My teacher was explaining preparation of alkyne by dehydrohalogenation of Alkane dihalide. Actually the preparation is carried out in two steps. In the first step an alkane dihalide is dehydrohalogenated with alcoholic KOH and then the formed alkene halide is again dehydrohalogenated with Sodamide (NaNH2). Given that sodamide is a more strong dehydrohalogenating agent than alcoholic potassium hydroxide, why can't we carry out it in a single step with sodamide itself ? Why is preparation of Alkynes by dehydrohalogenation is carried out in two steps ?

  • $\begingroup$ It may be a question of price. Sodamide is more expensive than $\ce{KOH}$ $\endgroup$
    – Maurice
    Mar 5, 2023 at 14:00
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    $\begingroup$ Consider the other pathways open to the dihalide starting material with sodamide and ammonia. $\endgroup$
    – Waylander
    Mar 5, 2023 at 15:08
  • $\begingroup$ What Waylander said - this wouldn't work, because sodium amide is nucleophilic. $\endgroup$
    – Mithoron
    Mar 5, 2023 at 15:30
  • $\begingroup$ @Mithoron So The starting material cannot be Sodamide ? $\endgroup$ Mar 5, 2023 at 15:53
  • $\begingroup$ It is not just that sodamide is nucleophilic. Any deprotonation that occurs generates ammonia which is more nucleophilic so you have two pathways that will give amines. $\endgroup$
    – Waylander
    Mar 5, 2023 at 16:06

2 Answers 2


Consider the possible reactions of sodamide with the dihalide:

1 - Sodamide acts as a base deprotonating the dihalide forming the double bond by elimination of halide anion giving the intermediate alkenyl halide that you want. This is stable to the reaction conditions, but the other product of this is a molecule of ammonia - we'll come back to this.

2 - Sodamide acts as a nucleophile displacing one of the halides (and then possibly going on to displace the other halide giving the aziridine). This pathway is irreversible, costing you yield and complicating your purification of the product. Remember that molecule of ammonia? It is a better nucleophile than sodamide and can do this pathway too, giving rise to multiple amine derivatives of the original dihalide.

  • $\begingroup$ Not sure about aziridine ring formation being "irreversible", especially in the presence of nucleophiles. $\endgroup$ Mar 5, 2023 at 19:04
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    $\begingroup$ Why is a ammonia a better nucleophile than NaNH2 ? $\endgroup$ Mar 5, 2023 at 19:49

The mechanism of the first dehalogenation is by KOH extracting a proton, a rearrangement to a vinyl chloride, and one of the chlorides as a leaving group (as a concerted E2 elimination).

Since the first proton extraction is stabilized by 2 electron withdrawing Cl, KOH is adequate for this step, less expensive, and more easily controlled$^1$.

The second proton extraction to form the alkyne requires a much stronger base, NaNH2, which is far more reactive.

One thing you don't want is something out of control, like a run-away polymerization or overheating of a large batch.

This is why sodamide could be added cautiously, in a second step.

There may also be a higher chemical cost to using 2 NaNH2 instead of KOH/NaNH2, but this would be off-set by saving a synthetic step and worth investigating if safe.

Alkyne is known to be synthesized in one step with 2 NaNH2 in anhydrous ammonia.

$^1$ which is helpful when they tell you, OK, go out into the plant and make 5000 gallons of it.

  • 1
    $\begingroup$ Do you have an example of the prep of an alkyne from a vicinal dihalide with only sodamide, rather than teaching materials claiming it works? $\endgroup$
    – Waylander
    Mar 5, 2023 at 17:36
  • $\begingroup$ Not to mention, since when E2 has carbanion? $\endgroup$
    – Mithoron
    Mar 5, 2023 at 18:52
  • $\begingroup$ @Waylander Here is another reference for vicinal dihalides. $\endgroup$ Mar 5, 2023 at 18:57
  • $\begingroup$ @Mithoron you are right. The carbanion would not form as a separate intermediate. E2. $\endgroup$ Mar 5, 2023 at 18:59
  • $\begingroup$ @RobertDiGiovanni Does the Chem. Rev. referenced discuss using NaNH2? I can only see the abstract $\endgroup$
    – Waylander
    Mar 5, 2023 at 22:09

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