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Alkynes: acidity, additions and polymerization: NEET MCQs with answers

50 questions on this topic of Hydrocarbons (NEET 2027 syllabus). Here are 12; tap one to see its answer and explanation.

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easy

1. Which of the following bases is strong enough to deprotonate a terminal alkyne like ethyne completely into its conjugate base?

  1. Aqueous sodium hydroxide (NaOH)
  2. Sodium amide in liquid ammonia (NaNH₂)
  3. Sodium bicarbonate solution (NaHCO₃)
  4. Sodium ethoxide in ethanol (NaOCH₂CH₃)
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Answer: (B) Sodium amide in liquid ammonia (NaNH₂)

Terminal alkynes have a pKa of approximately 25, which makes them less acidic than water (pKa ≈ 15.7) and ethanol (pKa ≈ 16). Therefore, hydroxide and alkoxide ions cannot completely deprotonate terminal alkynes. Only extremely strong bases like the amide ion (NH₂⁻, conjugate acid NH₃ pKa ≈ 38) can quantitatively deprotonate them.

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2. Ethyne reacts with sodium metal to form sodium acetylide and H₂, showing that:

  1. Ethyne is a strong base
  2. Ethyne is saturated
  3. Sodium adds across the triple bond
  4. Hydrogen atoms attached to triply bonded carbon are acidic
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Answer: (D) Hydrogen atoms attached to triply bonded carbon are acidic

The sp-hybridised carbon has 50% s character and is more electronegative, so the attached hydrogen is weakly acidic and can be replaced by sodium.

NCERT Class 11

hard

3. What is the volume of H₂ gas evolved at STP when 0.10 mol of but-1-yne reacts completely with excess sodium metal?

  1. 2.24 L
  2. 1.12 L
  3. 4.48 L
  4. 0.56 L
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Answer: (B) 1.12 L

The stoichiometric reaction is: CH₃CH₂C≡CH + Na → CH₃CH₂C≡C⁻Na⁺ + 0.5 H₂. Thus, 1 mol of but-1-yne produces 0.5 mol of H₂. For 0.10 mol of but-1-yne, 0.05 mol of H₂ is formed. Volume at STP = 0.05 × 22.4 L = 1.12 L.

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4. When ethyne is treated with sodium amide in liquid ammonia, compound P is formed. Subsequent reaction of P with bromoethane produces compound Q. What is compound Q?

  1. Prop-1-yne
  2. But-1-ene
  3. But-2-yne
  4. But-1-yne
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Answer: (D) But-1-yne

Ethyne reacts with NaNH₂ (a strong base) to undergo deprotonation yielding monosodium acetylide, HC≡C⁻Na⁺ (P). Nucleophilic substitution of bromoethane by acetylide ion (HC≡C⁻ + CH₃CH₂Br → HC≡C-CH₂CH₃ + Br⁻) produces but-1-yne (Q).

easy

5. Which of the following compounds has the highest pKa value?

  1. Ethyne
  2. Ethene
  3. Propyne
  4. Water
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Answer: (B) Ethene

pKa values reflect acid strength; a higher pKa corresponds to a weaker acid. The order of acidity is H₂O (pKa ≈ 15.7) > HC≡CH (pKa ≈ 25) > CH₃-C≡CH (pKa ≈ 26) > CH₂=CH₂ (pKa ≈ 44). Therefore, ethene has the highest pKa among the given compounds.

hard

6. Propyne reacts with hypochlorous acid (HOCl) in excess. What is the IUPAC name of the stable final organic product?

  1. 1,2-Dichloropropan-2-ol
  2. 1,1-Dichloropropan-2-one
  3. 1,3-Dichloropropan-2-one
  4. 2,2-Dichloropropanal
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Answer: (B) 1,1-Dichloropropan-2-one

HOCl adds as Cl⁺ and OH⁻ across the triple bond. Following Markovnikov's orientation, Cl⁺ adds to the terminal carbon and OH⁻ to C-2 to give 1-chloroprop-1-en-2-ol. Addition of a second HOCl gives 1,1-dichloropropane-2,2-diol, which spontaneously loses a water molecule to form 1,1-dichloropropan-2-one.

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7. Given below are two statements: One is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A): Terminal alkynes form a precipitate of copper acetylide with ammoniacal cuprous chloride, whereas internal alkynes do not. Reason (R): Terminal alkynes contain sp hybridized acetylenic hydrogen atoms which are weakly acidic and can be replaced by heavy metal cations. In the light of the above statements, choose the most appropriate answer:

  1. Both (A) and (R) are true and (R) is the correct explanation of (A)
  2. Both (A) and (R) are true but (R) is not the correct explanation of (A)
  3. (A) is true but (R) is false
  4. (A) is false but (R) is true
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Answer: (A) Both (A) and (R) are true and (R) is the correct explanation of (A)

Terminal alkynes possess a hydrogen attached to an sp-hybridized carbon atom (high s-character, 50%), which makes the C-H bond sufficiently polarized and acidic to undergo substitution with Cu⁺ ions from ammoniacal cuprous chloride, yielding a red precipitate. Internal alkynes lack such acidic terminal hydrogens and therefore do not form this precipitate. Thus, Assertion is true and Reason correctly explains it.

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8. How many total moles of bromine (Br₂) are completely consumed when a mixture containing 0.2 mol of ethyne and 0.3 mol of propyne is treated with excess bromine in CCl₄?

  1. 0.5 mol
  2. 1.0 mol
  3. 0.8 mol
  4. 1.2 mol
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Answer: (B) 1.0 mol

Each mole of an alkyne contains two carbon-carbon π bonds and therefore consumes exactly 2 moles of Br₂ to form a tetrahaloalkane. For 0.2 mol ethyne, moles of Br₂ required = 0.2 × 2 = 0.4 mol. For 0.3 mol propyne, moles of Br₂ required = 0.3 × 2 = 0.6 mol. Total Br₂ consumed = 0.4 + 0.6 = 1.0 mol.

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9. Passing ethyne through a red-hot iron tube at 873 K gives:

  1. Ethane
  2. Cyclohexane
  3. Benzene, by cyclic polymerisation
  4. Polythene
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Answer: (C) Benzene, by cyclic polymerisation

Three molecules of ethyne polymerise cyclically to form benzene under these conditions.

NCERT Class 11

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10. Given below are two statements: Statement I: Alkynes are generally less reactive than alkenes towards electrophilic addition reactions. Statement II: The intermediate carbocation formed by the addition of an electrophile to an alkyne is a vinylic carbocation, which is less stable than an alkyl carbocation. In light of the above statements, choose the correct answer:

  1. Both Statement I and Statement II are correct and Statement II is the correct explanation of Statement I.
  2. Both Statement I and Statement II are correct but Statement II is not the correct explanation of Statement I.
  3. Statement I is correct but Statement II is incorrect.
  4. Statement I is incorrect but Statement II is correct.
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Answer: (A) Both Statement I and Statement II are correct and Statement II is the correct explanation of Statement I.

Alkynes are less reactive towards electrophilic addition than alkenes despite having higher electron density because electrophilic attack on an alkyne generates a sp-hybridized vinylic carbocation (+C=C), which is considerably less stable than the alkyl carbocation formed from an alkene. Thus, Statement II is correct and correctly explains Statement I.

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11. Linear polymerization of ethyne under suitable conditions produces polyacetylene (polyethyne). Thin films of this polymer conduct electricity because:

  1. It contains free sp³ carbon atoms with lone pairs
  2. It forms a three-dimensional covalent network lattice like diamond
  3. It hydrolyzes in air to release conducting hydronium ions
  4. It possesses a conjugated system of alternating single and double bonds
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Answer: (D) It possesses a conjugated system of alternating single and double bonds

Polyacetylene is an unbranched polymer consisting of repeating -CH=CH- units. The continuous conjugation of alternating single and double bonds allows delocalization of π-electrons along the chain, making it electrically conductive upon doping.

easy

12. Which of the following alkynes will NOT react with ammoniacal silver nitrate solution (Tollens' reagent) to form a precipitate?

  1. Ethyne
  2. Propyne
  3. But-1-yne
  4. But-2-yne
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Answer: (D) But-2-yne

Only terminal alkynes possess acidic acetylenic hydrogen atoms bonded to sp-hybridized carbon. But-2-yne is an internal alkyne lacking acidic hydrogen, so it does not form a silver acetylide precipitate.

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