Haloalkanes and Haloarenes: JEE Main Previous Year Questions (2021-2026)

 Page 1


JEE Main Previous Year Questions (2021-2026): 
Haloalkanes and Haloarenes  
 
(January 2026) 
 
 
Q1:  
Consider the above reaction 
A. The reaction proceeds through a more stable radical intermediate. 
B. The role of peroxide is to generate H• (Hydrogen radical). 
C. During this reaction, benzene is formed as a byproduct. 
D. 1-Bromo-2-phenylethane is formed as the minor product. 
E. The same reaction in absence of peroxide proceeds via carbocation intermediate. 
Identify the correct statements. Choose the correct answer from the options given below: 
A: A & E Only 
B: A, B & D Only 
C: A, C & E Only 
D: C, D & E Only 
Answer: C 
Explanation: 
In presence of peroxide, addition of HBr to an alkene follows free radical mechanism (peroxide 
effect). 
For styrene Ph-CH=CH
2
: 
Major product (with peroxide) 
Br• adds in such a way that the more stable radical is formed. 
If Br• adds to terminal carbon (CH
2
), the radical comes on benzylic carbon: 
Ph-CH-CH
2
Br 
This is benzylic radical (resonance-stabilised), hence preferred. Then it abstracts H from HBr: 
Ph-CH
2
-CH
2
Br 
So 1-bromo-2-phenylethane is the major product. 
Checking each statement 
A. True 
The reaction proceeds via the more stable radical intermediate (benzylic radical). 
B. False 
Peroxide does not generate H•. 
It generates radicals that ultimately produce Br• (the chain-carrying radical), e.g. by abstraction 
of H from HBr. 
C. True 
Benzoyl peroxide can form phenyl radicals (Ph•), which can abstract H from HBr: 
Ph• + HBr ? PhH + Br• 
Page 2


JEE Main Previous Year Questions (2021-2026): 
Haloalkanes and Haloarenes  
 
(January 2026) 
 
 
Q1:  
Consider the above reaction 
A. The reaction proceeds through a more stable radical intermediate. 
B. The role of peroxide is to generate H• (Hydrogen radical). 
C. During this reaction, benzene is formed as a byproduct. 
D. 1-Bromo-2-phenylethane is formed as the minor product. 
E. The same reaction in absence of peroxide proceeds via carbocation intermediate. 
Identify the correct statements. Choose the correct answer from the options given below: 
A: A & E Only 
B: A, B & D Only 
C: A, C & E Only 
D: C, D & E Only 
Answer: C 
Explanation: 
In presence of peroxide, addition of HBr to an alkene follows free radical mechanism (peroxide 
effect). 
For styrene Ph-CH=CH
2
: 
Major product (with peroxide) 
Br• adds in such a way that the more stable radical is formed. 
If Br• adds to terminal carbon (CH
2
), the radical comes on benzylic carbon: 
Ph-CH-CH
2
Br 
This is benzylic radical (resonance-stabilised), hence preferred. Then it abstracts H from HBr: 
Ph-CH
2
-CH
2
Br 
So 1-bromo-2-phenylethane is the major product. 
Checking each statement 
A. True 
The reaction proceeds via the more stable radical intermediate (benzylic radical). 
B. False 
Peroxide does not generate H•. 
It generates radicals that ultimately produce Br• (the chain-carrying radical), e.g. by abstraction 
of H from HBr. 
C. True 
Benzoyl peroxide can form phenyl radicals (Ph•), which can abstract H from HBr: 
Ph• + HBr ? PhH + Br• 
So benzene (PhH) can be formed as a byproduct. 
D. False 
1-bromo-2-phenylethane (Ph-CH 2-CH 2Br) is the major, not minor, product in presence of 
peroxide. 
E. True 
In absence of peroxide, HBr adds by electrophilic addition via a carbocation intermediate (here 
benzylic carbocation), giving Markovnikov product. 
Correct option: Option C (A, C & E Only) 
 
Q2: Consider the following reactions giving major product. Identify the correct reaction. 
A:  
B:  
C:  
D:  
Answer: D 
Explanation: 
Page 3


JEE Main Previous Year Questions (2021-2026): 
Haloalkanes and Haloarenes  
 
(January 2026) 
 
 
Q1:  
Consider the above reaction 
A. The reaction proceeds through a more stable radical intermediate. 
B. The role of peroxide is to generate H• (Hydrogen radical). 
C. During this reaction, benzene is formed as a byproduct. 
D. 1-Bromo-2-phenylethane is formed as the minor product. 
E. The same reaction in absence of peroxide proceeds via carbocation intermediate. 
Identify the correct statements. Choose the correct answer from the options given below: 
A: A & E Only 
B: A, B & D Only 
C: A, C & E Only 
D: C, D & E Only 
Answer: C 
Explanation: 
In presence of peroxide, addition of HBr to an alkene follows free radical mechanism (peroxide 
effect). 
For styrene Ph-CH=CH
2
: 
Major product (with peroxide) 
Br• adds in such a way that the more stable radical is formed. 
If Br• adds to terminal carbon (CH
2
), the radical comes on benzylic carbon: 
Ph-CH-CH
2
Br 
This is benzylic radical (resonance-stabilised), hence preferred. Then it abstracts H from HBr: 
Ph-CH
2
-CH
2
Br 
So 1-bromo-2-phenylethane is the major product. 
Checking each statement 
A. True 
The reaction proceeds via the more stable radical intermediate (benzylic radical). 
B. False 
Peroxide does not generate H•. 
It generates radicals that ultimately produce Br• (the chain-carrying radical), e.g. by abstraction 
of H from HBr. 
C. True 
Benzoyl peroxide can form phenyl radicals (Ph•), which can abstract H from HBr: 
Ph• + HBr ? PhH + Br• 
So benzene (PhH) can be formed as a byproduct. 
D. False 
1-bromo-2-phenylethane (Ph-CH 2-CH 2Br) is the major, not minor, product in presence of 
peroxide. 
E. True 
In absence of peroxide, HBr adds by electrophilic addition via a carbocation intermediate (here 
benzylic carbocation), giving Markovnikov product. 
Correct option: Option C (A, C & E Only) 
 
Q2: Consider the following reactions giving major product. Identify the correct reaction. 
A:  
B:  
C:  
D:  
Answer: D 
Explanation: 
 
 
 
Q3: Given below are two statements: 
Statement II : The given optically active molecule, on hydrolysis 
gives a solution that can rotate the plane polarized light. 
In the light of the above statements, choose the correct answer from the options given 
below: 
A: Statement I is false but Statement II is true 
B: Statement I is true but Statement II is false 
C: Both Statement I and Statement II are true 
D: Both Statement I and Statement II are false 
Answer: B 
Page 4


JEE Main Previous Year Questions (2021-2026): 
Haloalkanes and Haloarenes  
 
(January 2026) 
 
 
Q1:  
Consider the above reaction 
A. The reaction proceeds through a more stable radical intermediate. 
B. The role of peroxide is to generate H• (Hydrogen radical). 
C. During this reaction, benzene is formed as a byproduct. 
D. 1-Bromo-2-phenylethane is formed as the minor product. 
E. The same reaction in absence of peroxide proceeds via carbocation intermediate. 
Identify the correct statements. Choose the correct answer from the options given below: 
A: A & E Only 
B: A, B & D Only 
C: A, C & E Only 
D: C, D & E Only 
Answer: C 
Explanation: 
In presence of peroxide, addition of HBr to an alkene follows free radical mechanism (peroxide 
effect). 
For styrene Ph-CH=CH
2
: 
Major product (with peroxide) 
Br• adds in such a way that the more stable radical is formed. 
If Br• adds to terminal carbon (CH
2
), the radical comes on benzylic carbon: 
Ph-CH-CH
2
Br 
This is benzylic radical (resonance-stabilised), hence preferred. Then it abstracts H from HBr: 
Ph-CH
2
-CH
2
Br 
So 1-bromo-2-phenylethane is the major product. 
Checking each statement 
A. True 
The reaction proceeds via the more stable radical intermediate (benzylic radical). 
B. False 
Peroxide does not generate H•. 
It generates radicals that ultimately produce Br• (the chain-carrying radical), e.g. by abstraction 
of H from HBr. 
C. True 
Benzoyl peroxide can form phenyl radicals (Ph•), which can abstract H from HBr: 
Ph• + HBr ? PhH + Br• 
So benzene (PhH) can be formed as a byproduct. 
D. False 
1-bromo-2-phenylethane (Ph-CH 2-CH 2Br) is the major, not minor, product in presence of 
peroxide. 
E. True 
In absence of peroxide, HBr adds by electrophilic addition via a carbocation intermediate (here 
benzylic carbocation), giving Markovnikov product. 
Correct option: Option C (A, C & E Only) 
 
Q2: Consider the following reactions giving major product. Identify the correct reaction. 
A:  
B:  
C:  
D:  
Answer: D 
Explanation: 
 
 
 
Q3: Given below are two statements: 
Statement II : The given optically active molecule, on hydrolysis 
gives a solution that can rotate the plane polarized light. 
In the light of the above statements, choose the correct answer from the options given 
below: 
A: Statement I is false but Statement II is true 
B: Statement I is true but Statement II is false 
C: Both Statement I and Statement II are true 
D: Both Statement I and Statement II are false 
Answer: B 
Explanation:  
 
 
 
Q4: Match the List-I with List-II 
 
Choose the correct answer from the options given below: 
A: A-III, B-IV, C-I, D-II 
Page 5


JEE Main Previous Year Questions (2021-2026): 
Haloalkanes and Haloarenes  
 
(January 2026) 
 
 
Q1:  
Consider the above reaction 
A. The reaction proceeds through a more stable radical intermediate. 
B. The role of peroxide is to generate H• (Hydrogen radical). 
C. During this reaction, benzene is formed as a byproduct. 
D. 1-Bromo-2-phenylethane is formed as the minor product. 
E. The same reaction in absence of peroxide proceeds via carbocation intermediate. 
Identify the correct statements. Choose the correct answer from the options given below: 
A: A & E Only 
B: A, B & D Only 
C: A, C & E Only 
D: C, D & E Only 
Answer: C 
Explanation: 
In presence of peroxide, addition of HBr to an alkene follows free radical mechanism (peroxide 
effect). 
For styrene Ph-CH=CH
2
: 
Major product (with peroxide) 
Br• adds in such a way that the more stable radical is formed. 
If Br• adds to terminal carbon (CH
2
), the radical comes on benzylic carbon: 
Ph-CH-CH
2
Br 
This is benzylic radical (resonance-stabilised), hence preferred. Then it abstracts H from HBr: 
Ph-CH
2
-CH
2
Br 
So 1-bromo-2-phenylethane is the major product. 
Checking each statement 
A. True 
The reaction proceeds via the more stable radical intermediate (benzylic radical). 
B. False 
Peroxide does not generate H•. 
It generates radicals that ultimately produce Br• (the chain-carrying radical), e.g. by abstraction 
of H from HBr. 
C. True 
Benzoyl peroxide can form phenyl radicals (Ph•), which can abstract H from HBr: 
Ph• + HBr ? PhH + Br• 
So benzene (PhH) can be formed as a byproduct. 
D. False 
1-bromo-2-phenylethane (Ph-CH 2-CH 2Br) is the major, not minor, product in presence of 
peroxide. 
E. True 
In absence of peroxide, HBr adds by electrophilic addition via a carbocation intermediate (here 
benzylic carbocation), giving Markovnikov product. 
Correct option: Option C (A, C & E Only) 
 
Q2: Consider the following reactions giving major product. Identify the correct reaction. 
A:  
B:  
C:  
D:  
Answer: D 
Explanation: 
 
 
 
Q3: Given below are two statements: 
Statement II : The given optically active molecule, on hydrolysis 
gives a solution that can rotate the plane polarized light. 
In the light of the above statements, choose the correct answer from the options given 
below: 
A: Statement I is false but Statement II is true 
B: Statement I is true but Statement II is false 
C: Both Statement I and Statement II are true 
D: Both Statement I and Statement II are false 
Answer: B 
Explanation:  
 
 
 
Q4: Match the List-I with List-II 
 
Choose the correct answer from the options given below: 
A: A-III, B-IV, C-I, D-II 
B: A-IV, B-I, C-III, D-II 
C: A-III, B-IV, C-II, D-I 
D: A-I, B-II, C-IV, D-III 
Answer: C 
Explanation:  
Vinyl, allyl, benzyl and alkyl chlorides are identified from the position of Cl: 
Step 1: Identify each example in List-II 
III. CH 2 = CHCl 
Here Cl is directly attached to a double-bond carbon ? vinyl chloride. 
I. CH 2 = CH — CH 2Cl 
Here Cl is on the carbon next to C = C (allylic position) ? allyl chloride. 
II. CH 3 — CH(Cl)CH 3 
This is a saturated alkyl group with Cl on an sp³ carbon ? alkyl chloride (2-chloropropane). 
IV. (structure shown is C 6H 5 — CH 2Cl) 
Cl is on the carbon next to benzene ring (benzylic position) ? benzyl chloride. 
Step 2: Match List-I with List-II 
A. Vinyl chloride ? III 
B. Benzyl chloride ? IV 
C. Alkyl chloride ? II 
D. Allyl chloride ? I 
So the correct option is: Option C: A-III, B-IV, C-II, D-I 
 
Q5: Consider the following compounds  
 
Arrange these compounds in the increasing order of reactivity with nitrating mixture. 
A: c < b < a 
B: b < c < a 
C: c < a < b 
D: b < a < c 
Answer: D 
Explanation: 
In Ph — OMe, the —OMe group is an electron-donating group and shows +M (plus mesomeric)