125 practice questions on Hydrocarbons, a chapter in the NCERT Class 11 Chemistry textbook, organised into 4 topics. Every question includes a written explanation. Pick a topic below, or try the samples first.
Propane is subjected to monochlorination. How many distinct monochlorination products can be obtained?
A1
B2
C3
D4
Show answer and explanation
Correct answer: B - 2
Monochlorination of propane can yield two distinct products: 1-chloropropane and 2-chloropropane, depending on whether the chlorine atom replaces a hydrogen atom bonded to the primary or secondary carbon.
Q2
easy
Which of the following statements is true regarding the Wurtz reaction for alkane preparation?
AIt involves the coupling of haloalkanes in the presence of sodium.
BIt is used for the preparation of branched alkanes exclusively.
CThe reaction is used to prepare higher alkanes from lower alkanes.
DIt involves the reaction of alkanes with sodium in dry ether.
Show answer and explanation
Correct answer: A - It involves the coupling of haloalkanes in the presence of sodium.
The Wurtz reaction involves the coupling of haloalkanes in the presence of sodium, typically in dry ether, to produce higher alkanes. It is not used to prepare alkanes from lower alkanes or specifically for branched alkanes.
Q3
easy
In the free-radical halogenation of methane, what is the first step in the mechanism?
AChain termination by combining radicals.
BFormation of chloromethane radicals.
CInitiation by homolytic cleavage of chlorine molecules.
DChlorine radicals abstract a hydrogen atom from methane.
Show answer and explanation
Correct answer: C - Initiation by homolytic cleavage of chlorine molecules.
The first step in the free-radical halogenation of methane is the initiation step, where chlorine molecules undergo homolytic cleavage to form chlorine radicals. Hydrogen abstraction and chloromethane radical formation occur in later steps.
Q4
easy
Benzene is known to have equal bond lengths due to which of the following phenomena?
AResonance
BInductive effect
CIonic bonding
DHyperconjugation
Show answer and explanation
Correct answer: A - Resonance
Resonance in benzene leads to the delocalization of electrons across all the carbon atoms, resulting in equal bond lengths. Ionic bonding, inductive effect, and hyperconjugation do not result in equal bond lengths in benzene.
Q5
hard
But-1-ene is to be converted into but-1-yne. Which sequence of reagents achieves this?
A$\text{HBr}$, then alcoholic KOH
B$\text{Br}_2$ in $\text{CCl}_4$, then zinc dust
CHot concentrated $\text{KMnO}_4$, then sodamide
D$\text{Br}_2$ in $\text{CCl}_4$, then excess sodamide
Show answer and explanation
Correct answer: D - $\text{Br}_2$ in $\text{CCl}_4$, then excess sodamide
Bromine adds across the double bond to give 1,2-dibromobutane, and excess sodamide then removes two molecules of HBr to leave a triple bond. Adding HBr and eliminating with alcoholic KOH simply returns an alkene. Bromine followed by zinc dust also regenerates but-1-ene, and hot $\text{KMnO}_4$ cleaves the double bond into carboxylate fragments instead.
Q6
medium
In Friedel-Crafts alkylation, why is a Lewis acid catalyst like AlCl3 necessary?
ATo reduce the alkyl halide
BTo generate the electrophile from the alkyl halide
CTo stabilize the benzene ring
DTo increase the reaction temperature
Show answer and explanation
Correct answer: B - To generate the electrophile from the alkyl halide
AlCl3 is necessary to generate the electrophile (carbocation) from the alkyl halide by accepting a chlorine atom. It does not increase temperature or stabilize the benzene ring, nor does it reduce the alkyl halide.
Practice the full Hydrocarbons bank free
All 125 questions with timed practice, instant scoring, and explanations. Free forever.