentane has a boiling point of approximately 309.1 K. 2 → c : Ethoxyethane has a boiling point of approximately 307.6 K. 3 → a : Butan-1-ol has a boiling point of approximately 390 K because of intermolecular hydrogen bonding. 4 → d : Water boils at 373 K under standard atmospheric pressure. Therefore, Option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → n-Pentane and butan-1-ol are incorrectly matched.
�� Option C → n-Pentane and ethoxyethane are interchanged.
�� Option D → Ethoxyethane and butan-1-ol are incorrectly matched.

🎯 Strategy

Used

Option Grouping

Application:

Match each compound with its known standard boiling point and compare intermolecular forces.

Final Logic:

Alcohol (H-bonding) > Water (H-bonding) > Alkane ≈ Ether (similar molecular masses).

🧠 Memory Trick

Alcohol Highest, Water 373 K

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

10 Comparing ethers and alkanes:
1. Ethers have a weak polarity.
2. Ethers have drastically higher boiling points than alkanes of comparable mass.
3. The weak polarity of ethers does not appreciably affect their boiling points compared to alkanes.
4. Hydrogen bonding in alkanes elevates their boiling point above ethers.

A (1), (2)

B (1), (3)

C (3), (4)

D (2), (4)

💡 Short Explanation

�� Ethers are weakly polar. �� Ethers do not form intermolecular hydrogen bonds. �� Their boiling points are close to comparable alkanes.

✅ Correct Answer Explanation

Statement 1 is correct because ethers possess a polar C–O–C linkage. Statement 2 is incorrect because ethers do not have drastically higher boiling points than alkanes of similar molecular mass. Statement 3 is correct because the weak polarity causes only a small difference in boiling point. Statement 4 is incorrect because alkanes do not exhibit hydrogen bonding. Therefore, option B is correct.

❓ Why Other Options Are Incorrect

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

🎯 Strategy

Used

Elimination

Application:

Evaluate each statement using intermolecular-force concepts.

Final Logic:

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

🧠 Memory Trick

Ether = Weak Polar, No H-Bond

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

11 The large difference in boiling points between alcohols and ethers of comparable molecular masses is fundamentally caused by:

A Ethers having a higher molecular mass

B Alcohols possessing intermolecular hydrogen bonding, which is absent in ethers

C Ethers forming stronger intermolecular hydrogen bonds than alcohols

D Alcohols having longer carbon chains

💡 Short Explanation

�� Alcohols form intermolecular hydrogen bonds. �� Ethers do not form intermolecular hydrogen bonds among themselves. �� Hydrogen bonding significantly raises boiling point.

✅ Correct Answer Explanation

Alcohol molecules contain an O–H bond and can form extensive intermolecular hydrogen bonding. These strong intermolecular attractions require more energy to overcome during boiling. Ethers lack an O–H bond and therefore cannot form intermolecular hydrogen bonds with other ether molecules. As a result, alcohols have much higher boiling points than ethers of comparable molecular masses. Therefore, option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Comparable molecular masses are assumed, so molecular mass is not the cause.
�� Option C → Ethers do not form stronger intermolecular hydrogen bonds than alcohols.
�� Option D → Carbon chain length is not the fundamental reason for the difference.

🎯 Strategy

Used

Elimination

Application:

Compare intermolecular forces present in alcohols and ethers.

Final Logic:

Hydrogen bonding in alcohols raises boiling points significantly.

🧠 Memory Trick

Alcohol = H-Bond = High BP

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

12 Arrange the following compounds in decreasing order of their boiling points based on their intermolecular forces (assume comparable molecular mass):
1. Butan-1-ol
2. n-Pentane
3. Ethoxyethane
4. Water

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

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

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

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

💡 Short Explanation

�� Alcohols exhibit intermolecular hydrogen bonding. �� Water also shows extensive hydrogen bonding. �� Ethers exhibit weak dipole-dipole interactions, while alkanes possess only van der Waals forces.

✅ Correct Answer Explanation

Compound 1 (Butan-1-ol) has the highest boiling point (≈390 K) because of strong intermolecular hydrogen bonding and its larger molecular mass. Compound 4 (Water) has a boiling point of 373 K due to its extensive hydrogen-bonding network. Compound 3 (Ethoxyethane) possesses a permanent dipole moment and exhibits dipole-dipole interactions but cannot form intermolecular hydrogen bonds with itself. Compound 2 (n-Pentane) is non-polar and experiences only weak van der Waals forces. Therefore, the decreasing order of boiling points is: Butan-1-ol > Water > Ethoxyethane > n-Pentane Hence, option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Completely reverses the correct boiling-point trend.
�� Option C → Places water below n-pentane, which is incorrect.
�� Option D → Incorrectly places water above butan-1-ol and n-pentane above ethoxyethane.

🎯 Strategy

Used

Option Grouping

Application:

Arrange the compounds by comparing the strength of their intermolecular forces and their known boiling points.

Final Logic:

Butan-1-ol > Water > Ether > Alkane.

🧠 Memory Trick

Alcohol > Water > Ether > Alkane

📚 Class XII – NCERT Chemistry Part II – Chapter: Alcohols, Phenols and Ethers – Topic: Physical Properties (Boiling Point Trends)

13 Ethoxyethane and water solubility properties:
1. Ethoxyethane is essentially immiscible with water.
2. Ethoxyethane is miscible to almost the same extent as butan-1-ol.
3. Oxygen in ether can form hydrogen bonds with water molecules.
4. Ethoxyethane dissolves only in non-polar solvents.

A (1), (4)

B (2), (4)

C (1), (3)

D (2), (3)

💡 Short Explanation

�� Ether oxygen can accept hydrogen bonds. �� Ethers show limited water solubility. �� Solubility is comparable to some lower alcohols.

✅ Correct Answer Explanation

Statement 1 is incorrect because ethoxyethane is not completely immiscible with water. Statement 2 is correct because ethoxyethane is soluble in water to an extent comparable to butan-1-ol. Statement 3 is correct because the oxygen atom in ether can accept hydrogen bonds from water molecules. Statement 4 is incorrect because ethoxyethane dissolves in both organic solvents and has limited solubility in water. Therefore, option D is correct.

❓ Why Other Options Are Incorrect

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

🎯 Strategy

Used

Elimination

Application:

Verify each statement using hydrogen-bonding and solubility concepts.

Final Logic:

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

🧠 Memory Trick

Ether O Accepts H-Bonds

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

14 The solubility of higher molecular mass alcohols in water is low because the alkyl group acts as a hydrophobic barrier. Conceptually, what does "hydrophobic" mean in this context?

A Water-attracting, enhancing hydrogen bond formation

B Water-repelling, decreasing interaction with polar water molecules

C Acid-donating, lowering the pH of the solution

D Electron-withdrawing, increasing acidity

💡 Short Explanation

�� Hydrophobic means water-repelling. �� Larger alkyl groups reduce water interaction. �� Solubility decreases as carbon chain length increases.

✅ Correct Answer Explanation

Hydrophobic groups are non-polar regions that interact poorly with polar water molecules. As the alkyl chain becomes larger, the hydrophobic character dominates over the hydrophilic hydroxyl group, reducing solubility in water. Therefore, option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Describes hydrophilic behavior.
�� Option C → Hydrophobicity is unrelated to acidity.
�� Option D → Electron-withdrawing effects do not define hydrophobicity.

🎯 Strategy

Used

Contextual/Tonal Matching

Application:

Interpret the scientific meaning of the term hydrophobic.

Final Logic:

Hydrophobic substances avoid interaction with water.

🧠 Memory Trick

Hydro + Phobic = Water Fear

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

15 While the C-O bonds in ethers are polar, their overall physical properties (like boiling point) closely resemble non-polar alkanes of comparable mass. This implies that:

A The dipole moment is extraordinarily large.

B Ethers form strong dipole-dipole networks akin to hydrogen bonds.

C The net dipole moment exists but is too weak to significantly increase intermolecular attraction.

D Ethers possess an ionic character.

💡 Short Explanation

�� Ethers possess weak polarity. �� No intermolecular hydrogen bonding occurs. �� Boiling points remain close to alkanes.

✅ Correct Answer Explanation

Although ethers have polar C–O bonds and possess a net dipole moment, the resulting intermolecular attractions are relatively weak compared with hydrogen bonding. Therefore, their boiling points are only slightly higher than those of comparable alkanes. Hence, option C is correct.

❓ Why Other Options Are Incorrect

�� Option A → Ether dipole moments are not extraordinarily large.
�� Option B → Ethers do not form hydrogen-bond networks among themselves.
�� Option D → Ethers are covalent, not ionic.

🎯 Strategy

Used

Elimination

Application:

Compare the strength of intermolecular attractions.

Final Logic:

Weak dipole interactions cause only modest boiling-point changes.

🧠 Memory Trick

Ether = Weak Dipole

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

16 Match List-I (Molecule) with List-II (Intermolecular Force/Polarity Characteristic)

List I	List II
1. Ethanol	a. Non-polar, purely van der Waals forces
2. Ethoxyethane	b. Weak polarity, possesses a net dipole moment but no intermolecular H-bonding
3. Propane	c. Intermolecular H-bonding, aliphatic chain
4. Phenol	d. Intermolecular H-bonding, aromatic ring

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

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

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

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

💡 Short Explanation

�� Ethanol exhibits hydrogen bonding. �� Ether has weak polarity. �� Propane is non-polar. �� Phenol contains aromatic-ring hydrogen bonding.

✅ Correct Answer Explanation

1 → c : Ethanol is an aliphatic alcohol exhibiting intermolecular hydrogen bonding. 2 → b : Ethoxyethane possesses a net dipole moment but lacks intermolecular hydrogen bonding. 3 → a : Propane is non-polar and experiences only van der Waals forces. 4 → d : Phenol exhibits intermolecular hydrogen bonding and contains an aromatic ring. Therefore, option D is correct.

❓ Why Other Options Are Incorrect

�� Option A → Multiple incorrect pairings are present.
�� Option B → Ethanol and phenol are interchanged.
�� Option C → Ethoxyethane and propane are incorrectly matched.

🎯 Strategy

Used

Option Grouping

Application:

Match each molecule with its dominant intermolecular-force characteristic.

Final Logic:

Alcohol → H-Bond, Ether → Weak Dipole, Alkane → van der Waals, Phenol → Aromatic H-Bond.

🧠 Memory Trick

Alcohol–H Bond, Ether–Dipole, Alkane–vdW

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

17 Methoxymethane and ethanol share the same molecular formula (C₂H₆O). Why does ethanol exist as a liquid with a high boiling point while methoxymethane is highly volatile?

A Ethanol has a higher molecular mass.

B Ethanol exhibits extensive intermolecular hydrogen bonding.

C Methoxymethane forms stronger hydrogen bonds.

D The C-O bond in ethanol is longer.

💡 Short Explanation

�� Ethanol and methoxymethane are functional isomers. �� Ethanol forms hydrogen bonds. �� Hydrogen bonding raises boiling point.

✅ Correct Answer Explanation

Both compounds have the same molecular formula and similar molecular masses. However, ethanol contains an O–H bond and forms strong intermolecular hydrogen bonds. Methoxymethane lacks an O–H bond and cannot form such hydrogen bonds among its own molecules. Consequently, ethanol has a much higher boiling point and exists as a liquid under ordinary conditions. Therefore, option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Molecular masses are identical.
�� Option C → Methoxymethane does not form stronger hydrogen bonds.
�� Option D → Bond length is not the determining factor.

🎯 Strategy

Used

Elimination

Application:

Compare intermolecular forces rather than molecular formula.

Final Logic:

Hydrogen bonding makes ethanol less volatile.

🧠 Memory Trick

Same Formula, Different Forces

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

18 Identify reaction type:
The conversion of a sodium alkoxide and an alkyl halide into an ether (e.g., sodium ethoxide + methyl bromide) is called Williamson synthesis. What type of mechanism does this primarily follow for primary alkyl halides?

A SN1

B SN2

C Electrophilic addition

D Free radical substitution

💡 Short Explanation

�� Williamson synthesis prepares ethers. �� Alkoxide acts as a nucleophile. �� Primary alkyl halides favour the SN2 mechanism.

✅ Correct Answer Explanation

In Williamson ether synthesis, the alkoxide ion attacks the carbon atom of a primary alkyl halide from the backside while the leaving group departs simultaneously. This one-step bimolecular nucleophilic substitution is characteristic of the SN2 mechanism. Therefore, option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → SN1 is unfavorable for primary alkyl halides.
�� Option C → No addition reaction occurs.
�� Option D → The reaction is not radical based.

🎯 Strategy

Used

Contextual/Tonal Matching

Application:

Recognize the standard mechanism associated with Williamson synthesis.

Final Logic:

Primary alkyl halides undergo backside attack via SN2.

🧠 Memory Trick

Williamson = Alkoxide + SN2

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

📄 Passage

Solubility of alcohols and phenols in water is primarily due to their ability to form hydrogen bonds with water molecules. The solubility decreases with an increase in the size of the alkyl or aryl groups, which are hydrophobic in nature. Because of this, several of the lower molecular mass alcohols are miscible with water in all proportions. In contrast, higher alcohols become increasingly insoluble. Furthermore, the boiling points of these compounds increase with molecular mass due to increasing van der Waals forces, while branching decreases the boiling point by reducing surface area.

Based on the passage, if you compare straight-chain butan-1-ol and a highly branched isomer like 2-methylpropan-2-ol, how would the branching affect their respective boiling points?

A Branching decreases the surface area, thus decreasing van der Waals forces and lowering the boiling point.

B Branching increases the surface area, increasing van der Waals forces and raising the boiling point.

C Branching completely eliminates hydrogen bonding, drastically lowering the boiling point.

D Branching has no effect on van der Waals forces.

💡 Short Explanation

�� Branching reduces surface area. �� Reduced surface area weakens van der Waals forces. �� Lower intermolecular attraction lowers boiling point.

✅ Correct Answer Explanation

The passage clearly states that branching decreases boiling point by reducing surface area. Smaller surface area reduces van der Waals attractions between molecules. Therefore, a highly branched alcohol generally has a lower boiling point than its straight-chain isomer. Hence, option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Branching decreases, not increases, surface area.
�� Option C → Branching does not eliminate hydrogen bonding.
�� Option D → Branching directly affects van der Waals interactions.

🎯 Strategy

Used

Contextual/Tonal Matching

Application:

Use the information explicitly stated in the passage.

Final Logic:

Less 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

Based on the passage, why is methanol completely miscible in water whereas heptan-1-ol is nearly insoluble?

A Methanol has a larger hydrophobic group.

B Heptan-1-ol lacks an –OH group.

C Heptan-1-ol has a much larger hydrophobic alkyl group which hinders water solubility.

D Methanol cannot form hydrogen bonds with water.

💡 Short Explanation

�� Both alcohols contain an –OH group. �� Long alkyl chains are hydrophobic. �� Larger hydrophobic portions reduce water solubility.

✅ Correct Answer Explanation

Methanol has a very small alkyl group, allowing hydrogen bonding with water to dominate. In heptan-1-ol, the large hydrophobic alkyl chain outweighs the effect of the hydroxyl group and significantly reduces interaction with water. Consequently, heptan-1-ol is nearly insoluble. Therefore, option C is correct.

❓ Why Other Options Are Incorrect

�� Option A → Methanol has the smaller hydrophobic group.
�� Option B → Heptan-1-ol contains an –OH group.
�� Option D → Methanol readily forms hydrogen bonds with water.

🎯 Strategy

Used

Elimination

Application:

Compare the size of hydrophobic alkyl groups.

Final Logic:

Larger hydrophobic 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 in Water

CUET UG Chemistry Booster Test – 2 Structure and Properties

CUET UG Chemistry Booster Test - 2 Structure and Properties

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