# Does a C=C double bond donate electron or withdraw it?

I have learned that a $$\ce{C=C}$$ double bond is electron-rich because of the p-p overlap and hence can easily donate an electron pair: I agree with this statement.

However, in a $$\ce{C=C}$$ double bond, the carbon is $$\ce{sp^2}$$ hybridized and I've also learned that more '$$\ce{s}$$-type' character increases electronegativity. Does this mean that the carbon atom - being more electronegative - tries to withdraw electrons?

For example consider the double bond connected to a positively charged Carbon, then will it donate an electron and stabilize it? or will it (being more electronegative) withdraw an electron?

• As you advance in chemistry, you tend to find that different factors often produce competing effects with different impacts in different situations, producing the complexity we observe in chemistry. – Zhe Jan 14 '19 at 18:11
• You need to specifically give us the context. Your question as of now is too general and unspecific. Where is the position of the double bond? Is it adjacent to a positively charged carbon (i.e. allylic to it)? Or is it adjacent to a negatively charged carbon? – Tan Yong Boon Jan 14 '19 at 22:53
• oh ok. I will edit my question then – The Jade Reaper Jan 15 '19 at 1:50

## 2 Answers

An allylic double bond can be said to be both electron-donating and electron-withdrawing, depending on the context. The simpler way to explain this is based on the idea of resonance. Both the allyl cation and allyl anion are stabilised by the presence of this double bond.

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The behaviour of allylic double bonds can also be rationalised based on molecular orbital (MO) theory. A molecule containing a $$\pi$$ bond has both an antibonding $$\pi^*$$ MO and a bonding $$\pi$$ MO. Due to the lower strength of a $$\pi$$ interaction, compared to $$\sigma$$ interactions, the $$\pi$$ MO is not too lowered in energy and the $$\pi^*$$ MO is not too raised in energy, relative to the energy level of the $$\ce {C}$$ p orbitals. The consequence of this is that, to attain greater stabilisation of the molecule, this filled $$\pi$$ MO can interact with empty orbitals at adjacent sites in the molecule and this empty $$\pi^*$$ MO can interact with filled orbitals at adjacent sites. The allyl cation and allyl anion are just two examples to illustrate the above concept.

Other systems illustrating these effects would be the conjugated enone, as well as allyl halides. For the case of the enone, the effect seems to be more of an electron donation as the $$\ce {C=C}$$ $$\pi$$ bonding MO interacts with the $$\ce {C=O}$$ $$\pi^*$$ MO. For the case of tha allyl halide, it also seems to be that there is electron donation of the allylic double bond to the partially positively-charged $$\ce {C}$$ bonded to the halogen atom as the $$\ce {C=C}$$ $$\pi$$ bonding MO interacts with the $$\ce {C-X}$$ $$\sigma^*$$ MO.

From these examples, we can see that the electronic effects of the allylic double bond is highly dependent on the system we are considering. We can also seem to qualitatively derive the rule of thumb that the allylic double bond donates or withdraws based on the "needs" of the adjacent site. If it is electron-deficient, there would be donation from the double bond. If there is a site of high electron density, there would be withdrawal towards the double bond.

I guess it depends on the way you quantify "electron donation/withdrawal" (because it is a useful concept, but not really a measurable observable). I'm sure someone will find a counterexample, but if you consider methane and ethylene as Brønsted acids, you find that methane is a much weaker acid. Thus, ethylene is better at stabilizing the negative charge. (This is in line with your sp2 argument.)

• True. But then again I would argue that in lots of compounds having C=C double bond, the pi electron cloud donates some negative charge in order to stabilize it. And that goes against my sp2 argument. – The Jade Reaper Jan 14 '19 at 19:26