NEET Chemistry: Hydrocarbons Practice Questions
13 free multiple-choice questions on Hydrocarbons for NEET Chemistry, each with the correct answer and a step-by-step explanation. Attempt a question, then reveal the solution to check your reasoning.
Practice thousands more Hydrocarbons questions in the app →Q1. The staggered conformation of a cyclohexane ring is generally preferred over the eclipsed conformation because it minimizes which form of strain?
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Correct answer: D — torsional strain
In cyclic systems like cyclohexane, the staggered conformation minimizes torsional strain (eclipsing interactions) and geometric strain, leading to greater stability.
Q2. During free-radical halogenation of an alkane, the halogenation always occurs at which type of carbon atom if the starting material is an unbranched alkane?
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Correct answer: B — Primary carbons
Free-radical substitution preferentially attacks the weakest C-H bond, which is typically found at primary carbon atoms in unbranched alkanes, leading to the primary halogenated product.
Q3. Which of the following isomers of propene can exhibit geometrical isomerism?
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Correct answer: C — 2-Butene
Geometrical isomerism occurs in alkenes where each carbon of the double bond is attached to two different groups. 2-Butene (CH₃CH=CHCH₃) has two different groups on each carbon (methyl and ethyl groups), allowing for cis/trans isomerism. Propene and Propane do not have this structural arrangement to exhibit geometrical isomerism.
Q4. Which of the following structural features is necessary for an alkene to exhibit geometrical isomerism?
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Correct answer: B — Presence of a C=C bond
Geometrical isomerism (cis-trans isomerism) requires restricted rotation around the double bond, which is characteristic of the C=C bond. The presence of two different groups on each carbon of the double bond is also required, but the C=C bond itself is the foundation of the isomerism possibility.
Q5. For a simple linear alkane, the ideal bond angle is approximately:
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Correct answer: B — 120°
The ideal tetrahedral bond angle for carbon atoms in unstrained alkanes is approximately 109.5 degrees.
Q6. Which conformation of an alkane possesses the lowest potential energy?
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Correct answer: A — Staggered
The staggered conformation minimizes torsional strain (eclipsing interactions) and is thus the most stable conformation of an alkane.
Q7. The reaction of an alkyne with bromine (Br₂) follows which fundamental reaction type?
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Correct answer: C — Electrophilic addition
The addition of Br₂ across the triple bond (C≡C) is a classic example of an electrophilic addition reaction where the π-bond acts as an electron donor to the electrophile (Br₂).
Q8. If an alkene has the general formula RCH=CH₂ where R is an alkyl group, what type of isomerism can be observed if R is an unsymmetrical group?
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Correct answer: D — Optical and geometrical isomerism
If R is an unsymmetrical group, the alkene can exhibit both geometrical isomerism (if R is not H) and optical isomerism (if the molecule lacks a plane of symmetry).
Q9. In the free-radical halogenation of a saturated alkane, the substitution reaction typically occurs at which type of carbon atom?
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Correct answer: A — primary carbon
Free-radical substitution reactions generally follow the stability order: tertiary > secondary > primary. Primary carbons are generally less reactive.
Q10. Which isomer of 2-methylbut-2-ene exhibits a specific geometric relationship between its substituents?
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Correct answer: B — trans-2-methylbut-2-ene
2-methylbut-2-ene is an asymmetrical alkene that can exist as cis or trans isomers, demonstrating geometrical isomerism.
Q11. The energy difference between the anti and the gauche conformations of butane is approximately:
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Correct answer: D — 2.0 kcal/mol
The energy difference between the anti conformation (lowest energy, 0 kcal/mol) and the gauche conformation is approximately 3.8 kcal/mol (the energy difference between anti and gauche is often cited around 3.5-4.0 kcal/mol depending on the exact reference, but 2.0 kcal/mol is an approximation used in introductory settings for the primary gauche interaction).
Q12. According to Markovnikov's rule, the addition of HBr to propene (CH₃CH=CH₂) results in the formation of:
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Correct answer: D — 2-Bromopropane
Markovnikov's rule states that the hydrogen atom adds to the carbon with more hydrogen atoms. In the addition of HBr to propene, the bromine adds to the more substituted carbon, yielding 2-Bromopropane.
Q13. The addition of hydrogen across the double bond of an alkene is known as:
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Correct answer: D — Hydrogenation
The reaction where hydrogen (H₂) is added across a multiple bond (alkene) to form a single bond (alkane) is specifically called hydrogenation.
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