Explaining Acidities of Ortho, Meta, & Para Fluorophenols

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In summary, the difference in acidity of ortho, meta, and para-fluorophenols can be explained by the inductive effect of fluorine, which is a stronger factor than the resonance effects. While the ortho and para positions have more stable resonance structures, the inductive effect of fluorine in these positions makes them less acidic compared to the meta position.
  • #1
sciontc03
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Homework Statement


How do you explain the difference in acidity of the ortho, meta, and para- fluorophenols.


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The Attempt at a Solution



I know that EWG enhance the acidity of phenols because they stabilize the delocalized negative charge.
I also drew out the resonance structures and saw that ortho and para positions have 4 resonance structures, whereas meta has only 3 structures. This means ortho and para would be more stable. However, the pKa show different.
pKa of phenol = 10.0
pKa of ortho-fluorophenol = 8.81
pKa of meta-fluorophenol = 9.28
pKa of para-fluorophenol = 9.81

How do you explain this?
 
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  • #2
If the fluorine were affecting the acidity via resonance, fluorine would act as an electron donating group, rather than an electron withdrawing group. The inductive effect is more important here than the resonance effects because resonance structures where the fluorine has a + charge are very small contributors.
 
  • #3


The difference in acidity of ortho, meta, and para-fluorophenols can be explained by a combination of electronic and steric effects. The presence of a fluorine atom in the ortho, meta, and para positions introduces an electron-withdrawing group (EWG) effect, which stabilizes the delocalized negative charge on the phenoxide ion. This makes the phenol more acidic compared to phenol without any substituents.

In addition, the position of the fluorine atom also plays a role. In ortho and para-fluorophenols, the fluorine atom is in close proximity to the hydroxyl group, leading to stronger EWG effects and therefore a greater stabilization of the phenoxide ion. This results in a lower pKa value for ortho and para-fluorophenols compared to meta-fluorophenol, where the fluorine atom is further away from the hydroxyl group.

However, the steric hindrance caused by the bulky fluorine atom in the ortho position can also play a role in decreasing the acidity. This is because the bulky substituent can hinder the approach of the base to the proton, making it more difficult to deprotonate the phenol. This steric hindrance is less pronounced in meta and para-fluorophenols, resulting in a slightly higher pKa value compared to ortho-fluorophenol.

In summary, the difference in acidity of ortho, meta, and para-fluorophenols can be attributed to a combination of electronic and steric effects. The ortho and para positions have stronger EWG effects, leading to a greater stabilization of the phenoxide ion, but the ortho position also has steric hindrance, which can decrease the acidity. This results in the observed pKa values for the different isomers of fluorophenol.
 

1. What is the difference between ortho, meta, and para fluorophenols?

Ortho, meta, and para fluorophenols are structural isomers of fluorophenols, meaning they have the same chemical formula but differ in the placement of the fluorine atom on the phenol ring. Ortho fluorophenols have the fluorine atom attached to the carbon atom adjacent to the hydroxyl group, meta fluorophenols have the fluorine atom attached to the carbon atom two positions away from the hydroxyl group, and para fluorophenols have the fluorine atom attached to the carbon atom opposite the hydroxyl group.

2. What is the acidity of ortho, meta, and para fluorophenols?

The acidity of ortho, meta, and para fluorophenols increases in the order of para < meta < ortho. This is due to the electron-withdrawing effect of the fluorine atom, which stabilizes the conjugate base of the phenol and makes it easier to remove a proton from the hydroxyl group.

3. How does the position of the fluorine atom affect the acidity of fluorophenols?

The position of the fluorine atom on the phenol ring affects the acidity of fluorophenols due to the different degrees of electron delocalization in the conjugate base. In ortho fluorophenols, the fluorine atom is in close proximity to the hydroxyl group, leading to strong electron delocalization and increased acidity. In meta fluorophenols, the fluorine atom is further away from the hydroxyl group, leading to weaker electron delocalization and lower acidity. In para fluorophenols, the fluorine atom is opposite the hydroxyl group, resulting in the weakest electron delocalization and lowest acidity.

4. How do fluorophenols compare to other phenols in terms of acidity?

Fluorophenols generally have a higher acidity compared to other phenols due to the electron-withdrawing effect of the fluorine atom. This makes it easier to remove a proton from the hydroxyl group and increases the stability of the conjugate base. However, the exact acidity of a fluorophenol will also depend on the position of the fluorine atom on the phenol ring.

5. How does the acidity of fluorophenols affect their reactivity?

The higher acidity of fluorophenols makes them more reactive compared to other phenols. This is because the more acidic a compound is, the more likely it is to donate a proton and participate in reactions. The presence of the electron-withdrawing fluorine atom also makes fluorophenols more susceptible to nucleophilic attack, resulting in a wider range of potential reactions.

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