* *
O
// (double Bond)
CH3-C
\*
*O-H
(*=unbonded electron pair)
Now resonance is when a pair of bonding electrons moves, either to an
nonbonding orbital, or to a different bonding orbital. Resonance tends
to occur when double bonds are present (I think because it results in a
high electron density, meaning a pair of electrons wouldn't mind popping
off for a bit for some freash air or something). In this case, before
Acetic acid has lost its hydrogen in a reaction, resonance occurs when an
nonbonding electron pair from the lower oxygen forms a double bond with
carbon while simutaniously one of the bonds to the upper oxygen moves to
a nonbonded state as a third nonbonding pair on the oxygen. That makes
the structure look like this:
*
*O*
/
CH3-C
\\
O-H
*
The two 'resonance structures' alternate between the two states.
However, if you were to calculate the formal charge on the oxygens in
both cases you'll find that in the first case, both oxygens have a formal
charge of 0 (it likes this). In the second case you'll find that the
oxygen on top now has a formal charge of -1, and the one on bottom has a
formal charge of +1. (It doesn't like this, heres why). This poses a
problem because to seperate the charges like this takes energy. When a
resonance structure like this occurs, it is called small resonance
stabilization (We'll explain the stabilization part in a bit). This is
because it prefers to stay in the first resonant structure due to the
energy it requires to change to and maintain the second structure. As a
side point, they also aren't equivalent because the OH group has the bond
now ( if the OH group was an O it would be equivalant because in both
cases the double bond is to an O, and thus are functionally the same.)
The reason it is called small resonance stabilization is because the
stabilization provided by the resonance is small.
So why does resonance stabilize structures? The basic reason is
a fundemental of kindergarden: Sharing. If two people carry two heavy
boards, each individually, it gets tiring, however if they both grab an
end of both boards, the task is easier. Why? Dunno, ask a physisist.
Anyway, that same idea is going on in a resonance structure. Contrary to
the idea that the molecule shifts from one resonance structure to the
other, the molecule actually finds a happy medium. Take this example:
* * * (-) * (-) * *
O *O* *O* O (-)
# ! ! |
C C C C
/ \ / # # \ / \
*O* *O* *O* O* *O *O* *O O*
* * * * * * (-)* *(-)
1 2 3 Resonant
Structure
*=Nonbonding Electron Pair (-)Partial - charge
#=Double Bond
! or / or \ =Single Bond
Unfortunately I forgot to include the Negative charge on the single
bonded O's, so pretend they are there on the first three charges. There
should be one on each single bonded Oxygen.
The first three structures (1,2,3) are the resonance structure stages,
however the fourth structure is what actually results. The bonds shown
in the fourth structure are not single bonds (but the media is limited so
I had to use single bond notation). They are actually 1 and 1/3 bonds
because the whole molecule evenly shares the bond throughout. Because of
this, the two extra full charges on the oxygens in the resonant stages
are also evenly shared (2/3 a charge on each). See SHARING!! This is
more stable because the charges aren't seperated. Also, with the bond
and two electron pairs roving indescrimanantly amongst the atom, it is in
a sence more random, and thus the free energy change is more negative
when in this stage (when randomness goes up, f.e.c. goes down). This
also stabilizes it. This is the factor that gives small resonance
stabilization its 'stability'. Now lets look at larger resonance stability.
In the Acetate Ion (the Acetate Acid after it gives up it's H+) this
stability occurs.
* * * * *
O *O* (-) O (~)
// / %
CH3-C CH3-C CH3-C
\ * \\ %
O O O*
* * (-) * * * (~)
R.S.1 R.S.2 Actual
*=Nonbonding electron pairs
-,\,/= Bonds
(-)= Full negative charge
%=Partial bond (1 1/2)
(~)=Partial Negative Charge (~)
Here is the two Resonant structures, and the resulting actual structure.
The resonanance provides larger stability because not only does it have
the increased sharing of bonds and extra electron pairs, but also the
charge is evenly distributed. This makes the Acetate Ion more stable,
and have a more negative free energy change (because in the decrease of
orderlyness) than the Acetic Acid state.
Now that that's out of the way, why does that make it more acidic than
an alcohol? First understand that an alcohol does not have resonance
stabilization in either the acid form or the ionic form. Thus it does
have the resulting negative effect on the free energy change. This makes
it less willing than a caroboxylic acid to give up the H+. Since it is
less willing, it is a less powerful acid.
Ok, that was dauntingly vague...