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Organic Chemistry 1 · Alkene and Alkyne Chemistry

Preparation of Alkynes

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On this page 7 sections
  1. In 30 seconds
  2. Why this matters
  3. The college version
  4. Eli explains
  5. Worked example
  6. Key takeaway
  7. Study tools

In 30 seconds

Alkynes are most often prepared by double dehydrohalogenation — removing two equivalents of HX from a dihalide with a . Both vicinal dihalides (halogens on adjacent carbons) and geminal dihalides (both halogens on the same carbon) work, because each route passes through a vinyl (alkenyl) halide intermediate: dihalide → (first elimination) → alkyne (second elimination). The second elimination is harder, so an excess of a very strong base such as sodium amide (NaNH₂) is used. The terminal alkyne product can then be deprotonated to an and alkylated to extend the carbon skeleton (topics 35 and 42).

Why this matters

The alkyne-forming is a foundational route to acetylenic building blocks used in medicinal and materials chemistry — rigid linear "spacers" and click-chemistry handles often begin as terminal alkynes made this way. The strong bases involved (sodium amide and similar) are highly reactive, air- and moisture-sensitive, and hazardous; they must be handled only by qualified personnel following approved institutional safety documentation. No operational, quantity, or disposal instructions are provided here.

The college version

1. Vicinal and Geminal Dihalides as Starting Materials

A has halogens on adjacent carbons (Br–CH₂–CH₂–Br); a has both on the same carbon (CH₃–CHBr₂). Both converge on the same alkyne because the first elimination converts either one into a vinyl halide — a halogen on a double-bond carbon (CH₂=CHBr) — and the second elimination forms the triple bond. These dihalides are often made by adding halogen or HX to an alkene, so alkynes can be reached from alkenes via a dihalide "detour."

2. The Double Elimination Sequence

Each step is an E2 elimination requiring a strong base, which removes a β-hydrogen while the C–X bond breaks and X⁻ leaves as the electrons form a new π bond.

  • First elimination: dihalide → vinyl halide (one C=C forms).
  • Second elimination: vinyl halide → alkyne (a second π bond forms). The second step is slower because vinylic C–H and C–X bonds are held more tightly, so a strong base, excess base, and heat are required.

3. Synthesis Planning and Compatibility Limits

For , the product ends in C≡C–H (pKa ≈ 25), so any strong base present deprotonates it to an acetylide as it forms. In practice excess NaNH₂ does the double elimination and then deprotonates the product; water is added at the end to regenerate the neutral terminal alkyne. This deprotonation is the gateway to alkylation — reacting the acetylide with a methyl or primary alkyl halide to extend the chain (topics 35 and 42). Substrate-compatibility caution: the alkyl halide partner must be methyl or primary (unhindered) for a clean SN2 reaction; secondary and tertiary halides favor elimination instead.

How it works

  1. Start with a vicinal or geminal dihalide (often made from an alkene).
  2. Add a strong base (excess NaNH₂) to remove the first HX and form a vinyl halide.
  3. The base removes a second HX to form the alkyne.
  4. For terminal alkynes, the base deprotonates the product to an acetylide; protonate with water at the end.
  5. Optionally alkylate the acetylide with a methyl/primary halide to extend the chain.

Common confusions

Do not confuseWithDifference
Vicinal dihalideGeminal dihalideVicinal = adjacent carbons; geminal = same carbon
First eliminationSecond eliminationFirst makes a vinyl halide; second makes the alkyne (harder)
Vinyl halideAlkyl halideVinyl has X on a double-bond carbon (tighter, slower E2)
Double eliminationOne E2One E2 makes an alkene; two E2 steps make an alkyne

Memory aids

"V and G both go through V-H." Whether you start with a Vicinal or Geminal dihalide, the route runs through a Vinyl Halide before the alkyne forms. Add "strong base ×2, plus one more to deprotonate" to remember the excess NaNH₂.

Quick review

Topic Recap

Alkynes are prepared by double dehydrohalogenation of vicinal or geminal dihalides using a strong base such as excess sodium amide. The two-step sequence passes through a vinyl halide, and the second elimination is harder. Terminal alkynes are deprotonated to acetylides under the reaction conditions, then protonated on workup; that acetylide is the nucleophile used to extend the carbon chain by SN2 alkylation with methyl or primary halides.

Knowledge Check

  1. What two classes of dihalide can be converted to an alkyne by double elimination?
  2. What intermediate forms after the first E2 elimination of 1,2-dibromoethane?
  3. Why is excess strong base (rather than one equivalent) needed?
  4. When a terminal alkyne is made in excess NaNH₂, what species is present before water is added?
  5. Which alkyl halides are suitable partners for alkylating the acetylide made here, and why?

Answers and Rationales

  1. Vicinal and geminal dihalides. Both give a vinyl halide after the first elimination, then the alkyne.
  2. Bromoethene (a vinyl halide). The first E2 removes one HBr, leaving one C=C and one Br.
  3. The second elimination is harder, and a terminal alkyne is deprotonated as it forms. Extra base both pushes the reaction and removes the acidic C≡C–H.
  4. The acetylide ion (HC≡C⁻). The terminal alkyne is deprotonated; water protonates it back to ethyne.
  5. Methyl and primary halides. They are unhindered and undergo clean SN2 substitution; secondary/tertiary halides eliminate.
Eli, the EliExplains learning guide

Eli explains

The same idea, in plain words

Explain it like I’m 10

Think of building a triple bond as removing two barriers from a two-carbon chain. If two halogens are already attached to those carbons (a "dihalide"), a strong base peels off a hydrogen and a halogen, one from each end, squeezing the single bond into a double bond. Repeat that peeling a second time and the double bond is squeezed into a triple bond.

Compare this with making a double bond (topic 27), which is one peeling pass; making a triple bond is the same idea done twice.

Where this stops being exact: the "squeezing" image hides that each pass is its own elimination with a specific proton and leaving group that must line up, and the second pass is genuinely harder — it needs the strongest bases and extra base (often with heat), and it will not conveniently stop at the double-bond stage.

Simple Example

  • 1,2-Dibromoethane (vicinal dihalide) + excess NaNH₂ → ethyne (via bromoethene).
  • 1,1-Dibromoethane (geminal dihalide) + excess NaNH₂ → ethyne (same intermediate).

Worked example

Trace the double elimination of 1,2-dibromoethane with excess NaNH₂, moving electrons before naming products.

  1. First E2. A double-headed curved arrow shows NH₂⁻ (strong base) removing a β-hydrogen from one carbon of Br–CH₂–CH₂–Br while the C–Br bond on the other carbon breaks and Br⁻ departs; the C–H electron pair moves in to form a π bond. Product: bromoethene (CH₂=CHBr), a vinyl halide.
  2. Accounting after step 1. One HBr is lost; the two carbons now share four electrons (σ + π), the departing Br holds a full octet, and charge balances (neutral product + Br⁻ + Na⁺).
  3. Second E2. Another NH₂⁻ removes the vinylic hydrogen from CH₂=CHBr; Br⁻ leaves and the former C–H pair becomes the second π bond. Product: ethyne (HC≡CH).
  4. Deprotonation. A third NH₂⁻ removes the acidic C≡C–H proton to give acetylide HC≡C⁻; water is added last to protonate it back to neutral ethyne.
  5. Overall. Two HBr equivalents and one C–H proton are removed by three equivalents of base; atom, charge, and octet counts balance throughout.

Key takeaways

  • High yield: Vicinal and geminal dihalides both give the same alkyne via a vinyl halide.
  • High yield: Making an alkyne requires two E2 eliminations and a strong base (excess NaNH₂).
  • High yield: The second elimination (vinyl halide → alkyne) is harder and needs excess base, often with heat.
  • High yield: A terminal alkyne product is deprotonated by excess base to an acetylide; water protonates it back.
  • Methyl/primary alkyl halides are required for acetylide alkylation (SN2); secondary/tertiary eliminate.

Keep learning

Ready to build on this? Continue to the next lesson.

Practice Organic Chemistry 1

This lesson has no separate scored set. Practice draws from the subject’s question bank.

Study tools & related lessonsYou’ll learn to · Key vocabulary · Related

You’ll learn to

  • Explain how alkynes are prepared by double elimination from vicinal and geminal dihalides.
  • Describe the role of a strong base in driving two sequential eliminations through a vinyl-halide intermediate.
  • Plan a synthesis that produces a terminal alkyne and recognize where the product can be deprotonated and alkylated.
  • Identify substrate-compatibility limitations and the safety boundaries around strong-base elimination reactions.

Key vocabulary

Vicinal dihalide
Two halogens on adjacent carbons
Geminal dihalide
Two halogens on the same carbon
Double elimination
Two sequential E2 steps removing 2 HX
Strong base
Species like NaNH₂ that removes a proton readily
Vinyl halide
Halogen on a double-bond carbon
Terminal alkyne formation
Making an alkyne that ends in C≡C–H
Acetylide
Deprotonated terminal alkyne (R–C≡C⁻)
Elimination sequence
dihalide → vinyl halide → alkyne
Substrate compatibility
Which partners react without side reactions

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