Partial reduction of alkynes to trans alkenes using sodium and ammonia
Description: Alkynes treated with sodium in ammonia are reduced to trans alkenes. Alternatively, potassium in ammonia can also be used.
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Would this not lead to a low yield of conversion? Could the hydrocarbons not form polymers when in their radical state?
Yields can be very good (95%!) so this doesn’t really happen in solution. Electron transfer to give the radical anion from the vinyl radical is fast, relative to addition to another alkyne. (Remember that we’re dealing with 2 equiv of Na relative to the alkyne, here. so there is an adequate supply of electrons).
It’s likely *possible* to do a polymerization of alkynes via some kind of one-electron reduction to give a vinyl radical that then adds to another equivalent of alkyne to give a new vinyl radical and so on. This would require a catalytic amount of reductant, and then the chain gets propagated.
is this reaction the same with benzene Birch reduction??
It is very similar, yes, but the Birch reduction uses t-BuOH as a proton source.
Why is the radical anion able to undergo flipping when there is a pi bond? Does that require a great amount of energy to first break the pi bond, flip and then reform the pi bond again?
I was wondering the same thing….? Prehaps the alkyne pi bond cleavage actually takes electrons from each of the separate pi bonds which allows temporary alkane rotation to form the more stable trans intermediate?
I think it’s just illustrating that either cis or trans could have been produced. However, only the trans product will end up being be produced because it is more stable.
As I understand it, the pi bond doesn’t “block” rotation like a rigid beam, but renders rotation unfavorable because there would be no stabilizing overlap in the intermediate: “if it’s not broken, don’t fix it”. If the radical system destabilizes the cis isomer enough, maybe breaking the remaining pi bond would be less unstable by comparison, and no longer inhibitive?