Acidity of substituted phenols


by maverick280857
Tags: acidity, phenols, substituted
maverick280857
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#1
Oct30-04, 10:20 AM
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Hello again

My question:

Rank the following compounds in order of decreasing acidity:

(a) phenol
(b) o-nitrophenol
(c) m-nitrophenol
(d) p-nitrophenol

I know that (d) will be most acidic. But here's the problem: in orthonitrophenol the hydrogen of the -OH group satisfies the conditions for intramolecular hydrogen bonding and should in theory be bound to the nitrogen of the nitro substituent making difficult its loss as H+. So (c) should be more acidic than (b).

1. Is this reasoning correct?
2. Is o-nitrophenol more acidic than phenol despite the above reasoning (if at all the above reasoning is correct)?
3. What is the actual order of acidity.

I would be grateful if someone could help me out with this because I have two contradicting answers.

Thanks and cheers
Vivek
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chem_tr
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#2
Oct30-04, 11:24 AM
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Quote Quote by maverick280857
Rank the following compounds in order of decreasing acidity:

(a) phenol
(b) o-nitrophenol
(c) m-nitrophenol
(d) p-nitrophenol
Hello, I will turn the question upside down again Don't worry, somebody will help us

If we want a phenol-based acid to be strong, we must decrease the electronic density in the benzene ring. Thus, o- and p-nitrophenols have some "relief" because of their positions; however, m-nitrophenol is sufficiently electron deficient, so I presume that this one should be the most acidic one. Later comes phenol, as the others possess increased electron flow through the ring. The p-derivative is sterically very relieved, so this should be the least acidic one.

According to my reasoning, the final order is likely to be as follows:

[tex]m-nitrophenol > phenol > o-nitrophenol > p-nitrophenol[/tex]
GCT
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Oct30-04, 01:09 PM
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m-nitro phenol is actually lower is acidity than p or 0 nitro-phenol. Based on this, I'm sure you can finish up this question (which is more acidic, phenol or m-nitrophenol?). I apologize if I sounded condenscending.

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Oct30-04, 01:09 PM
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Acidity of substituted phenols


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chem_tr
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Oct30-04, 02:16 PM
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Well, it seems that everything goes upside down again

Therefore we decide the acidity whether delocalization meets with the functional group; an ortho- and para-directing group and a deactivating one at the meta-position isn't good. This is because of the fact that with delocalization, the blocking force to the acidity (electron pair) cannot be "wasted" in another group in the proper position. In meta-isomer, phenolate oxygen cannot be used, so it cannot leave the proximity of OH group, hence the acidity decreases.

However, if we place a nitro group at para-position of hydroxyl, delocalization can be done effectively and the electron pair around OH group can leave it for a considerable time, so the hydrogen can be more dissociated. The same thing can be said for ortho-substituted one, but a more pronounced steric hindrance will be present.

Well, someone has better explain this phenomena, I'm confused
slepsta
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#6
Oct30-04, 10:39 PM
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I have not read the previous posts so perhaps this was already answered. The addition of a Nitro group, regardless of position, will make the phenol a stronger acid due to its inductive effect. However, nitro groups also add a resonance effect. It is impossible to show the reasonance effect through here but any organic text should explain this. Bascially the ortho compound will give the strongest acid since its inductive effect is closest to the acidic phenol. What is slightly counter intuitive is that the para compound is more acidic than the meta compound. This is due to the resonance effect. The acid strenght goes liek this:

Ortho > Para > Meta > non-substituted Phenol
movies
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Oct31-04, 12:03 PM
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As Slepsta said, I would expect the o-nitro compound to be the most acidic. However, the H-bonding Maverick mentioned may have an effect as well. There might also be a small steric contribution: if the nitro group cannot be co-planar with the aromatic ring then the resonance withdrawing effect will be reduced because the orbitals cannot overlap as effectively.

I found a pKa table that lists all three nitrophenols and gives pKa values of 7.14 (para), 7.23 (ortho), and 8.35 (meta). The pKa of phenol itself is 9.95. All these values have water as solvent and reference.

Here is a link to the page that has the table:
http://www.chem.wisc.edu/areas/organic/index-chem.htm
slepsta
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#8
Oct31-04, 12:52 PM
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That is interesting because the pKa values of nitro benzoic acids are as follows: Ortho 2.2, meta 3.5, and para 3.4. You might expect the ortho group to experience the same steric hinderance as in the phenol but the resonance contribution more than makes up for it.
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#9
Oct31-04, 01:38 PM
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The benzoic acid case is a little different though because you can't draw a resonance structure to delocalize the negative charge into the nitro group. I think that the effect is more inductive than resonance in that case. There must be some contribution of resonance effects or else you would expect the para to have a higher pKa than the meta. Surprising that ortho is so much lower than the other two though.
maverick280857
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#10
Nov1-04, 02:59 AM
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Hello everyone

Thanks for your help. As far as the benzoic acids are concerned, I would like to believe that the ortho offect is operative in the ortho substitution product so it increases the acidity over phenol and the other two isomers. Let me get this right: are you trying to tell me that the effect of intramolecular hydrogen bonding is little in comparison with the other effects that have been mentioned?

Certainly, the pKas given by movies indicate the para derivative to be most acidic followed by ortho and then meta. So there is no question about the acidities. Now I am thinking this: the hydrogen of the hydroxy group can (and theoretically will as I pointed out in my first post) participate in chelation with the oxygen of the nitro group. This weak bond needs to be overcome for the hydrogen to be appreciably acidic in my opinion. So now if the ortho isomer is known to be more acidic, this intramolecular hydrogen bonding effect is negligible isn't it? For otherwise, it would hinder the loss of the acidic protion making the ortho isomer less acidic than para (I understand why the para derivative is more acidic than meta).

Additionally, all groups which withdraw electrons by resonance or donate electrons by resonance will withdraw electrons inductively (except the ones that have a negative charge on a key atom which would enable them to give electrons inductively as well as by resonance). So inductive effect is a common factor at every position in the phenol ring. Why should we consider it? Small variations in the inductive effect shouldn't after all be affected by more pronounced effects due to resonance.

Thanks and cheers
vivek
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#11
Nov1-04, 12:31 PM
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The hydrogen bonding may still be operative in the o-nitrobenzoic acid case, but the chelate requires a 7-membered ring, which isn't as favorable as the o-nitrophenol case, which has a 6-membered ring chelate. Hydrogen bonds are worth somewhere around 3 kcal/mol and the actual bond is probably worth a little less in this case because the oxygens of a nitro group aren't very basic (relatively). 3 kcal/mol is a significant amount, but in aqueous solution (the pKa's were reported in water) these hydrogen bonds are most likely broken up by water molecules.

Some experiments have suggested that inductive effects are about as powerful as resonance effects. The difficulty is separating the two effects, as there aren't many groups that are purely inductive or purely resonace withdrawing/donating (CF3 and R3+ are generally considered purely inductive, however). The generality about inductive effects is that the decrease over distance. For example, an ortho inductive withdrawing group will have a greater effect than a para inductive withdrawing group (assuming no resonance contribution). Two examples: 1) a phenol substituted with Me3N+ has pKa's of 7.42 (ortho), ~8 (meta), and ~8 (para); 2) a bromophenol has pKa's of 8.42 (ortho), 9.11 (meta), and 9.34 (para). [Same source as above]

I hope this addresses some of your questions, Maverick.


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