Formation of alkynes through double elimination of vicinal dibromides

by James

Description: Sodium amide will convert 1,2-dihalides (“vicinal dihalides”) into alkynes through two consecutive elimination reactions.

Notes: The reaction is greatly assisted by heat. Also note that vicinal dihalides are formed through bromination of alkenes. So this is an important step on the pathway from converting an alkene into an alkyne.

Bromides, chlorides, and iodides all work fine.


Notes: The reaction works well for iodides, bromides, and chlorides. Note that the byproducts of this reaction are NH3 and the sodium salt of the respective halide. Example 4 shows how this reaction can be applied to form alkynes from alkenes in two steps.

Mechanism: Strong base (NaNH2) removes a proton from carbon, leading to formation of a C–C double bond and loss of halide (Step 1, arrows A, B and C). [This goes through an E2 mechanism]. A second equivalent of NaNH2 removes a second proton from the alkene carbon, leading to formation of the triple bond (Step 2, arrows D, E and F).

Notes: It’s also reasonable to show removal of proton from C-2 first and then from C-1 (i.e. to reverse the order of the eliminations).


{ 4 comments… read them below or add one }

Quinn Haynie

My course taught us that we need three molar equivalents of NaNH2 for this reaction, as it prevents the formation of the alkynyl salt.



True, for terminal alkynes. Once the alkyne is formed, it can then be deprotonated by NaNH2 in solution. For practical purposes, three equivalents for the formation of terminal alkynes is best.


Jason Mathias

Does this work the same way for Geminal Dihalides?



Yes it does!


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