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Hydrocarbons – Class 11 Chemistry notes, important questions and MCQs

NCERT Class 11 Chemistry, chapter 9. Notes, key terms, important questions with answers and MCQs.

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Chapter overview

Hydrocarbons contain only carbon and hydrogen. Alkanes, alkenes, alkynes and aromatic compounds differ in bonding and reactions.

Aliphatic hydrocarbons

  • Alkanes (CₙH₂ₙ₊₂) undergo free-radical substitution and combustion; Wurtz reaction joins alkyl halides.
  • Alkenes (CₙH₂ₙ) undergo electrophilic addition; Markovnikov's rule; peroxide effect (anti-Markovnikov) only with HBr.
  • Alkynes (CₙH₂ₙ₋₂); terminal alkynes are weakly acidic; ozonolysis locates double bonds.

Aromatic hydrocarbons

  • Benzene: planar, resonance-stabilised; Hückel's rule (4n + 2) π electrons for aromaticity.
  • Electrophilic substitution: nitration, halogenation, sulphonation, Friedel–Crafts alkylation and acylation.
  • Directive influence: –OH, –CH₃ are ortho/para directing; –NO₂, –COOH are meta directing.

Key terms

Markovnikov's rule
in HX addition, H goes to the carbon with more hydrogens
Aromaticity
extra stability of planar cyclic conjugated systems with (4n + 2) π electrons
Ozonolysis
reaction of an alkene with ozone followed by Zn/H₂O that splits the double bond into two carbonyl compounds, locating its position

Important questions with answers

Why is benzene highly stable though it has three double bonds?2 marks

Its six π electrons are delocalised over the ring (resonance), giving extra stability (aromaticity) and equal C–C bond lengths; so it prefers substitution to addition.

Why are terminal alkynes acidic? Show with a reaction.3 marks

The C–H hydrogen of ethyne is attached to an sp carbon, which has 50% s-character and is more electronegative, so the H can be removed as H⁺. HC≡CH + 2Na → NaC≡CNa + H₂ (or with NaNH₂).

Explain the mechanism of addition of HBr to propene with and without peroxide, giving the major product in each case.5 marks

Without peroxide (ionic, Markovnikov): H⁺ adds to the CH₂ end to form the more stable secondary carbocation CH₃–C⁺H–CH₃; Br⁻ attacks it to give 2-bromopropane. With peroxide (free-radical, anti-Markovnikov): the peroxide forms Br• radicals; Br• adds to CH₂ to give the more stable secondary radical CH₃–C•H–CH₂Br, which takes H from HBr to give 1-bromopropane. The peroxide effect works only with HBr, because H–Cl is too strong and the I• radical too unreactive.

Quick revision

  • Peroxide effect only with HBr.
  • Benzene prefers substitution over addition.
  • –NO₂ is meta directing.

Exam tips

  • Conformations of ethane: staggered is more stable than eclipsed.
  • Conformations of ethane (staggered, eclipsed) interconvert freely; geometrical (cis–trans) isomers of alkenes do not, because the C=C cannot rotate.

Practice MCQs

  1. General formula of alkenes is

    1. CₙH₂ₙ₊₂
    2. CₙH₂ₙ₋₂
    3. CₙH₂ₙ
    4. CₙHₙ
    Show answer

    Answer: (c) CₙH₂ₙ. One double bond.

  2. Benzene mainly undergoes

    1. addition
    2. electrophilic substitution
    3. elimination
    4. polymerisation
    Show answer

    Answer: (b) electrophilic substitution. Keeps aromaticity.

  3. Addition of HBr to propene (no peroxide) gives mainly

    1. 1-bromopropane
    2. propane
    3. 2-bromopropane
    4. 1,2-dibromopropane
    Show answer

    Answer: (c) 2-bromopropane. Markovnikov.

  4. Number of π electrons in benzene is

    1. 4
    2. 2
    3. 8
    4. 6
    Show answer

    Answer: (d) 6. Three double bonds.

  5. Which shows geometrical (cis–trans) isomerism

    1. propene
    2. but-1-ene
    3. but-2-ene
    4. ethene
    Show answer

    Answer: (c) but-2-ene. Each double-bond carbon has two different groups.

  6. Which group is meta directing

    1. –OH
    2. –NO₂
    3. –CH₃
    4. –NH₂
    Show answer

    Answer: (b) –NO₂. Electron withdrawing.

  7. Anti-Markovnikov addition (peroxide effect) is seen with

    1. HBr
    2. HCl
    3. HI
    4. HF
    Show answer

    Answer: (a) HBr. Only HBr.

  8. Ozonolysis of an alkene gives only propanone (acetone). The alkene is

    1. but-2-ene
    2. 2,3-dimethylbut-2-ene
    3. propene
    4. 2-methylpropene
    Show answer

    Answer: (b) 2,3-dimethylbut-2-ene. Symmetrical, (CH₃)₂C=C(CH₃)₂.

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Written by Kundali IQ from standard, well-established facts to help with revision. This is not an official NCERT publication; always read your textbook too.