Alkynes Via Elimination Reactions

by James

in Alkyl Halides, Alkynes, Organic Chemistry 1, Organic Reactions

We’ve gone through elimination reactions before – treatment of alkyl halides with base gives alkenes.

Some time ago we discussed elimination reactions here. 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.

1-alkene-elim

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 one?

2-alkyne-elim

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. 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.

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 [NaNH2].

Here’s an illustration of how it works.

2-alkyne-vicinal

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 me writing 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.

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].

3-alkyne-geminal

In total, the elimination reactions described above represent a second way of making alkynes, other than through SN2 of acetylides with alkyl halides.

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

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

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.

5-bromine-alkene-alkyne

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

1_alkynebanner_orange

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{ 4 comments… read them below or add one }

Rajendra Samant

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.

Reply

James

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. : – )

Reply

B

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?

Reply

XYZ

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

Reply

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