Alkyne Reactions

By James Ashenhurst

Alkenes To Alkynes Via Halogenation And Elimination Reactions

Last updated: November 15th, 2022 |

Formation of Alkynes Via Double Elimination Of Halides

  • Alkynes can be produced from vicinal or geminal dihalides through double elimination reactions.
  • The usual choice of base for these reactions is sodium amide (NaNH2)
  • The first equivalent of strong base forms an alkenyl halide. The second equivalent forms the alkyne.
  • If a terminal alkyne is formed, a third equivalent of base will be consumed, since the resulting alkyne proton is relatively acidic (pKa 25)
  • Overall, this process can be used for the synthesis of alkynes from alkenes, through 1) halogenation of alkene 2) double elimination of dihalide

Table of Contents

  1. Elimination Reactions: Form C–C (pi), Break C–H And C–LG
  2. Elimination Of A Vinyl Halide To Give An Alkyne
  3. Formation of Alkynes From Double Elimination Of Vicinal Dihalides
  4. Formation Of Alkynes From Double Elimination Of Geminal Dihalides
  5. Why Is This Useful? Going From An Alkene To An Alkyne
  6. Notes
  7. (Advanced) References and Further Reading

1. Alkenes Via Elimination Reactions Of Alkyl Halides: Form C–C (pi), Break C–H And C–LG

We’ve gone through elimination reactions before – treatment of alkyl halides with base gives alkenes. Today we’ll discuss a pathway to get to alkynes from alkyl dibromides (which in turn can be made from alkenes) via double elimination.

Some time ago we discussed elimination reactions (See post: Elimination reactions) follow the general pattern below, where two adjacent bonds to carbon are broken – usually C-H and C-X, where X is a leaving group – and in place we form a new π bond.

elimination of alkyl halide with base form c c pi break c x and c h giving new pi bond

2. Alkynes From Alkenyl Halides: Elimination Of A Vinyl Halide To Give An Alkyne

If we can form alkenes through elimination reactions of alkyl halides, it’s natural to ask: can we also use elimination reactions to form alkynes? Perhaps through a reaction like this elimination of an alkenyl (aka vinyl) halide?

elimination of alkenyl halide with base giving alkyne forms c c pi breaks c h c br requires more harsh conditions alkoxide amide

The answer is yes! Although to be fair, these types of molecules [alkenyl halides]  are perhaps not the most familiar to us so far, We’ve only seen them once, and that was a post about how they can be synthesized from alkynes. (See post: Halogenation of Alkynes)

For our purposes, in order to synthesize them it’s common practice to start with an alkyl di halide.

In other words, we start with an alkyl di-halide, and then do two elimination reactions – one to form the alkenyl halide, and a second to form the alkyne.

Let’s have a look.

3. Alkynes From Double Elimination Of Vicinal Dihalides

There are two types of alkyl dihalides we’ve met so far. Vicinal dihalides have halogens on adjacent carbons – “in the vicinity”, if you will. Treatment of vicinal dihalides with strong base can lead to an elimination reaction [through the E2 mechanism] giving an alkenyl halide.

Treatment of this alkenyl halide with a second equivalent of base then gives the alkyne.

In previous posts we saw that a common base used for elimination reactions are alkoxide bases such as sodium ethoxide. But here, we typically go for a more powerful base, sodium amide (See post: Sodium Amide NaNH2)

Here’s an illustration of how it works.

elimination of vicinal dihalides with nanh2 gives alkenyl halides further elimination gives alkynes 3 equiv required alkyne acidic

In the first step, NaNH2 is the base in an elimination reaction [E2] to give the alkenyl bromide. In the second reaction, likewise a second equivalent of  NaNH2 performs a second elimination reaction to form the alkyne.

This is one example – a rare example, I may add – of an elimination reaction that works on an sp2 hybridized carbon. You might recall seeing at some point that reactions of the SN1/SN2/E1/E2 types typically don’t occur on carbons other than sp3 hybridized systems. This is a rare exception!

Finally, what’s often not mentioned in this reaction is that the product alkyne, if terminal [i.e. has a C-H on the end] is acidic [pKa 25] – any excess NaNH2 will thus remove the alkyne C-H and give the alkynyl anion. So if a terminal alkyne is formed, three equivalents of NaNH2 will be consumed; the alkyne is protonated upon workup, usually by adding water.

4. Alkynes From Double Elimination Of Geminal Dihalides

Geminal dihalides contain two halogen atoms attached to the same carbon.

Treatment of geminal dihalides with NaNH2 likewise gives alkynes through two successive elimination reactions. [We haven’t really covered any reactions that form geminal dihalides, except for the di-addition of HX to alkynes. So if you happened to start off with an alkyne and made a geminal dihalide with it, this would be a way of getting the alkyne back].

elimination of geminal dihalide to give alkyne with nanh2 gives alkyne via alkenyl halide 3 equiv required if terminal alkyne formed

In total, the elimination reactions described above represent a second way of making alkynes, other than through SN2 of acetylides with alkyl halides (See post: SN2 of Acetylides)

Let’s bring it  back.  What does it matter that we can do this?

OK. Let’s start putting some reactions together that show why this might be important.

5. Alkenes To Alkynes, Via Halogenation And Double Elimination

Let’s say we wanted to make an alkyne. But all we have to start with is an alkene. How might we get there?

  • We’ve just learned how to make alkynes from vicinal dihalides.
  • How might we make a vicinal dihalide from an alkene?
  • Bromination! 

So we can start off with an alkene… and brominate… and then add two equivalents of strong base to give ourselves the alkyne.

how to convert alkene to alkyne start with halogenation br2 and double elimination with nanh2 to give alkyne synthesis problem

This is a quick example of multi-step organic synthesis. We will soon see much more of this in the context of alkynes!

Next Post: Alkynes Are A Blank Canvas


(Advanced) References and Further Reading

  1. Eliminations from Olefins
    Angew. Chem. Int. Ed. 1965 4 (1), 49
    DOI: 10.1002/anie196500491
    A review describing various types of a- and b-elimination reactions of alkenes to give alkynes.
  2. Elimination Strategy for Aromatic Acetylenes
    Orita, H.; Otera, J.
    Chem. Rev. 2006, 106, 5387
    DOI: 10.1021/cr050560m
    Section 3.3 in this review covers the synthesis of alkynes by double dehydrobromination reactions from vic-dibromoalkanes.
  3. Unsaturated eight-membered ring compounds. XI. Synthesis of sym-dibenzo-1,5-cyclooctadiene-3,7-diyne and sym-dibenzo-1,3,5-cyclooctatrien-7-yne, presumably planar conjugated eight-membered ring compounds
    Henry N. C. Wong, Peter J. Garratt, and Franz Sondheimer
    Journal of the American Chemical Society 1974 96 (17), 5604-5605
    The reaction used to synthesize the strained cyclooctyne is the double dehydrobromination reaction, using fairly standard conditions (KOtBu in THF).
  4. Proton NMR study of two tetradehydrocyclodecabiphenylenes
    Charles F. Wilcox Jr. and Karl A. Weber
    The Journal of Organic Chemistry 1986 51 (7), 1088-1094
    DOI: 10.1021/jo00357a028
    The authors also use a double dehydrobromination reaction to obtain cyclic dialkynes, also using standard conditions (KOtBu in THF). The experimental section has detailed procedures.


Comment section

13 thoughts on “Alkenes To Alkynes Via Halogenation And Elimination Reactions

  1. Should not the dehydrohalogenation of alkenyl halide be E1CB instead of E2?
    I am saying so because the leaving group will become poor due to partial double bond character.

  2. Hey brother, I’m totally confused. Please reply anyone if you know solution of my query!
    First of all, talking about – ve charged species. It’s nucleophile cum base”(depending upon its attacking group). You treated alkane alkenes with NaNH2. Here this acts NaNH2 acts as an base, which deprotonates the substrate giving elimination product. Also, using NaOH(OH^-) when treated with substrate (mostly alkane derivatives) gives substitution reactions, whereas the same NaOH, acts as a base for the substrate, like in above bromoalkene did (when treated with NaOH gives alkynes… an elimination product). So I totally don’t understand where this types of reagents are going to act as BASE and when NUCLEOPHILE.
    Please anyone help me, I’ve asked this question many a times to other, but none of them were able. This was the reason for my hate for organic. But, I want to grow up once again. Please help… This is my building organic concept for JEE ADVANCED.

  3. We can also use alcoholic KOH for dehydrohalogenation of vicinal dihalide right?But from what I have seen sodamide is preferred why is that so?

  4. When NaNH2 is added in excess is the reaction E1cB as there is carbanion? ? Am I right? Please clear up my misunderstanding. Thank You.

  5. I thought for E2 to happen on a primary substrate, you need a strong, bulky (hindered) base. NH2 is not a bulky base. How does E2 still happen to the terminal halogen when making a terminal alkyne?

    1. For E2 reaction you need only a strong base not a bulky one. A bulky base can lead to Hoffman product instead of zaitsev product.

  6. Bromination of an alkene gives vicinal dibromide. Because of the bulky bromo group, should we get a pair of stereoisomers ? If yes, will they differ in the ease of elimination? Please elaborate. Thanks.

    1. Whether you get stereoisomers depends on the structure of the alkene, but in most cases, yes, you will get a pair of stereoisomers upon bromination. The first elimination to give the vinyl bromide will be the easiest. Which hydrogen is removed by base will depend on the structure of the compound – the least hindered will likely be removed first. I’m sorry to give you an “it depends” answer, but, it depends. : – )

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