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When a vicinal dibromide undergoes elimination reactions in vigorous basic conditions, say in ammonia with $\ce {NaNH2}$, is the product an alkyne or is it a conjugated diene? I would think that the answer depends on thermodynamic and kinetic factors. Considering bond energies of the $\ce {C-C}$ $\pi$ bonds, perhaps the conjugated diene would be more favourably formed?

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  • $\begingroup$ That reaction isn't reversible, so it's not going to dependent on thermodynamic factors. $\endgroup$ – Zhe Jan 10 at 16:09
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After the first elimination, you have a vinylic bromide, so there are two options for the second elimination. The two acidic hydrogens that the base can abstract are one attached to sp2 carbon and one to sp3. Since sp2 hybridized carbon is more electronegative, the base abstracts the hydrogen attached to it, and then bromine leaves being a good leaving group, which gives alkyne as the major product.

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