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Chapter 23: Problem 585

(a) Why is phenol \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{OH}\) a stronger acid than an alcohol? (b) Would para-acetylphenol be a stronger or weaker acid than phenol? (c) Would 2, 4 -dinitrophenol be a stronger or weaker acid than \(p\) -nitrophenol? Draw contributing structures to support the answers.

Short Answer

Expert verified
(a) Phenol is a stronger acid than an alcohol because its conjugate base, the phenoxide ion, is resonance stabilized, making it more stable than the alkoxide ion (alcohol's conjugate base). (b) Para-acetylphenol is a stronger acid than phenol as its conjugate base is more stable due to the electron-withdrawing properties of the methyl carbonyl group. (c) 2,4-dinitrophenol is a stronger acid than p-nitrophenol because its conjugate base has more resonance-contributing structures due to the presence of two nitro groups, making it more stable.

Step by step solution

01

Determine the conjugate bases of phenol and alcohol

When phenol loses a proton (H+), it forms the phenoxide ion, and when an alcohol loses a proton, it forms an alkoxide ion. Phenol: \(\mathrm{C}_{6}\mathrm{H}_{5}\mathrm{OH} \rightarrow \mathrm{C}_{6}\mathrm{H}_{5}\mathrm{O}^{-} + \mathrm{H}^{+}\) (phenoxide ion) Alcohol: \(\mathrm{R}\mathrm{OH} \rightarrow \mathrm{R}\mathrm{O}^{-} + \mathrm{H}^{+}\) (alkoxide ion)
02

Compare the stability of the conjugate bases

The phenoxide ion is resonance stabilized, meaning the negative charge is delocalized over the aromatic ring, which increases its stability. The alkoxide ion doesn't have any resonance stabilization, so it's less stable than the phenoxide ion. A more stable conjugate base means a stronger acid, so phenol is a stronger acid than an alcohol. #b) Comparing acidity of phenol and para-acetylphenol#
03

Determine the conjugate base of para-acetylphenol

When para-acetylphenol loses a proton (H+), it forms its conjugate base: Para-acetylphenol: \(\mathrm{C}_{6}\mathrm{H}_{4}\mathrm{O}(\mathrm{C}\mathrm{H}_{3})\mathrm{OH} \rightarrow \mathrm{C}_{6}\mathrm{H}_{4}\mathrm{O}(\mathrm{C}\mathrm{H}_{3})\mathrm{O}^{-} + \mathrm{H}^{+}\) (para-acetylphenoxide ion)
04

Compare the stability of the conjugate bases

In the para-acetylphenoxide ion, the methyl carbonyl group introduces electron-withdrawing properties, which can help delocalize the negative charge in the ion via resonance. This makes the para-acetylphenoxide ion more stable than the phenoxide ion. A more stable conjugate base means a stronger acid, so para-acetylphenol is a stronger acid than phenol. #c) Comparing acidity of 2, 4-dinitrophenol and p-nitrophenol#
05

Determine the conjugate bases of 2, 4-dinitrophenol and p-nitrophenol

When 2, 4-dinitrophenol loses a proton (H+), it forms its conjugate base, and when p-nitrophenol loses a proton, it forms its conjugate base: 2,4-dinitrophenol: \(\mathrm{C}_{6}\mathrm{H}_{3}(\mathrm{NO}_{2})_{2}\mathrm{OH} \rightarrow \mathrm{C}_{6}\mathrm{H}_{3}(\mathrm{NO}_{2})_{2}\mathrm{O}^{-} + \mathrm{H}^{+}\) (2, 4-dinitrophenoxide ion) p-nitrophenol: \(\mathrm{C}_{6}\mathrm{H}_{4}\mathrm{NO}_{2}\mathrm{OH} \rightarrow \mathrm{C}_{6}\mathrm{H}_{4}\mathrm{NO}_{2}\mathrm{O}^{-} + \mathrm{H}^{+}\) (p-nitrophenoxide ion)
06

Draw resonance contributors

Draw resonance contributors for 2, 4-dinitrophenoxide ion and p-nitrophenoxide ion. For 2,4-dinitrophenoxide ion, we can see that there are more resonance-contributing structures due to the two nitro groups present in the molecule as compared to p-nitrophenoxide ion, which has only one nitro group.
07

Compare the stability of the conjugate bases

The higher number of resonance-contributing structures in the 2, 4-dinitrophenoxide ion (compared to the p-nitrophenoxide ion) indicates that it's more stable. A more stable conjugate base means a stronger acid, so 2, 4-dinitrophenol is a stronger acid than p-nitrophenol.

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Most popular questions from this chapter

Outline a synthesis of cumene from cheap, readily available hydrocarbons. Then synthesize phenol from cumene.

How much difference in physical properties would you expect for o-and \(p\) -cyanophenol isomers? Explain.

A student attempted the following synthesis of - methoxy- benzyl alcohol from o-cresol, but got essentially no yield. What went wrong?

Tropolone \(\left(\mathrm{I}, \mathrm{C}_{7} \mathrm{H}_{7} \mathrm{O}_{2}\right)\) has a flat molecule with all carbon-carbon bonds of the same length (1.40 A). The measured heat of combustion is \(20 \mathrm{kcal}\) lower than that calculated by the method of summing bond energies. Its dipole moment is \(3.71 \mathrm{D}\); that of 5-bromotropolone is \(2.07 \mathrm{D}\). Tropolone undergoes the Reimer-Tiemann reaction, couples with diazonium ions, and is nitrated by dilute nitric acid. It gives a green color with ferric chloride, and does not react with 2,4 -dinitrophenylhydrazine. Tropolone is both acidic \(\left(\mathrm{K}_{\mathrm{a}}=10^{-7}\right)\) and weakly basic, forming a hydrochloride in ether. (a) What class of compounds does tropolone resemble? Is it adequately represented by formula I? (b) Using both valence-bond and orbital structures, account for the properties of tropolone. (c) In what direction is the dipole moment of tropolone? Is this consistent with the structure you have proposed? (d) The infrared spectrum of tropolone shows a broad band at about \(3150 \mathrm{~cm}^{-1}\) that changes only slightly upon dilution. What does this tell you about the structure of tropolone?

How do you account for the fact that, unlike most phenols, \(2,4-\) dinitrophenol and \(2,4,6\) -trinitrophenol are soluble in aqueous sodium bicarbonate?
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