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So I was randomly looking at some really random exercises and I stumbled across synthesis of propyne from given compounds. There was $\ce{CH3CHBrCH2Br}$ as my substrate and I had to propose a way of synthesis of propyne. It is obvious, isn't it? Just treat it with base and I'll be alright. Then I started thinking... can't allene form also? What is stopping one Br from getting eliminated on one side and second Br on the second side.

I am almost certainly sure that alkyne is the major product, but shouldn't I be "scared" of allyne? Could somoene explain to me if it actually forms and if it is significant in amount.

And for why it is forming i figured out that E2 eliminations favour the s-trans (or just trans in double bonds) configuration and the firstly acquired alkene has already 1 hydrogen in trans configuration. Also because the double bond hydrogen is more acidic so it is more easily removed by base. I don't really know why this thought came to my mind so randomly but I just cant stop thinking about it, thanks in advance for answers.

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You may get some allene, but the alykene would be by some measure the major product. This sort of elimination is typically carried out with a particuarly strong base such as NaNH2. Once you have eliminated your first bromine, you have a choice of deprotonating an an alkane proton or an alkene one. The alkene protons are considerably more acidic as the carbon to which they bond is sp2 hybridized as opposed to sp3. This means the carbon has more s character which is closer to the nucleus and hence the carbon is more electronegative, so the bond to hydrogen is more electronegative, weaker and hence the alkene proton is more acidic and will be preferntially deprotonated. There is no need to consider carbanion stability in this case since this would always be an E2. So as deprotonation of the alkene would lead to the alkyne, the alkyne will be formed in much greater yield.

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