Hydrocarbons — Class 11 MCQs with Answers

Hydrocarbons — important questions & MCQs with answers (Class 11 Chemistry)

14 solved questions from this chapter's difficulty levels, each with its answer and explanation. The other 76 are timed and scored when you sign in.

1. Q1Easy

   General formula of alkenes:

   A.CₙH₂ₙ₊₂

   B.CₙH₂ₙ✓ Correct

   C.CₙH₂ₙ₋₂

   D.CₙHₙ

   Answer: B. CₙH₂ₙ

   Explanation: One C=C bond removes 2 H from the alkane formula: CₙH₂ₙ₊₂ → CₙH₂ₙ, e.g. propene C₃H₆. Adding the extra +2 gives the alkane formula instead, a common mix-up.

2. Q2Easy

   Molecular formula of benzene:

   A.C₂H₆

   B.C₆H₁₂

   C.CH₄

   D.C₆H₆✓ Correct

   Answer: D. C₆H₆

   Explanation: Benzene's ring has 6 C and 6 H — degree of unsaturation = (2×6+2−6)/2 = 4, matching 1 ring + 3 C=C. Mistaking it for cyclohexane (C₆H₁₂) skips the unsaturation entirely.

3. Q3Easy

   General formula of alkanes:

   A.CₙH₂ₙ₊₂✓ Correct

   B.CₙH₂ₙ

   C.CₙH₂ₙ₋₂

   D.CₙH₂ₙ₊₁

   Answer: A. CₙH₂ₙ₊₂

   Explanation: Saturated chains have no π bonds, so every carbon is sp³: degree of unsaturation = 0 gives CₙH₂ₙ₊₂, e.g. C₄H₁₀. Dropping the +2 gives the alkene formula CₙH₂ₙ instead.

4. Q4Easy

   Nitration of benzene uses:

   A.Conc. HNO₃ + conc. H₂SO₄✓ Correct

   B.Dilute HNO₃

   C.NaNO₂

   D.HNO₂

   Answer: A. Conc. HNO₃ + conc. H₂SO₄

   Explanation: Conc. HNO₃ + conc. H₂SO₄ generates the electrophile NO₂⁺ (H₂SO₄ protonates HNO₃, which loses H₂O). Dilute HNO₃ alone can't ionise enough to form NO₂⁺ for attack.

5. Q5Easy

   Hybridisation of carbon in ethyne (C₂H₂) is:

   A.sp³d

   B.sp²

   C.sp³

   D.sp✓ Correct

   Answer: D. sp

   Explanation: Each carbon in HC≡CH forms 2 σ bonds (to C and H), so it needs only 2 hybrid orbitals: sp, leaving 2 p orbitals for the 2 π bonds. Picking sp² confuses it with ethene's geometry.

6. Q6Easy

   Friedel-Crafts alkylation catalyst is:

   A.KMnO₄

   B.NaOH

   C.H₂SO₄

   D.Anhydrous AlCl₃✓ Correct

   Answer: D. Anhydrous AlCl₃

   Explanation: Anhydrous AlCl₃ is a Lewis acid: it pulls Cl⁻ from R−Cl to generate the R⁺ electrophile needed for electrophilic substitution. KMnO₄ and NaOH can't generate a carbocation this way.

7. Q7Medium

   Alkynes have ___ bonds between carbons.

   A.Single

   B.Double

   C.Triple✓ Correct

   D.None

   Answer: C. Triple

   Explanation: The −yne suffix specifically denotes a C≡C triple bond (2π + 1σ), giving sp hybridised carbons, unlike −ane (single) or −ene (double). Picking 'double' confuses alkynes with alkenes.

8. Q8Medium

   Hückel's rule for aromaticity: (4n+2) π electrons. For benzene, n =

   A.1✓ Correct

   B.0

   C.2

   D.3

   Answer: A. 1

   Explanation: Benzene has 6 π electrons; solving 4n+2 = 6 gives n = 1. Dividing 6 by 4 directly (instead of solving the equation) is the mistake that gives a non-integer or wrong n.

9. Q9Medium

   Addition of HBr to propene in presence of peroxide gives mainly:

   A.Propanol

   B.2-Bromopropane

   C.1,2-Dibromopropane

   D.1-Bromopropane✓ Correct

   Answer: D. 1-Bromopropane

   Explanation: Peroxides generate Br· radicals that add first to the terminal, less hindered carbon, placing the radical (and later H) on C2 — anti-Markovnikov: 1-bromopropane. Without peroxide, ionic addition gives the opposite regiochemistry.

10. Q10Medium

    −NH₂ group on benzene is:

    A.Meta directing, activating

    B.Meta directing, deactivating

    C.Para directing, deactivating

    D.Ortho/para directing, activating✓ Correct

    Answer: D. Ortho/para directing, activating

    Explanation: −NH₂'s N lone pair delocalises into the ring by resonance, raising electron density most at ortho/para — strongly activating and o/p-directing. Confusing it with −NO₂ flips both labels.

11. Q11Medium

    Ozonolysis of 2-butene followed by Zn/H₂O gives:

    A.Acetone + formaldehyde

    B.Two molecules of acetaldehyde✓ Correct

    C.Two molecules of formaldehyde

    D.Butanal

    Answer: B. Two molecules of acetaldehyde

    Explanation: O₃ cleaves the C2=C3 bond of CH₃CH=CHCH₃ symmetrically; Zn/H₂O reduces each ozonide fragment to CH₃CHO. Both halves are identical, so the product is 2 mol acetaldehyde, not a mixed pair.

12. Q12Hard

    Markovnikov's rule applies to addition of HX to:

    A.Esters

    B.Alkanes

    C.Alkenes (asymmetric)✓ Correct

    D.Amines

    Answer: C. Alkenes (asymmetric)

    Explanation: Markovnikov's rule governs HX addition to unsymmetrical (asymmetric) alkenes: H adds to the carbon already bearing more H, via the more stable carbocation. Symmetric alkenes give only one product.

13. Q13Hard

    Benzene typically undergoes:

    A.Electrophilic substitution✓ Correct

    B.Nucleophilic addition

    C.Free radical addition

    D.None

    Answer: A. Electrophilic substitution

    Explanation: Benzene reacts by electrophilic substitution, replacing one ring H while keeping all 3 π bonds and the resonance stabilisation intact. Addition, as with alkenes, would destroy that delocalisation.

14. Q14Hard

    Addition of HCl to 3-methyl-1-butene gives major product:

    A.1-Chloro-3-methylbutane

    B.3-Chloro-2-methylbutane

    C.2-Chloro-2-methylbutane (via 1,2-H shift)✓ Correct

    D.2-Chloro-3-methylbutane

    Answer: C. 2-Chloro-2-methylbutane (via 1,2-H shift)

    Explanation: Protonation of CH₂=CH−CH(CH₃)−CH₃ gives a 2° carbocation at C2; a 1,2-hydride shift from C3 converts it to the more stable 3° cation, which Cl⁻ attacks: 2-chloro-2-methylbutane, not the unrearranged product.

Hydrocarbons — FAQs

More Class 11 Chemistry chapters

• 01Some Basic Concepts of Chemistry
• 02Structure of Atom
• 03Classification of Elements and Periodicity
• 04Chemical Bonding and Molecular Structure
• 05Thermodynamics
• 06Equilibrium

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