Phenol reacts with three equivalents of bromine in CCl 4 (in the dark) to give a product of formula C 6 H 3 OBr 3 . When this product is added to bromine water, a yellow solid of molecular formula C 6 H 2 OBr 4 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.

All right. Hi everyone. So for this question it says that when three equivalents of grooming are added 235 dimethyl fino in CCL four, A product with the molecular formula C. Eight H seven B. R 30 is obtained. This product is further treated with running water which results in another product with the molecular formula C. Eight H six BR four oh. The Ir spectrum of the second product shows a strong absorption of around 85 in verse, centimeters similar to that of Quinault determine the structures of the two products. So let's get started right? And let's first get started by drawing out our starting material which is going to be our 35 dimethyl phenol. So I'm going to draw a benzene ring with a final group, in other words an O. H. And two methyl groups on position three. And then on position five. So we're treating it, We're treating our 35 dimethyl fino with um browning in CCL four. And we get a product with a formula of C. Eight H seven B. R three oh, so that be our three sticks out to me. Right? So that must mean that when we react elemental roaming with CCL four then that is going to be a hallucination of our ring here in three positions. In other words, in the three free positions that we have within our starting material. So if I go ahead and I redraw our starting material like. So then we are going to have three bro means in the empty spaces that we have, so to speak. So these are our bro means and this is going to be our C. eight H 7 B. R. three. Oh so now we are further reacting this tri substituted um starting material with roaming water to generate a product with C eight H six BR 40 as our molecular formula. So now because we are adding 1/4 grooming, that must mean that our aromatics city within this fennel has been lost, right? Because where else would we be able to add our roaming? So if I just react this with B. R. Two once again, then like I said before, we're going to have um 1/4 browning added to this molecule which is going to it's gonna end our aromatic city. Pretty much so now, what that must mean since also. Right. What's interesting about the IR spectrum of the second product is that we have a strong absorption at around 16 85 inverse centimeters. Now recall for a second that a C double bond oh, stretch typically occurs at around inverse centimeters. So 1685 is pretty close to 1700. Right, and recall once again that a lower wave number or other words a lower number, inverse centimeters. Then you would expect then the 1700 that we would expect indicates that we have a conjugated system. So that would be um why we have our C double bond O absorption at a smaller number than we expect. So if we have oxidation of our final O. H, then that means we're going to have a car bottle here, and if it's conjugated, we must have our pi bonds, exactly one sigma bond away from our carbonnel. So that must mean that our browning added to the para position and everything else is still intact. So let me just draw that. And this is going to be your final answer. This is going to be your final answer, and thank you very much for watching. I hope you found this helpful.

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19. Reactions of Aromatics: EAS and Beyond

EAS:Halogenation Mechanism

Organic Chemistry

19. Reactions of Aromatics: EAS and Beyond

EAS:Halogenation Mechanism

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Problem 71

Textbook Question

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.

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.

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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