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JEE Main Previous Year Questions (2025): Alcohols, Phenols and Ethers

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JEE Main Previous Year Questions 
(2025): Alcohols, Phenols and Ethers 
Q1: What amount of bromine will be required to convert 2 g of phenol into 
2,4,6tribromophenol? 
(Given molar mass in ???? ?? ?? - ?? of ?? , ?? , ?? , ???? are ???? , ?? , ???? , ???? respectively) 
JEE Main 2025 (Online) 23rd January Morning Shift 
Options: 
A. 6.0 g 
B. 10.22 g 
C. 20.44 g 
D. 4.0 g 
Ans: B 
Solution: 
 
Moles of phenol =
2
94
= 0 . 021 
? Moles of bromine = 0 . 021 × 3 = 0 . 064 
? Mass of bromine = 0 . 064 × 160 = 10 . 22 g 
Q2: Given below are two statements : 
Statement (I): The boiling points of alcohols and phenols increase with increase in the 
number of ?? -atoms. 
Statement (II): The boiling points of alcohols and phenols are higher in comparison to 
other class of compounds such as ethers, haloalkanes. 
In the light of the above statements, choose the correct answer from the options 
given below : 
JEE Main 2025 (Online) 23rd January Evening Shift 
Options: 
A. Statement I is false but Statement II is true 
B. Both Statement I and Statement II are true 
C. Statement I is true but Statement II is false 
D. Both Statement I and Statement II are false 
Page 2


JEE Main Previous Year Questions 
(2025): Alcohols, Phenols and Ethers 
Q1: What amount of bromine will be required to convert 2 g of phenol into 
2,4,6tribromophenol? 
(Given molar mass in ???? ?? ?? - ?? of ?? , ?? , ?? , ???? are ???? , ?? , ???? , ???? respectively) 
JEE Main 2025 (Online) 23rd January Morning Shift 
Options: 
A. 6.0 g 
B. 10.22 g 
C. 20.44 g 
D. 4.0 g 
Ans: B 
Solution: 
 
Moles of phenol =
2
94
= 0 . 021 
? Moles of bromine = 0 . 021 × 3 = 0 . 064 
? Mass of bromine = 0 . 064 × 160 = 10 . 22 g 
Q2: Given below are two statements : 
Statement (I): The boiling points of alcohols and phenols increase with increase in the 
number of ?? -atoms. 
Statement (II): The boiling points of alcohols and phenols are higher in comparison to 
other class of compounds such as ethers, haloalkanes. 
In the light of the above statements, choose the correct answer from the options 
given below : 
JEE Main 2025 (Online) 23rd January Evening Shift 
Options: 
A. Statement I is false but Statement II is true 
B. Both Statement I and Statement II are true 
C. Statement I is true but Statement II is false 
D. Both Statement I and Statement II are false 
Ans: B 
Solution: 
Statement (I): 
As the number of carbon atoms in alcohols or phenols increases, the molecule becomes larger, 
which in turn enhances the van der Waals (dispersion) forces between the molecules. 
Stronger intermolecular forces require more energy (i.e., higher temperature) to overcome, 
leading to higher boiling points. 
Therefore, Statement (I) is true. 
Statement (II): 
Alcohols and phenols have a hydroxyl group ( -OH ) that can form hydrogen bonds, which are 
significantly stronger than the dipole-dipole interactions typically seen in ethers or the 
interactions in haloalkanes. 
This stronger hydrogen bonding means that alcohols and phenols have higher boiling points 
compared to ethers and haloalkanes of similar molecular weight. 
Hence, Statement (II) is also true. 
Based on the above points, the correct answer is: 
Option B: Both Statement I and Statement II are true. 
Q3: Given below are two statements : 
Statement I : 
 
Statement II : In 
 
, intramolecular substitution takes place first by involving lone pair of electrons on 
nitrogen. 
In the light of the above statements, choose the most appropriate answer from the 
options given below 
JEE Main 2025 (Online) 28th January Morning Shift 
Options: 
A. Both Statement I and Statement II are correct 
Page 3


JEE Main Previous Year Questions 
(2025): Alcohols, Phenols and Ethers 
Q1: What amount of bromine will be required to convert 2 g of phenol into 
2,4,6tribromophenol? 
(Given molar mass in ???? ?? ?? - ?? of ?? , ?? , ?? , ???? are ???? , ?? , ???? , ???? respectively) 
JEE Main 2025 (Online) 23rd January Morning Shift 
Options: 
A. 6.0 g 
B. 10.22 g 
C. 20.44 g 
D. 4.0 g 
Ans: B 
Solution: 
 
Moles of phenol =
2
94
= 0 . 021 
? Moles of bromine = 0 . 021 × 3 = 0 . 064 
? Mass of bromine = 0 . 064 × 160 = 10 . 22 g 
Q2: Given below are two statements : 
Statement (I): The boiling points of alcohols and phenols increase with increase in the 
number of ?? -atoms. 
Statement (II): The boiling points of alcohols and phenols are higher in comparison to 
other class of compounds such as ethers, haloalkanes. 
In the light of the above statements, choose the correct answer from the options 
given below : 
JEE Main 2025 (Online) 23rd January Evening Shift 
Options: 
A. Statement I is false but Statement II is true 
B. Both Statement I and Statement II are true 
C. Statement I is true but Statement II is false 
D. Both Statement I and Statement II are false 
Ans: B 
Solution: 
Statement (I): 
As the number of carbon atoms in alcohols or phenols increases, the molecule becomes larger, 
which in turn enhances the van der Waals (dispersion) forces between the molecules. 
Stronger intermolecular forces require more energy (i.e., higher temperature) to overcome, 
leading to higher boiling points. 
Therefore, Statement (I) is true. 
Statement (II): 
Alcohols and phenols have a hydroxyl group ( -OH ) that can form hydrogen bonds, which are 
significantly stronger than the dipole-dipole interactions typically seen in ethers or the 
interactions in haloalkanes. 
This stronger hydrogen bonding means that alcohols and phenols have higher boiling points 
compared to ethers and haloalkanes of similar molecular weight. 
Hence, Statement (II) is also true. 
Based on the above points, the correct answer is: 
Option B: Both Statement I and Statement II are true. 
Q3: Given below are two statements : 
Statement I : 
 
Statement II : In 
 
, intramolecular substitution takes place first by involving lone pair of electrons on 
nitrogen. 
In the light of the above statements, choose the most appropriate answer from the 
options given below 
JEE Main 2025 (Online) 28th January Morning Shift 
Options: 
A. Both Statement I and Statement II are correct 
B. Statement I is incorrect but Statement II is correct 
C. Both Statement I and Statement II are incorrect 
D. Statement I is correct but Statement II is incorrect 
Ans: A 
Solution: 
 
Rate of (a) is faster than (b) because of its intramolecular substitution. 
Q4: Which one of the following, with HBr will give a phenol? 
JEE Main 2025 (Online) 29th January Evening Shift 
Options: 
A.  
B.  
Page 4


JEE Main Previous Year Questions 
(2025): Alcohols, Phenols and Ethers 
Q1: What amount of bromine will be required to convert 2 g of phenol into 
2,4,6tribromophenol? 
(Given molar mass in ???? ?? ?? - ?? of ?? , ?? , ?? , ???? are ???? , ?? , ???? , ???? respectively) 
JEE Main 2025 (Online) 23rd January Morning Shift 
Options: 
A. 6.0 g 
B. 10.22 g 
C. 20.44 g 
D. 4.0 g 
Ans: B 
Solution: 
 
Moles of phenol =
2
94
= 0 . 021 
? Moles of bromine = 0 . 021 × 3 = 0 . 064 
? Mass of bromine = 0 . 064 × 160 = 10 . 22 g 
Q2: Given below are two statements : 
Statement (I): The boiling points of alcohols and phenols increase with increase in the 
number of ?? -atoms. 
Statement (II): The boiling points of alcohols and phenols are higher in comparison to 
other class of compounds such as ethers, haloalkanes. 
In the light of the above statements, choose the correct answer from the options 
given below : 
JEE Main 2025 (Online) 23rd January Evening Shift 
Options: 
A. Statement I is false but Statement II is true 
B. Both Statement I and Statement II are true 
C. Statement I is true but Statement II is false 
D. Both Statement I and Statement II are false 
Ans: B 
Solution: 
Statement (I): 
As the number of carbon atoms in alcohols or phenols increases, the molecule becomes larger, 
which in turn enhances the van der Waals (dispersion) forces between the molecules. 
Stronger intermolecular forces require more energy (i.e., higher temperature) to overcome, 
leading to higher boiling points. 
Therefore, Statement (I) is true. 
Statement (II): 
Alcohols and phenols have a hydroxyl group ( -OH ) that can form hydrogen bonds, which are 
significantly stronger than the dipole-dipole interactions typically seen in ethers or the 
interactions in haloalkanes. 
This stronger hydrogen bonding means that alcohols and phenols have higher boiling points 
compared to ethers and haloalkanes of similar molecular weight. 
Hence, Statement (II) is also true. 
Based on the above points, the correct answer is: 
Option B: Both Statement I and Statement II are true. 
Q3: Given below are two statements : 
Statement I : 
 
Statement II : In 
 
, intramolecular substitution takes place first by involving lone pair of electrons on 
nitrogen. 
In the light of the above statements, choose the most appropriate answer from the 
options given below 
JEE Main 2025 (Online) 28th January Morning Shift 
Options: 
A. Both Statement I and Statement II are correct 
B. Statement I is incorrect but Statement II is correct 
C. Both Statement I and Statement II are incorrect 
D. Statement I is correct but Statement II is incorrect 
Ans: A 
Solution: 
 
Rate of (a) is faster than (b) because of its intramolecular substitution. 
Q4: Which one of the following, with HBr will give a phenol? 
JEE Main 2025 (Online) 29th January Evening Shift 
Options: 
A.  
B.  
C.  
D.  
Ans: A 
Solution: 
Relation with HBr to give phenol 
(1) 
 
The reaction involves two steps, 
? ) The lone pair of oxygen atom attacks on the hydrogen of HBr . 
? ) The nucleophile Br- attacks on the methyl group which lead to the formation of phenol along 
with methyl bromide. 
Phenol is formed from this compound on reaction with HBr . 
(2) 
Page 5


JEE Main Previous Year Questions 
(2025): Alcohols, Phenols and Ethers 
Q1: What amount of bromine will be required to convert 2 g of phenol into 
2,4,6tribromophenol? 
(Given molar mass in ???? ?? ?? - ?? of ?? , ?? , ?? , ???? are ???? , ?? , ???? , ???? respectively) 
JEE Main 2025 (Online) 23rd January Morning Shift 
Options: 
A. 6.0 g 
B. 10.22 g 
C. 20.44 g 
D. 4.0 g 
Ans: B 
Solution: 
 
Moles of phenol =
2
94
= 0 . 021 
? Moles of bromine = 0 . 021 × 3 = 0 . 064 
? Mass of bromine = 0 . 064 × 160 = 10 . 22 g 
Q2: Given below are two statements : 
Statement (I): The boiling points of alcohols and phenols increase with increase in the 
number of ?? -atoms. 
Statement (II): The boiling points of alcohols and phenols are higher in comparison to 
other class of compounds such as ethers, haloalkanes. 
In the light of the above statements, choose the correct answer from the options 
given below : 
JEE Main 2025 (Online) 23rd January Evening Shift 
Options: 
A. Statement I is false but Statement II is true 
B. Both Statement I and Statement II are true 
C. Statement I is true but Statement II is false 
D. Both Statement I and Statement II are false 
Ans: B 
Solution: 
Statement (I): 
As the number of carbon atoms in alcohols or phenols increases, the molecule becomes larger, 
which in turn enhances the van der Waals (dispersion) forces between the molecules. 
Stronger intermolecular forces require more energy (i.e., higher temperature) to overcome, 
leading to higher boiling points. 
Therefore, Statement (I) is true. 
Statement (II): 
Alcohols and phenols have a hydroxyl group ( -OH ) that can form hydrogen bonds, which are 
significantly stronger than the dipole-dipole interactions typically seen in ethers or the 
interactions in haloalkanes. 
This stronger hydrogen bonding means that alcohols and phenols have higher boiling points 
compared to ethers and haloalkanes of similar molecular weight. 
Hence, Statement (II) is also true. 
Based on the above points, the correct answer is: 
Option B: Both Statement I and Statement II are true. 
Q3: Given below are two statements : 
Statement I : 
 
Statement II : In 
 
, intramolecular substitution takes place first by involving lone pair of electrons on 
nitrogen. 
In the light of the above statements, choose the most appropriate answer from the 
options given below 
JEE Main 2025 (Online) 28th January Morning Shift 
Options: 
A. Both Statement I and Statement II are correct 
B. Statement I is incorrect but Statement II is correct 
C. Both Statement I and Statement II are incorrect 
D. Statement I is correct but Statement II is incorrect 
Ans: A 
Solution: 
 
Rate of (a) is faster than (b) because of its intramolecular substitution. 
Q4: Which one of the following, with HBr will give a phenol? 
JEE Main 2025 (Online) 29th January Evening Shift 
Options: 
A.  
B.  
C.  
D.  
Ans: A 
Solution: 
Relation with HBr to give phenol 
(1) 
 
The reaction involves two steps, 
? ) The lone pair of oxygen atom attacks on the hydrogen of HBr . 
? ) The nucleophile Br- attacks on the methyl group which lead to the formation of phenol along 
with methyl bromide. 
Phenol is formed from this compound on reaction with HBr . 
(2) 
 
No phenol formation in this case because alkyl chain is attached to the benzene ring and on 
addition of H - Br, either H + or Br - get attached to this chain and no - OH group is formed as 
it is directly bonded to the benzene ring. 
(3)  
 
 
No phenol formation in this case. -OH group is formed as 
(4)
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FAQs on JEE Main Previous Year Questions (2025): Alcohols, Phenols and Ethers

1. What are the main types of alcohols, and how are they classified?
Ans. Alcohols are primarily classified into three categories: primary, secondary, and tertiary alcohols. A primary alcohol (1°) has the hydroxyl (-OH) group attached to a carbon atom that is connected to only one other carbon atom. A secondary alcohol (2°) has the -OH group attached to a carbon that is connected to two other carbon atoms. A tertiary alcohol (3°) has the -OH group attached to a carbon connected to three other carbon atoms. This classification is important for understanding their chemical reactivity and properties.
2. What are phenols, and how do they differ from alcohols?
Ans. Phenols are organic compounds that contain a hydroxyl (-OH) group attached to an aromatic hydrocarbon ring. The key difference between phenols and alcohols is that in phenols, the -OH group is directly bonded to a carbon atom of an aromatic ring, whereas in alcohols, it is bonded to a saturated carbon atom. This structural difference results in distinct chemical properties, such as phenols being more acidic than typical alcohols.
3. What are ethers, and what is their general structure?
Ans. Ethers are organic compounds characterized by the presence of an oxygen atom connected to two alkyl or aryl groups. Their general structure can be represented as R-O-R', where R and R' are hydrocarbon chains. Ethers are known for their relatively low reactivity compared to alcohols and phenols, making them useful as solvents in organic reactions.
4. How do alcohols undergo oxidation, and what are the possible products?
Ans. Alcohols can undergo oxidation to form aldehydes, ketones, or carboxylic acids, depending on their classification. Primary alcohols oxidize to form aldehydes, which can further oxidize to carboxylic acids. Secondary alcohols are oxidized to form ketones. Tertiary alcohols do not oxidize easily because there is no hydrogen atom bonded to the carbon with the -OH group. The oxidizing agents commonly used include potassium dichromate (K₂Cr₂O₇) and potassium permanganate (KMnO₄).
5. What is the significance of the dehydration of alcohols, and what products are formed?
Ans. The dehydration of alcohols is a significant reaction in organic chemistry, where alcohols lose a molecule of water (H₂O) to form alkenes. This reaction typically occurs under acidic conditions and can be used to synthesize alkenes from alcohols efficiently. The reaction can be represented as R-OH → R-CH=CH₂ + H₂O, where R represents the hydrocarbon chain. This process is important in the production of various industrial chemicals and fuels.
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