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NEET Hydrocarbons MCQs with answers

379 questions on Hydrocarbons for NEET 2027. Below are 15 to try; tap a question to see its answer and explanation.

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15 sample questions

easy

1. In the study of conformations of alkanes, intermediate conformations between the staggered and eclipsed forms are collectively known as:

  1. Gauche conformations
  2. Skew conformations
  3. Anti conformations
  4. Enantiomeric conformations
Show answer and explanation

Answer: (B) Skew conformations

NCERT specifically states that between the staggered and eclipsed conformations of an alkane, an infinite number of possible intermediate conformations exist, and these are known as skew conformations. Gauche and anti are specific conformations with fixed dihedral angles.

medium

2. Arrange the following compounds in decreasing order of reactivity towards electrophilic aromatic substitution: I. Benzene II. Nitrobenzene III. Benzaldehyde IV. Benzoic acid

  1. I > III > IV > II
  2. I > IV > III > II
  3. I > III > II > IV
  4. IV > III > II > I
Show answer and explanation

Answer: (A) I > III > IV > II

Benzene has no deactivating groups and is the most reactive among the four. The groups –CHO, –COOH, and –NO₂ are all electron-withdrawing (-M and -I), deactivating the benzene ring in the order: –NO₂ > –COOH > –CHO. Therefore, the reactivity order towards electrophiles is Benzene > Benzaldehyde > Benzoic acid > Nitrobenzene.

hard

3. In the cyclic polymerization of ethyne, 3 moles of ethyne trimerize to form 1 mole of benzene. What is the percentage change in the hybridization of carbon atoms having sp character transitioning to sp² character in this process?

  1. 100% of the carbons change hybridization
  2. 50% of the carbons change hybridization
  3. 0% (hybridization remains identical)
  4. 75% of the carbons change hybridization
Show answer and explanation

Answer: (A) 100% of the carbons change hybridization

In ethyne (HC≡CH), all carbon atoms are sp-hybridized. In benzene (C₆H₆), all six carbon atoms are sp²-hybridized. Therefore, every carbon atom (100%) undergoes a transition from sp hybridization to sp² hybridization during the cyclic trimerization reaction.

medium

4. Consider the following statements regarding the addition of hydrogen halides to alkenes: Statement I: The peroxide effect is observed with HBr, HCl, and HI. Statement II: The C-Cl bond is too strong to be broken homolytically by the peroxy radical, while the addition of iodine radical to an alkene is endothermic and iodine radicals preferentially recombine. In light of the above statements, choose the correct answer:

  1. Both Statement I and Statement II are correct
  2. Statement I is correct but Statement II is incorrect
  3. Statement I is incorrect but Statement II is correct
  4. Both Statement I and Statement II are incorrect
Show answer and explanation

Answer: (C) Statement I is incorrect but Statement II is correct

Statement I is incorrect because the peroxide effect (anti-Markovnikov addition) is observed only with HBr, not with HCl or HI. Statement II is correct because the H-Cl bond is too strong (energetically unfavourable to break homolytically), and the addition of iodine free radical is endothermic, causing iodine radicals to combine to form I₂ instead.

easy

5. When but-2-yne is hydrogenated over Lindlar's catalyst (Pd/CaCO₃ partially poisoned with quinoline or lead salts), the predominant product is:

  1. cis-But-2-ene
  2. trans-But-2-ene
  3. Butane
  4. But-1-ene
Show answer and explanation

Answer: (A) cis-But-2-ene

Hydrogenation of non-terminal alkynes using Lindlar's catalyst undergoes stereoselective syn-addition of hydrogen, yielding the cis-alkene (cis-but-2-ene) as the major product.

hard

6. When m-xylene (1,3-dimethylbenzene) undergoes mononitration under standard conditions, which position on the ring is least substituted by the incoming –NO₂ group?

  1. C-4
  2. C-5
  3. C-6
  4. C-2
Show answer and explanation

Answer: (D) C-2

Although position 2 lies ortho to both methyl groups where their activating effects reinforce each other electronically, it is severely sterically hindered by the two adjacent methyl groups. Hence, substitution at C-2 occurs in negligible amounts, making it the least substituted position compared to C-4 (or C-6).

medium

7. Which of the following reagent combinations can be successfully used to separate a mixture of hex-1-yne and hex-2-yne?

  1. Ammoniacal cuprous chloride solution followed by dilute HCl
  2. Cold dilute alkaline KMnO₄
  3. Bromine water in CCl₄
  4. Concentrated H₂SO₄
Show answer and explanation

Answer: (A) Ammoniacal cuprous chloride solution followed by dilute HCl

Hex-1-yne is a terminal alkyne and forms a red precipitate of copper acetylide with ammoniacal Cu₂Cl₂ solution, whereas hex-2-yne does not react. The filtered precipitate can then be treated with dilute HCl to regenerate pure hex-1-yne.

easy

8. When 2-methylbut-2-ene is subjected to ozonolysis followed by treatment with Zn dust and H₂O, the products formed are:

  1. Propanal and ethanal
  2. Propan-2-one and ethanal
  3. Butan-2-one and methanal
  4. Propan-2-one and methanal
Show answer and explanation

Answer: (B) Propan-2-one and ethanal

2-Methylbut-2-ene has the structure (CH₃)₂C=CH-CH₃. Cleaving the double bond via reductive ozonolysis produces propan-2-one ((CH₃)₂C=O) and ethanal (CH₃CHO).

medium

9. The addition of bromine (Br₂) in CCl₄ to trans-but-2-ene proceeds via an anti-addition mechanism. The stereochemical outcome of this reaction is:

  1. A racemic mixture of (2R, 3R) and (2S, 3S)-2,3-dibromobutane
  2. A mixture of diastereomers in equal amounts
  3. An optically inactive meso-2,3-dibromobutane
  4. A chiral product with net optical rotation
Show answer and explanation

Answer: (C) An optically inactive meso-2,3-dibromobutane

Anti-addition of bromine across a trans-alkene (like trans-but-2-ene) leads to a meso compound due to internal compensation of optical activity. The resulting meso-2,3-dibromobutane has a plane of symmetry and is optically inactive.

medium

10. During the conformational transition of n-butane about the C2–C3 bond, which conformation represents a local (relative) potential energy maximum, but NOT the absolute maximum?

  1. Anti conformation
  2. Gauche conformation
  3. Partially eclipsed conformation
  4. Fully eclipsed conformation
Show answer and explanation

Answer: (C) Partially eclipsed conformation

The fully eclipsed conformation (dihedral angle 0°) is the absolute maximum in potential energy. The partially eclipsed conformations (dihedral angle 120° and 240°), where CH₃ eclipses H, represent local energy maxima along the rotational coordinate.

easy

11. Regarding the dipole moments of cis- and trans-isomers of 1,2-dichloroethene, which of the following statements is correct?

  1. Both cis and trans isomers have zero dipole moment.
  2. The cis-isomer has a higher dipole moment than the trans-isomer.
  3. The trans-isomer has a higher dipole moment than the cis-isomer.
  4. Both isomers have equal, non-zero dipole moments.
Show answer and explanation

Answer: (B) The cis-isomer has a higher dipole moment than the trans-isomer.

In trans-1,2-dichloroethene, the two opposed C-Cl bond dipoles cancel out symmetrically, resulting in a net dipole moment of zero. In cis-1,2-dichloroethene, the bond dipoles reinforce each other vectorially, yielding a significant net dipole moment (μ > 0).

hard

12. Identify the final product 'Z' in the following reaction sequence: Propyne + HBr → X (major) X + HBr → Y (major) Y + aqueous KOH → Z

  1. Propane-1,2-diol
  2. Propan-2-one
  3. Propanal
  4. Propane-2,2-diol
Show answer and explanation

Answer: (B) Propan-2-one

Addition of HBr to propyne follows Markovnikov's rule to give 2-bromopropene (X). Second addition of HBr gives 2,2-dibromopropane (Y). Reaction of Y with aqueous KOH replaces both -Br groups with -OH to give an unstable gem-diol, which loses water to form propan-2-one (Z).

medium

13. Given that the relative rates of abstraction of 3°, 2°, and 1° hydrogens by a chlorine radical at room temperature are 5.0 : 3.8 : 1.0 respectively, calculate the approximate percentage of 2-chloropropane formed during the monochlorination of propane.

  1. 44%
  2. 79%
  3. 56%
  4. 21%
Show answer and explanation

Answer: (C) 56%

Propane has six 1° hydrogens and two 2° hydrogens. Relative reactivity: for 1° = 6 × 1.0 = 6.0; for 2° = 2 × 3.8 = 7.6. Total reactivity = 6.0 + 7.6 = 13.6. Percentage of 2-chloropropane = (7.6 / 13.6) × 100 ≈ 55.88% ≈ 56%.

medium

14. Given below are two statements: Statement I: cis-But-2-ene has a higher boiling point than trans-but-2-ene. Statement II: cis-But-2-ene has a higher melting point than trans-but-2-ene. In light of the above statements, choose the correct answer:

  1. Both Statement I and Statement II are correct
  2. Statement I is correct but Statement II is incorrect
  3. Statement I is incorrect but Statement II is correct
  4. Both Statement I and Statement II are incorrect
Show answer and explanation

Answer: (B) Statement I is correct but Statement II is incorrect

cis-But-2-ene possesses a net dipole moment, giving rise to stronger dipole-dipole attractions and hence a higher boiling point (approx. 4 °C) than trans-but-2-ene (approx. 1 °C). However, trans-but-2-ene is more symmetrical, packs better into a crystal lattice, and thus has a higher melting point (-106 °C) compared to cis-but-2-ene (-139 °C).

easy

15. Which of the following compounds has a non-zero dipole moment?

  1. trans-1-chloroprop-1-ene
  2. trans-hex-3-ene
  3. trans-1,2-dichloroethene
  4. trans-but-2-ene
Show answer and explanation

Answer: (A) trans-1-chloroprop-1-ene

In trans-1-chloroprop-1-ene, the two bond dipoles (C-Cl and C-CH₃) do not cancel each other out because the groups have different electronegativities and vector magnitudes. In the other three molecules, the symmetrical trans groups result in complete cancellation of bond dipoles (μ = 0).

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