Why this preparation of Alkyne by dehydrohalogenation is done in two steps? [closed]








**Closed.** This question is [off-topic](/help/closed-questions). It is not currently accepting answers.
 
 






---








 **Homework questions** must demonstrate some effort to understand the underlying concepts. For help asking a good homework question, see: [How do I ask homework questions on Chemistry Stack Exchange?](https://chemistry.meta.stackexchange.com/questions/141/how-do-i-ask-homework-questions-on-chemistry-stack-exchange)




Closed 6 months ago.








[Improve this question](/posts/171928/edit) 



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 ?




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.


 3

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](https://www.jove.com/science-education/11835/preparation-of-alkynes-dehydrohalogenation) in one step with 2 [NaNH2](https://en.m.wikipedia.org/wiki/Sodium_amide) in anhydrous ammonia.


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


 0