[nextpauling]

So we may prepare an alkyne from a dihalide by two successive E2 reactions.  Suppose however that in the second reaction we may remove a beta proton from a different carbon than in the first reaction to form two adjacent double bonds instead of the triple bond (i hope i'm being clear without a way to show a drawing).  Will the major product be the one with two double bonds or one triple bond?  My guess would be the product with two double bonds because the formation of the triple bond requires the abstraction of a proton on an sp2 carbon whereas the formation of the second double bond requires removal of H+ from an sp3 carbon. 

[sjb]

I've only really seen the conversion RCH=CXR' to yield alkynes in the literature where allenes are not doable, for instance when R is aromatic or tertiary e.g. PhCHBr₂CH₃ to PhC#CH.

However, sp³ bonded protons are in general less acidic than sp², and similarly through to sp protons (consider the pKas of ethane, ethylene and acetylene), so you may be on to something. It may be that you do initially form allenes (when you can), but then these reduce strain or something tautomerising to the alkyne.

S