I am not able to decide which among 1,3-butadiene and 2-butyne is the major product of this reaction. Will there be even a major product out of the two or will there be a mixture?

According to me, the product formed from the 1st mechanism should be the major product as it is more stable (two double bonds in a conjugate alkene are more stable than a triple bond). But I also think that product formed from mechanism II can be formed faster as the hydrogen being picked up in the first step of mechanism II is more acidic (due to the electron withdrawing effect of electronegative chlorine atom) as compared to the hydrogen being picked up in mechanism I.

Another factor, which seems to favor the formation of product by mechanism I, is the picking of hydrogen atom in the 2nd step which is easier from an sp3 carbon than from an sp2 carbon (in mechanism II).

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2 Answers 2


These are both reasonable mechanisms, and the question outlines well the factors favoring each. In favor of mechanism I:

  • Low temperature suggests kinetic deprotonation
  • Statistically more terminal hydrogens than internal hydrogens

In favor of mechanism II:

  • Small base suggests thermodynamic deprotonation

In these cases where there are conflicting factors, I agree with @Lighthart that the best answer is to actually look at the experimental results. That will give an indication of how to weigh the various factors.

In this case, I do not find an exact hit in SciFinder, but the reaction of 2,3-dibromobutane does come up in two older reviews by Bergstrom and Franke. I do not have access to Franke's review, although SciFinder has a note that "Bromine represents any halogen."

The SciFinder result for the Bergstrom review is deceptive. There is a comment "generalized reaction, halogen and R-groups can vary," and I don't find the exact result (or reasonable analogs) in the review.

More broadly, there are seven results for an internal, vicinal (1,2) dibromide reacting with sodium amide to give an alkyne. One reliable example is the synthesis of stearolic acid from Organic Syntheses.

Similar SciFinder searches for the 1,3-diene product come up empty.

Based on this research, I believe the alkyne would be the predominant product.

  • 2
    $\begingroup$ A further conclusion is that I wouldn't try to make either of the possible compounds by this route. $\endgroup$
    – jerepierre
    Mar 22, 2016 at 18:01
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    $\begingroup$ I agree. These kinds of questions are interesting from a teaching standpoint to compare various factors that influence the outcome of a reaction that can occur by competing pathways. The experimental version would be an interesting lab exercise (except for the liquid ammonia) to demonstrate the principal. If I needed a conjugated diene or an internal alkyne in high yield, I would something else. $\endgroup$
    – Ben Norris
    Mar 22, 2016 at 18:55
  • $\begingroup$ It would be a good experiment to run because teaching how to handle liquid ammonia is a worthwhile exercise in itself $\endgroup$
    – Waylander
    May 20, 2018 at 21:25

Liquid ammonia is cold, this means you are going to form the 'kinetic' product.

Kinetic products are the ones that are most likely to occur via molecule collision, generally from the least sterically hindered deprotonations. There is some hinderance for forming the first intermediate on the way to product II.

I would expect product I to form.

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    $\begingroup$ Consider however, that in very strong base, product II can isomerize into an acidic alkyne: $$\ce{CH3 -C#C -CH3 <=>[\ce{NH2-}]CH3CH2 -C#C -H} \\ \ce{CH3CH2 -C#C -H + NH2- -> CH3CH2 -C#C- + NH3}$$ $\endgroup$
    – Ben Norris
    Mar 22, 2016 at 16:20
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    $\begingroup$ True, under thermodynamic conditions.(read: not liquid ammonia temperatures). Less sure about this under kinetic conditions. $\endgroup$
    – Lighthart
    Mar 22, 2016 at 16:25
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    $\begingroup$ On the other hand, NaNH2 is an extremely small base so the steric effect will be minimized. $\endgroup$
    – jerepierre
    Mar 22, 2016 at 17:11
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    $\begingroup$ Agree on this point except that the likely molecular collisions are more likely to happen on the terminal hydrogens, of which there are three times as many. My answer is a weighted guess, I would defer to an actual experiment to resolve this. $\endgroup$
    – Lighthart
    Mar 22, 2016 at 17:22
  • 2
    $\begingroup$ Then I shall defer to the experiment. $\endgroup$
    – Lighthart
    Mar 22, 2016 at 19:34

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