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[FONT=Verdana, Arial, Helvetica, sans-serif]31) Phenols are soluble in a strongly basic sodium hydroxide solution, and insoluble in dilute sodium bicarbonate. Phenol has a pKa = 10.0. The introduction of an ortho bromine atom into the phenol would have the effect of
- lowering the pKa and thus decreasing the acidity of the phenol.
- lowering the pKa and thus increasing the acidity of the phenol.
- [FONT=Verdana, Arial, Helvetica, sans-serif][FONT=Verdana, Arial, Helvetica, sans-serif]increasing the pKa and thus decreasing the acidity of the phenol...
- increasing the pKa and thus increasing the acidity of the phenol
Can anyone help me out with this question? I picked C but supposedly the answer is B.
..
There's two things to take into consideration:
A. Halogens are electron donating groups (via pi bonds ~~ resonance).
B. Halogens are electronegative (via sigma bonds ~~ so there's an inductive effect).
Prior to the addition of Halogen, consider the acidity of the phenol. The hydrogen of a phenol (a benzene ring with an attached -OH group), is moderately acidic because it's attached to an aromatic benzene ring. The negative charge on the oxygen (after deprotonation) can be delocalized along the benzene ring via resonance. This stability is what enables the hydrogen to be deprotonated in the presence of a weak base.
The second thing to ask is if it's possible to stabilize the conjugate base of the phenol even further (increasing the acidity of the phenol group) and the answer is yes. In order to understand how, you need to consider the resonance structures of the deprotonated phenol. The now negatively charged oxygen (with 3 lone pairs) can reach a "happy state" (2 bonds, 2 lone pairs) by donating one of it's lone pairs to the carbon below it. Meanwhile the pi bond connected to the carbon below it must move it's electron pair to it's next door neighbor (a neighboring carbon atom), which results in a negative charge on that atom. Even though resonance increases stability, these particar resonance structures aren't that great because carbon atoms don't want a negative charge (because they're not very electronegative).
It turns out that if you drew out all the resonance structures, this negative charge (on carbon) would reside in 3 positions of the benzene ring (2 on ortho and 1 on para).
My next question is, can you make these resonance structures even MORE stable? The answer is yes... by simply adding a halogen (OR an Electron Withdrawing Group). Remember, increasing the stability of the conjugate base, increases the acidity of the acid form. Halogens are notorious for their electronegativity. As such, they are able to pull electron density away from neighboring atoms to themselves. This is why adding a halogen in the ortho or para positions stabilize the phenol group.
One other important thing to clarify. Halogens, just like the -OH of the phenol group, are also electrondonating (they can donate an electron pair via a pi bond). Earlier, the stabalizing effect was via inductive effect. However, the electron DONATING ability of halogens are considered to have poor resonance ability (even though they are EDG's ...which is why it's usually ignored). 95% of the time, resonance is considered first before inductive effects, but for halogens the resonance contribution is so little that we can consider it negligible. In most other cases, the resonance effect would win out. This is why the answer choice is B and not C.
Hope this helps.
