[shehri]

In my book it's said:

The reaction of alkynes with certain metyals are due to acidic Hydrogen
present in terminal alkynes.In alkynes the Carbon atom is sp hybridized.It has 50% s-character so
it is strongly electron withdrawing.I'm unable to unerstan this reason of acidity.Plz. explain?Thanks.

[Dan]

Which is further reaching, an sp orbital or an sp3 orbital?

[english]

Think about electronegativity.  The more s character you have, how will polarity in the carbon hydrogen bond change?  Think about what Dan said and apply this to that reasoning.  Electrons in what orbitals experience greater penetration towards the nucleus, s or p?

For all intents and purposes you usually disregard the polarity of a C—H bond because the difference is quite small.

If you're comparing an alkyne hydrogen to an alkene hydrogen, or to an alkane hydrogen, this difference becomes significant.

[movies]

I like to think of it this way:

Which orbital is lower in energy, s or p?  A lower energy means that the electrons in that orbital are at lower energy.

Remember that an anion is a pair of electrons.

If you think of hybrid orbitals in terms of % s-character, then you can easily rationalize the difference between sp³ and sp based on the % contribution for the lower energy orbital.

[english]

I like to think of it this way:

Which orbital is lower in energy, s or p?  A lower energy means that the electrons in that orbital are at lower energy.

Remember that an anion is a pair of electrons.

If you think of hybrid orbitals in terms of % s-character, then you can easily rationalize the difference between sp³ and sp based on the % contribution for the lower energy orbital.

Never thought of it that way.

So you're saying that the more s character you have, the more stable the anion, and thus the stronger the conjugate acid?

[Dan]

Yeah, it's a good way to aproach the idea. The sp orbital is lower in energy because the orbital is more penetrating (ie. electron closer to the nucleus, which is what I was getting at).

[movies]

Yup.

I think that the "penetrating" argument is just a little harder to follow, although it is certainly the effect that is responsible for the lowering in energy.  Everyone knows that s is lower than p, so then it just becomes a question of which orbital has more % s.

[shehri]

Thank U all for providing worthwhile infos.This forum's doing a great job of helping chemistry students all over the globe,undoubtedly.