Alkyne Acidic Character

8 MCQs9-step worked example

Source: NCERT Aldehydes, Ketones and Carboxylic AcidsPYQ coverage: NEET 2021, 2022, 2023, 2024, 2025Official key: NTA-verifiedLast reviewed: May 2026

Lesson

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 22: 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.

Practice 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:

MCQ 2Easy RecallPractice

Which of the following reacts with sodamide (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:

MCQ 4Direct ApplicationPractice

Which of the following alkynes will NOT react with sodium metal to liberate H₂?

MCQ 5Direct ApplicationPractice

Propyne reacts with NaNH₂ in liquid ammonia. The organic product is:

MCQ 6Direct ApplicationPractice

Among the following, the most acidic hydrogen is present in:

MCQ 7Concept TrapPractice

Terminal alkynes are more acidic than alkenes because:

MCQ 8Concept TrapPractice

An unknown hydrocarbon X does not react with NaNH₂ but decolourises bromine water. X is most likely:

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, Part 2, page 22).
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. ---

Before solving, remember these

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, p. 22

Formulas

Markovnikov's rule (and anti-Markovnikov)

H^{+} to C with more H

Without peroxide: ionic mechanism — H goes to carbon with MORE hydrogens (carbocation stability rule). With peroxide (HBr only, Kharasch): radical mechanism — anti-Markovnikov.

Symbol	Quantity	SI Unit
H,X	added atoms	-

Valid when

Asymmetric alkene
H-X with X = Cl, Br, I
Without peroxide for Markovnikov

Exam Traps & Common Mistakes

These are the exact patterns that cause wrong answers in NEET. Each trap includes when it triggers and how to avoid it.

Trap

Markovnikov reverses with peroxide

Category: Organic Reaction Conditions

HBr addition to alkene: WITHOUT peroxide → Markovnikov (H to C with more H). WITH peroxide (Kharasch effect) → anti-Markovnikov (radical mechanism). Specific to HBr only — not HCl, HI.

When it triggers

Question gives HX addition to alkene with explicit peroxide condition or hints (e.g. ROOR, light).

How to avoid

Without peroxide: ionic mechanism, carbocation stability → Markovnikov. With peroxide: radical mechanism, radical stability → anti-Markovnikov. Effect ONLY for HBr (HCl too strong, HI too weak).

Trap

Solvent changes product distribution

Category: Organic Reaction Conditions

Same starting materials give different products depending on solvent. Polar protic (water, alcohols): SN1/E1 favoured. Polar aprotic (DMSO, DMF): SN2 favoured. Affects substitution vs elimination.

When it triggers

Question contrasts product when solvent is changed; or specifies solvent type.

How to avoid

Polar protic stabilises carbocation → SN1/E1 (3° preferred). Polar aprotic doesn't solvate nucleophile → strong SN2 nucleophile (1°/2° preferred). Bulky base (t-BuOK) favours E2 over SN2.

Mistake

Confuses ortho/para directors with meta directors.

Root cause: concept gap

Correction

o,p-directors (activators except halogens): -OH, -OR, -NH₂, -NHR, alkyl. m-directors (deactivators): -NO₂, -CN, -COOH, -CHO. Halogens: o,p-directors but DEACTIVATORS.

Mistake