Alkyne Synthesis

Core Concepts

In this tutorial, you will learn about the reagents used and mechanisms involved in synthesizing alkynes.

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The main method by which alkynes are synthesized are via dehydrohalogenation reactions. These reactions are a type of elimination which rely on the deprotonation of a carbon which is adjacent to a carbon bonded to a halogen. The halogen serves as a leaving group and the two carbons gain a pi bond between one another. This proceeds mainly via the E2 mechanism which is a concerted-step elimination that requires a strong base to remove a hydrogen. To create an alkyne from an alkane, either a vicinal (one carbon apart) or geminal (same carbon) dihalide is needed and a double dehydrohalogenation needs to proceed. This reaction using a strong base such as hydroxide, lithium diisopropylamide (LDA), or any alkoxide dissolved in polar aprotic solvent.

An example of an Alkyne Synthesis using a vicinal dihalide

One key requirement in the E2 mechanism is the need for leaving constituents to be antiperiplanar to one another before the elimination can occur. While the E1 mechanism does not have this stereospecificity, the concerted mechanism of the E2 requires the removed hydrogen and halogen leaving group to be aligned anti to each other before the reaction can occur.

As stated prior, this means that a candidate for E2 reaction can either be vicinal or geminal alkyl dihalide.

To undergo E2, the carbon adjacent to the carbon bonded to the halogens must have at least as many hydrogens as there are halogens. Without a hydrogen on the carbon adjacent to the halogenated, there is no possibility of an E2 because the molecule cannot be deprotonated.

E2 Mechanism

Double dehydrohalogenation to form alkynes requires two elimination reactions to occur. Usually, these elimination reactions are both an E2 mechanism, however, it is possible to perform an E1 to form an alkene, and a subsequent E2 to form an alkyne.

Double Dehydrohalogenation via two E2 reactions

An E2 reaction requires a strong base which will deprotonate the carbon adjacent to the halogenated carbon. Strong basicity is important as to not undergo a substitution with the halogen, which will serve as a good leaving group for an SN1 or SN2 reaction. Therefore reagents such as CN should not be used as they are more apt to undergo substitution than elimination. Although alkoxides can undergo substitutions such as SN2, these usually are only done at terminal carbons. Therefore, the secondary halogenated carbons usually involved in double dehydrohalogenation are safe from substitution. Another way to avoid nucleophilic substitution is to use a bulky base such as LDA(Lithium Diisopropylamide) or DBU(1,8-Diazabicycloundec-7-ene).

Example of a successful alkyne synthesis via two E2 steps.


Synthesis of an alkane requires carbons to rehybridize, forming different molecular geometries than their original configurations. In the case of an alkane, an originally sp³ hybrid carbon will change its geometry from tetrahedral to trigonal planar as it transitions instantaneously to become an sp² hybridized carbon in the formation of an alkene. A second elimination will change the trigonal planar sp² carbon to a linear sp hybrid carbon. These unhybridized p orbitals are what comprise the new double and triple bonds in the molecule. Therefore, it is essential that the carbons adjacent two each other both have constituents which leave simultaneously. This ensures that a rehybridization and new unsaturated bond formation is thermodynamically favorable.

Reaction Specifics

As previously stated, strong bases are generally the key to undergoing E2 elimination reactions. They will often be placed in a polar aprotic solvent. The polar aprotic solvent is used to thoroughly solvate the nucleophile involved in the E2 reaction. Solvents such as DMF, DMSO, or THF have a dipole, and this polarity helps ensure adequate solvation of the nucleophile. Another very important feature of double dehydrohalogenation reactions is their need for two equivalents of the same base to be used in the reaction, so that the E2 reaction can happen twice.

Typical polar aprotic solvents

Overall, the strong base, and polar aprotic solvent are what comprise the E2 reaction. To denote dehydrohalogenation they should be used together with a specification for 2 eq. of base used.

Practice Problems

Question 1

Question 2

The intermediate in a double dehydrohalogenation looks like the alkene shown below. Why does it experience no reaction when exposed to an excess of a strong base?

Question 3

Will the following molecule undergo alkyne synthesis via double dehydrohalogenation?


Question 1

The following will undergo double dehydrohalogenation to yield this product.

Question 2

The molecule shown will not undergo dehydrohalogenation because the hydrogen is not anti to the halogen. Therefore, it cannot proceed via the E2 mechanism.

Question 3

The molecule will not undergo alkyne synthesis. Despite having both vicinal and geminal halogens, the carbons involved have insufficient hydrogens on adjacent carbons to undergo the E2 mechanism twice. The only outcome possible with a strong base would be an alkene.