What is anti-bond?

[UIUCstudent]

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The BR says "An anti-bond is a molecular orbital that results in bonding breaking when coupled with a bonding orbital." Which doesn't clarify things for me.

Also what does it mean when a carbon is highly substituted?

 

[BerkReviewTeach]

What is an anti-Bond? It's the guy in Quantum Solace trying to kill James of course. 😀

Seriously, it's best described as an excited state for a molecular orbital. The basic idea is that if you have a compound like an alkene for instance, if you add exactly the right amount of energy, you can excite one of the electron in the bonding orbital (pi-bond) into and anti-bonding orbital (where the wto p-orbitals don't overlap, so the electrons cannot be shared with neighboring p-orbitals). Once that electron is excited into an anti-bonding orbital, the system has one electron in a bonding orbital and another in an anti-bonding orbital, resulting in no pi-bond.

It's basically a theory to explain why an exact amount of energy is necessary to break a bond (in this case a pi-bond).

Highly substituted means that the carbons forming the double bond have R-groups rather than Hs attached. For instance, 2,3-dimethyl-2-butene is tetrasubstituted, meaning it has for methyl groups on the double bond carbons.

 

[loveoforganic]

To just cover one little gap, antibonding orbitals lie at specific energy levels, just like bonding orbitals. There's a n=1 sigma bonding, n=1 sigma s antibonding, n=2 sigma s bonding, n=2 sigma s antibonding, etc. The n=1 sigma s antibonding falls before the n=2 sigma s bonding, etc. So antibonding orbitals do get filled in nonexcited molecules and can even be filled in the valence level. The relative numbers of bonding and antibonding orbitals is what determines how many bonds will be present in that molecule (bonding orbital electrons - antibonding orbital electrons)/2. So, by exciting electrons from bonding to antibonding orbitals, you can cause bonds to break, like BRT was saying.

I don't remember which of this is MCAT level and not.

 

[UIUCstudent]

Thanks for the explanations! I'm definitely starting to recall this from orgo class now.

Edit: I had another question, but the explanation in BR fixed that right up!

 

[UIUCstudent]

So BR says the order of highest to lowest electronegativity is F>O>N>Cl>Br>I> and fluorine would have the most inductive effect.

Instead, this is the chart in my book with fluorine as the least EWG.
http://www.screencast.com/users/trinhn812/folders/Jing/media/974e394b-73ff-40dd-b034-999507011ee2

I'm a bit confused on this. Is EWG trend and inductive effect two different concepts?

 

[BerkReviewTeach]

So BR says the order of highest to lowest electronegativity is F>O>N>Cl>Br>I> and fluorine would have the most inductive effect.

Instead, this is the chart in my book with fluorine as the least EWG.
http://www.screencast.com/users/trinhn812/folders/Jing/media/974e394b-73ff-40dd-b034-999507011ee2

I'm a bit confused on this. Is EWG trend and inductive effect two different concepts?

The list you are considering deals with benzene activation, which takes both the inductive effect and resonance into account. Although it seems like it shouldn't work this way, halogens are ortho-para directors because they are slightly electron donating through resonance.

So here is the deal on that chart. Fluorine has a 2p orbital, which overlaps with the pi-system of benzene better than the other halogens. As a result, it is the best resonance donator of the four halogens. So while it is the most electronegative of the four halogens in question, which would make it the most electron withdrawing by the inductive effect, it is also the best resonance donator, making it the best electron donating group of the four by resonance. Because benzene activation takes both resonance and inductive effect into account, F is overall a more donating group to benzene than Cl, Br, and I.

And FYI, they don't test benzene chemistry on the MCAT any longer, so that should alleviate possible confusion in this area.