125 practice questions covering Hydrocarbons, organised into 4 topics. 125 questions include a written explanation. Pick a topic below, or try the samples first.
alkenes
35 questions with answers
aromatic-hydrocarbons
34 questions with answers
alkynes
29 questions with answers
alkanes
27 questions with answers
Propane is subjected to monochlorination. How many distinct monochlorination products can be obtained?
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.
Which of the following statements is true regarding the Wurtz reaction for alkane preparation?
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.
In the free-radical halogenation of methane, what is the first step in the mechanism?
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.
Benzene is known to have equal bond lengths due to which of the following phenomena?
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.
But-1-ene is to be converted into but-1-yne. Which sequence of reagents achieves this?
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.
In Friedel-Crafts alkylation, why is a Lewis acid catalyst like AlCl3 necessary?
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.
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