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Why do vinylic carbocations generally not undergo hydride rearrangement from neighbouring sp3 carbon to get more stability?
According to me, a rearrangement would lead to an allylic carbocation, which is more stable than vinylic, hence the rearrangement should be favorable.
If you think about protonating 1-propyne on the terminal acetylenic carbon, you get a vinylic cation that could apparently form an allylic cation by a hydride shift. The bond to the hydride that would shift has to be lined up with the empty orbital to which it is migrating.
So far, no problem. But after the shift, the empty $\mathrm{2p}$ orbital of the new carbocation is orthogonal (at right angles) to the $\mathrm{2p}$ orbitals of the adjacent double bond. Therefore, what looks like an allylic carbocation is actually a twisted allylic carbocation.
The $\mathrm{2p}$ orbitals of the double bond can't stabilize the new cation by resonance. Worse yet, the $\mathrm{2p^2}$ carbons of the double bond are relatively electronegative, so they actually destabilize the new carbocation. In other words, the migration is forming another unstable carbocation.
This is why rearrangements in vinylic carbocations don't occur. (There is a Russian Chemical Reviews article [1] that explains all this, but it's hard to get a copy of this, and their PDFs seem to be corrupted.)
(Wikipedia has a very extensively detailed section on vinyl cation rearrangement, so be sure to give it a read as well for a more in-depth view. I'll base my answer on that section)
1,2-hydride shifts are very uncommon in vinyl cations. "The interacting orbitals during the conversion of a linear vinyl cation to a non-linear allyl cation are orthogonal and passes through a non-planar transition state". This leads to a very high activation energy for such a rearrangement. 1,2-methyl shifts too have a very high activation energy.
Hence, even though such a rearrangement would lead to a thermodynamically favorable product, it is uncommon.
