I've come across the following problem in Klein's Organic Chemistry, 2nd edition:

Transformation of (1,1-dibromoethyl)cyclohexane to ethynylcyclohexane with excess sodium amide (NaNH2)

However, I'm a little bit confused about why the terminal alkyne is produced. I believe that an allene should be formed via stepwise elimination of HBr, using the strong base NaNH2. The first elimination should give the more substituted, more thermodynamically stable alkene (Zaitsev's rule):

Proposed elimination to form vinylidenecyclohexane

Why does the former occur instead of the latter?

  • 10
    $\begingroup$ monosubstituted acetilenes are deprotonated by sodium amide, allenes are not. Since alkenes and allenes can converse to each other in amides via deprotonation, typically even if you put a nonterminal allene or alkyne into sodium amide solution, it transforms into terminal alkyne. $\endgroup$
    – permeakra
    Commented Dec 8, 2015 at 22:10
  • $\begingroup$ I think the first proton is abstracted from the primary carbon due to the following reasons: Primary is less sterically hindered and hence is easily accessible by $\ce{NH2-}$ ion. Also, even a partial negative charge on carbon in the transition state of the reaction is better off on primary carbon. $\endgroup$
    – ShankRam
    Commented Feb 2, 2016 at 4:04

1 Answer 1


Sodamide is bloody dangerous. If your bottle is too old, it likely has begun to oxidize some and you have potential azides or other polynitrogen compounds that may lead to an explosion.

Even with good stuff, generally the reagent is not heated to avoid similar complications. This means normally this reagent works at low temperatures (often times dry ice/acetone). Under these conditions kinetic products are favored over thermodynamic ones.

The exomethyl group is much more accessible for deprotonation, and so kinetically, I would expect the amide ion to abstract the terminal hydrogen.

First elimination of HBr

After that, the resulting bromoalkene still has the possibility of forming the allene product, but the transition state resulting in elimination to the alkyne is favoured as the vinylic hydrogen and the bromine are always locked in an anti conformation due to alkene geometry, as indicated in the above diagram. On the other hand, the allylic hydrogen may not necessarily be anti to the bromine due to C–C bond rotation.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.