Skip to main content
Back to HYDROCARBONS

Your Progress

0 / 118 Answered
0
Correct
0
Incorrect
0%
Accuracy
JEE • CHEMISTRY • HYDROCARBONS

Hydrocarbons practice · Medium

Try a question, then open “View answer” to check it. Sign in to submit answers and save progress.

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 91-118 of 118 questions
91 Medium
91 of 118
Analyzing +4 | -1
Which of the following will not produce ethane
View answer and explanation
Correct option: B
reduction of CH3COCH3C{H_3}COC{H_3} with HI/P4HI/{P_4}
Option B (reduction of CH3COCH3C{H_3}COC{H_3} with HI/P4HI/{P_4}) will produce propane, not ethane.
92 Medium
92 of 118
Remembering +4 | -1
The most stable conformational isomer of cyclohexane is:
View answer and explanation
Correct option: A
Chair form
Option A (Chair form) is correct because the chair conformation of cyclohexane is the most stable due to minimal torsional and steric strain.
93 Medium
93 of 118
Remembering +4 | -1
Among the following number of meta directing groups are:
−CN,−NO2,−COOH,−CH3,−SO3H,−N+H3,−F-\mathrm{CN},-\mathrm{NO}_2,-\mathrm{COOH},-\mathrm{CH}_3,-\mathrm{SO}_3 \mathrm{H},-\stackrel{+}{\mathrm{N}} \mathrm{H}_3,-\mathrm{F}
View answer and explanation
Correct option: A
5
Option A (5) is correct because there are five meta-directing groups among the listed substituents: -CN, -NO2, -COOH, -SO3H, and -NH3+.
94 Medium
94 of 118
Remembering +4 | -1
Which of the following has the least octane number?
View answer and explanation
Correct option: D
nn -heptane
Option D (nn -heptane) is correct because n-heptane is assigned an octane number of 0, representing very poor antiknock properties.
95 Medium
95 of 118
Understanding +4 | -1
The common name and IUPAC name of CH3CH2−C≡CH\mathrm{CH}_3 \mathrm{CH}_2-\mathrm{C} \equiv \mathrm{CH}, respectively are
View answer and explanation
Correct option: B
ethylacetylene and but-1-yne
The correct answer is A (Liquid solvent, solid solvent) because during the depression of freezing point, the liquid solvent and solid solvent are in equilibrium.
96 Medium
96 of 118
Analyzing +4 | -1
In the reaction C2H6→(CH3COO)2MnA→NaOHB,{C_2}{H_6}\xrightarrow{{{{\left( {C{H_3}COO} \right)}_2}Mn}}A\xrightarrow{{NaOH}}B, the correct statement about BB is
View answer and explanation
Correct option: D
Both 1 and 2
Option D is correct because after the given reactions, A is acetic acid and B is sodium acetate, which on decarboxylation gives methane and on Kolbe's electrolysis gives ethane.
97 Medium
97 of 118
Understanding +4 | -1
Which of the following will decolourise alkaline KMnO4\mathrm{KMnO}_4 solution?
View answer and explanation
Correct option: D
C2H4\mathrm{C}_2 \mathrm{H}_4
Option D (C2H4\mathrm{C}_2 \mathrm{H}_4) is correct because it is an unsaturated compound (alkene) that can react with and decolorize alkaline KMnO4\mathrm{KMnO}_4 solution via oxidation.
98 Medium
98 of 118
Analyzing +4 | -1
In the following sequence the product D is,

CH≡CH→HBrA→HBrB→KOH(alc.)C→NaNH2D:\mathrm{CH} \equiv \mathrm{CH}\xrightarrow{{HBr}}A\xrightarrow{{HBr}}B\xrightarrow{{\mathrm{KOH}(\mathrm{alc}\mathrm{.})}}C\xrightarrow{{\mathrm{NaN}{\mathrm{H}_2}}}D:
View answer and explanation
Correct option: C
Ethyne
Option C (Ethyne) is correct because the sequence of reactions leads back to the starting material, ethyne.
99 Medium
99 of 118
Analyzing +4 | -1
An unknown compound A has a molecular formula C4H6{\mathrm{C}_4}{\mathrm{H}_6}. When A is treated with excess of Br2\mathrm{B}{\mathrm{r}_2} a new substance B with formula C4H6Br4  {\mathrm{C}_4}{\mathrm{H}_6}\mathrm{B}{\mathrm{r}_4}\mathrm{\;} is formed. A forms a white ppt. with ammoniacal silver nitrate solution. A may be :
View answer and explanation
Correct option: A
But-1-yne
The correct answer is But-1-yne because it has a terminal alkyne which reacts with ammoniacal silver nitrate to form a white precipitate, and the formula C4H6 suggests two degrees of unsaturation.
100 Medium
100 of 118
Understanding +4 | -1
Which of the following can be sulphonated most easily
View answer and explanation
Correct option: B
toluene
Option B (toluene) is correct because the methyl group on toluene is an electron-donating group, activating the ring towards electrophilic substitution, such as sulfonation.
101 Medium
101 of 118
Understanding +4 | -1
Which of the following reagent is not involved in the halogenation of benzene
View answer and explanation
Correct option: C
Br2/CCl4B{r_2}/CC{l_4}
Option C (Br2/CCl4B{r_2}/CC{l_4}) is correct because CCl4CC{l_4} acts as a solvent but not as a halogen carrier to generate an electrophile, which is necessary for the halogenation of benzene.
102 Medium
102 of 118
Analyzing +4 | -1
CH3−C≡CH→H2SO4HgSO4A    CH3−CH=CH2→2⋅H2O2/OH−1.BH3,THFX→PCCB{\text{C}}{{\text{H}}_3} - {\text{C}} \equiv {\text{CH}}\xrightarrow[{{{\text{H}}_2}{\text{S}}{{\text{O}}_4}}]{{{\text{HgS}}{{\text{O}}_4}}}{\text{A}}\,\,\,\,{\text{C}}{{\text{H}}_3} - {\text{CH}} = {\text{C}}{{\text{H}}_2}\xrightarrow[{2 \cdot {{\text{H}}_2}{{\text{O}}_2}/{\text{O}}{{\text{H}}^ - }}]{{1.{\text{B}}{{\text{H}}_3},{\text{THF}}}}{\text{X}}\xrightarrow{{{\text{PCC}}}}{\text{B}}
Products A\mathrm{A} and B\mathrm{B} are differentiated by
View answer and explanation
Correct option: B
Tollen’s reagent
Option B is correct because Tollen's reagent can differentiate between a ketone (A) and an aldehyde (B).
103 Medium
103 of 118
Remembering +4 | -1
Ethylene reacts with 1% alkaline KMnO4\mathrm{KMn}{\mathrm{O}_4} to form
View answer and explanation
Correct option: B
Ethylene glycol
Option B (Ethylene glycol) is correct because the reaction of ethylene with 1% alkaline KMnO4\mathrm{KMn}{\mathrm{O}_4} (Baeyer's reagent) produces ethylene glycol.
104 Medium
104 of 118
Analyzing +4 | -1
Which of the following hydrocarbon on reductive ozonolysis will produce CO2\mathrm{CO}_2 gas?
View answer and explanation
Correct option: C
CH2=C=CH2\mathrm{CH}_2=\mathrm{C}=\mathrm{CH}_2
Option C is correct because reductive ozonolysis of allene (CH₂=C=CH₂) produces two moles of formaldehyde (HCHO) and one mole of carbon dioxide (CO₂).
105 Medium
105 of 118
Application +4 | -1
In a nitration experiment, 10.0 g10.0 \mathrm{~g} of benzene gave 13.2 g13.2 \mathrm{~g} of nitrobenzene. The percentage yield for nitration of benzene is:
View answer and explanation
Correct option: A
83.7%83.7 \%
Option A (83.7%83.7 \%) is correct because it is the percentage yield calculated from the actual yield of nitrobenzene compared to the theoretical yield based on the amount of benzene used.
106 Medium
106 of 118
Analyzing +4 | -1
Assertion (A): Acidity of C−H\mathrm{C}-\mathrm{H} bond decreases in the order 3CH≡CH→Red hot iron tube3{\text{CH}} \equiv {\text{CH}}\xrightarrow{{{\text{Red hot iron tube}}}}
Reason (R): Acidity of the C−H\mathrm{C}-\mathrm{H} bond decreases as the electronegativity of the carbon to which it is attached decreases.
View answer and explanation
Correct option: A
Both A &\& R are correct and R is correct explanation of A
Both A & R are correct and R is correct explanation of A because the acidity of C-H bond is directly related to the s-character of the carbon atom, which also affects its electronegativity.
107 Medium
107 of 118
Understanding +4 | -1
Butyne on oxidation with hot. alk KMnO4KMn{O_4} gives
View answer and explanation
Correct option: C
CH3CH2COOH+CO2+H2OC{H_3}C{H_2}COOH + C{O_2} + {H_2}O
Option C (CH3CH2COOH+CO2+H2OC{H_3}C{H_2}COOH + C{O_2} + {H_2}O) is correct because butyne oxidation with hot alkaline KMnO4 cleaves the triple bond, yielding propanoic acid and carbon dioxide.
108 Medium
108 of 118
Understanding +4 | -1
Which of the following show Acidic Character of Alkyne?
View answer and explanation
Correct option: C
NaNH2{NaNH}_2
Option C (NaNH2{NaNH}_2) is correct because sodium amide is a strong base that can deprotonate terminal alkynes, demonstrating their acidic character.
109 Medium
109 of 118
Understanding +4 | -1
Which of the following statement(s) is/are correct?

(i) Alkanes can have infinite number of conformations by rotation around a C−C{C}-{C} single bonds.

(ii) Rotation around C−C{C}-{C} single bond is completely free.

(iii) Rotation is hindered by a small energy barrier of 1−20 kJ mol−11-20 {~kJ} {~mol}^{-1} due to torsional strain.
View answer and explanation
Correct option: B
(i) and (iii)
Option B ((i) and (iii)) is correct because alkanes can have infinite conformations due to rotation around C-C single bonds, and this rotation is hindered by a small energy barrier due to torsional strain.
110 Medium
110 of 118
Remembering +4 | -1
The neutral wax called Ozokerite found near petroleum well is a mixture of:
View answer and explanation
Correct option: B
Solid hydrocarbons
Option B (Solid hydrocarbons) is correct because Ozokerite is a naturally occurring mineral wax composed of various solid hydrocarbons.
111 Medium
111 of 118
Application +4 | -1
Which of the compounds with molecular formula C5H10{C_5}{H_{10}} yields acetone on ozonalysis
View answer and explanation
Correct option: B
2 – methyl – 2 – butene
Option B (2 – methyl – 2 – butene) is correct because ozonolysis of 2-methyl-2-butene yields acetone and another aldehyde.
112 Medium
112 of 118
Understanding +4 | -1
Iodination of alkanes is always carried out in the presence of some oxidising agents like HIO3{HIO}_3. This is because
View answer and explanation
Correct option: B
it is a reversible reaction
Option B (it is a reversible reaction) is correct because iodination of alkanes is a reversible reaction, and the HI formed can react with the iodinated alkane to regenerate the alkane and iodine. Oxidizing agents are used to consume the HI and drive the reaction forward.
113 Medium
113 of 118
Analyzing +4 | -1
Which among the following is very strong o−, p−o - ,\,p - directing group?
View answer and explanation
Correct option: D
−NHR- \mathrm{NHR}
Option D (−NHR-\mathrm{NHR}) is correct because alkylamino groups are stronger activating and directing groups than amino groups due to the electron-donating effect of the alkyl group.
114 Medium
114 of 118
Analyzing +4 | -1
Assertion :- Neopentane forms only one monosubstituted compound.

Reason :- It has only one type of carbon atoms.
View answer and explanation
Correct option: C
If Assertion is True but the Reason is False
Option C is correct because while the assertion is true (neopentane forms only one monosubstituted product), the reason is false (all 12 hydrogens are equivalent).
115 Medium
115 of 118
Understanding +4 | -1
When benzene is heated in air, it burns with sooty flame producing XX and YY, as shown in the given reaction: C6H6+152O2⟶X+Y.\mathrm{C}_6 \mathrm{H}_6+\frac{15}{2} \mathrm{O}_2 \longrightarrow X+Y.\quadHere, XX and YY refer to
View answer and explanation
Correct option: B
CO2,H2O\mathrm{CO}_2, \mathrm{H}_2 \mathrm{O}
Option B is correct because the complete combustion of hydrocarbons like benzene in the presence of sufficient oxygen produces carbon dioxide and water.
116 Medium
116 of 118
Analyzing +4 | -1
S-I: Both Benzene &\& ethyne give same product on ozonolysis.
S-II: Ethyne &\& benzene possess same emperical formula.
View answer and explanation
Correct option: A
S-I &\& S-II are true
Option A (S-I & S-II are true) is correct because both benzene and ethyne give glyoxal on ozonolysis, and they have the same empirical formula (CH).
117 Medium
117 of 118
Analyzing +4 | -1
Which of the following reaction do not give alkyne :-
View answer and explanation
Correct option: C
CH3−CH∣       Br        −CH∣        Br         −CH3C{H_3} - \mathop {CH}\limits_{\begin{array}{cccccccccccccccccccc} {|\,\,\,\,\,\,\,} \\ {Br\,\,\,\,\,\,\,\,} \end{array}} - \mathop {CH}\limits_{\begin{array}{cccccccccccccccccccc} {|\,\,\,\,\,\,\,\,} \\ {Br\,\,\,\,\,\,\,\,\,} \end{array}} - C{H_3}
Option C is correct because the reaction with Zn/Alcohol yields an alkene, not an alkyne, while the other reactions can lead to alkyne formation.
118 Medium
118 of 118
Analyzing +4 | -1
Which alkane (molecular mass = 72) would yield only three mono chlorinated derivates
View answer and explanation
Correct option: A
n - pentane
Option A (n - pentane) is correct because n-pentane yields only three mono chlorinated derivatives due to its symmetry.
Revision notes and key takeaways

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

Showing 91-118 of 118 questions
0 /118
0✓ 0✗
0% Accuracy
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.