Hydrocarbons

44 NEET-UG practice questions on Hydrocarbons, part of Chemistry. The first 25 are below; the first 5 with answers and explanations (tap Show answer after you have chosen), the rest in Learn with a free account.

1. Question 1 · Easy

   Bromoethane is treated with sodium metal in dry ether. Identify the main hydrocarbon formed.

   1. AEthene
   2. BHexane
   3. CEthane
   4. DButane

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   Answer: D · Butane

   This is the Wurtz reaction: 2C₂H₅Br + 2Na → C₄H₁₀ + 2NaBr. Two ethyl groups join, so the carbon count doubles.

   • A: Assumes elimination, which needs alcoholic KOH rather than sodium.
   • B: Joins three ethyl groups instead of two.
   • C: Keeps two carbons, as though bromine were simply replaced by hydrogen.

   Remember this

   The Wurtz reaction doubles the carbon chain and gives a symmetrical alkane with an even number of carbons.

2. Question 2 · Medium

   Propyne reacts with excess hydrogen bromide in the dark, with no peroxide present. The final product is

   1. A1,2-dibromopropane
   2. B1,3-dibromopropane
   3. C1,1-dibromopropane
   4. D2,2-dibromopropane

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   Answer: D · 2,2-dibromopropane

   The first HBr adds by Markovnikov's rule to give 2-bromopropene. The second H goes to the CH₂ end, leaving the cation at C-2, which is stabilised by bromine's lone pair, so Br again adds to C-2.

   • A: This vicinal dihalide comes from adding Br₂ to propene.
   • B: Puts bromine on non-adjacent carbons, which addition across a triple bond cannot do.
   • C: Places both bromines on the terminal carbon, the anti-Markovnikov direction.

   Remember this

   Two molecules of HX add to a terminal alkyne to give a gem-dihalide, with both halogens on one carbon.

3. Question 3 · Medium

   Match List-I with List-II. List-I (reaction) A. Sodium propanoate heated with soda lime B. Ethyne passed through a red-hot iron tube at 873 K C. Ethene shaken with cold, dilute alkaline KMnO₄ D. But-2-yne with H₂ over Lindlar's catalyst List-II (main organic product) I. Ethane-1,2-diol II. cis-But-2-ene III. Ethane IV. Benzene

   1. AA-III, B-IV, C-II, D-I
   2. BA-IV, B-III, C-I, D-II
   3. CA-III, B-IV, C-I, D-II
   4. DA-I, B-II, C-III, D-IV

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   Answer: C · A-III, B-IV, C-I, D-II

   Soda lime removes CO₂, so propanoate gives ethane. Ethyne trimerises to benzene, cold Baeyer's reagent gives the 1,2-diol, and Lindlar's catalyst delivers both H atoms to one face of the triple bond, giving the cis alkene.

   • A: Swaps the products of the KMnO₄ and Lindlar reactions.
   • B: Swaps the products of the soda lime and red-hot tube reactions.
   • D: Pairs each reaction with a product of an unrelated substrate.

   Remember this

   Decarboxylation with soda lime gives an alkane with one carbon fewer than the salt.

4. Question 4 · Easy

   During the nitration of benzene with a mixture of concentrated HNO₃ and concentrated H₂SO₄, the species that attacks the ring is

   1. ANO₃⁻
   2. BNO₂⁻
   3. CHSO₄⁻
   4. DNO₂⁺

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   Answer: D · NO₂⁺

   Sulphuric acid protonates nitric acid, which then loses water to give the nitronium ion, NO₂⁺. This electrophile attacks the π electrons of the ring.

   • A: The nitrate ion is an anion and does not attack the ring.
   • B: The nitrite ion is an anion and cannot act as an electrophile.
   • C: Hydrogen sulphate is the by-product anion, not the electrophile.

   Remember this

   Benzene undergoes electrophilic substitution, so the attacking species must be electron-seeking; in nitration it is NO₂⁺.

5. Question 5 · Hard

   Which route starting from benzene gives m-chloronitrobenzene as the main product?

   1. ACl₂ in UV light, then nitration with HNO₃/H₂SO₄
   2. BNitration with HNO₃/H₂SO₄, then Cl₂ with anhydrous FeCl₃
   3. CCl₂ with anhydrous FeCl₃, then nitration with HNO₃/H₂SO₄
   4. DNitration with HNO₃/H₂SO₄, then Cl₂ in UV light

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   Answer: B · Nitration with HNO₃/H₂SO₄, then Cl₂ with anhydrous FeCl₃

   –NO₂ is meta-directing, so it must be in place before the second substitution. FeCl₃ then generates Cl⁺, which substitutes at the meta position.

   • A: UV chlorination of benzene gives the addition product benzene hexachloride.
   • C: Chlorine is ortho/para-directing, so this route gives mainly the para and ortho isomers.
   • D: UV light promotes radical addition, not electrophilic ring substitution.

   Remember this

   Plan a disubstituted benzene by introducing the group that directs the next substituent to the required position first.

6. Question 6 · Medium

   Arrange these compounds in increasing order of reactivity towards electrophilic nitration. A. Benzene B. Anisole C. Nitrobenzene D. Chlorobenzene

   1. AC, A, D, B
   2. BB, A, D, C
   3. CC, D, A, B
   4. DD, C, A, B

   Answer and explanation in LearnFree account

7. Question 7 · Medium

   Given below are two statements: Statement I: The cyclopentadienyl anion is aromatic because it is cyclic and planar with six delocalised π electrons. Statement II: Cycloocta-1,3,5,7-tetraene is aromatic because it is a planar ring with eight π electrons. Choose the correct answer.

   1. AStatement I is correct but Statement II is incorrect
   2. BBoth Statement I and Statement II are correct
   3. CBoth Statement I and Statement II are incorrect
   4. DStatement I is incorrect but Statement II is correct

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8. Question 8 · Hard

   An alkene X (C₆H₁₂) gives propanone as the only organic product on ozonolysis followed by Zn/H₂O. What is the main product when X reacts with HBr in the absence of peroxide?

   1. A2-Bromo-2,3-dimethylbutane
   2. B3-Bromohexane
   3. C1-Bromo-2,3-dimethylbutane
   4. D3-Bromo-2,2-dimethylbutane

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9. Question 9 · Medium

   When 2-methylpropene reacts with HBr in the presence of benzoyl peroxide, what is the main product?

   1. A1-Bromobutane
   2. B2-Bromo-2-methylpropane
   3. C2-Bromobutane
   4. D1-Bromo-2-methylpropane

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10. Question 10 · Medium

    Complete hydrogenation of 1.4 g of an alkene consumed 560 mL of H₂ measured at STP. What is the molecular formula of the alkene? (Take the molar volume of a gas at STP = 22.4 L and C = 12, H = 1.)

    1. AC₄H₈
    2. BC₈H₁₆
    3. CC₂H₄
    4. DC₅H₁₀

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11. Question 11 · Medium

    Assertion A: Ethene is a stronger acid than ethyne. Reason R: An sp hybridised carbon has more s-character than an sp² carbon and is more electronegative. Choose the correct answer.

    1. ABoth A and R are correct but R is not the correct explanation of A
    2. BBoth A and R are correct and R is the correct explanation of A
    3. CA is correct but R is not correct
    4. DA is not correct but R is correct

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12. Question 12 · Easy

    Refer to species X and Y in the figure. Assertion A: X is aromatic but Y is not. Reason R: X is planar and cyclically conjugated with 6 π electrons, while Y has an sp³ carbon that breaks the conjugation. Choose the correct answer.

    

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    1. AA is correct but R is not correct
    2. BA is not correct but R is correct
    3. CBoth A and R are correct but R is not the correct explanation of A
    4. DBoth A and R are correct and R is the correct explanation of A

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13. Question 13 · Hard

    The cycloalkene in the figure is ozonised and the ozonide is reduced as shown. Name the product P.

    

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    1. A6-Oxoheptanoic acid
    2. B6-Oxoheptanal
    3. CHexanedial
    4. DHeptane-2,6-dione

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14. Question 14 · Medium

    Identify the major product P of the reaction shown.

    

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    1. A2-Methylpropan-1-ol
    2. B2-Bromo-2-methylpropane
    3. C1,2-Dibromo-2-methylpropane
    4. D1-Bromo-2-methylpropane

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15. Question 15 · Easy

    Which reagent converts but-2-yne mainly into cis-but-2-ene?

    1. ASodium in liquid ammonia
    2. BH₂ with Lindlar's catalyst
    3. CZinc and dilute HCl
    4. DExcess H₂ over platinum

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16. Question 16 · Medium

    Propene is treated with HCl in the presence of benzoyl peroxide. The major product is

    1. A2-chloropropane
    2. B1,2-dichloropropane
    3. C1-chloropropane
    4. D3-chloroprop-1-ene

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17. Question 17 · Medium

    Consider the statements about electrophilic substitution in benzene derivatives: A. Toluene undergoes nitration faster than benzene. B. The nitro group directs an incoming electrophile to the ortho and para positions. C. Chlorine deactivates the benzene ring yet directs an incoming electrophile to the ortho and para positions. D. Nitrobenzene readily undergoes Friedel-Crafts alkylation. Choose the correct answer from the options given below.

    1. AA and C only
    2. BC and D only
    3. CA, B and C only
    4. DA, C and D only

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18. Question 18 · Medium

    Sodium butanoate is heated with soda lime. The main hydrocarbon formed is

    1. Apropene
    2. Bpropane
    3. Cethane
    4. Dbutane

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19. Question 19 · Easy

    Ozonolysis of but-2-ene, followed by treatment with zinc and water, gives

    1. Apropanal and methanal
    2. Bethanoic acid only
    3. Cpropanone and methanal
    4. Dethanal only

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20. Question 20 · Medium

    Which hydrocarbon reacts with sodium metal to give hydrogen gas?

    1. AEthane
    2. BBut-2-yne
    3. CPropyne
    4. DPropene

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21. Question 21 · Hard

    Benzene is converted by the sequence: (i) CH₃Cl, anhydrous AlCl₃ → P (ii) alkaline KMnO₄ and heat, then dilute H₂SO₄ → Q (iii) concentrated HNO₃ and concentrated H₂SO₄ → R (major) Identify R.

    1. A2-Nitrobenzoic acid
    2. B4-Nitrotoluene
    3. C4-Nitrobenzoic acid
    4. D3-Nitrobenzoic acid

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22. Question 22 · Medium

    Which C₅H₁₀ alkene gives propanone and ethanal as the only products when treated with ozone and then with zinc dust and water?

    1. A2-Methylbut-2-ene
    2. BPent-2-ene
    3. C2-Methylbut-1-ene
    4. D3-Methylbut-1-ene

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23. Question 23 · Hard

    A hydrocarbon P (C₅H₈) liberates hydrogen gas when treated with sodium metal. On catalytic hydrogenation it takes up 2 mol of H₂ per mole and gives 2-methylbutane. What is P?

    1. APent-2-yne
    2. BPent-1-yne
    3. C3-Methylbut-1-yne
    4. D2-Methylbuta-1,3-diene

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24. Question 24 · Medium

    Propyne is passed into dilute sulphuric acid containing mercury(II) sulphate at 333 K. The main organic product is

    1. Apropanone
    2. Bpropanal
    3. Cpropan-2-ol
    4. Dpropan-1-ol

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25. Question 25 · Medium

    Arrange the carbon–carbon bonds in the following hydrocarbons in increasing order of bond length. A. Ethyne B. Benzene C. Ethene D. Ethane Choose the correct answer from the options given below.

    1. AA, C, B, D
    2. BC, A, B, D
    3. CD, B, C, A
    4. DA, B, C, D

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