CUET UG Chemistry Booster Test -2 Classification and Basics -

🚀

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

Alcohol structural properties:
1. Contain one or more hydroxyl groups
2. The -OH group is directly attached to a carbon of an aromatic system
3. Formed by replacing a hydrogen atom in an aliphatic hydrocarbon
4. Examples include CH₃OH

Answer Review

1 Alcohol structural properties:
1. Contain one or more hydroxyl groups
2. The -OH group is directly attached to a carbon of an aromatic system
3. Formed by replacing a hydrogen atom in an aliphatic hydrocarbon
4. Examples include CH₃OH

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

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

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

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

💡 Short Explanation

�� Alcohols contain hydroxyl groups. �� Alcohols are derived from aliphatic hydrocarbons. �� Methanol is the simplest alcohol.

✅ Correct Answer Explanation

Statement 1 is correct because alcohols contain one or more hydroxyl (-OH) groups. Statement 2 is incorrect because an -OH group directly attached to an aromatic ring forms a phenol, not an alcohol. Statement 3 is correct because alcohols are formed by replacing a hydrogen atom in an aliphatic hydrocarbon with an -OH group. Statement 4 is correct because CH₃OH (methanol) is an alcohol. Therefore, option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Includes statement 2, which is incorrect.
�� Option C → Includes statement 2 and omits statement 1.
�� Option D → Includes statement 2, which is incorrect.

🎯 Strategy

Used

Elimination

Application:

Check each statement against the NCERT definition of alcohols.

Final Logic:

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

🧠 Memory Trick

Alcohol = Aliphatic + OH

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

2 Which of the following compounds accurately represents the fundamental definition of a phenol?

A C₆H₅CH₂OH

B C₆H₅OH

C CH₃OH

D CH₃OCH₃

💡 Short Explanation

�� Phenol contains a benzene ring. �� OH group is directly attached to the ring. �� Formula of phenol is C₆H₅OH.

✅ Correct Answer Explanation

Phenol is defined as a compound in which the hydroxyl group is directly attached to an aromatic ring. C₆H₅OH satisfies this condition and is therefore phenol. Option A is benzyl alcohol, option C is methanol, and option D is dimethyl ether.

❓ Why Other Options Are Incorrect

�� Option A → Benzyl alcohol; OH is attached to a side-chain carbon.
�� Option C → Methanol is an alcohol, not a phenol.
�� Option D → Dimethyl ether is an ether.

🎯 Strategy

Used

Odd One Out

Application:

Identify the structure with OH directly attached to a benzene ring.

Final Logic:

Phenol = C₆H₅OH.

🧠 Memory Trick

Phenol = Phenyl + OH

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

3 Match List I with List II.

List I	List II
1. CH₃OCH₃	a. Methyl phenyl ether
2. C₂H₅OC₂H₅	b. Diethyl ether
3. C₆H₅OCH₃	c. Ethyl phenyl ether
4. C₆H₅OCH₂CH₃	d. Dimethyl ether

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

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

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

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

💡 Short Explanation

�� Common names depend on groups attached to oxygen. �� Same alkyl groups form simple ethers. �� Aromatic ethers contain phenyl groups.

✅ Correct Answer Explanation

1 → d : CH₃OCH₃ is dimethyl ether. 2 → b : C₂H₅OC₂H₅ is diethyl ether. 3 → a : C₆H₅OCH₃ is methyl phenyl ether (anisole). 4 → c : C₆H₅OCH₂CH₃ is ethyl phenyl ether. Therefore, option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → CH₃OCH₃ and C₆H₅OCH₂CH₃ are incorrectly matched.
�� Option C → Diethyl ether and methyl phenyl ether are interchanged.
�� Option D → CH₃OCH₃ and C₂H₅OC₂H₅ are incorrectly matched.

🎯 Strategy

Used

Option Grouping

Application:

Match ether structures with their standard common names.

Final Logic:

Dimethyl, Diethyl, Methyl Phenyl and Ethyl Phenyl correspond respectively.

🧠 Memory Trick

Di-Me, Di-Et, Anisole, Phenetole

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

4 The bond angle (C-O-C) in ethers is slightly greater than the tetrahedral angle due to the repulsive interaction between the two bulky groups. The value of this angle in methoxymethane is:

A 108.9°

B 109°

C 111.7°

D 120°

💡 Short Explanation

�� Ether bond angle exceeds tetrahedral angle. �� Lone pair and bulky group repulsions affect geometry. �� Methoxymethane has a bond angle of 111.7°.

✅ Correct Answer Explanation

According to NCERT, the C–O–C bond angle in methoxymethane is approximately 111.7°. This is slightly greater than the tetrahedral angle because of repulsion between the bulky alkyl groups attached to oxygen. Therefore, option C is correct.

❓ Why Other Options Are Incorrect

�� Option A → Not the reported NCERT value.
�� Option B → Approximately tetrahedral but not the actual ether angle.
�� Option D → Much larger than the observed value.

🎯 Strategy

Used

Memory-Based Elimination

Application:

Recall the NCERT structural data for methoxymethane.

Final Logic:

Methoxymethane has a C–O–C bond angle of 111.7°.

🧠 Memory Trick

Ether Angle ≈ 112°

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

5 Arrange the following primary alcohols in decreasing order of their carbon chain length:
1. Methanol
2. Butan-1-ol
3. Propan-1-ol
4. Ethanol

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

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

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

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

💡 Short Explanation

�� Carbon chain length decreases from four to one carbon. �� Butanol has the longest chain. �� Methanol has the shortest chain.

✅ Correct Answer Explanation

Butan-1-ol contains four carbon atoms, propan-1-ol contains three, ethanol contains two and methanol contains one. Therefore, the decreasing order of carbon chain length is Butan-1-ol > Propan-1-ol > Ethanol > Methanol.

❓ Why Other Options Are Incorrect

�� Option B → Gives increasing order instead of decreasing order.
�� Option C → Ethanol and propan-1-ol are incorrectly arranged.
�� Option D → Butan-1-ol is not placed first.

🎯 Strategy

Used

Option Grouping

Application:

Count carbon atoms in each alcohol.

Final Logic:

4C > 3C > 2C > 1C.

🧠 Memory Trick

But > Prop > Eth > Meth

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

6 IUPAC name of the secondary alcohol derived from butane is:

A Butan-1-ol

B Butan-2-ol

C 2-Methylpropan-1-ol

D 2-Methylpropan-2-ol

💡 Short Explanation

�� Secondary alcohol has OH on a secondary carbon. �� Butane gives butan-2-ol as the secondary isomer. �� OH-bearing carbon is attached to two carbons.

✅ Correct Answer Explanation

In butan-2-ol, the hydroxyl group is attached to carbon-2, which is bonded to two other carbon atoms. Hence it is a secondary alcohol. Therefore, option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Primary alcohol.
�� Option C → Branched primary alcohol.
�� Option D → Tertiary alcohol.

🎯 Strategy

Used

Substitution

Application:

Locate the hydroxyl group and determine the carbon type.

Final Logic:

OH on secondary carbon = Butan-2-ol.

🧠 Memory Trick

2-ol = Secondary

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

📄 Passage

Monohydric alcohols may be further classified according to the hybridization of the carbon atom to which the hydroxyl group is attached. In compounds containing C(sp³)-OH bonds, the -OH group is attached to an sp³ hybridized carbon. These include primary, secondary, and tertiary alcohols. In allylic alcohols, the -OH group is attached to an sp³ hybridized carbon adjacent to the carbon-carbon double bond. Allylic alcohols may themselves be primary, secondary, or tertiary depending on the number of carbon atoms directly attached to the sp³ carbon bearing the -OH group.

According to the passage, a tertiary alcohol must possess an -OH group attached to:

A An sp² hybridized carbon

B An sp³ hybridized carbon bonded to three other carbons

C A carbon-carbon double bond directly

D An aromatic ring directly

💡 Short Explanation

�� Tertiary alcohols contain C(sp³)-OH bonds. �� OH-bearing carbon is attached to three carbons. �� Classification depends on the carbon attached to OH.

✅ Correct Answer Explanation

The passage states that primary, secondary and tertiary alcohols contain C(sp³)-OH bonds. In a tertiary alcohol, the carbon bearing the hydroxyl group is attached to three other carbon atoms. Hence option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Describes vinylic systems.
�� Option C → OH directly attached to a double-bond carbon is not a tertiary alcohol.
�� Option D → Describes phenolic structures.

🎯 Strategy

Used

Contextual/Tonal Matching

Application:

Use the classification information provided in the passage.

Final Logic:

Tertiary alcohol = OH-bearing sp³ carbon attached to three carbons.

🧠 Memory Trick

Tertiary = Three Carbon Neighbours

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

📄 Passage

Based on the passage, which statement about allylic alcohols is correct?

A They only exist as primary alcohols

B The hydroxyl group is attached to an sp² hybridized carbon

C The carbon attached to the -OH is sp³ hybridized and adjacent to a double bond

D They are classified as aromatic alcohols

💡 Short Explanation

�� Allylic alcohols contain an sp³ carbon. �� The carbon is adjacent to a double bond. �� They may be primary, secondary or tertiary.

✅ Correct Answer Explanation

The passage clearly states that in allylic alcohols, the hydroxyl group is attached to an sp³ hybridized carbon adjacent to a carbon-carbon double bond. Therefore, option C is correct.

❓ Why Other Options Are Incorrect

�� Option A → Allylic alcohols may be primary, secondary or tertiary.
�� Option B → The OH-bearing carbon is sp³, not sp².
�� Option D → Allylic alcohols are not aromatic alcohols.

🎯 Strategy

Used

Contextual/Tonal Matching

Application:

Directly identify the definition given in the passage.

Final Logic:

Allylic alcohol = sp³ carbon adjacent to C=C.

🧠 Memory Trick

Ally = Adjacent to Alkene

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

9 Identify the classification type of the compound where an -OH group is attached to a -CH(CH₃)- group which is directly attached to a benzene ring.

A Primary benzylic alcohol

B Secondary benzylic alcohol

C Tertiary benzylic alcohol

D Vinylic alcohol

💡 Short Explanation

�� OH-bearing carbon is adjacent to a benzene ring. �� That carbon is attached to two carbons. �� Therefore it is secondary benzylic.

✅ Correct Answer Explanation

The carbon bearing the hydroxyl group is benzylic because it is directly attached to a benzene ring. It is secondary because that carbon is attached to two carbon groups. Hence the compound is a secondary benzylic alcohol.

❓ Why Other Options Are Incorrect

�� Option A → OH-bearing carbon is not attached to only one carbon.
�� Option C → OH-bearing carbon is not attached to three carbons.
�� Option D → No carbon-carbon double bond carbon bears the OH group.

🎯 Strategy

Used

Option Grouping

Application:

Determine both benzylic position and degree of substitution.

Final Logic:

Benzylic + two carbon attachments = Secondary benzylic alcohol.

🧠 Memory Trick

Benzylic + Two = Secondary

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

10 In vinylic alcohols, the hybridization of the carbon atom directly bonded to the hydroxyl oxygen is:

A sp³

B dsp²

C sp

D sp²

💡 Short Explanation

�� Vinylic carbon participates in a double bond. �� Double-bond carbons are sp² hybridized. �� OH is directly attached to that carbon.

✅ Correct Answer Explanation

In a vinylic alcohol, the hydroxyl group is directly attached to a carbon atom involved in a carbon-carbon double bond. Such carbons are sp² hybridized. Therefore, option D is correct.

❓ Why Other Options Are Incorrect

�� Option A → Represents saturated tetrahedral carbons.
�� Option B → Not applicable to vinylic carbon atoms.
�� Option C → Represents triple-bond carbons.

🎯 Strategy

Used

Elimination

Application:

Determine hybridization from the presence of a carbon-carbon double bond.

Final Logic:

Vinylic carbon = sp² hybridization.

🧠 Memory Trick

Vinyl = Double Bond = sp²

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

11 IUPAC name of the monohydric alcohol containing three carbon atoms with the -OH on the second carbon is:

A Propan-1-ol

B Propan-2-ol

C Propene-1-ol

D Butan-2-ol

💡 Short Explanation

�� The parent chain contains three carbon atoms. �� The hydroxyl group is on carbon 2. �� It is a secondary alcohol.

✅ Correct Answer Explanation

A three-carbon alcohol is derived from propane. Since the hydroxyl group is attached to carbon 2, the correct IUPAC name is propan-2-ol. Therefore, option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Hydroxyl group is attached to carbon 1.
�� Option C → Contains a double bond and is not derived from propane.
�� Option D → Contains four carbon atoms.

🎯 Strategy

Used

Substitution

Application:

Identify the parent chain and locate the hydroxyl group.

Final Logic:

Three carbons + OH on carbon 2 = Propan-2-ol.

🧠 Memory Trick

3C + OH at 2 = Propan-2-ol

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

12 Match List I with List II.

List I	List II
1. Methanol	a. Dihydric
2. Ethylene glycol	b. Trihydric
3. Glycerol	c. Monohydric
4. Resorcinol	d. Polyhydric

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

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

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

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

💡 Short Explanation

�� Methanol contains one hydroxyl group. �� Ethylene glycol and resorcinol contain two hydroxyl groups. �� Glycerol contains three hydroxyl groups.

✅ Correct Answer Explanation

1 → c : Methanol contains one hydroxyl group and is monohydric. 2 → a : Ethylene glycol contains two hydroxyl groups and is dihydric. 3 → b : Glycerol contains three hydroxyl groups and is trihydric. 4 → a : Resorcinol is benzene-1,3-diol and is dihydric. Therefore, option D is correct.

❓ Why Other Options Are Incorrect

�� Option A → Incorrectly classifies methanol and resorcinol.
�� Option B → Incorrectly classifies methanol and glycerol.
�� Option C → Interchanges dihydric and trihydric classifications.

🎯 Strategy

Used

Option Grouping

Application:

Count the number of hydroxyl groups present in each compound.

Final Logic:

1 OH = Monohydric, 2 OH = Dihydric, 3 OH = Trihydric.

🧠 Memory Trick

Meth-1, Glycol-2, Glycerol-3

📚 Class XII – NCERT Chemistry Part II – Chapter: Alcohols, Phenols and Ethers – Topic: Classification Based on Hydroxyl Count

13 Trihydric alcohol properties:
1. Propane-1,2,3-triol is an example
2. Contain three -OH groups
3. Retain the 'e' of alkane in IUPAC naming
4. Also called catechols

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

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

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

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

💡 Short Explanation

�� Trihydric alcohols contain three hydroxyl groups. �� Glycerol is a common example. �� The parent alkane name retains the 'e'.

✅ Correct Answer Explanation

Statement 1 is correct because propane-1,2,3-triol (glycerol) is a trihydric alcohol. Statement 2 is correct because trihydric alcohols contain three hydroxyl groups. Statement 3 is correct because NCERT nomenclature retains the terminal 'e' of the alkane name in polyhydric alcohols. Statement 4 is incorrect because catechol is a dihydric phenol, not a trihydric alcohol. Therefore, option C is correct.

❓ Why Other Options Are Incorrect

�� Option A → Includes statement 4, which is incorrect.
�� Option B → Includes statement 4, which is incorrect.
�� Option D → Includes statement 4 and omits statement 2.

🎯 Strategy

Used

Elimination

Application:

Verify each statement using definitions and nomenclature rules.

Final Logic:

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

🧠 Memory Trick

Trihydric = Triol = 3 OH

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

14 The IUPAC naming convention for polyhydric alcohols retains the 'e' of the parent alkane name. Which of the following is correct for a dihydric alcohol of ethane?

A Ethanediol

B Ethane-1,2-diol

C Ethanol-1,2

D Ethylenediol

💡 Short Explanation

�� Polyhydric alcohols retain the 'e' of the alkane. �� Two hydroxyl groups require the suffix "diol". �� Locants must be specified.

✅ Correct Answer Explanation

The correct IUPAC name is ethane-1,2-diol because the parent hydrocarbon is ethane and hydroxyl groups are present on carbons 1 and 2. NCERT specifies retention of the terminal 'e' in polyhydric alcohol nomenclature. Therefore, option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Incomplete IUPAC name without locants.
�� Option C → Incorrect naming format.
�� Option D → Common name, not the preferred IUPAC name.

🎯 Strategy

Used

Elimination

Application:

Apply IUPAC nomenclature rules for polyhydric alcohols.

Final Logic:

Ethane + OH on C-1 and C-2 = Ethane-1,2-diol.

🧠 Memory Trick

Diol → Keep the "e"

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

15 IUPAC name of the monohydric phenol with a methyl group at the para position is:

A 2-Methylphenol

B 3-Methylphenol

C 4-Methylphenol

D Benzene-1,4-diol

💡 Short Explanation

�� Para position corresponds to carbon 4. �� Phenol is the parent compound. �� Methyl group occupies the para position.

✅ Correct Answer Explanation

In phenol, the hydroxyl-bearing carbon is numbered as carbon 1. The para position corresponds to carbon 4. Therefore, a methyl substituent at the para position gives the IUPAC name 4-methylphenol. Hence, option C is correct.

❓ Why Other Options Are Incorrect

�� Option A → Represents ortho-methylphenol.
�� Option B → Represents meta-methylphenol.
�� Option D → A dihydric phenol.

🎯 Strategy

Used

Substitution

Application:

Determine the position of the methyl group relative to the hydroxyl group.

Final Logic:

Para = Position 4.

🧠 Memory Trick

Ortho-2, Meta-3, Para-4

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

16 Arrange the following dihydric phenols in decreasing order of the positional numbers of their hydroxyl groups (e.g., 1,4 > 1,3 > 1,2):
1. Catechol
2. Resorcinol
3. Hydroquinone
4. Phenol

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

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

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

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

💡 Short Explanation

�� Hydroquinone is benzene-1,4-diol. �� Resorcinol is benzene-1,3-diol. �� Catechol is benzene-1,2-diol. �� Phenol contains only one hydroxyl group.

✅ Correct Answer Explanation

3 → Hydroquinone corresponds to benzene-1,4-diol. 2 → Resorcinol corresponds to benzene-1,3-diol. 1 → Catechol corresponds to benzene-1,2-diol. 4 → Phenol contains only one hydroxyl group and therefore comes after the dihydric phenols when arranged according to the positional numbering sequence given. Therefore, the correct arrangement is: Hydroquinone > Resorcinol > Catechol > Phenol Hence, option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Gives the reverse order of the dihydric phenols.
�� Option C → Places resorcinol before hydroquinone incorrectly.
�� Option D → Interchanges catechol and resorcinol.

🎯 Strategy

Used

Option Grouping

Application:

Recall the common names and corresponding IUPAC positions of the hydroxyl groups.

Final Logic:

Hydroquinone (1,4) > Resorcinol (1,3) > Catechol (1,2) > Phenol.

🧠 Memory Trick

Hydro-4, Res-3, Cate-2, Phenol-1

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

17 Identify the classification type of a compound having a benzene ring directly attached to three -OH groups.

A Monohydric phenol

B Dihydric phenol

C Trihydric phenol

D Substituted ether

💡 Short Explanation

�� Classification depends on hydroxyl count. �� Three hydroxyl groups indicate trihydric nature. �� All hydroxyl groups are attached to the aromatic ring.

✅ Correct Answer Explanation

A phenol containing three hydroxyl groups attached directly to a benzene ring is classified as a trihydric phenol. Therefore, option C is correct.

❓ Why Other Options Are Incorrect

�� Option A → Contains only one hydroxyl group.
�� Option B → Contains two hydroxyl groups.
�� Option D → Does not contain the phenolic hydroxyl arrangement.

🎯 Strategy

Used

Odd One Out

Application:

Count the number of hydroxyl groups attached to the ring.

Final Logic:

Three hydroxyl groups = Trihydric phenol.

🧠 Memory Trick

Trihydric = Three OH

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

18 When phenol is substituted with a methyl group at the ortho position, the common name of the resulting compound is:

A o-Cresol

B m-Cresol

C p-Cresol

D Anisole

💡 Short Explanation

�� Cresols are methyl-substituted phenols. �� Ortho position corresponds to carbon 2. �� o-Cresol contains adjacent OH and CH₃ groups.

✅ Correct Answer Explanation

Phenol containing a methyl group at the ortho position is called o-cresol. The methyl group is adjacent to the hydroxyl group on the benzene ring. Therefore, option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Represents meta substitution.
�� Option C → Represents para substitution.
�� Option D → Anisole is methoxybenzene.

🎯 Strategy

Used

Memory-Based Elimination

Application:

Recall the common names of methyl-substituted phenols.

Final Logic:

Ortho methyl phenol = o-Cresol.

🧠 Memory Trick

Ortho = o-Cresol

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

19 Symmetrical ether properties:
1. The alkyl or aryl groups attached to the oxygen are identical
2. Diethyl ether is an example
3. C₂H₅OCH₃ is an example
4. Also known as simple ethers

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

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

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

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

💡 Short Explanation

�� Symmetrical ethers contain identical groups. �� Diethyl ether is symmetrical. �� They are called simple ethers.

✅ Correct Answer Explanation

Statement 1 is correct because symmetrical ethers contain identical alkyl or aryl groups on both sides of oxygen. Statement 2 is correct because diethyl ether (C₂H₅OC₂H₅) is symmetrical. Statement 3 is incorrect because C₂H₅OCH₃ contains two different groups and is an unsymmetrical ether. Statement 4 is correct because symmetrical ethers are also called simple ethers. Therefore, option B is correct.

❓ Why Other Options Are Incorrect

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

🎯 Strategy

Used

Elimination

Application:

Compare the groups attached on both sides of oxygen.

Final Logic:

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

🧠 Memory Trick

Same Groups = Simple Ether

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

20 IUPAC name of the unsymmetrical ether CH₃OCH₂CH₂CH₃ is:

A 1-Methoxypropane

B Methyl propyl ether

C Methoxyethane

D Propoxymethane

💡 Short Explanation

�� Ethers are named as alkoxyalkanes. �� Propane is the longer carbon chain. �� Methoxy is the substituent.

✅ Correct Answer Explanation

The longer chain is propane and the smaller group attached through oxygen is methoxy. The methoxy group is attached to carbon 1 of propane. Therefore, the correct IUPAC name is 1-methoxypropane. Hence, option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Common name, not IUPAC name.
�� Option C → Represents a different ether.
�� Option D → Incorrect parent chain selection.

🎯 Strategy

Used

Substitution

Application:

Select the longest carbon chain and treat the smaller group as an alkoxy substituent.

Final Logic:

Propane parent chain + methoxy substituent = 1-Methoxypropane.

🧠 Memory Trick

Long Chain = Parent

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

CUET UG Chemistry Booster Test - 1 Classification and Basics

Test Complete!

Answer Review