Skip to main content
Back to HYDROCARBONS

Your Progress

0 / 269 Answered
0
Correct
0
Incorrect
0%
Accuracy
JEE • CHEMISTRY • HYDROCARBONS

Hydrocarbons practice · Mixed difficulty

Choose an answer to check it, or open “View answer” to study the explanation. Sign in to save progress to your account.

Practice format and chapter quiz

These sets use single-correct-answer MCQs. They do not cover every JEE Main or JEE Advanced question format and are not full-length mock papers.

Take a chapter quiz →
Showing 91-120 of 269 questions

Preparing practice and saved progress…

91 Medium
91 of 269
Understanding +4 | -1
Select the organic compounds aliphatic in nature but burn with smoky flame :
View answer and explanation
Correct option: D
Both (a) and (b)
Option D (Both (a) and (b)) is correct because both carbon tetrachloride (CCl4\mathrm{CCl_4}) and chloroform (CHCl3\mathrm{CHCl_3}) are aliphatic compounds that burn with a smoky flame due to incomplete combustion.
92 Medium
92 of 269
Analyzing +4 | -1
The compound that will react most readily with gaseous bromine has the formula:
View answer and explanation
Correct option: D
C4H10{C_4}{H_{10}}
Option A (C2H4) is correct because alkenes react most readily with bromine due to the presence of a pi bond, which is more reactive than sigma bonds in alkanes.
93 Medium
93 of 269
Understanding +4 | -1
Acetylenic hydrogens are acidic because
View answer and explanation
Correct option: A
Sigma electron density of C−H\mathrm{C}-\mathrm{H} bond in acetylene is nearer to carbon, which has 50%s50 \% s-character
Option A is correct because the sigma electron density of the C-H bond in acetylene is closer to the carbon, which has 50% s-character, making the hydrogen relatively acidic.
94 Easy
94 of 269
Remembering +4 | -1
Lindlar’s catalyst is:
View answer and explanation
Correct option: A
Pd−CaCO3\mathrm{Pd} - \mathrm{CaC}{\mathrm{O}_3} deactivated by lead acetate
Option A (Pd−CaCO3\mathrm{Pd} - \mathrm{CaC}{\mathrm{O}_3} deactivated by lead acetate) is the correct definition of Lindlar's catalyst, used for the partial hydrogenation of alkynes to alkenes.
95 Easy
95 of 269
Remembering +4 | -1
Which among the following are used as catalyst in cracking?
View answer and explanation
Correct option: D
All of these
Option D (All of these) is correct because oxides of Al, Cr, and Mo can be used as catalysts in cracking.
96 Easy
96 of 269
Understanding +4 | -1
The product obtained at anode upon Kolbe's electrolysis of sodium ethanoate is
View answer and explanation
Correct option: B
C2H6\mathrm{C}_2 \mathrm{H}_6
Option B (C2H6\mathrm{C}_2 \mathrm{H}_6) is correct because the product obtained at the anode during Kolbe's electrolysis of sodium ethanoate is ethane.
97 Medium
97 of 269
Analyzing +4 | -1
Compound 'X' on treatment with O3\mathrm{O}_3 followed by Zn/H2O\mathrm{Zn} / \mathrm{H}_2 \mathrm{O} gives methanal and cyclohexyl methanal. The compound 'X' is:
View answer and explanation
Correct option: D
3-cyclohexyl-1-propene
Option D (3-cyclohexyl-1-propene) is correct because its ozonolysis followed by reductive workup yields methanal and cyclohexyl methanal.
98 Hard
98 of 269
Analyzing +4 | -1
Consider the following statements : A hydrocarbon of molecular formula C5H10{C}_5 {H}_{10} is a

I. monosubstituted alkene

II. disubstituted alkene

III. trisubstituted alkene

Which of the following statement(s) is(are) correct?
View answer and explanation
Correct option: A
I, II and III
Option A (I, II and III) is correct because a hydrocarbon with the formula C5H10{C}_5 {H}_{10} can be mono-, di-, or trisubstituted.
99 Medium
99 of 269
Analyzing +4 | -1
The olefin which on ozonolysis gives CH3CH2CHO  and  CH3CHO\mathrm{C}{\mathrm{H}_3}\mathrm{C}{\mathrm{H}_2}\mathrm{CHO}\;and\;\mathrm{C}{\mathrm{H}_3}\mathrm{CHO} is:
View answer and explanation
Correct option: D
2-pentene
Option D (2-pentene) is correct because ozonolysis of 2-pentene yields propanal (CH3CH2CHO) and acetaldehyde (CH3CHO).
100 Easy
100 of 269
Analyzing +4 | -1
An isomer of C6H14{C}_6 {H}_{14} forms three monochloro derivatives. The isomer may be (Excluding stereo isomer)
View answer and explanation
Correct option: B
nn-hexane
The correct answer is C (2, 3-dimethylbutane) because it forms three monochloro derivatives due to its symmetry.
101 Medium
101 of 269
Analyzing +4 | -1
Assertion :- In the following reaction CH2 = CH2+NaNH2→CH2=C−HNa++NH3{\text{C}}{{\text{H}}_2}{\text{ = C}}{{\text{H}}_{\text{2}}} + {\text{NaN}}{{\text{H}}_{\text{2}}} \to {\text{C}}{{\text{H}}_{\text{2}}} = \mathop {\text{C}}\limits^ - {\text{H}}\mathop {{\text{Na}}}\limits^ + + {\text{N}}{{\text{H}}_{\text{3}}}
reaction is always in forward direction
Reason :-Ethene is stronger acid than ammonia.
View answer and explanation
Correct option: D
If both Assertion & Reason are false
Option D (If both Assertion & Reason are false) is correct because ethene is a weaker acid than ammonia, and the reaction is not always in the forward direction.
102 Easy
102 of 269
Application +4 | -1
The compound (CH3)2CH−CHCl−CH3{\left( {\mathrm{C}{\mathrm{H}_3}} \right)_2}\mathrm{CH}-\mathrm{CHCl}-\mathrm{C}{\mathrm{H}_3} reacts with alcoholic KOH to give the following alkene:
View answer and explanation
Correct option: D
(CH3)2C=CH−CH3{\left( {\mathrm{C}{\mathrm{H}_3}} \right)_2}\mathrm{C} = \mathrm{CH}-\mathrm{C}{\mathrm{H}_3}
The correct answer is D. The major product is determined by Saytzeff's rule, which favors the more substituted alkene.
103 Medium
103 of 269
Understanding +4 | -1
Butyne-2 contains:
View answer and explanation
Correct option: D
Both sps p and sp3s p^3 hybridised carbon atoms
Option D (Both sps p and sp3s p^3 hybridised carbon atoms) is correct because Butyne-2 has two spsp hybridized carbons (the ones in the triple bond) and two sp3sp^3 hybridized carbons (the methyl groups).
104 Medium
104 of 269
Understanding +4 | -1
Which of the following compounds will show cis-trans isomerism?
View answer and explanation
Correct option: C
C6H5CH=CH−CH3\mathrm{C}_6 \mathrm{H}_5 \mathrm{CH}=\mathrm{CH}-\mathrm{CH}_3
Option C (C6H5CH=CH−CH3\mathrm{C}_6 \mathrm{H}_5 \mathrm{CH}=\mathrm{CH}-\mathrm{CH}_3) is correct because for cis-trans isomerism to occur, each carbon of the double bond must have two different groups attached to it.
105 Medium
105 of 269
Analyzing +4 | -1
Which of the following statement is incorrect regarding nitration of aromatic compounds?

I. The rate of nitration of benzene decreases at 323-333 K.

II. The rate of nitration of toluene is greater than that of benzene.

III. The rate of nitration of chlorobenzene is greater than that of benzene.

IV. Nitration is an electrophilic substitution reaction. Choose the incorrect statement.
View answer and explanation
Correct option: B
Both I and III
Option B (Both I and III) is correct because the rate of nitration of benzene increases with temperature, and the rate of nitration of chlorobenzene is lower than that of benzene due to the -I effect of chlorine.
106 Easy
106 of 269
Application +4 | -1
A fuel contains 25% nn-heptane and 75% iso-octane. Its octane number is:
View answer and explanation
Correct option: B
75
Option B (75) is correct because the octane number is defined as the percentage of iso-octane in the fuel mixture.
107 Easy
107 of 269
Understanding +4 | -1
The number of isomeric xylenes is:
View answer and explanation
Correct option: B
3
Option B (3) is correct because there are three possible isomers of xylene: ortho-xylene, meta-xylene, and para-xylene.
108 Medium
108 of 269
Remembering +4 | -1
2-Hexyne can be converted into trans-2-hexene
by the action of:
View answer and explanation
Correct option: B
Na in liq. NH3\mathrm{NH}_3
Option B (Na in liq. NH3\mathrm{NH}_3) is correct because sodium in liquid ammonia reduces alkynes to trans-alkenes.
109 Medium
109 of 269
Understanding +4 | -1
The alkane that gives only one mono-chloro product on chlorination with Cl2C{l_2} in presence of diffused sunlight is
View answer and explanation
Correct option: B
Ethane
Option B (Ethane) is correct because ethane (CH3−CH3C{H_3} - C{H_3}) has only one type of hydrogen atom, leading to only one monochloro product upon chlorination.
110 Medium
110 of 269
Understanding +4 | -1
Pure methane can be produced by
View answer and explanation
Correct option: C
Soda lime decarboxylation
Option C (Soda lime decarboxylation) is correct because it's the only method listed that directly produces methane from sodium acetate.
111 Easy
111 of 269
Understanding +4 | -1
Which of the following does not decolourise bromine solution in carbon disulphide?
View answer and explanation
Correct option: C
Ethane
Option C (Ethane) is correct because it is a saturated compound with no pi bonds and therefore does not react with bromine solution.
112 Medium
112 of 269
Understanding +4 | -1
Statement I : The boiling point of alkanes increases steadily with increase in size of chain
Statement II : Greater the length of the chain, greater is the surface area and hence, larger the vander Waal's attractive forces
View answer and explanation
Correct option: C
Both statements are true
Option C is correct because both statements are true regarding the boiling point of alkanes and the influence of chain length on van der Waals forces.
113 Medium
113 of 269
Remembering +4 | -1
CH≡CH→Cu2Cl2NH4Clproduct\mathrm{CH} \equiv \mathrm{CH} \xrightarrow[\mathrm{Cu}_2 \mathrm{Cl}_2]{\mathrm{NH}_4 \mathrm{Cl}} \text{product}

Product is:
View answer and explanation
Correct option: A
Cu−C≡C−Cu\mathrm{Cu}-\mathrm{C} \equiv \mathrm{C}-\mathrm{Cu}
Option A (Cu−C≡C−Cu\mathrm{Cu}-\mathrm{C} \equiv \mathrm{C}-\mathrm{Cu}) is correct because the reaction of acetylene with Cu2Cl2\mathrm{Cu}_2 \mathrm{Cl}_2 in NH4Cl\mathrm{NH}_4 \mathrm{Cl} forms copper acetylide.
114 Easy
114 of 269
Remembering +4 | -1
The electrophile involved in chlorination of benzene is
View answer and explanation
Correct option: B
chloronium ion
Option B (chloronium ion) is correct because the chlorination of benzene involves the electrophilic attack of the chloronium ion (Cl+Cl^+) on the benzene ring.
115 Medium
115 of 269
Analyzing +4 | -1
The alkyne which gives pyruvic acid (CH3COCOOH)\left( {\mathrm{C}{\mathrm{H}_3}\mathrm{COCOOH}} \right) on oxidation with alk. KMnO4\mathrm{KMn}{\mathrm{O}_4} is:
View answer and explanation
Correct option: B
CH3C≡CH\mathrm{C}{\mathrm{H}_3}\mathrm{C} \equiv \mathrm{CH}
Option B (CH₃C≡CH) is correct because terminal alkynes with a methyl group adjacent to the triple bond yield pyruvic acid upon oxidation with alkaline KMnO₄.
116 Medium
116 of 269
Analyzing +4 | -1
Assertion (A): In the electrophilic substitution of aryl halides, the incoming electrophile gets attached to the meta position.

Reason (R): Aryl halides are moderately deactivating.
View answer and explanation
Correct option: D
R is correct; A is incorrect
Option D is correct because aryl halides are ortho/para directing due to resonance, not meta directing, and the reason that aryl halides are moderately deactivating is correct.
117 Easy
117 of 269
Analyzing +4 | -1
C6H12{\mathrm{C}_6}{\mathrm{H}_{12}} on addition of HBr in presence and in absence of peroxide gives some product.

It is :
View answer and explanation
Correct option: D
All of these
Option D is correct because C6H12{\mathrm{C}_6}{\mathrm{H}_{12}} can be hexene-3 or 2,3-dimethylbut-2-ene (symmetrical alkenes), and addition of HBr can give different products depending on the presence or absence of peroxide.
118 Medium
118 of 269
Analyzing +4 | -1
(A): Ethene is more reactive than ethane
(R): C=C\mathrm{C}=\mathrm{C} bond energy in ethene is more than that in ethane
View answer and explanation
Correct option: B
Both A and R are correct, R is not the correct explanation of A
Option B (Both A and R are correct, R is not the correct explanation of A) is correct because ethene is indeed more reactive than ethane, but the reason is due to the presence of the pi bond, not the overall bond energy.
119 Easy
119 of 269
Remembering +4 | -1
A chlorohydrocarbon, named chlorodane is used especially as:
View answer and explanation
Correct option: D
Anti-termite
Option D (Anti-termite) is correct because chlorodane is a chlorohydrocarbon specifically used as an anti-termite agent.
120 Medium
120 of 269
Understanding +4 | -1
Propyne does not react with
View answer and explanation
Correct option: D
NaOH
Option D (NaOH) is correct because propyne, being a weak acid, does not react with a strong base like NaOH as the reverse reaction is favored.
Revision notes and key takeaways

Review HYDROCARBONS in your textbook or class notes, then use this set to check your understanding.

Showing 91-120 of 269 questions

Your practice

0 / 269 answered · 269 remaining · 0% accuracy

Progress is stored on this device. Sign in to save to your account.

✓ Correct · ✗ Incorrect · ○ Unanswered

Answer feedback and difficulty

Type at least 2 characters or jump directly:
🔓

Save your learning progress

Sign in with a free account to save your practice answers and quiz results, then return to them when you revise.