Hydrocarbons — NEET Chemistry MCQs with answers

Hydrocarbons is a Class 11 Chemistry chapter in the NEET (UG) syllabus. NEET720 has 992 reviewed practice questions on it, each with a quick answer and a step-by-step explanation. The 8 questions below are free and fixed, so you can bookmark this page; the full chapter, plus mistake tracking and spaced revision, is in the app.

Topics covered

Alkanes · Conformations · Alkenes · Alkynes · Aromatic hydrocarbons · Nomenclature and isomerism · Hydrocarbons · Arenes · Alkanes/Alkenes/Alkynes · Alkyl halides / Alkenes · Dienes · Cycloalkanes · Nomenclature · Isomerism in hydrocarbons · Hydrocarbons - general · Aromaticity · Aromatic compounds · Alkenes and Alkynes · General hydrocarbons · Mixed hydrocarbons · Conformations and Alkenes · Aromaticity and Alkanes · Physical properties of alkanes · Heat of hydrogenation · Aromatic ring stability · Alkyne physical properties · Free radical halogenation kinetics · Conformational analysis · Combustion enthalpy trend · Electrophilic aromatic substitution rate · Combustion analysis · Ozonolysis product identification · Wurtz reaction limitation · Baeyer's test / bromine water test · Relative rates of monochlorination · Alkyne hydration regioselectivity · Ozone/bromine titration data · Alkyne acidity · Peroxide effect · Side-chain oxidation of arenes

8 free Hydrocarbons practice questions with answers

Choose an answer in your head before opening it. Each explanation says why the correct option is right and, where relevant, why the tempting wrong option is wrong.

1. Question 1 · medium · Alkanes

   A student claims: "Since branched alkanes are more compact and have stronger internal C-C bonds packed closer together, branching should INCREASE the boiling point compared to the straight-chain isomer." Where does this reasoning go wrong?

   • A.The student is correct, but only for alkanes with more than 10 carbons
   • B.Boiling point does not depend on branching at all — it depends only on molecular weight
   • C.It confuses intramolecular bond strength (unaffected by branching) with intermolecular attractive forces (which decrease with branching due to reduced surface contact); branching actually lowers the boiling point
   • D.The student is correct — branching does increase boiling point because compact molecules have less empty space and pack more efficiently as liquids
   Show answer and explanation

   Answer: C. It confuses intramolecular bond strength (unaffected by branching) with intermolecular attractive forces (which decrease with branching due to reduced surface contact); branching actually lowers the boiling point

   Boiling point is governed by intermolecular (van der Waals) forces between separate molecules, which depend on surface area of contact, not by intramolecular C-C bond strength (which is essentially unchanged by branching); branching makes a molecule more compact/spherical, reducing intermolecular contact area and hence lowering the boiling point, the opposite of the student's claim.

   The student conflates two unrelated ideas: (1) intramolecular bond strength — the C-C and C-H covalent bond strengths within a molecule, which do not meaningfully change with branching pattern, and (2) intermolecular forces — the weak London dispersion forces between neighboring molecules that must be overcome for boiling, which strongly depend on how much surface area two molecules can bring into contact. A branched, more spherical molecule has less surface area available for contact with neighbors compared to an elongated straight chain of the same molecular formula, so its dispersion forces (and hence boiling point) are lower, not higher. This is empirically confirmed: n-pentane (36°C) > isopentane (28°C) > neopentane (9.5°C).

   Common mistake: Conflating intramolecular bond strength/compactness with the intermolecular forces that actually govern boiling point

2. Question 2 · medium · Alkanes

   Consider the following statements about alkane preparation methods: (I) The Wurtz reaction must be carried out in dry ether because sodium and organosodium intermediates react violently with any trace of water. (II) Kolbe electrolysis is carried out on a concentrated AQUEOUS solution of the sodium/potassium salt of a carboxylic acid, unlike Wurtz which uses a dry, non-aqueous solvent. (III) Decarboxylation of a sodium carboxylate salt requires plain NaOH alone (without CaO) for a clean, controlled reaction. (IV) Catalytic hydrogenation of an alkene requires a finely divided metal catalyst (Ni, Pd, or Pt) to proceed at a practical rate under moderate conditions. Which of these statements are correct?

   • A.I, II, III, and IV
   • B.II, III, and IV only
   • C.I and IV only
   • D.I, II, and IV only
   Show answer and explanation

   Answer: D. I, II, and IV only

   Statements I, II, and IV are standard, correct facts about reaction conditions for these preparation methods; statement III is false because decarboxylation is carried out with soda lime (NaOH + CaO), not plain NaOH, specifically to protect the glass apparatus from corrosive molten NaOH.

   I: True — sodium metal and the organosodium intermediate in Wurtz reaction react violently/explosively with water, so strictly anhydrous (dry) ether is essential. II: True — Kolbe electrolysis uses a concentrated aqueous solution of the carboxylate salt as the electrolyte, in clear contrast to Wurtz's non-aqueous dry ether medium. III: False — decarboxylation with 'soda lime' specifically uses a mixture of NaOH and CaO; CaO is added because it reduces the corrosiveness of fused/molten NaOH toward the glass reaction vessel, making the reaction practically safer, not because plain NaOH alone gives a 'cleaner' reaction (plain NaOH heated in glass is actually more likely to damage the apparatus). IV: True — catalytic hydrogenation requires a metal catalyst (Ni typically, sometimes Pd or Pt) to adsorb H2 and the alkene and enable practical reaction rates; without a catalyst, H2 does not add across a C=C bond under ordinary conditions.

   Common mistake: Assuming plain NaOH (without CaO) is used for decarboxylation, missing the practical role of CaO

3. Question 3 · medium · Alkanes

   A student, when predicting the major product of free radical monochlorination of isobutane, argues: "By Markovnikov's rule, the halogen should go to the carbon with fewer hydrogens (the more substituted one), so the tertiary C-H is chlorinated preferentially." Is this reasoning sound?

   • A.Yes — Markovnikov's rule is a completely general principle that applies to any reaction forming a new C-X bond, including radical substitution on alkanes
   • B.No — the tertiary C-H is actually LESS reactive, so both the reasoning and the conclusion are wrong
   • C.Yes — because Markovnikov's rule and free radical stability considerations are, in this specific case, mathematically identical formulas
   • D.No — the conclusion (tertiary C-H is preferred) happens to be correct, but Markovnikov's rule is the wrong justification; it applies to electrophilic addition across a C=C double bond, not to radical substitution on a saturated alkane, which is governed instead by relative radical stability/reactivity
   Show answer and explanation

   Answer: D. No — the conclusion (tertiary C-H is preferred) happens to be correct, but Markovnikov's rule is the wrong justification; it applies to electrophilic addition across a C=C double bond, not to radical substitution on a saturated alkane, which is governed instead by relative radical stability/reactivity

   Isobutane's tertiary C-H is indeed preferentially chlorinated, but the correct reason is that H-abstraction there forms the more stable tertiary radical (lower activation energy), NOT Markovnikov's rule, which is specific to electrophilic addition across alkene double bonds and does not apply to a saturated alkane substrate with no double bond at all.

   Markovnikov's rule addresses regiochemistry when an unsymmetrical reagent (like HX) adds across an unsymmetrical alkene's C=C double bond, based on the relative stability of the possible carbocation intermediates. Isobutane has no double bond, so there is no possible carbocation intermediate and Markovnikov's rule simply does not apply as a mechanism. Free radical halogenation instead proceeds by homolytic C-H bond cleavage, and the site selectivity (tertiary > secondary > primary) is explained by the relative stability of the resulting free RADICAL (not carbocation), governed by hyperconjugation/induction, exactly as discussed in radical-stability-focused questions. The student reaches the right answer (tertiary preferred) through an incorrect mechanistic justification — a common but conceptually serious error that NEET tests to check genuine mechanistic understanding rather than rote pattern matching.

   Common mistake: Applying Markovnikov's rule outside its proper mechanistic context (electrophilic addition to alkenes)

4. Question 4 · easy · Alkenes

   According to Markovnikov's rule, when an unsymmetrical alkene reacts with HX, the hydrogen atom of HX adds to the carbon of the double bond that:

   • A.is the terminal carbon regardless of substitution
   • B.already carries the fewer number of hydrogen atoms
   • C.is attached to the more electronegative substituent
   • D.already carries the greater number of hydrogen atoms
   Show answer and explanation

   Answer: D. already carries the greater number of hydrogen atoms

   Markovnikov's rule (empirical form): H adds to the carbon already bearing more H atoms, so X adds to the more substituted carbon, forming the more stable carbocation intermediate.

   In electrophilic addition of HX to an unsymmetrical alkene, H+ adds first to give the more stable carbocation. The carbon that already has more hydrogens is the less substituted one; protonating there leaves the positive charge on the more substituted carbon, which is stabilized by more alkyl (hyperconjugation/inductive) groups. Hence 'rich gets richer': H goes to the carbon already richer in H.

   Common mistake: Reversing which carbon gets H vs X

5. Question 5 · medium · Alkenes

   In the acid-catalysed addition of HX to propene, the rate-determining step generates a carbocation. The regiochemistry (Markovnikov orientation) is ultimately controlled by:

   • A.the polarity of the H–X bond only
   • B.the size of the halide ion, which attacks the less hindered carbon
   • C.the relative stability of the secondary vs primary carbocation formed on protonation
   • D.the boiling point of the hydrogen halide used
   Show answer and explanation

   Answer: C. the relative stability of the secondary vs primary carbocation formed on protonation

   Protonation of propene can give either a primary or secondary carbocation; the secondary cation is more stable (more hyperconjugation/+I from two alkyl groups), so it forms preferentially and dictates that Br- attacks C2, giving 2-bromopropane.

   H+ can add to C1 (terminal CH2) giving a secondary carbocation at C2, or to C2 giving a primary carbocation at C1. The secondary carbocation is markedly more stable due to greater hyperconjugative and inductive electron donation from two alkyl groups. The reaction proceeds through the lower-energy transition state leading to this more stable intermediate, so X- then bonds to C2, giving the Markovnikov product 2-halopropane.

   Common mistake: Attributing regiochemistry to steric bulk of the nucleophile instead of cation stability

6. Question 6 · easy · Alkenes

   Saytzeff's rule states that in a beta-elimination reaction giving a mixture of alkenes, the major product is the one in which:

   • A.the double bond is between the two least substituted carbons
   • B.the double bond is terminal, giving the least hindered alkene
   • C.the alkene formed is the most stable, generally the most highly substituted one
   • D.the alkene retains the original halogen's position as a substituent
   Show answer and explanation

   Answer: C. the alkene formed is the most stable, generally the most highly substituted one

   Saytzeff's rule: the more substituted (thermodynamically more stable) alkene is the major elimination product, because more alkyl substitution stabilizes the C=C via hyperconjugation.

   In E1/E2 eliminations with unhindered bases, the transition state leading to the more substituted double bond is lower in energy, mirroring the greater thermodynamic stability of more substituted alkenes (more hyperconjugating C–H bonds at the alkene carbons). This more-substituted alkene is termed the Saytzeff product and is normally the major product unless a bulky base favours the Hofmann (less substituted) alkene.

   Common mistake: Confusing Saytzeff orientation with Hofmann (bulky-base) orientation

7. Question 7 · medium · Alkenes

   Reductive ozonolysis of an alkene (O3 followed by Zn/H2O or Zn/CH3COOH) is used in preference to plain hydrolytic workup of the ozonide because the reducing agent:

   • A.prevents further oxidation of the aldehyde products to carboxylic acids
   • B.prevents the ozonide from decomposing back to the starting alkene
   • C.is required to cleave the C=C bond in the first place
   • D.converts the alkene to an epoxide before cleavage
   Show answer and explanation

   Answer: A. prevents further oxidation of the aldehyde products to carboxylic acids

   Zn (or dimethyl sulfide) reduces the intermediate ozonide/peroxidic species and prevents H2O2 generated during workup from oxidizing any aldehyde product further to a carboxylic acid, so aldehydes are obtained cleanly (reductive/'mild' ozonolysis).

   Ozone adds across the C=C bond to form a molozonide, which rearranges to the ozonide. Aqueous hydrolysis of the ozonide alone would release H2O2 as a by-product, which can oxidise any aldehyde formed to a carboxylic acid, contaminating the product mixture. Adding a reducing agent such as Zn dust (or Me2S) destroys this H2O2 as it forms, so aldehydes survive unoxidised — this is why 'Zn/H2O' or 'Zn/AcOH' workup is specified for clean aldehyde/ketone isolation, whereas an oxidative workup (H2O2) deliberately converts any aldehyde product to acid.

   Common mistake: Believing Zn cleaves the double bond rather than ozone

8. Question 8 · medium · Alkenes

   Among 1-butene, cis-2-butene and 2-methylpropene, the relative thermodynamic stability (based on heat of hydrogenation and degree of substitution) follows the order:

   • A.1-butene > cis-2-butene > 2-methylpropene
   • B.2-methylpropene ≈ cis-2-butene > 1-butene, since both are disubstituted alkenes while 1-butene is monosubstituted
   • C.cis-2-butene > 2-methylpropene > 1-butene by a large, clearly resolved margin
   • D.all three have identical stability since they are all C4H8 isomers
   Show answer and explanation

   Answer: B. 2-methylpropene ≈ cis-2-butene > 1-butene, since both are disubstituted alkenes while 1-butene is monosubstituted

   Alkene stability increases with the number of alkyl groups on the double-bond carbons (more hyperconjugation); 2-methylpropene and cis-2-butene are both disubstituted and close in stability, while 1-butene (monosubstituted) is less stable, shown by its larger heat of hydrogenation.

   Heats of hydrogenation decrease as alkene stability increases. 1-Butene (CH2=CH-CH2-CH3) has only one alkyl group on the double bond (monosubstituted) and the highest heat of hydrogenation (~126 kJ/mol) among these. cis-2-Butene (CH3-CH=CH-CH3) and 2-methylpropene ((CH3)2C=CH2) are both disubstituted alkenes with comparable, lower heats of hydrogenation (~120 and ~118 kJ/mol respectively), reflecting greater hyperconjugative stabilization from the extra alkyl substituents. Hence stability: disubstituted pair > monosubstituted 1-butene.

   Common mistake: Treating all isomeric alkenes as equally stable or ranking by chain length instead of substitution

Questions about Hydrocarbons for NEET

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Questions are original NEET720 compositions reviewed for correctness, syllabus fit and option quality. Counts update as the bank grows (992 active practice questions in this chapter today).