CUET UG Chemistry Booster Test - 3 Classification and Basics -

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CUET UG Chemistry Booster Test - 1 Classification and Basics

⏱️ 30 Minutes

❓ 20 Questions

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

CUET UG Chemistry Booster Test - 1 Classification and Basics

1 / 20

QUESTION 1 OF 20

Match List I with List II.

List I	List II
1. Alcohols	a. -OH attached directly to an aromatic ring
2. Phenols	b. Oxygen bridging two alkyl/aryl groups
3. Ethers	c. -OH attached to an aliphatic system
4. Hydrocarbons	d. Parent compounds containing only C and H

Answer Review

1 Match List I with List II.

List I	List II
1. Alcohols	a. -OH attached directly to an aromatic ring
2. Phenols	b. Oxygen bridging two alkyl/aryl groups
3. Ethers	c. -OH attached to an aliphatic system
4. Hydrocarbons	d. Parent compounds containing only C and H

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

💡 Short Explanation

�� Alcohols contain OH on aliphatic carbon. �� Phenols contain OH on aromatic carbon. �� Ethers contain an oxygen bridge.

✅ Correct Answer Explanation

1 → c : Alcohols contain an OH group attached to an aliphatic carbon. 2 → a : Phenols contain an OH group directly attached to an aromatic ring. 3 → b : Ethers contain an oxygen atom bridging two alkyl or aryl groups. 4 → d : Hydrocarbons contain only carbon and hydrogen atoms. Therefore, option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Alcohols and phenols are interchanged.
�� Option C → Ethers and phenols are incorrectly matched.
�� Option D → Multiple functional groups are mismatched.

🎯 Strategy

Used

Option Grouping

Application:

Match each compound class with its defining functional characteristic.

Final Logic:

Alcohol–Aliphatic OH, Phenol–Aromatic OH, Ether–Oxygen Bridge.

🧠 Memory Trick

Alcohol–Aliphatic, Phenol–Phenyl, Ether–Bridge

📚 Class XII – NCERT Chemistry Part II – Chapter: Alcohols, Phenols and Ethers – Topic: Classification of Oxygen-Containing Organic Compounds

2 Identify the reaction/derivation type that conceptually produces a phenol from a hydrocarbon.

A Substitution of hydrogen by an alkoxy group

B Substitution of hydrogen in an aromatic hydrocarbon by a hydroxyl group

C Dehydration of an alcohol

D Hydration of an alkene

💡 Short Explanation

�� Phenols are aromatic hydroxy compounds. �� OH replaces a hydrogen atom on an aromatic ring. �� Benzene gives phenol conceptually.

✅ Correct Answer Explanation

Phenols are conceptually derived when a hydroxyl group replaces a hydrogen atom in an aromatic hydrocarbon. This is the fundamental definition of phenols according to NCERT. Therefore, option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Produces ether-type structures.
�� Option C → Removes water from alcohols.
�� Option D → Produces alcohols from alkenes.

🎯 Strategy

Used

Elimination

Application:

Identify the definition that specifically describes phenol formation.

Final Logic:

Phenol = Aromatic hydrocarbon + OH substitution.

🧠 Memory Trick

Phenol = Phenyl + OH

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

3 Ethers structural properties:
1. Regarded as hydrocarbon derivatives where hydrogen is replaced by -OR or -OAr
2. The central oxygen is sp³ hybridized
3. The C-O bond length is almost the same as in alcohols
4. The larger alkyl group is chosen as the parent hydrocarbon in IUPAC naming

A (1), (3)

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

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

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

💡 Short Explanation

�� Ethers contain an oxygen bridge. �� Oxygen is sp³ hybridized. �� IUPAC names use alkoxyalkane nomenclature.

✅ Correct Answer Explanation

Statement 1 is correct because ethers may be viewed as hydrocarbon derivatives containing -OR or -OAr groups. Statement 2 is correct because the oxygen atom in ethers is sp³ hybridized. Statement 3 is correct because the C-O bond length in ethers is nearly the same as that in alcohols. Statement 4 is correct because in IUPAC nomenclature the larger alkyl group is selected as the parent hydrocarbon. Therefore, all four statements are correct.

❓ Why Other Options Are Incorrect

�� Option A → Omits statements 2 and 4, which are correct.
�� Option B → Omits statement 3, which is correct.
�� Option C → Omits statement 1, which is correct.

🎯 Strategy

Used

Elimination

Application:

Verify each structural and nomenclature property of ethers.

Final Logic:

All four statements are correct.

🧠 Memory Trick

Ether = OR + sp³ O

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

4 Arrange the following carbon-oxygen (C-O) bond lengths in decreasing order:
1. C-O bond length in methanol
2. C-O bond length in methoxymethane
3. C-O bond length in phenol
4. C=O bond length in methanal

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

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

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

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

💡 Short Explanation

�� Phenol has a shorter C-O bond because of resonance. �� Methanol and methoxymethane have single C-O bonds. �� Carbonyl C=O bonds are shortest due to double-bond character.

✅ Correct Answer Explanation

1 → Methanol has a C-O bond length of approximately 143 pm. 2 → Methoxymethane has a C-O bond length of approximately 141 pm. 3 → Phenol has a C-O bond length of approximately 136 pm because conjugation with the aromatic ring gives the bond partial double-bond character. 4 → Methanal contains a carbonyl (C=O) bond, which is significantly shorter than all C-O single bonds because of its full double-bond character. Therefore, the decreasing order is: Methanol > Methoxymethane > Phenol > Methanal Hence, option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Interchanges methanol and methoxymethane.
�� Option C → Gives the reverse trend.
�� Option D → Places phenol before methoxymethane incorrectly.

🎯 Strategy

Used

Option Grouping

Application:

Compare NCERT bond-length values and the effect of resonance and bond order.

Final Logic:

C-O (alcohol) > C-O (ether) > C-O (phenol) > C=O.

🧠 Memory Trick

Alcohol > Ether > Phenol > Carbonyl

📚 Class XII – NCERT Chemistry Part II – Chapter: Alcohols, Phenols and Ethers – Topic: Bond Lengths and Resonance Effects

5 Why is 2-methylpropan-1-ol classified as a primary alcohol despite having a branched carbon chain?

A The -OH group is attached to an sp² hybridized carbon

B The carbon bearing the -OH group is directly attached to only one other carbon atom

C The carbon bearing the -OH group is directly attached to three other carbon atoms

D It contains only one hydroxyl group

💡 Short Explanation

�� Alcohol classification depends on the OH-bearing carbon. �� Branching elsewhere does not affect classification. �� The OH-bearing carbon is primary.

✅ Correct Answer Explanation

In 2-methylpropan-1-ol, the carbon attached to the hydroxyl group is bonded to only one other carbon atom. Therefore, it is classified as a primary alcohol regardless of branching in the carbon chain. Hence, option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → The OH-bearing carbon is sp³ hybridized.
�� Option C → Describes a tertiary alcohol.
�� Option D → Indicates monohydric nature, not primary classification.

🎯 Strategy

Used

Elimination

Application:

Focus on the carbon directly bonded to OH.

Final Logic:

One carbon neighbour = Primary alcohol.

🧠 Memory Trick

Primary = One Carbon Attached

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

6 IUPAC name of the secondary cyclic alcohol with a methyl group at the second position is:

A 2-Methylcyclopentanol

B 1-Methylcyclopentan-2-ol

C Cyclopentylmethanol

D 2-Methylcyclohexanol

💡 Short Explanation

�� The hydroxyl group gets priority in numbering. �� Cyclopentanol is the parent structure. �� Methyl is located at carbon 2.

✅ Correct Answer Explanation

In cycloalkanols, the carbon bearing the hydroxyl group is assigned position 1. A methyl group at the adjacent carbon gives the name 2-methylcyclopentanol. Therefore, option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Incorrect numbering system.
�� Option C → Different structural arrangement.
�� Option D → Contains six carbons in the ring.

🎯 Strategy

Used

Substitution

Application:

Apply IUPAC numbering rules for cycloalkanols.

Final Logic:

Cyclopentanol parent + methyl at C-2.

🧠 Memory Trick

OH Gets Position 1

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

7 Tertiary alcohol properties:
1. The -OH is attached to a tertiary carbon atom
2. An example is 2-methylpropan-2-ol
3. They contain a C(sp³)-OH bond
4. They form turbidity immediately with Lucas reagent at room temperature

A (1), (2)

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

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

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

💡 Short Explanation

�� Tertiary alcohols contain tertiary carbon atoms. �� tert-Butyl alcohol is a common example. �� They react fastest with Lucas reagent.

✅ Correct Answer Explanation

Statement 1 is correct because the hydroxyl group is attached to a tertiary carbon. Statement 2 is correct because 2-methylpropan-2-ol is a tertiary alcohol. Statement 3 is correct because tertiary alcohols contain a C(sp³)-OH bond. Statement 4 is correct because tertiary alcohols produce immediate turbidity with Lucas reagent at room temperature. Therefore, all four statements are correct.

❓ Why Other Options Are Incorrect

�� Option A → Omits statements 3 and 4.
�� Option B → Omits statement 4.
�� Option C → Omits statement 1.

🎯 Strategy

Used

Elimination

Application:

Verify each defining property of tertiary alcohols.

Final Logic:

All four statements are correct.

🧠 Memory Trick

Tertiary = Instant Lucas Test

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

8 Match List I with List II.

List I	List II
1. 2-Methylpropan-2-ol	a. Primary allylic alcohol
2. Propan-2-ol	b. Primary benzylic alcohol
3. CH₂=CH-CH₂-OH	c. Tertiary alcohol
4. C₆H₅CH₂OH	d. Secondary alcohol

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

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

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

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

💡 Short Explanation

�� Classification depends on OH-bearing carbon. �� Allylic alcohols are adjacent to double bonds. �� Benzylic alcohols are adjacent to aromatic rings.

✅ Correct Answer Explanation

1 → c : 2-Methylpropan-2-ol is a tertiary alcohol. 2 → d : Propan-2-ol is a secondary alcohol. 3 → a : CH₂=CH-CH₂OH is a primary allylic alcohol. 4 → b : C₆H₅CH₂OH is a primary benzylic alcohol. Therefore, option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Tertiary and secondary alcohols are interchanged.
�� Option C → Multiple classifications are incorrect.
�� Option D → Propan-2-ol is not an allylic alcohol.

🎯 Strategy

Used

Option Grouping

Application:

Classify each compound using structural features.

Final Logic:

tert-Butyl → Tertiary, Propan-2-ol → Secondary, Allyl alcohol → Allylic, Benzyl alcohol → Benzylic.

🧠 Memory Trick

Allyl–Double Bond, Benzyl–Benzene

📚 Class XII – NCERT Chemistry Part II – Chapter: Alcohols, Phenols and Ethers – Topic: Allylic, Benzylic and Alcohol Classification

📄 Passage

Hydroxy compounds containing aromatic rings display different properties depending on the exact point of attachment of the hydroxyl group. In benzylic alcohols, the -OH group is attached to an sp³-hybridized carbon atom next to an aromatic ring. Benzylic alcohols can be further classified as primary, secondary, or tertiary. On the other hand, in compounds containing a C(sp²)-OH bond, such as vinylic alcohols and phenols, the -OH group is bonded to a carbon-carbon double bond or to an aryl carbon. The carbon-oxygen bond length in phenol (136 pm) is slightly less than that in methanol due to partial double bond character resulting from conjugation, and the sp² hybridized state of the carbon.

According to the passage, benzylic alcohols differ from phenols primarily because:

A Benzylic alcohols have an -OH attached to an sp² carbon

B Benzylic alcohols have an -OH attached to an sp³ carbon next to an aromatic ring

C Phenols do not have an aromatic ring

D Benzylic alcohols exhibit stronger partial double bond character

💡 Short Explanation

�� Benzylic alcohols contain C(sp³)-OH bonds. �� Phenols contain C(sp²)-OH bonds. �� Position of attachment determines classification.

✅ Correct Answer Explanation

The passage states that benzylic alcohols contain an OH group attached to an sp³-hybridized carbon adjacent to an aromatic ring, whereas phenols contain an OH group attached directly to an aromatic sp² carbon. Therefore, option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Describes phenols, not benzylic alcohols.
�� Option C → Phenols are aromatic compounds.
�� Option D → Partial double-bond character is associated with phenols.

🎯 Strategy

Used

Contextual/Tonal Matching

Application:

Identify the exact distinction given in the passage.

Final Logic:

Benzylic = sp³ adjacent to ring; Phenol = sp² on ring.

🧠 Memory Trick

Benzylic = Beside Ring

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

📄 Passage

Based on the passage, the C-O bond in phenol is shorter than in methanol because:

A The carbon is sp³ hybridized in phenol

B The conjugation of unshared electron pairs of oxygen with the aromatic ring gives it partial double bond character

C Vinylic alcohols have completely triple bonds

D Methanol lacks any hydrogen atoms

💡 Short Explanation

�� Phenol exhibits resonance. �� Conjugation shortens the C-O bond. �� Partial double-bond character develops.

✅ Correct Answer Explanation

The passage states that conjugation between the lone pair of oxygen and the aromatic ring gives the C-O bond in phenol partial double-bond character. This decreases the bond length relative to methanol. Therefore, option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Phenolic carbon is sp² hybridized, not sp³.
�� Option C → Vinylic alcohols contain double bonds, not triple bonds.
�� Option D → Methanol contains hydrogen atoms.

🎯 Strategy

Used

Contextual/Tonal Matching

Application:

Use the exact explanation given in the passage.

Final Logic:

Resonance → Partial double bond → Shorter C-O bond.

🧠 Memory Trick

Resonance Shortens Bonds

📚 Class XII – NCERT Chemistry Part II – Chapter: Alcohols, Phenols and Ethers – Topic: Resonance and Bond Length in Phenol

11 Arrange the following in decreasing order of standard boiling point as indicated by their increasing molecular mass and number of hydroxyl groups (assume straight chains):
1. Ethane-1,2-diol
2. Propane-1,2,3-triol
3. Ethanol
4. Methanol

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

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

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

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

💡 Short Explanation

�� Boiling point increases with hydrogen bonding. �� More hydroxyl groups increase intermolecular attraction. �� Glycerol has the highest boiling point.

✅ Correct Answer Explanation

Propane-1,2,3-triol contains three hydroxyl groups and exhibits extensive hydrogen bonding, giving it the highest boiling point. Ethane-1,2-diol contains two hydroxyl groups and has the next highest boiling point. Ethanol contains one hydroxyl group and has a lower boiling point than ethylene glycol. Methanol has the smallest molecular mass and only one hydroxyl group, resulting in the lowest boiling point among the given compounds. Therefore, the decreasing order is: Propane-1,2,3-triol > Ethane-1,2-diol > Ethanol > Methanol

❓ Why Other Options Are Incorrect

�� Option A → Ethylene glycol cannot have a higher boiling point than glycerol.
�� Option C → Gives an incorrect reverse trend.
�� Option D → Places ethanol ahead of ethylene glycol incorrectly.

🎯 Strategy

Used

Option Grouping

Application:

Compare hydroxyl-group count and molecular mass.

Final Logic:

3 OH > 2 OH > 1 OH (higher mass) > 1 OH (lower mass).

🧠 Memory Trick

Triol > Diol > Ethanol > Methanol

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

12 Identify the classification type of the compound represented by the formula HO-CH₂-CH(OH)-CH₃.

A Monohydric alcohol

B Dihydric alcohol

C Trihydric alcohol

D Polyhydric phenol

💡 Short Explanation

�� The compound contains two hydroxyl groups. �� Both hydroxyl groups are attached to an aliphatic chain. �� It is a glycol-type alcohol.

✅ Correct Answer Explanation

The structure HO-CH₂-CH(OH)-CH₃ contains two hydroxyl groups. Alcohols containing two hydroxyl groups are classified as dihydric alcohols. Therefore, option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Contains only one hydroxyl group.
�� Option C → Requires three hydroxyl groups.
�� Option D → The compound is an alcohol, not a phenol.

🎯 Strategy

Used

Odd One Out

Application:

Count the number of hydroxyl groups present.

Final Logic:

Two hydroxyl groups = Dihydric alcohol.

🧠 Memory Trick

2 OH = Dihydric

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

13 The correct IUPAC methodology for naming a trihydric alcohol, such as propane-1,2,3-triol, specifies that:

A The 'e' of the parent alkane name is dropped

B The multiplicative prefix 'tri' is placed before 'ol' and the 'e' of alkane is retained

C The suffix is replaced entirely by 'triol' without locants

D The prefix 'hydroxyl' must be used three times

💡 Short Explanation

�� Polyhydric alcohols retain the terminal "e". �� Trihydric alcohols use the suffix "triol". �� Locants must be specified.

✅ Correct Answer Explanation

According to IUPAC nomenclature, polyhydric alcohols retain the terminal "e" of the parent alkane. The multiplicative prefix "tri" is added before "ol" to form the suffix "triol". Thus, propane-1,2,3-triol is the correct name. Therefore, option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → The terminal "e" is retained in polyhydric alcohol nomenclature.
�� Option C → Locants are essential in IUPAC names.
�� Option D → Hydroxyl is not repeatedly used as a prefix in this case.

🎯 Strategy

Used

Elimination

Application:

Apply NCERT nomenclature rules for polyhydric alcohols.

Final Logic:

Retain "e" + use triol + specify locants.

🧠 Memory Trick

Polyhydric = Keep the "e"

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

14 IUPAC name of the dihydric alcohol with two carbons is:

A Ethylene glycol

B Methanediol

C Dimethyl ether

D Ethane-1,2-diol

💡 Short Explanation

�� The parent chain contains two carbon atoms. �� Two hydroxyl groups are present. �� It is the IUPAC name of ethylene glycol.

✅ Correct Answer Explanation

A two-carbon alcohol containing hydroxyl groups at carbon 1 and carbon 2 is named ethane-1,2-diol according to IUPAC nomenclature. Therefore, option D is correct.

❓ Why Other Options Are Incorrect

�� Option A → Common name, not IUPAC name.
�� Option B → Contains only one carbon atom.
�� Option C → An ether, not an alcohol.

🎯 Strategy

Used

Substitution

Application:

Determine the parent chain and hydroxyl positions.

Final Logic:

Two carbons + two OH groups = Ethane-1,2-diol.

🧠 Memory Trick

Ethane + Diol = Ethane-1,2-diol

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

15 Match List I with List II.

List I	List II
1. o-Nitrophenol	a. 10.0
2. Phenol	b. 10.1
3. m-Cresol	c. 7.2
4. p-Nitrophenol	d. 7.1

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

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

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

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

💡 Short Explanation

�� Nitro groups increase acidity. �� Cresol is less acidic than phenol. �� Lower pKa indicates stronger acidity.

✅ Correct Answer Explanation

1 → c : o-Nitrophenol has a pKa of approximately 7.2. 2 → a : Phenol has a pKa of approximately 10.0. 3 → b : m-Cresol has a pKa of approximately 10.1. 4 → d : p-Nitrophenol has a pKa of approximately 7.1. Therefore, option C is correct.

❓ Why Other Options Are Incorrect

�� Option A → Multiple pKa values are interchanged.
�� Option B → Nitro-substituted phenols and phenol are incorrectly matched.
�� Option D → Incorrect assignment of acidic strengths.

🎯 Strategy

Used

Option Grouping

Application:

Match each phenol with its NCERT pKa value.

Final Logic:

Nitrophenols have lower pKa than phenol and cresol.

🧠 Memory Trick

Nitro = Strong Acid = Low pKa

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

16 Dihydric phenol properties:
1. They are designated as 1,2-, 1,3-, and 1,4-benzenediols
2. Resorcinol is an example of a 1,3-benzenediol
3. They possess exactly two hydroxyl groups attached to the aromatic ring
4. Hydroquinone is considered a monohydric phenol

A (1), (2)

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

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

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

💡 Short Explanation

�� Dihydric phenols contain two hydroxyl groups. �� Catechol, resorcinol and hydroquinone are examples. �� Hydroquinone is not monohydric.

✅ Correct Answer Explanation

Statement 1 is correct because dihydric phenols are named as benzene-1,2-diol, benzene-1,3-diol and benzene-1,4-diol. Statement 2 is correct because resorcinol is benzene-1,3-diol. Statement 3 is correct because dihydric phenols contain exactly two hydroxyl groups attached to the aromatic ring. Statement 4 is incorrect because hydroquinone is benzene-1,4-diol and therefore a dihydric phenol. Therefore, option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Omits statement 3, which is correct.
�� Option C → Includes statement 4, which is incorrect.
�� Option D → Includes statement 4, which is incorrect and omits statement 2.

🎯 Strategy

Used

Elimination

Application:

Verify each statement using definitions and examples of dihydric phenols.

Final Logic:

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

🧠 Memory Trick

Cate-2, Res-3, Hydro-4

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

17 In the nomenclature of trihydric phenols (e.g., benzene-1,2,3-triol), the unit used to specify the exact positions of the substituent groups on the benzene ring is:

A Greek letters (alpha, beta, gamma)

B Arabic numerals (1, 2, 3)

C Roman numerals (I, II, III)

D Alphabetical letters (a, b, c)

💡 Short Explanation

�� IUPAC nomenclature uses locants. �� Locants are written as Arabic numerals. �� Positions identify substituent locations.

✅ Correct Answer Explanation

In IUPAC nomenclature, the positions of substituents on the benzene ring are indicated using Arabic numerals such as 1, 2 and 3. Hence, benzene-1,2,3-triol specifies the exact positions of the hydroxyl groups. Therefore, option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Not used in modern IUPAC nomenclature.
�� Option C → Roman numerals are not used as locants.
�� Option D → Alphabetical letters do not indicate positions.

🎯 Strategy

Used

Odd One Out

Application:

Identify the numbering system used in IUPAC nomenclature.

Final Logic:

IUPAC locants are Arabic numerals.

🧠 Memory Trick

IUPAC = 1, 2, 3

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

18 Arrange the following substituted phenols in decreasing order of their acidic strength (highest to lowest, based on pKa values):
1. p-Cresol
2. Phenol
3. p-Nitrophenol
4. 2,4,6-Trinitrophenol

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

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

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

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

💡 Short Explanation

�� Nitro groups increase acidity. �� Methyl groups decrease acidity. �� Lower pKa means stronger acid.

✅ Correct Answer Explanation

2,4,6-Trinitrophenol contains three strongly electron-withdrawing nitro groups and is the most acidic. p-Nitrophenol is more acidic than phenol due to the nitro group. Phenol is more acidic than p-cresol because the methyl group donates electron density and decreases acidity. Therefore, the decreasing order is: 2,4,6-Trinitrophenol > p-Nitrophenol > Phenol > p-Cresol

❓ Why Other Options Are Incorrect

�� Option A → Gives the reverse acidity trend.
�� Option B → Places p-nitrophenol above trinitrophenol incorrectly.
�� Option D → Places phenol above p-nitrophenol incorrectly.

🎯 Strategy

Used

Option Grouping

Application:

Compare electron-withdrawing and electron-donating substituents.

Final Logic:

More nitro groups = Higher acidity.

🧠 Memory Trick

Tri-Nitro > Nitro > Phenol > Cresol

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

19 IUPAC name of the symmetrical aromatic ether C₆H₅OC₆H₅ is:

A Diphenyl ether

B Phenoxybenzene

C Ethoxybenzene

D Methoxybenzene

💡 Short Explanation

�� The ether contains two phenyl groups. �� IUPAC uses alkoxy/aroxy nomenclature. �� Phenoxybenzene is the preferred IUPAC name.

✅ Correct Answer Explanation

The compound C₆H₅OC₆H₅ consists of two phenyl groups connected through an oxygen atom. The preferred IUPAC name is phenoxybenzene, while diphenyl ether is the common name. Therefore, option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Common name, not the preferred IUPAC name.
�� Option C → Refers to ethoxybenzene.
�� Option D → Refers to methoxybenzene (anisole).

🎯 Strategy

Used

Elimination

Application:

Differentiate between common and IUPAC nomenclature.

Final Logic:

C₆H₅OC₆H₅ = Phenoxybenzene.

🧠 Memory Trick

Phenoxy + Benzene

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

20 Which of the following best explains why C₂H₅OCH₃ is classified as an unsymmetrical ether?

A It contains two identical ethyl groups

B It contains an ethyl group and a methyl group bonded to the same oxygen atom

C It contains an aromatic ring bonded to an oxygen

D The C-O-C bond angle is exactly 180 degrees

💡 Short Explanation

�� Unsymmetrical ethers contain different groups. �� One side contains methyl. �� The other side contains ethyl.

✅ Correct Answer Explanation

C₂H₅OCH₃ contains two different alkyl groups attached to the oxygen atom: an ethyl group and a methyl group. Ethers with different groups on either side of oxygen are classified as unsymmetrical or mixed ethers. Therefore, option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Identical groups would form a symmetrical ether.
�� Option C → No aromatic ring is present.
�� Option D → Ether bond angles are not 180°.

🎯 Strategy

Used

Option Grouping

Application:

Compare the groups attached on both sides of oxygen.

Final Logic:

Different alkyl groups = Unsymmetrical ether.

🧠 Memory Trick

Different Sides = Mixed Ether

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

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