CUET UG Chemistry Booster Test

🚀

CUET UG Chemistry Booster Test - 3 Nomenclature

⏱️ 30 Minutes

❓ 20 Questions

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

CUET UG Chemistry Booster Test - 3 Nomenclature

1 / 20

QUESTION 1 OF 20

When distinguishing common naming structures, what conceptual difference separates an isobutyl group from a sec-butyl group in alcohols?

Answer Review

1 When distinguishing common naming structures, what conceptual difference separates an isobutyl group from a sec-butyl group in alcohols?

A Isobutyl has a linear carbon chain, while sec-butyl is branched.

B Isobutyl is a primary alcohol with a branched chain, while sec-butyl is a secondary alcohol.

C Isobutyl refers to a tertiary carbon, while sec-butyl refers to a primary carbon.

D Both are completely synonymous in IUPAC nomenclature.

💡 Short Explanation

�� Isobutyl alcohol is 2-methylpropan-1-ol. �� sec-Butyl alcohol is butan-2-ol. �� Primary and secondary classifications differ.

✅ Correct Answer Explanation

Isobutyl alcohol (2-methylpropan-1-ol) contains a branched carbon skeleton, but the hydroxyl-bearing carbon is attached to only one other carbon atom, making it a primary alcohol. sec-Butyl alcohol (butan-2-ol) has the hydroxyl-bearing carbon attached to two other carbon atoms, making it a secondary alcohol. Therefore, option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Isobutyl is also branched, not linear.
�� Option C → Incorrect alcohol classifications.
�� Option D → They are different compounds.

🎯 Strategy

Used

Elimination

Application:

Compare the carbon attached directly to the hydroxyl group.

Final Logic:

Primary branched alcohol vs secondary alcohol.

🧠 Memory Trick

Iso = Primary, Sec = Secondary

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

2 Which of the following strictly adhere to the IUPAC rules for naming alcohols?
1. In polyhydric alcohols, the 'e' of alkane is retained and ending 'ol' added.
2. The position of substituents are indicated using numerals determined by proximity to the hydroxyl group.
3. 'Cyclo' is added before the parent chain of any alcohol containing a benzene ring.
4. Multiplicative prefixes do not alter the retention of the 'e' in polyhydric species.

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

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

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

D (1), (3)

💡 Short Explanation

�� Polyhydric alcohols retain the terminal "e". �� OH group gets numbering priority. �� Cyclo is used only for cyclic aliphatic systems.

✅ Correct Answer Explanation

Statement 1 is correct because names such as ethane-1,2-diol retain the terminal "e". Statement 2 is correct because numbering is based on giving the hydroxyl group the lowest possible locant. Statement 3 is incorrect because benzene-containing compounds are not named using the prefix "cyclo". Statement 4 is correct because the use of diol, triol, etc. retains the "e" of the parent alkane. Therefore, statements 1, 2 and 4 are correct.

❓ Why Other Options Are Incorrect

�� Option B → Includes statement 3, which is incorrect.
�� Option C → Includes statement 3, which is incorrect.
�� Option D → Omits statements 2 and 4.

🎯 Strategy

Used

Elimination

Application:

Check each statement against NCERT nomenclature rules.

Final Logic:

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

🧠 Memory Trick

Poly = Keep e

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

3 What is the IUPAC name of the complex unsaturated alcohol CH₂=CH–CH(OH)–CH₃?

A But-3-en-2-ol

B But-1-en-3-ol

C 2-Hydroxybut-3-ene

D 3-Hydroxybut-1-ene

💡 Short Explanation

�� OH group receives numbering priority. �� Double bond is also included in the parent chain. �� Alcohol suffix has priority over alkene numbering.

✅ Correct Answer Explanation

Numbering is carried out from the end nearest to the hydroxyl group. This gives the hydroxyl group locant 2 and the double bond locant 3. Therefore, the correct IUPAC name is But-3-en-2-ol.

❓ Why Other Options Are Incorrect

�� Option B → Gives lower priority to OH.
�� Option C → Uses hydroxy as a prefix instead of the principal suffix.
�� Option D → Incorrect numbering convention.

🎯 Strategy

Used

Substitution

Application:

Assign the lowest possible number to the hydroxyl group.

Final Logic:

OH priority → But-3-en-2-ol.

🧠 Memory Trick

OH Before EN

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

4 Arrange the sequence of the substituent locants in ascending numerical order for the molecule 4-Chloro-3-ethyl-2-(1-methylethyl)-butan-1-ol:
1. Chloro
2. (1-methylethyl)
3. Ethyl
4. Hydroxyl (ol)

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

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

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

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

💡 Short Explanation

�� OH group determines numbering. �� Remaining substituents receive locants accordingly. �� Ascending order follows the assigned positions.

✅ Correct Answer Explanation

From the name: 4 → Hydroxyl group at carbon 1. 2 → (1-methylethyl) group at carbon 2. 3 → Ethyl group at carbon 3. 1 → Chloro group at carbon 4. Therefore, the ascending locant order is: Hydroxyl (1) < (1-methylethyl) (2) < Ethyl (3) < Chloro (4).

❓ Why Other Options Are Incorrect

�� Option B → Assigns incorrect locants.
�� Option C → Incorrect ordering.
�� Option D → Gives descending order.

🎯 Strategy

Used

Option Grouping

Application:

Read the locants directly from the IUPAC name.

Final Logic:

1 < 2 < 3 < 4.

🧠 Memory Trick

OH Starts the Count

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

5 In compounds containing both a double bond and a hydroxyl group (e.g., Hex-1-en-3-ol):
1. The alkene gets numbering priority over the hydroxyl group.
2. The hydroxyl group gets numbering priority, dictating the end from which the chain is numbered.
3. The suffix 'en' comes before 'ol' in the final name.
4. The parent chain must contain both the double bond and the hydroxyl carbon.

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

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

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

D (2), (3)

💡 Short Explanation

�� OH has higher priority than C=C. �� Both functional features remain in the parent chain. �� Names contain both "en" and "ol".

✅ Correct Answer Explanation

Statement 1 is incorrect because the hydroxyl group has higher priority than the double bond. Statement 2 is correct because numbering is chosen to give the hydroxyl group the lowest locant. Statement 3 is correct because names are written in the form en-ol. Statement 4 is correct because the selected parent chain must contain both the double bond and the carbon bearing the hydroxyl group. Therefore, statements 2, 3 and 4 are correct.

❓ Why Other Options Are Incorrect

�� Option A → Includes statement 1, which is incorrect.
�� Option C → Includes statement 1, which is incorrect.
�� Option D → Omits statement 4.

🎯 Strategy

Used

Elimination

Application:

Apply priority rules between hydroxyl and alkene groups.

Final Logic:

OH > C=C for numbering priority.

🧠 Memory Trick

OH Wins Over EN

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

6 Match List I with List II.

List I	List II
1. 4-Chloro-2,3-dimethylpentan-1-ol	a. A diol with methyls at C-2 and C-5
2. 2-Ethoxy-1,1-dimethylcyclohexane	b. An ether with a nitro group at C-2
3. 1-Ethoxy-2-nitrocyclohexane	c. A primary alcohol with chloro at C-4
4. 2,5-Dimethylhexane-1,3-diol	d. A cyclic ether with gem-dimethyls at C-1

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

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

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

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

💡 Short Explanation

�� Names directly describe structural features. �� Functional groups determine classification. �� Locants identify positions precisely.

✅ Correct Answer Explanation

1 → c : Primary alcohol with chloro at C-4. 2 → d : Cyclohexane containing ethoxy and gem-dimethyl substitution. 3 → b : Ether with a nitro substituent at carbon 2. 4 → a : Diol with methyl groups at carbons 2 and 5. Therefore, option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Multiple mismatches.
�� Option C → Ether structures incorrectly paired.
�� Option D → Primary alcohol incorrectly matched.

🎯 Strategy

Used

Option Grouping

Application:

Match structural descriptions with their nomenclature.

Final Logic:

Interpret locants and functional groups directly.

🧠 Memory Trick

Name → Structure Match

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

7 The IUPAC name of the heavily substituted diol structure CH₃–CH(CH₃)–CH₂–CH(OH)–CH(CH₃)–CH₂–OH is:

A 2-Methyl-5-methylhexane-1,3-diol

B 2,5-Dimethylhexane-4,6-diol

C 1,3-Dihydroxy-2,5-dimethylhexane

D 2,5-Dimethylhexane-1,3-diol

💡 Short Explanation

�� Longest chain contains six carbons. �� Two hydroxyl groups are present. �� Lowest locants are assigned to OH groups.

✅ Correct Answer Explanation

Numbering from the hydroxyl end gives hydroxyl positions 1 and 3 and methyl substituents at positions 2 and 5. Therefore, the correct IUPAC name is 2,5-Dimethylhexane-1,3-diol.

❓ Why Other Options Are Incorrect

�� Option A → Improper substituent naming.
�� Option B → Incorrect hydroxyl numbering.
�� Option C → Not the preferred NCERT naming style.

🎯 Strategy

Used

Substitution

Application:

Assign lowest locants to hydroxyl groups.

Final Logic:

OH positions = 1,3; methyls = 2,5.

🧠 Memory Trick

OH First, Then Methyl

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

8 Why is the formal IUPAC name of glycerol "Propane-1,2,3-triol" rather than simply "Propanetriol"?

A Because 'triol' implies the presence of three distinct carbon chains.

B Locants are always strictly required in polyhydric systems to prevent isomer ambiguity, regardless of the carbon count.

C The locants must explicitly prove the molecule isn't Propane-1,1,1-triol.

D Because the common name must be overridden by numbers.

💡 Short Explanation

�� Polyhydric alcohols require locants. �� Positions of OH groups must be specified. �� Prevents ambiguity.

✅ Correct Answer Explanation

IUPAC nomenclature requires explicit locants for hydroxyl groups in polyhydric alcohols. The name Propane-1,2,3-triol clearly specifies the locations of all three hydroxyl groups and avoids ambiguity. Therefore, option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Triol does not imply multiple carbon chains.
�� Option C → Too specific and not the general rule.
�� Option D → Not the reason for locants.

🎯 Strategy

Used

Elimination

Application:

Apply IUPAC rules for naming polyhydric alcohols.

Final Logic:

Locants ensure unambiguous structure identification.

🧠 Memory Trick

Polyhydric = Position Required

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

9 What is the proper IUPAC name of a cyclohexane ring bearing a hydroxyl group at C-1 and a bromine atom at C-3?

A 3-Bromocyclohexanol

B 1-Bromo-3-cyclohexanol

C Cyclohexyl-3-bromide

D 3-Bromophenol

💡 Short Explanation

�� OH-bearing carbon is C-1. �� Bromine occupies carbon 3. �� Parent compound is cyclohexanol.

✅ Correct Answer Explanation

In cyclic alcohols, the carbon bearing the hydroxyl group is automatically assigned position 1. Bromine is located at carbon 3. Therefore, the correct name is 3-Bromocyclohexanol.

❓ Why Other Options Are Incorrect

�� Option B → Incorrect numbering style.
�� Option C → Incorrect parent structure.
�� Option D → Phenol contains a benzene ring.

🎯 Strategy

Used

Substitution

Application:

Assign OH-bearing carbon as C-1.

Final Logic:

Cyclohexanol + Br at C-3.

🧠 Memory Trick

OH = 1 in Rings

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

10 Identify the structural classification of Cyclopent-3-en-1-ol:

A Saturated cyclic primary alcohol

B Unsaturated cyclic secondary alcohol

C Aromatic dihydric phenol

D Unsymmetrical cyclic ether

💡 Short Explanation

�� Contains a ring structure. �� Contains a carbon-carbon double bond. �� Hydroxyl-bearing carbon is secondary.

✅ Correct Answer Explanation

Cyclopent-3-en-1-ol contains a cyclopentane ring with a double bond, making it unsaturated. The carbon bearing the hydroxyl group is attached to two other ring carbons and is therefore secondary. Hence, the compound is classified as an unsaturated cyclic secondary alcohol.

❓ Why Other Options Are Incorrect

�� Option A → Compound is unsaturated.
�� Option C → Not aromatic and contains only one OH group.
�� Option D → Not an ether.

🎯 Strategy

Used

Elimination

Application:

Identify ring type, saturation and alcohol classification.

Final Logic:

Cyclic + C=C + Secondary OH carbon.

🧠 Memory Trick

Ring + EN + OH = Unsaturated Alcohol

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

11 Match List I with List II.

List I	List II
1. Catechol	a. Benzene-1,3-diol
2. Resorcinol	b. Benzene-1,4-diol
3. Hydroquinone	c. Benzene-1,2-diol
4. Phenol	d. Hydroxybenzene

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

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

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

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

💡 Short Explanation

�� Catechol is the ortho dihydric phenol. �� Resorcinol is the meta dihydric phenol. �� Hydroquinone is the para dihydric phenol.

✅ Correct Answer Explanation

1 → c : Catechol = Benzene-1,2-diol. 2 → a : Resorcinol = Benzene-1,3-diol. 3 → b : Hydroquinone = Benzene-1,4-diol. 4 → d : Phenol = Hydroxybenzene. Therefore, option C is correct.

❓ Why Other Options Are Incorrect

�� Option A → Catechol and hydroquinone are interchanged.
�� Option B → All three dihydric phenols are incorrectly matched.
�� Option D → Resorcinol and hydroquinone are interchanged.

🎯 Strategy

Used

Option Grouping

Application:

Match common names of dihydric phenols with their IUPAC names.

Final Logic:

Cate = 1,2; Res = 1,3; Hydro = 1,4.

🧠 Memory Trick

Cate-2, Res-3, Hydro-4

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

12 The IUPAC name of a phenol molecule with two methyl substituents placed at both ortho positions relative to the hydroxyl group is:

A 2,4-Dimethylphenol

B 1,3-Dimethylbenzene

C 3,5-Dimethylphenol

D 2,6-Dimethylphenol

💡 Short Explanation

�� Hydroxyl-bearing carbon is C-1. �� Ortho positions are C-2 and C-6. �� Two methyl groups occupy these positions.

✅ Correct Answer Explanation

In phenol, the carbon carrying the hydroxyl group is assigned position 1. The two ortho positions relative to the hydroxyl group are carbons 2 and 6. Therefore, the correct IUPAC name is 2,6-Dimethylphenol.

❓ Why Other Options Are Incorrect

�� Option A → Contains one ortho and one para methyl group.
�� Option B → Not a phenol.
�� Option C → Represents meta substitution.

🎯 Strategy

Used

Substitution

Application:

Assign carbon 1 to OH and identify ortho positions.

Final Logic:

Ortho to OH = 2 and 6.

🧠 Memory Trick

OH = 1 → Ortho = 2 & 6

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

13 The locant unit values corresponding to 'meta' substitution in a disubstituted benzene system are:

A 1,2

B 1,3

C 1,4

D 1,5

💡 Short Explanation

�� Meta means one carbon separation. �� Standard aromatic nomenclature. �� Corresponds to 1,3-substitution.

✅ Correct Answer Explanation

In benzene nomenclature: Ortho = 1,2 Meta = 1,3 Para = 1,4 Therefore, meta substitution corresponds to positions 1 and 3.

❓ Why Other Options Are Incorrect

�� Option A → Ortho substitution.
�� Option C → Para substitution.
�� Option D → Not represented by meta nomenclature.

🎯 Strategy

Used

Memory-Based Elimination

Application:

Recall ortho-meta-para positions.

Final Logic:

Meta = 1,3.

🧠 Memory Trick

O = 2, M = 3, P = 4

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

14 Regarding the structure 2,4,6-Trinitrophenol (Picric acid):
1. It is an extensively substituted phenol.
2. It contains nitro groups at both ortho positions and the para position.
3. It has a nitro group at the meta position.
4. The hydroxyl group is at C-1.

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

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

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

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

💡 Short Explanation

�� Picric acid is 2,4,6-trinitrophenol. �� Nitro groups occupy ortho and para positions. �� OH-bearing carbon is carbon 1.

✅ Correct Answer Explanation

Statement 1 is correct because picric acid is a highly substituted phenol. Statement 2 is correct because nitro groups are present at positions 2 and 6 (ortho) and position 4 (para). Statement 3 is incorrect because no nitro group is present at a meta position. Statement 4 is correct because numbering starts from the hydroxyl-bearing carbon, which is C-1. Therefore, statements 1, 2 and 4 are correct.

❓ Why Other Options Are Incorrect

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

🎯 Strategy

Used

Elimination

Application:

Verify each substituent position from the IUPAC name.

Final Logic:

2,4,6 = Ortho, Para, Ortho.

🧠 Memory Trick

Picric = 2,4,6 Nitro

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

15 The common name of C₆H₅O(CH₂)₆CH₃ is:

A Phenyl heptyl ether

B Heptyl phenyl ether

C Benzyl heptyl ether

D Phenoxyheptane

💡 Short Explanation

�� Oxygen joins a phenyl group and a heptyl group. �� Common names list groups followed by ether. �� This is a mixed ether.

✅ Correct Answer Explanation

The ether contains a phenyl group (C₆H₅–) and a heptyl group ((CH₂)₆CH₃). According to common nomenclature, the two groups are written as separate words followed by "ether". Therefore, the common name is heptyl phenyl ether.

❓ Why Other Options Are Incorrect

�� Option A → Not the NCERT naming order.
�� Option C → Contains a benzyl group, not phenyl.
�� Option D → IUPAC-style nomenclature, not common nomenclature.

🎯 Strategy

Used

Substitution

Application:

Identify the groups attached to oxygen.

Final Logic:

Heptyl + Phenyl + Ether.

🧠 Memory Trick

Phenyl + Alkyl = Alkyl Phenyl Ether

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

16 According to the alphabetical order rule for common names, what is the common name for the molecule 2-Methoxypropane (CH₃–O–CH(CH₃)₂)?

A Propyl methyl ether

B Methyl isopropyl ether

C Isopropyl methyl ether

D Dimethyl ethyl ether

💡 Short Explanation

�� Ether contains methyl and isopropyl groups. �� Common names use alphabetical order. �� Ether is written at the end.

✅ Correct Answer Explanation

The molecule contains a methyl group and an isopropyl group attached to oxygen. In common nomenclature, these groups are written as separate words followed by "ether". NCERT lists this compound as methyl isopropyl ether.

❓ Why Other Options Are Incorrect

�� Option A → Uses propyl instead of isopropyl.
�� Option C → Not the convention used in NCERT examples.
�� Option D → Incorrect structure.

🎯 Strategy

Used

Elimination

Application:

Identify the alkyl groups attached to oxygen.

Final Logic:

Methyl + Isopropyl + Ether.

🧠 Memory Trick

Me + Iso = Ether

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

17 What is the IUPAC name of a cyclohexane ring bearing an ethoxy group at C-1 and a nitro group at C-2?

A 2-Nitro-1-ethoxycyclohexane

B 1-Nitro-2-ethoxycyclohexane

C Ethoxy-nitrocyclohexyl ether

D 1-Ethoxy-2-nitrocyclohexane

💡 Short Explanation

�� Cyclohexane is the parent ring. �� Ethoxy group is at carbon 1. �� Nitro group is at carbon 2.

✅ Correct Answer Explanation

The parent structure is cyclohexane. The substituents are ethoxy at C-1 and nitro at C-2. Therefore, the correct IUPAC name is 1-Ethoxy-2-nitrocyclohexane.

❓ Why Other Options Are Incorrect

�� Option A → Incorrect locant assignment.
�� Option B → Incorrect numbering sequence.
�� Option C → Not a valid IUPAC name.

🎯 Strategy

Used

Substitution

Application:

Assign substituent positions according to the given structure.

Final Logic:

Ethoxy at 1, Nitro at 2.

🧠 Memory Trick

Ethoxy First, Nitro Next

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

18 Arrange the following groups in decreasing priority to serve as the parent chain based on carbon count:
1. Isopentyl (5 carbons)
2. Phenyl (6 carbons)
3. Heptyl (7 carbons)
4. Methyl (1 carbon)

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

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

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

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

💡 Short Explanation

�� Larger carbon frameworks receive parent priority. �� Carbon count determines the order. �� Heptyl contains the highest carbon count.

✅ Correct Answer Explanation

Carbon counts are: 3 → Heptyl = 7 carbons 2 → Phenyl = 6 carbons 1 → Isopentyl = 5 carbons 4 → Methyl = 1 carbon Therefore, decreasing priority is: Heptyl > Phenyl > Isopentyl > Methyl.

❓ Why Other Options Are Incorrect

�� Option B → Places heptyl after smaller groups.
�� Option C → Gives increasing order.
�� Option D → Phenyl should precede isopentyl.

🎯 Strategy

Used

Option Grouping

Application:

Arrange groups according to decreasing carbon count.

Final Logic:

7C > 6C > 5C > 1C.

🧠 Memory Trick

H > P > I > M

📚 Class XII – NCERT Chemistry Part II – Chapter: Alcohols, Phenols and Ethers – Topic: Parent Chain Selection in Ethers

📄 Passage

According to IUPAC system of nomenclature, ethers are regarded as hydrocarbon derivatives in which a hydrogen atom is replaced by an –OR or –OAr group. The larger (R) group is chosen as the parent hydrocarbon. Cyclic alcohols are named using the prefix cyclo and considering the —OH group attached to C-1. For polyhydric alcohols, the number of -OH groups is indicated by adding the multiplicative prefix, di, tri, etc., before 'ol'.

Based on the rules for cyclic alcohols outlined in the passage, what is the IUPAC name for a five-membered carbon ring containing an –OH group and a methyl group on the adjacent carbon?

A 2-Methylcyclopentanol

B 1-Methylcyclopentan-2-ol

C Cyclopentylmethanol

D 2-Methylcyclohexanol

💡 Short Explanation

�� The hydroxyl-bearing carbon is C-1. �� The methyl group is on the adjacent carbon. �� The ring contains five carbons.

✅ Correct Answer Explanation

The parent structure is cyclopentanol. The carbon carrying the hydroxyl group is assigned position 1. The adjacent carbon receives position 2 and bears the methyl substituent. Hence, the correct IUPAC name is 2-Methylcyclopentanol.

❓ Why Other Options Are Incorrect

�� Option B → Incorrect numbering convention.
�� Option C → Different structural arrangement.
�� Option D → Six-membered ring.

🎯 Strategy

Used

Contextual/Tonal Matching

Application:

Apply the cyclic alcohol naming rule from the passage.

Final Logic:

Cyclopentanol + Methyl at C-2.

🧠 Memory Trick

OH = 1 in Rings

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

📄 Passage

Applying the parent chain rule for ethers from the passage, what is the correct IUPAC name for C₆H₅OCH₃ (methyl phenyl ether)?

A Phenoxymethane

B Methoxybenzene

C Benzene methyl ether

D Phenylmethoxide

💡 Short Explanation

�� Benzene is the larger parent structure. �� Methoxy acts as the substituent. �� Ether is named as an alkoxy derivative.

✅ Correct Answer Explanation

The larger group is the benzene ring, which becomes the parent hydrocarbon. The smaller group attached through oxygen is the methoxy substituent. Therefore, the correct IUPAC name is Methoxybenzene.

❓ Why Other Options Are Incorrect

�� Option A → Incorrect parent-chain selection.
�� Option C → Not the accepted IUPAC name.
�� Option D → Not a valid IUPAC name.

🎯 Strategy

Used

Substitution

Application:

Select the larger group as parent and convert the smaller group into an alkoxy substituent.

Final Logic:

Benzene parent + Methoxy substituent = Methoxybenzene.

🧠 Memory Trick

Anisole = Methoxybenzene

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

CUET UG Chemistry Booster Test – 3 Nomenclature

CUET UG Chemistry Booster Test - 3 Nomenclature

Test Complete!

Answer Review