As we will learn in Section 15.12, many antioxidants—compounds that prevent unwanted radical oxidation reactions from occurring—are phenols, compounds that contain an OH group bonded directly to a benzene ring.

  1. Explain why homolysis of the O-H bond in phenol requires considerably less energy than homolysis of the O-H bond in ethanol (362 kJ/mol vs 438 kJ/mol).
  2. Why is the C-O bond in phenol shorter than C-O bond in ethanol?

Short Answer

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Answer

a. Homolysis of the O-H bond in phenol requires considerably less energy than homolysis of the O-H bond in ethanol as the phenoxide radical is resonance stabilized.

b. The C-O bond in phenol is shorter than the C-O bond in ethanol as the carbon atom involved in bond formation with oxygen in phenol is sp2 hybridized while the one in ethanol is sp3hybridized.

Step by step solution

01

Step-by-Step SolutionStep 1: Homolysis

The cleavage of a bond in which both the atoms involved in bonding retain their shared electron is named homolysis.

02

Comparing the O-H bond dissociation energy of phenol and ethanol

The homolysis of the oxygen-hydrogen (O-H) bond in phenol requires less energy than the homolysis of the oxygen-hydrogen (O-H) bond in ethanol as the phenoxide radical is resonance stabilized.

Generation of phenoxide radical

The resonance structures of phenoxide radical are provided below.

Resonance structures of phenoxide radical

03

Comparison of carbon-oxygen (C-O) bond length

The carbon-oxygen (C-O) bond in phenol is shorter than the carbon-oxygen (C-O) bond in ethanol, as, in the case of phenol, the carbon atom involved in bonding with oxygen is sp2hybridized.

While in the case of ethanol, the carbon atom is sp3hybridized.

As the percent of s character of the atoms involved in bonding increases, the bond length decreases.

Carbon-oxygen bond in the given compounds

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Most popular questions from this chapter

PGF2α (Section 4.15) is synthesized in cells using a cyclooxygenase enzyme that catalyzesa multistep radical pathway. Two steps in the pathway are depicted in the accompanying equations.

(a) Draw in curved arrows to illustrate how C is converted to D in Step [1].

(b) Identify Y, the product of Step [2], using the curved arrows that are drawn on compound D.

The equilibrium constant for the conversion of the axial to the equatorial conformation of methoxycyclohexane is 2.7.


  1. Given these data, which conformation is present in the larger amount at equilibrium?
  2. Is ΔG° for this process positive or negative?
  3. From the values in Table 6.3, approximate the size of ΔG°.

Esterification is the reaction of a carboxylic acid (RCOOH) with an alcohol (R'OH) to form an ester (RCOOR') with a loss of water. Equation [1] is an example of an intermolecular esterification reaction. Equation [2] is an example of an intramolecular esterification reaction; that is, the carboxylic acid and alcohol are contained in the same starting material, forming a cyclic ester as the product. The equilibrium constants for both reactions are given. Explain why Keqis different for these two apparently similar reactions.

[1]

[2]

As we learned in Chapter 4, propane (CH3CH2CH3)has both 1°and 2°hydrogens.

  1. Draw the carbon radical formed by homolysis of each type of C-H bond.
  2. Use the values in Table 6.2 to determine which C-H bond is stronger.
  3. Explain how this information can be used to determine the relative stability of the two radicals formed. Which radical formed from propane is more stable?

Given each of the following values, is the starting material or product favored at equilibrium?

a.Keq=5.5

b.ΔG°=40kJ/mol


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