I've got a burning orgo question for which I can't seem to find the answer. Consider the addition of a hydrogen halide to an alkyne. The addition of one equivalent merely involves the pi electrons of the triple bond attacking the hydrogen in H-X, thereby displacing the halide. Thus, one forms a vinyl cation on the more substituted carbon (Markovnikov addition). X- then adds to the carbocation, giving both E and Z isomers. So now you have an alkene with a halogen substituent on one of the double-bond carbons. When you have an alkyne whose triple-bond carbons are equally substituted, you get a mixture of products. This is all pretty simple. The problem comes when you add another equivalent of H-X. So, let's say you started with but-2-yne, a symmetrical alkyne. Addition of, say, hydrobromic acid (HBr) yields 2-bromo-but-2-ene: H3CBrC=CHCH3 When you add another equivalent of HBr, experiments demonstrate that one obtains only the geminal dihalide: H3CBr2C-CH2CH3 In other words, both bromides add to the same carbon. This suggests that the carbocation in the second addition formed on the carbon that already has the bromine. The bromine is electron-withdrawing and it should, therefore, destabilize the carbocation. If the alkyne were unsymmetrical, then maybe the argument could be made that the electron-donating effect of an additional alkyl group would outweigh the electron-withdrawing effect of the bromine, although I am even skeptical of this. However, in the case of the symmetrical alkyne, carbocation formation on the non-halogenated carbon would still provide the electron-donating benefits of an alkyl while avoiding the electron-withdrawing effect of the halogen. Why, then, does the geminal dihalide form? I'd really appreciate any insight anyone could provide.