Hydrocarbons

14 NEET-UG practice questions on Hydrocarbons, part of Chemistry. The first 14 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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