JEE Main Previous Year Questions (2025): Hydrocarbons

 Page 1


JEE Main Previous Year Questions 
(2025): Hydrocarbons 
Q1: The compound with molecular formula ?? ?? ?? ?? , which gives only one monobromo 
derivative and takes up four moles of hydrogen per mole for complete hydrogenation 
has ____ ?? electrons. 
JEE Main 2025 (Online) 22nd January Evening Shift 
Ans: 6 
Solution: 
 
 
No. of ?? e = 8 
 
 
Q2: The sum of sigma ( ?? ) and pi ( ?? ) bonds in Hex-1,3-dien-5-yne is ____ . 
JEE Main 2025 (Online) 29th January Morning Shift 
Ans: 15 
Solution: 
The compound given is hex-1,3-dien-5-yne structure is 
1
H
2
C = CH- CH = CH- C = CH
 
A doble bond has one ?? bond and one sigma bond. 
A triple bond has one sigma bond and two ?? bonds. 
For the one triple bond, number of ?? bonds = 2 
For one double bond, number of ?? bond = 1 
Page 2


JEE Main Previous Year Questions 
(2025): Hydrocarbons 
Q1: The compound with molecular formula ?? ?? ?? ?? , which gives only one monobromo 
derivative and takes up four moles of hydrogen per mole for complete hydrogenation 
has ____ ?? electrons. 
JEE Main 2025 (Online) 22nd January Evening Shift 
Ans: 6 
Solution: 
 
 
No. of ?? e = 8 
 
 
Q2: The sum of sigma ( ?? ) and pi ( ?? ) bonds in Hex-1,3-dien-5-yne is ____ . 
JEE Main 2025 (Online) 29th January Morning Shift 
Ans: 15 
Solution: 
The compound given is hex-1,3-dien-5-yne structure is 
1
H
2
C = CH- CH = CH- C = CH
 
A doble bond has one ?? bond and one sigma bond. 
A triple bond has one sigma bond and two ?? bonds. 
For the one triple bond, number of ?? bonds = 2 
For one double bond, number of ?? bond = 1 
For two double bonds, number of ?? bonds = 2 
So, total number of ?? bonds = 2 + 2 = 4 
For sigma bonds, count all carbon - carbon bonds and carbon - hydrogen bonds except the ?? 
bonds in double bonds and triple bond. 
 
 
The total number of sigma bonds = 11 
So, total number of bonds: 
?? = 11 
?? = 4 
Sum of ?? and ?? bonds 
= 11 + 4 = 15 
Q3: Isomeric hydrocarbons ? negative Baeyer's test 
(Molecular formula ?? ?? ?? ????
 ) 
The total number of isomers from above with four different nonaliphatic substitution 
sites is - 
JEE Main 2025 (Online) 29th January Evening Shift 
Ans: 2 
Solution: 
Molecular formula of isomeric hydrocarbon ? C
9
H
12
 - Negative Baeyer's test 
Baeyers test is given by compounds that contain readily active carbon-carbon double bond. 
So, in a negative Baeyer's test, there is no readily active c = c. 
Compounds that give negative Baeyer's test are alkanes and aromatic compounds. 
C
9
H
12
 compound is not an alkane 
(Alkane general formula C
?? H
2?? +2
 ) 
Hydrocarbons with the formula C
9
H
12
 are considered as aromatic and isomers of substituted 
benzene rings. 
The possible isomers of C
9
H
12
 are 
Page 3


JEE Main Previous Year Questions 
(2025): Hydrocarbons 
Q1: The compound with molecular formula ?? ?? ?? ?? , which gives only one monobromo 
derivative and takes up four moles of hydrogen per mole for complete hydrogenation 
has ____ ?? electrons. 
JEE Main 2025 (Online) 22nd January Evening Shift 
Ans: 6 
Solution: 
 
 
No. of ?? e = 8 
 
 
Q2: The sum of sigma ( ?? ) and pi ( ?? ) bonds in Hex-1,3-dien-5-yne is ____ . 
JEE Main 2025 (Online) 29th January Morning Shift 
Ans: 15 
Solution: 
The compound given is hex-1,3-dien-5-yne structure is 
1
H
2
C = CH- CH = CH- C = CH
 
A doble bond has one ?? bond and one sigma bond. 
A triple bond has one sigma bond and two ?? bonds. 
For the one triple bond, number of ?? bonds = 2 
For one double bond, number of ?? bond = 1 
For two double bonds, number of ?? bonds = 2 
So, total number of ?? bonds = 2 + 2 = 4 
For sigma bonds, count all carbon - carbon bonds and carbon - hydrogen bonds except the ?? 
bonds in double bonds and triple bond. 
 
 
The total number of sigma bonds = 11 
So, total number of bonds: 
?? = 11 
?? = 4 
Sum of ?? and ?? bonds 
= 11 + 4 = 15 
Q3: Isomeric hydrocarbons ? negative Baeyer's test 
(Molecular formula ?? ?? ?? ????
 ) 
The total number of isomers from above with four different nonaliphatic substitution 
sites is - 
JEE Main 2025 (Online) 29th January Evening Shift 
Ans: 2 
Solution: 
Molecular formula of isomeric hydrocarbon ? C
9
H
12
 - Negative Baeyer's test 
Baeyers test is given by compounds that contain readily active carbon-carbon double bond. 
So, in a negative Baeyer's test, there is no readily active c = c. 
Compounds that give negative Baeyer's test are alkanes and aromatic compounds. 
C
9
H
12
 compound is not an alkane 
(Alkane general formula C
?? H
2?? +2
 ) 
Hydrocarbons with the formula C
9
H
12
 are considered as aromatic and isomers of substituted 
benzene rings. 
The possible isomers of C
9
H
12
 are 
 
 
From these, total number of the isomers with four different non-aliphatic substitution sites are 2 
 
 
Q4: Given below are two statements : 
Statement I: One mole of propyne reacts with excess of sodium to liberate half a mole 
of ?? ?? gas. 
Statement II: Four g of propyne reacts with ???????? ?? to liberate ????
?? gas which 
occupies 224 mL at STP. 
In the light of the above statements, choose the most appropriate answer from the 
options given below : 
Page 4


JEE Main Previous Year Questions 
(2025): Hydrocarbons 
Q1: The compound with molecular formula ?? ?? ?? ?? , which gives only one monobromo 
derivative and takes up four moles of hydrogen per mole for complete hydrogenation 
has ____ ?? electrons. 
JEE Main 2025 (Online) 22nd January Evening Shift 
Ans: 6 
Solution: 
 
 
No. of ?? e = 8 
 
 
Q2: The sum of sigma ( ?? ) and pi ( ?? ) bonds in Hex-1,3-dien-5-yne is ____ . 
JEE Main 2025 (Online) 29th January Morning Shift 
Ans: 15 
Solution: 
The compound given is hex-1,3-dien-5-yne structure is 
1
H
2
C = CH- CH = CH- C = CH
 
A doble bond has one ?? bond and one sigma bond. 
A triple bond has one sigma bond and two ?? bonds. 
For the one triple bond, number of ?? bonds = 2 
For one double bond, number of ?? bond = 1 
For two double bonds, number of ?? bonds = 2 
So, total number of ?? bonds = 2 + 2 = 4 
For sigma bonds, count all carbon - carbon bonds and carbon - hydrogen bonds except the ?? 
bonds in double bonds and triple bond. 
 
 
The total number of sigma bonds = 11 
So, total number of bonds: 
?? = 11 
?? = 4 
Sum of ?? and ?? bonds 
= 11 + 4 = 15 
Q3: Isomeric hydrocarbons ? negative Baeyer's test 
(Molecular formula ?? ?? ?? ????
 ) 
The total number of isomers from above with four different nonaliphatic substitution 
sites is - 
JEE Main 2025 (Online) 29th January Evening Shift 
Ans: 2 
Solution: 
Molecular formula of isomeric hydrocarbon ? C
9
H
12
 - Negative Baeyer's test 
Baeyers test is given by compounds that contain readily active carbon-carbon double bond. 
So, in a negative Baeyer's test, there is no readily active c = c. 
Compounds that give negative Baeyer's test are alkanes and aromatic compounds. 
C
9
H
12
 compound is not an alkane 
(Alkane general formula C
?? H
2?? +2
 ) 
Hydrocarbons with the formula C
9
H
12
 are considered as aromatic and isomers of substituted 
benzene rings. 
The possible isomers of C
9
H
12
 are 
 
 
From these, total number of the isomers with four different non-aliphatic substitution sites are 2 
 
 
Q4: Given below are two statements : 
Statement I: One mole of propyne reacts with excess of sodium to liberate half a mole 
of ?? ?? gas. 
Statement II: Four g of propyne reacts with ???????? ?? to liberate ????
?? gas which 
occupies 224 mL at STP. 
In the light of the above statements, choose the most appropriate answer from the 
options given below : 
JEE Main 2025 (Online) 22nd January Morning Shift 
Options: 
A. Both Statement I and Statement II are incorrect 
B. Statement I is incorrect but Statement II is correct 
C. Both Statement I and Statement II are correct 
D. Statement I is correct but Statement II is incorrect 
Ans: D 
Solution: 
Statement I 
"One mole of propyne reacts with excess of sodium to liberate half a mole of H
2
 gas." 
Reaction and Stoichiometry 
Propyne (terminal alkyne): CH
3
C = CH . 
Reaction with sodium: 
2CH
3
C = CH+ 2Na? 2(CH
3
C = C
-
Na
+
) + H
2
. 
From this balanced equation, 2 moles of propyne produce 1 mole of H
2
. 
Hence, 1 mole of propyne will produce 
1
2
 mole of H
2
. 
Statement I is correct. 
Statement II 
"Four grams of propyne reacts with NaNH
2
 to liberate NH
3
 gas which occupies 224 mL at STP." 
Analysis 
Moles of propyne 
Molecular mass of propyne (C
3
H
4
) : 
3 × 12 + 4 × 1 = 36 + 4 = 40 g/mol . 
Four grams of propyne is: 
4 g
40 g/mol
= 0.1 mol . 
Reaction with NaNH
2
 
For a terminal alkyne: 
CH
3
C = CH+ NaNH
2
? CH
3
C = C
-
Na
+
+ NH
3
. 
?? mole of propyne produces ?? mole of NH
3
. 
Moles of NH
3
 produced 
With 0.1 mol of propyne, we get 0.1 mol of NH
3
. 
Volume of NH
3
 at STP 
Page 5


JEE Main Previous Year Questions 
(2025): Hydrocarbons 
Q1: The compound with molecular formula ?? ?? ?? ?? , which gives only one monobromo 
derivative and takes up four moles of hydrogen per mole for complete hydrogenation 
has ____ ?? electrons. 
JEE Main 2025 (Online) 22nd January Evening Shift 
Ans: 6 
Solution: 
 
 
No. of ?? e = 8 
 
 
Q2: The sum of sigma ( ?? ) and pi ( ?? ) bonds in Hex-1,3-dien-5-yne is ____ . 
JEE Main 2025 (Online) 29th January Morning Shift 
Ans: 15 
Solution: 
The compound given is hex-1,3-dien-5-yne structure is 
1
H
2
C = CH- CH = CH- C = CH
 
A doble bond has one ?? bond and one sigma bond. 
A triple bond has one sigma bond and two ?? bonds. 
For the one triple bond, number of ?? bonds = 2 
For one double bond, number of ?? bond = 1 
For two double bonds, number of ?? bonds = 2 
So, total number of ?? bonds = 2 + 2 = 4 
For sigma bonds, count all carbon - carbon bonds and carbon - hydrogen bonds except the ?? 
bonds in double bonds and triple bond. 
 
 
The total number of sigma bonds = 11 
So, total number of bonds: 
?? = 11 
?? = 4 
Sum of ?? and ?? bonds 
= 11 + 4 = 15 
Q3: Isomeric hydrocarbons ? negative Baeyer's test 
(Molecular formula ?? ?? ?? ????
 ) 
The total number of isomers from above with four different nonaliphatic substitution 
sites is - 
JEE Main 2025 (Online) 29th January Evening Shift 
Ans: 2 
Solution: 
Molecular formula of isomeric hydrocarbon ? C
9
H
12
 - Negative Baeyer's test 
Baeyers test is given by compounds that contain readily active carbon-carbon double bond. 
So, in a negative Baeyer's test, there is no readily active c = c. 
Compounds that give negative Baeyer's test are alkanes and aromatic compounds. 
C
9
H
12
 compound is not an alkane 
(Alkane general formula C
?? H
2?? +2
 ) 
Hydrocarbons with the formula C
9
H
12
 are considered as aromatic and isomers of substituted 
benzene rings. 
The possible isomers of C
9
H
12
 are 
 
 
From these, total number of the isomers with four different non-aliphatic substitution sites are 2 
 
 
Q4: Given below are two statements : 
Statement I: One mole of propyne reacts with excess of sodium to liberate half a mole 
of ?? ?? gas. 
Statement II: Four g of propyne reacts with ???????? ?? to liberate ????
?? gas which 
occupies 224 mL at STP. 
In the light of the above statements, choose the most appropriate answer from the 
options given below : 
JEE Main 2025 (Online) 22nd January Morning Shift 
Options: 
A. Both Statement I and Statement II are incorrect 
B. Statement I is incorrect but Statement II is correct 
C. Both Statement I and Statement II are correct 
D. Statement I is correct but Statement II is incorrect 
Ans: D 
Solution: 
Statement I 
"One mole of propyne reacts with excess of sodium to liberate half a mole of H
2
 gas." 
Reaction and Stoichiometry 
Propyne (terminal alkyne): CH
3
C = CH . 
Reaction with sodium: 
2CH
3
C = CH+ 2Na? 2(CH
3
C = C
-
Na
+
) + H
2
. 
From this balanced equation, 2 moles of propyne produce 1 mole of H
2
. 
Hence, 1 mole of propyne will produce 
1
2
 mole of H
2
. 
Statement I is correct. 
Statement II 
"Four grams of propyne reacts with NaNH
2
 to liberate NH
3
 gas which occupies 224 mL at STP." 
Analysis 
Moles of propyne 
Molecular mass of propyne (C
3
H
4
) : 
3 × 12 + 4 × 1 = 36 + 4 = 40 g/mol . 
Four grams of propyne is: 
4 g
40 g/mol
= 0.1 mol . 
Reaction with NaNH
2
 
For a terminal alkyne: 
CH
3
C = CH+ NaNH
2
? CH
3
C = C
-
Na
+
+ NH
3
. 
?? mole of propyne produces ?? mole of NH
3
. 
Moles of NH
3
 produced 
With 0.1 mol of propyne, we get 0.1 mol of NH
3
. 
Volume of NH
3
 at STP 
1 mole of any ideal gas at STP ˜ 22.4 L = 22400 mL. 
0.1 mol of NH
3
 occupies 0.1 × 22.4 L = 2.24 L = 2240 mL. 
However, Statement II says the liberated NH
3
 occupies only ?????? ???? at STP, which corresponds 
to 0.01 mol of NH
3
, not 0.1 mol . Therefore, the statement's volume is off by a factor of 10 and 
is thus incorrect if the reaction goes to completion in a typical way. 
Statement II is incorrect. 
Conclusion 
Statement I is correct. 
Statement II is incorrect. 
Hence, the best choice is: 
Option D: Statement I is correct but Statement II is incorrect. 
Q5: The alkane from below having two secondary hydrogens is : 
JEE Main 2025 (Online) 22nd January Evening Shift 
Options: 
A. 2,2,4,5-Tetramethylheptane 
B. 2,2,4,4-Tetramethylhexane 
C. 4-Ethyl-3,4-dimethyloctane 
D. 2,2,3,3-Tetramethylpentane 
Ans: D 
Solution: