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Organic Chemistry · Alkynes: An Introduction to Organic Synthesis

Preparation of Alkynes: Elimination Reactions of Dihalides

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General lab-safety guidance only (strong bases and liquid ammonia are hazardous; proper PPE/ventilation required) — no specific experimental procedures are provided.
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On this page 9 sections
  1. In 30 seconds
  2. Why this matters
  3. The college version
  4. Eli explains
  5. Worked example
  6. Key takeaway
  7. Check yourself
  8. Study tools
  9. Sources & references

In 30 seconds

Alkynes are rarely found in nature, so synthetic chemists must make them. The classic laboratory route is a double : take a dihalide (two C–X bonds on adjacent carbons, or on the same carbon), remove two equivalents of HX, and the two lost H's and two lost X's come from neighboring carbons, leaving a triple bond behind.

The overall transformation, for a (halogens on adjacent carbons):

R-CHX-CH2X → [2 equiv strong base] R-C#CH + 2 HX

Mechanistically this is two consecutive E2 eliminations. The first E2 removes one HX and makes an alkene that still carries one halogen — a (halogen on a double-bond carbon). The second E2 removes the remaining HX, and the C=C becomes C#C. Because vinyl halides are sluggish toward elimination, the second step needs a very strong base — typically sodium amide, NaNH2, dissolved in liquid ammonia — whereas the first elimination can often be done with the milder alcoholic potassium hydroxide, KOH/ethanol.

The same two-step logic works for geminal dihalides (both halogens on the same carbon, R-CH2-CHX2), which also give alkynes after double elimination. Because elimination is an anti (or E2 anti-periplanar) process in the usual staggered conformations, the stereochemistry of the starting dihalide matters in detail, but the practical message is simple: heat + strong base + dihalide → alkyne.

Why this matters

  • It's the standard alkyne synthesis. If a target molecule contains a triple bond, the retrosynthetic question is usually "which dihalide (or which alkene) do I start from?" This reaction is the answer, and it connects directly to Chapter 8: bromination of an alkene gives a vicinal dibromide, and double elimination of that dibromide gives the alkyne — a two-step alkene → alkyne route.
  • It teaches E2 in stereo and in series. You get to see elimination chemistry twice in one molecule: the first E2 is normal (alkyl halide), the second is special (vinyl halide). Understanding why the second needs a stronger base is a favorite exam question.
  • It feeds everything else in this chapter. The alkynes made here are the substrates for hydration (Topic 4), reduction (Topic 5), oxidative cleavage (Topic 6), and — if terminal — acetylide chemistry (Topics 7–8). You cannot do the rest of the chapter without a reliable way to make the starting material.
  • Industrial relevance. Acetylene and substituted alkynes are industrial feedstocks (acetylene in welding; alkynes in fine chemicals). The dehydrohalogenation route is also the historical basis for large-scale alkyne production.

The college version

Core Concepts

Step 1: E2 to the vinyl halide

A vicinal dihalide such as CH3CH2-CHBr-CH2Br (1,2-dibromobutane) is treated with base. The base abstracts the β-hydrogen while the C–Br bond breaks, forming a C=C double bond — this is a standard E2, exactly like the dehydrohalogenation you learned in Chapter 8:

CH3CH2-CHBr-CH2Br + KOH → [EtOH, heat] CH3CH2-CBr=CH2 + KBr + H2O

The product, CH3CH2-CBr=CH2, is a vinyl halide — a bromide on an sp2 carbon of an alkene. (Regiochemistry follows the usual E2 preference for the more substituted alkene, so the major product is the one with the double bond between the two more-substituted carbons, as drawn.)

Step 2: E2 on the vinyl halide — why NaNH2?

Removing the second HX from the vinyl halide is harder. Two factors fight the reaction:

  1. The leaving group is on an sp2 carbon. The C–Br bond of a vinyl halide is stronger than an alkyl C–Br bond because the sp2 carbon holds its electrons closer to the nucleus. Vinyl halides are famously unreactive in SN2 and SN1 — but E2 is still possible with a strong enough base.
  2. The abstracted H is on a vinylic position, attached to the double-bond carbon bearing the halide.

So the second elimination needs a much stronger base: NaNH2 in liquid NH3 (sodium amide is a powerful, non-nucleophilic-but-basic reagent) or potassium tert-butoxide. The reaction:

CH3CH2-CBr=CH2 + NaNH2 → [liq NH3] CH3CH2-C#CH + NaBr + NH3

Watch the stoichiometry: a terminal alkyne product still has its acidic ≡C–H, and excess NaNH2 will deprotonate it to the CH3CH2-C#C−. That's not a failure of the synthesis — it's the standard situation. The reaction is usually run with  ≥ 2 equivalents of NaNH2, and the acetylide is simply protonated back to the neutral alkyne on aqueous workup (adding water at the end). Expect exam problems that ask: "Why is the final step a water quench?" — because the product was isolated as its acetylide.

Vicinal vs. geminal dihalides

  • Vicinal (X–C–C–X, adjacent): obtained by adding X2 across an alkene (Chapter 8). Double elimination gives the alkyne.
  • Geminal (CX2 on one carbon): obtained by adding two equivalents of HX across an alkyne (or from ketones via PCl5/SOCl2 chemistry). Double elimination also gives the alkyne.

Both routes converge on the same C#C. The practical difference: which starting material is available in your synthesis.

The alkene → dibromide → alkyne strategy

Because bromination of an alkene (Chapter 8, Topic 2) gives a vicinal dibromide cleanly, the standard two-step alkene-to-alkyne sequence is:

  1. alkene + Br2 → vicinal dibromide (anti addition, no rearrangements)
  2. vicinal dibromide + 2 NaNH2 → alkyne

This works especially well for terminal alkynes because the starting alkene can be made by any alkene synthesis. It is a classic "how do you make that?" answer on synthesis exams.

Common Confusions

Do Not ConfuseWithDifference
First elimination (KOH/EtOH)Second elimination (NaNH2)The first removes HX from an alkyl halide (mild base suffices); the second must eliminate from a vinyl halide (needs NaNH2).
Acetylide formationUnwanted side reactionDeprotonating the terminal alkyne product is expected; aqueous workup restores the neutral alkyne.
Vicinal dihalideGeminal dihalideAdjacent carbons vs. same carbon; both eliminate to alkynes, but they come from different starting materials.
Elimination making a double bondElimination making a triple bondOne E2 → C=C; two E2s in the same molecule → C#C.
NaNH2 as baseNaNH2 as nucleophileIn this reaction amide acts as a base (abstracts H); it is too basic to be a practical nucleophile here.
E2 "anti" requirement"Any geometry works"The H and leaving group must be anti-periplanar in the E2 transition state; this controls which conformer reacts.
Eli, the EliExplains learning guide

Eli explains

The same idea, in plain words

Explain it like I’m 10

Imagine a chain of building blocks where two blocks each have a little knob (a bromine atom) and the chain has two loose hydrogen ends. A strong base comes along like a magnet: it grabs a hydrogen from one block, the knob pops off the neighboring block, and the two blocks snap together with a double bond. Now one knob is still left. A much stronger magnet (sodium amide) grabs the next hydrogen, pops the last knob, and the blocks snap together with a triple bond — the strongest link of all. Two yanks, two pops, one triple bond.

Worked example

Example 1: 1,2-dibromobutane → 1-butyne (mechanism walkthrough)

Setup: Convert CH3CH2-CHBr-CH2Br to CH3CH2-C#CH and describe each step's mechanism in words.

Step 1 — first elimination (E2): In ethanol with KOH and heat, the base abstracts a β-hydrogen from the CH2Br carbon while the C–Br bond on the adjacent carbon breaks. A curved arrow from the C–H bond forms the C=C. Product: CH3CH2-CBr=CH2 (2-bromobut-1-ene, a vinyl bromide).

Step 2 — second elimination (E2, strong base): In liquid ammonia, NaNH2 abstracts the vinylic hydrogen on the CH2 group, the C–Br bond breaks, and the π bond forms a triple bond. Product: CH3CH2-C#CH (1-butyne), which immediately loses its terminal proton to excess amide, giving the acetylide CH3CH2-C#C−.

Step 3 — workup: Adding water protonates the acetylide back to neutral 1-butyne.

Answer: The two eliminations are both E2; the second requires the much stronger base NaNH2 because the leaving group sits on an sp2 carbon.

Example 2: Mole-ratio planning (dimensional analysis)

Setup: How many moles of NaNH2 are required to convert 0.40 mol of 1,2-dibromobutane into 1-butyne, accounting for the terminal-alkyne deprotonation? (Assume the reaction is run without a separate quenching base.)

Formula first (reaction stoichiometry: 1 mol dihalide needs 2 mol NaNH2 for the two eliminations, and the resulting terminal alkyne consumes a further 1 mol NaNH2 as it forms the acetylide):

n(NaNH2) = n(dihalide) × 3 mol NaNH21 mol dihalide

Substitute:

n(NaNH2) = 0.40 mol × 31 = 1.2 mol

Answer: 1.2 mol of NaNH2 (units: mol dihalide × mol NaNH2/mol dihalide → mol NaNH2). If the textbook problem states the amide is used in excess, "≥2 equiv" is the minimum for elimination and 3 equiv covers acetylide formation — check the problem's wording.

Example 3: Retrosynthesis — "make 1-hexyne from an alkene"

Setup: Propose a synthesis of CH3CH2CH2CH2-C#CH (1-hexyne) starting from 1-hexene.

Forward plan:

  1. CH3CH2CH2CH=CH2 + Br2 → CH3CH2CH2CHBr-CH2Br (vicinal dibromide; anti addition, no rearrangement).
  2. CH3CH2CH2CHBr-CH2Br + 2 NaNH2 (liq NH3) → CH3CH2CH2CH2-C#C−, then H2O workup → CH3CH2CH2CH2-C#CH.

Answer: Two steps: bromination, then double dehydrohalogenation with sodium amide followed by aqueous workup. (The alternative — starting from a geminal dibromide — is also valid but requires making the dibromide first.)

Key takeaways

  • Double dehydrohalogenation: dihalide → (1st E2) → vinyl halide → (2nd E2, NaNH2) → alkyne.
  • First elimination: KOH/EtOH, heat. Second elimination: NaNH2 in liquid NH3 — vinyl halides need a much stronger base.
  • Vicinal dihalides (from alkene + X2) and geminal dihalides both give alkynes.
  • Use ≥2 equivalents of NaNH2 for terminal-alkyne targets; the product forms as the acetylide and is protonated on aqueous workup.
  • The alkene → Br2 → dibromide → NaNH2 → alkyne sequence is a standard retrosynthetic move.
  • E2 requires the H and X to be anti-periplanar in the transition state — the reason conformational analysis matters.
  • Safety note (general principle): strong bases and liquid ammonia are hazardous; these reactions are run with proper PPE, ventilation, and thermal control — never improvise lab conditions.

Check yourself

6 review questions from the chapter. Try each one, then open the answer.

  1. What two E2 eliminations convert a vicinal dihalide into an alkyne, and why does the second need a stronger base?

    Show answer

    First E2 (alkyl halide → vinyl halide) with KOH/EtOH; second E2 (vinyl halide → alkyne) with NaNH2 in liquid ammonia. The second is harder because the C–X bond on an sp2 carbon is stronger, so a much stronger base is required.

  2. Why is aqueous workup the final step when making a terminal alkyne from a dihalide with excess NaNH2?

    Show answer

    The terminal alkyne's acidic hydrogen is removed by excess amide, forming the acetylide anion; adding water (aqueous workup) protonates it back to the neutral R-C#CH.

  3. How many equivalents of NaNH2 are needed (minimum) to convert 1,1-dibromobutane to 1-butyne, if the acetylide is the isolated intermediate?

    Show answer

    Three: two for the two eliminations plus one to deprotonate the terminal alkyne to its acetylide (2 equiv minimum for elimination alone).

  4. Give the two-step sequence that converts 1-pentene into 1-pentyne, naming the reagents.

    Show answer

    (1) 1-pentene + Br2 → 1,2-dibromopentane; (2) 1,2-dibromopentane + 2 NaNH2 (liq NH3), then H2O → 1-pentyne.

  5. A student claims vinyl halides are unreactive toward substitution. Is that consistent with them undergoing elimination here? Explain.

    Show answer

    Yes. Vinyl halides are poor substrates for substitution (the sp2 carbon resists SN2 and SN1), but E2 elimination still proceeds when a strong enough base abstracts the adjacent hydrogen — elimination does not require a nucleophilic substitution step.

  6. Name one difference between making an alkyne from a vicinal vs. a .

    Show answer

    Vicinal dihalides come from X2 addition to alkenes (halogens on adjacent carbons); geminal dihalides come from two HX additions to an alkyne or carbonyl-derived chemistry (halogens on the same carbon). Both give the same alkyne after double elimination.

Keep learning

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Study tools & related lessonsKey vocabulary · Related

Key vocabulary

Dehydrohalogenation
Loss of HX from a molecule, making a π bond.
Vicinal dihalide
Two halogens on adjacent carbons.
Geminal dihalide
Two halogens on the same carbon.
Vinyl halide
A halide on an sp2 (alkene) carbon.
E2 elimination
One-step bimolecular elimination: base removes H, leaving group departs, π bond forms.
Sodium amide (ceNaNH2)
Very strong base (amide ion, NH2−).
Acetylide anion
R-C#C−, the conjugate base of a terminal alkyne.

Sources & references

  1. openstax.org — Organic Chemistry

This lesson was adapted from the open educational references above; their licenses and attributions are preserved. See Copyright & Licensing.

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