Key Fact
Acidic character of alkynes
Terminal alkynes (≡CH) acidic (pKa ~25); H replaceable with metals (Na/NaNH₂ → metal acetylide). Due to high s-character of sp C-H bond.
-- NCERT Class 11 Chemistry, Ch. 9, p. 316Alkyne Acidic Character
8 MCQs9-step worked example
Source: NCERT HydrocarbonsOfficial key: NTA-verifiedLast updated: 24 Sep 2026
Alkyne Acidic Character, explained for NEET
The trap: aspirants remember that alkynes have a triple bond and are unsaturated — but when asked why terminal alkynes are acidic, they freeze or confuse acidity with reactivity toward electrophilic addition. The acidic character of alkynes is about the C–H bond on a terminal sp carbon, not about the π electrons.
Why terminal alkynes are acidic. In a terminal alkyne (R–C≡C–H), the carbon bearing the hydrogen is sp-hybridised. An sp orbital has 50% s-character, compared with 33% for sp² and 25% for sp³. Greater s-character means the electron pair in the C–H bond is held closer to the carbon nucleus. When the hydrogen departs as H⁺, the resulting carbanion (acetylide ion, R–C≡C⁻) retains the electron pair in a high-s-character orbital — this stabilises the negative charge, making the proton easier to lose.
The acidity order follows s-character:
| Hybridisation | s-character | Example | Relative acidity |
|---|---|---|---|
| sp | 50% | HC≡CH (ethyne) | Most acidic |
| sp² | 33% | H₂C=CH₂ | Intermediate |
| sp³ | 25% | CH₃–CH₃ | Least acidic |
As stated in NCERT Class 11 Chemistry Chapter 9 (Hydrocarbons), Part 2, page 316: terminal alkynes are weakly acidic and react with strong bases such as sodamide (NaNH₂) to form sodium acetylides. Ethene and ethane do not undergo this reaction — their C–H hydrogens are not acidic enough.
Chemical evidence. Ethyne reacts with sodium metal and with NaNH₂ in liquid ammonia to give sodium acetylide (HC≡C⁻Na⁺) and H₂ (or NH₃). This reaction is the diagnostic test for a terminal alkyne — internal alkynes (R–C≡C–R′) lack the terminal C–H and do not react.
Watch-out for NEET: questions may ask you to arrange hydrocarbons in order of acidity or to identify which compound reacts with NaNH₂. The key is s-character → acidity. Don't confuse this with nucleophilicity of the triple bond.
Can you answer these Alkyne Acidic Character MCQs?
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
MCQ 1Easy RecallPractice
The acidic character of hydrocarbons follows the order:
Show answer and why every option is right or wrong
Answer: C. C is correct. Acidity increases with increasing s-character of the C–H bond: sp (50%, ethyne) > sp² (33%, ethene) > sp³ (25%, ethane). This is directly stated in NCERT Class 11 Chemistry, Chapter 9, Part 2.
Why A is wrong: A places ethene above ethyne. sp² has only 33% s-character versus 50% for sp, so ethene is less acidic than ethyne.
Why B is wrong: B reverses the order. Greater s-character stabilises the conjugate base (carbanion), so sp³ carbon (ethane) is the LEAST acidic, not most.
Why D is wrong: D places ethane above ethene. Ethane (sp³, 25% s-character) is the least acidic of the three.
MCQ 2Easy RecallPractice
Which of the following reacts with sodamide (NaNH₂) to form a sodium salt?
Show answer and why every option is right or wrong
Answer: D. D is correct. Ethyne (HC≡CH) is a terminal alkyne with an acidic sp C–H bond. It reacts with NaNH₂ to form sodium acetylide (HC≡C⁻Na⁺) and NH₃, as described in NCERT Class 11 Chemistry, Chapter 9, page 316.
Why A is wrong: A is wrong. Ethane has only sp³ C–H bonds (25% s-character), which are far too weakly acidic to react with NaNH₂.
Why B is wrong: B is wrong. Ethene has sp² C–H bonds (33% s-character), still insufficient acidity to react with NaNH₂ under normal conditions.
Why C is wrong: C is wrong. Propene has sp² and sp³ C–H bonds but no sp C–H bond. It does not react with NaNH₂ to form a sodium salt.
MCQ 3Easy RecallPractice
The percentage s-character in the hybrid orbital of the carbon bonded to hydrogen in a terminal alkyne is:
Show answer and why every option is right or wrong
Answer: D. D is correct. The terminal carbon of an alkyne is sp-hybridised. An sp hybrid orbital is formed from one s and one p orbital, giving 50% s-character (1 s out of 2 orbitals used).
Why A is wrong: A corresponds to sp³ hybridisation (1 s out of 4 orbitals = 25%), which is the case for alkanes, not alkynes.
Why B is wrong: B corresponds to sp² hybridisation (1 s out of 3 orbitals ≈ 33%), which is the case for alkenes, not alkynes.
Why C is wrong: C (75%) does not correspond to any standard hybridisation state used in NEET organic chemistry.
MCQ 4Direct ApplicationPractice
Which of the following alkynes will NOT react with sodium metal to liberate H₂?
Show answer and why every option is right or wrong
Answer: A. A is correct. But-2-yne (CH₃–C≡C–CH₃) is an internal alkyne — it has no terminal C–H bond. Only terminal alkynes possess the acidic sp C–H that reacts with Na to release H₂.
Why B is wrong: B is wrong. Propyne (CH₃–C≡CH) has a terminal C–H on an sp carbon. It reacts with Na to form sodium propynide and H₂.
Why C is wrong: C is wrong. Ethyne (HC≡CH) is a terminal alkyne with two acidic terminal C–H bonds. It readily reacts with sodium metal.
Why D is wrong: D is wrong. But-1-yne (HC≡C–CH₂–CH₃) has a terminal C–H bond and reacts with sodium metal.
MCQ 5Direct ApplicationPractice
Propyne reacts with NaNH₂ in liquid ammonia. The organic product is:
Show answer and why every option is right or wrong
Answer: B. B is correct. NaNH₂ is a strong enough base to abstract the terminal hydrogen of propyne (pKa ≈ 25), forming sodium propynide (CH₃–C≡C⁻Na⁺) and NH₃. This is an acid-base reaction, not an addition or substitution.
Why A is wrong: A is wrong. CH₃–CH=CH₂ (propene) would result from reduction or rearrangement, not from treatment with NaNH₂. NaNH₂ acts as a base here, not a reducing agent.
Why C is wrong: C is wrong. Propane would require hydrogenation (H₂/catalyst). NaNH₂ does not reduce alkynes.
Why D is wrong: D is wrong. The NH₂⁻ ion acts as a base (proton abstractor), not a nucleophile adding to the triple bond under these conditions.
MCQ 6Direct ApplicationPractice
Among the following, the most acidic hydrogen is present in:
Show answer and why every option is right or wrong
Answer: B. B is correct. The O–H bond in methanol (pKa ≈ 15.5) is significantly more acidic than the sp C–H of ethyne (pKa ≈ 25). Oxygen is more electronegative than carbon, stabilising the resulting alkoxide anion far more effectively than s-character alone stabilises an acetylide.
Why A is wrong: A is wrong. Ethane (sp³ C–H, pKa ≈ 50) has the least acidic hydrogens among these options.
Why C is wrong: C is wrong. Ethyne (sp C–H, pKa ≈ 25) is the most acidic hydrocarbon listed, but methanol's O–H (pKa ≈ 15.5) is still about 10 pKa units more acidic. The question asks about the most acidic hydrogen overall, not just among hydrocarbons.
Why D is wrong: D is wrong. Ethene (sp² C–H, pKa ≈ 44) is more acidic than ethane but far less than methanol or ethyne.
MCQ 7Concept TrapPractice
Terminal alkynes are more acidic than alkenes because:
Show answer and why every option is right or wrong
Answer: A. A is correct. The acetylide anion (conjugate base) holds its lone pair in an sp orbital with 50% s-character. Greater s-character means the electron pair is closer to the nucleus and better stabilised, making the conjugate base more stable and the parent acid stronger.
Why B is wrong: B is wrong. The π electrons of the triple bond relate to nucleophilicity and electrophilic addition, not to the acidity of the terminal C–H bond. Acidity depends on stabilisation of the conjugate base, not on electron donation.
Why C is wrong: C is wrong. C–C bond strength in the triple bond is actually greater than in a double bond (839 vs 614 kJ/mol). Bond strength of C–C has no direct bearing on C–H acidity.
Why D is wrong: D is wrong. While the triple bond is shorter than a double bond, molecular polarity is not the reason for terminal alkyne acidity. The relevant factor is the s-character of the orbital holding the departing electron pair.
MCQ 8Concept TrapPractice
An unknown hydrocarbon X does not react with NaNH₂ but decolourises bromine water. X is most likely:
Show answer and why every option is right or wrong
Answer: C. C is correct. But-2-yne is an internal alkyne — no terminal C–H, so no reaction with NaNH₂. However, it still contains a triple bond (unsaturation) and decolourises bromine water through electrophilic addition. This combination uniquely identifies an internal alkyne.
Why A is wrong: A is wrong. Ethyne IS a terminal alkyne — it would react with NaNH₂ to form sodium acetylide. Both conditions (reacts with NaNH₂ AND decolourises Br₂ water) would be satisfied, not just one.
Why B is wrong: B is wrong. Ethane is a saturated hydrocarbon. It does NOT decolourise bromine water (no π bonds for addition). The observation states X decolourises Br₂ water, ruling out ethane.
Why D is wrong: D is wrong. Propyne has a terminal C–H (sp) and WOULD react with NaNH₂. The observation states X does not react with NaNH₂, ruling out any terminal alkyne.
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How do you solve a Alkyne Acidic Character question? A worked example
- 1
Given
Four hydrocarbons with C–H bonds on carbons of different hybridisation:• (i) Ethyne: sp C–H• (ii) Ethene: sp² C–H• (iii) Ethane: sp³ C–H• (iv) Benzene: sp² C–H
- 2
Required
Arrange in decreasing order of acidity of the indicated C–H bond and explain the basis.
- 3
Concept
Acidity of a C–H bond depends on the stability of the conjugate base (carbanion). Greater s-character in the hybrid orbital holding the electron pair → electrons held closer to the carbon nucleus → more stable carbanion → stronger acid (NCERT Class 11 Chemistry, Chapter 9, page 316).
- 4
Formula / Principle
s-character: sp = 50%, sp² = 33%, sp³ = 25%. Higher s-character → lower pKa → higher acidity.
For two carbons with the same hybridisation (ethene vs benzene, both sp²): aromatic C–H in benzene is slightly more acidic because the phenyl anion is stabilised by the aromatic ring's electron-withdrawing inductive effect and the orbital is in the plane of the ring. - 5
Substitution / Analysis
| Compound | Hybridisation | s-character | Approximate pKa |
|----------|---------------|-------------|-----------------|
| HC≡CH | sp | 50% | ~25 |
| C₆H₅–H | sp² | 33% | ~43 |
| H₂C=CH₂ | sp² | 33% | ~44 |
| H₃C–CH₃ | sp³ | 25% | ~50 | - 6
Ordering
Decreasing acidity (increasing pKa):
HC≡CH > C₆H₅–H > H₂C=CH₂ > H₃C–CH₃ - 7
Final answer
HC≡CH > C₆H₅–H > H₂C=CH₂ > H₃C–CH₃
The ordering follows s-character: sp (50%) > sp² (33%) > sp³ (25%). Between the two sp² cases, benzene's C–H is slightly more acidic than ethene's because of additional stabilisation of the phenyl carbanion. - 8
Common trap
Confusing acidity (thermodynamic, pKa-based) with reactivity toward electrophilic addition. Alkynes are more acidic than alkenes, but both undergo electrophilic addition — those are separate properties. A question asking "which is more reactive toward Br₂?" has a different answer from "which is more acidic?"
Also: forgetting that benzene (sp²) fits between ethyne (sp) and ethene (sp²) — some students place all sp² compounds at the same acidity, ignoring ring stabilisation. - 9
Similar NEET-style question
"Among ethane, ethene, ethyne, and benzene, which can react with NaNH₂ in liquid ammonia to form a sodium salt? Justify."
Answer: Only ethyne — its pKa (~25) is low enough for the strong base NH₂⁻ (conjugate acid NH₃, pKa ~38) to deprotonate it. The others have pKa values well above 38.
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What to remember before solving Alkyne Acidic Character questions
More in Hydrocarbons: 4 exam traps and mistakes · 1 formula · 2 question patterns from its other lessons.
Alkyne Acidic Character questions from past NEET papers
No question in our NEET 2020–2025 set targets this topic directly.
Sources
Page numbers are the ones printed in the current NCERT textbook (2023 rationalised edition), unless marked pre-2023. The books are free at ncert.nic.in.
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