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Ch. 17 - Reactions of Aromatic Compounds
Wade - Organic Chemistry 9th Edition
Wade9th EditionOrganic ChemistryISBN: 9780135213728Not the one you use?Change textbook
All textbooksWade 9th EditionCh. 17 - Reactions of Aromatic CompoundsProblem 71
Chapter 17, Problem 71

Phenol reacts with three equivalents of bromine in CCl4 (in the dark) to give a product of formula C6H3OBr3. When this product is added to bromine water, a yellow solid of molecular formula C6H2OBr4 precipitates out of the solution. The IR spectrum of the yellow precipitate shows a strong absorption (much like that of a quinone) around 1680 cm–1. Propose structures for the two products.

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Step 1: Analyze the reaction of phenol with bromine in CCl4. Phenol is an aromatic compound with an -OH group attached to the benzene ring. The -OH group is an activating group, making the ortho and para positions more reactive toward electrophilic substitution. When phenol reacts with three equivalents of bromine in CCl4, bromine atoms will substitute at the ortho and para positions relative to the -OH group, forming 2,4,6-tribromophenol (C6H3OBr3).
Step 2: Examine the reaction of 2,4,6-tribromophenol with bromine water. Bromine water is an aqueous solution of bromine, which can act as both an electrophile and an oxidizing agent. The reaction leads to the formation of a yellow precipitate with the molecular formula C6H2OBr4. This suggests that one additional bromine atom is introduced into the molecule.
Step 3: Consider the IR spectrum of the yellow precipitate. The strong absorption around 1680 cm⁻¹ indicates the presence of a carbonyl group (C=O). This suggests that the product is a quinone derivative, as quinones typically exhibit such IR absorptions.
Step 4: Propose the structure of the yellow precipitate. The formation of a quinone derivative implies that the -OH group of 2,4,6-tribromophenol is oxidized to a carbonyl group, and an additional bromine atom is introduced at the para position relative to the carbonyl group. The resulting structure is 2,4,6,4'-tetrabromo-1,4-benzoquinone (C6H2OBr4).
Step 5: Summarize the structures of the two products. The first product is 2,4,6-tribromophenol (C6H3OBr3), formed by electrophilic substitution of bromine at the ortho and para positions of phenol. The second product is 2,4,6,4'-tetrabromo-1,4-benzoquinone (C6H2OBr4), formed by oxidation of the -OH group to a carbonyl group and bromination at the para position relative to the carbonyl group.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Electrophilic Aromatic Substitution

Electrophilic aromatic substitution (EAS) is a fundamental reaction in organic chemistry where an electrophile replaces a hydrogen atom on an aromatic ring. In the case of phenol reacting with bromine, the hydroxyl group (-OH) activates the ring, making it more susceptible to electrophilic attack. This reaction leads to the formation of brominated phenol derivatives, which is crucial for understanding the products formed in this question.
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Bromination of Phenol

Bromination of phenol involves the addition of bromine to the aromatic ring, resulting in the substitution of hydrogen atoms with bromine atoms. The presence of three equivalents of bromine indicates that multiple bromination occurs, leading to a tri-brominated product. The product C6H3OBr3 suggests that the bromine atoms are likely positioned ortho and para to the hydroxyl group, which is a directing effect of the -OH group.
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Infrared Spectroscopy and Functional Groups

Infrared (IR) spectroscopy is a technique used to identify functional groups in organic compounds based on their characteristic absorption bands. The strong absorption around 1680 cm-1 in the IR spectrum of the yellow precipitate suggests the presence of a carbonyl group, typical of quinones. This information is essential for deducing the structure of the final product, C6H2OBr4, which likely contains a quinone structure due to the oxidation of the brominated phenol.
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Related Practice
Textbook Question

In Chapter 14, we saw that Agent Orange contains (2,4,5-trichlorophenoxy) acetic acid, called 2,4,5-T. This compound is synthesized by the partial reaction of 1,2,4,5-tetrachlorobenzene with sodium hydroxide, followed by reaction with sodium chloroacetate, ClCH2CO2Na.

a. Draw the structures of these compounds, and write equations for these reactions.

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Textbook Question

Show how you would use a Suzuki reaction to synthesize the following biaryl compound. As starting materials, you may use the two indicated compounds, plus any additional reagents you need.

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Textbook Question

Unlike most other electrophilic aromatic substitutions, sulfonation is often reversible (see Section 17-4). When one sample of toluene is sulfonated at 0 °C and another sample is sulfonated at 100 °C, the following ratios of substitution products result:

a. Explain the change in the product ratios when the temperature is increased.

b. Predict what will happen when the product mixture from the reaction at 0 °C is heated to 100 °C.

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Textbook Question

Starting with benzene and any other reagents you need, show how you would synthesize the compound shown here. (Hint: Consider a Pd-catalyzed coupling for the final step.)

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Textbook Question

A common illicit synthesis of methamphetamine involves an interesting variation of the Birch reduction. A solution of ephedrine in alcohol is added to liquid ammonia, followed by several pieces of lithium metal. The Birch reduction usually reduces the aromatic ring (Section 17-14C), but in this case it eliminates the hydroxy group of ephedrine to give methamphetamine. Propose a mechanism, similar to that for the Birch reduction, to explain this unusual course of the reaction.

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Textbook Question

Unlike most other electrophilic aromatic substitutions, sulfonation is often reversible (see Section 17-4). When one sample of toluene is sulfonated at 0 °C and another sample is sulfonated at 100 °C, the following ratios of substitution products result:

c. Because the SO3H group can be added to a benzene ring and removed later, it is sometimes called a blocking group. Show how 2,6-dibromotoluene can be made from toluene using sulfonation and desulfonation as intermediate steps in the synthesis.