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Aspirin is acetylsalicylic acid (o-acetoxybenzoic acid, \(\left.0-\mathrm{CH}_{3} \mathrm{COO}-\mathrm{C}_{6} \mathrm{H}_{4} \mathrm{COOH}\right)\); oil of wintergreen is the ester, methyl salicylate. Outline the synthesis of these two compounds from phenol.

Short Answer

Expert verified
To synthesize acetylsalicylic acid (aspirin) and methyl salicylate (oil of wintergreen) from phenol, follow these steps: 1. Convert phenol to salicylic acid via the Kolbe-Schmitt reaction with NaOH and CO2, followed by acidification with HCl. 2. Synthesize aspirin by reacting salicylic acid with acetic anhydride in the presence of a strong acid catalyst. $$\text{C}_{6}\text{H}_{4}(\text{OH})\text{COOH} + (\text{CH}_{3}\text{CO})_{2}\text{O} \rightarrow \text{CH}_{3}\text{COOC}_{6}\text{H}_{4}\text{COOH} + \text{CH}_{3}\text{COOH}$$ 3. Synthesize oil of wintergreen by reacting salicylic acid with methanol in the presence of a strong acid catalyst. $$\text{C}_{6}\text{H}_{4}(\text{OH})\text{COOH} + \text{CH}_{3}\text{OH} \rightarrow \text{CH}_{3}\text{OC}_{6}\text{H}_{4}\text{COOH} + \text{H}_{2}\text{O}$$

Step by step solution

01

Convert phenol to salicylic acid

First, we need to convert phenol into salicylic acid. This can be done by the Kolbe-Schmitt reaction, in which phenol reacts with sodium hydroxide (NaOH) and carbon dioxide (CO2) to form sodium salicylate, followed by acidification to give salicylic acid. Reactions: 1. Phenol + NaOH -> Sodium phenoxide (NaOC6H5) 2. Sodium phenoxide + CO2 -> Sodium salicylate (NaOC6H4COO-) 3. Sodium salicylate + HCl -> Salicylic acid (C6H4(OH)COOH) + NaCl
02

Synthesis of Acetylsalicylic Acid (Aspirin)

To synthesize aspirin, we need to convert the salicylic acid obtained in step 1 into its acetyl derivative, acetylsalicylic acid. This can be done via an esterification reaction with acetic anhydride in the presence of or using a strong acid catalyst like sulfuric acid. Reaction: Salicylic acid (C6H4(OH)COOH) + Acetic anhydride ((CH3CO)2O) -> Acetylsalicylic acid (Aspirin) (CH3COOC6H4COOH) + Acetic acid (CH3COOH)
03

Synthesis of Methyl Salicylate (Oil of Wintergreen)

To synthesize methyl salicylate, we need to convert the salicylic acid obtained in step 1 into its methyl ester. This can be done by an esterification reaction with methanol in the presence of a strong acid catalyst like sulfuric acid or hydrochloric acid. Reaction: Salicylic acid (C6H4(OH)COOH) + Methanol (CH3OH) -> Methyl salicylate (Oil of Wintergreen) ( CH3OC6H4COOH) + Water (H2O) These steps outline the synthesis of both acetylsalicylic acid (aspirin) and methyl salicylate (oil of wintergreen) from phenol.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Acetylsalicylic Acid
Acetylsalicylic acid, commonly known as aspirin, is a widely used medication that alleviates pain and reduces inflammation. It has the chemical formula \( ext{CH}_3 ext{COOC}_6 ext{H}_4 ext{COOH}\). This compound is synthesized through a process known as esterification. This involves salicylic acid reacting with acetic anhydride in the presence of a catalyst.

Here's a closer look at how this happens:
  • Salicylic acid acts as the substrate, while acetic anhydride provides the acetyl group that forms the ester linkage in aspirin.
  • A strong acid catalyst, such as sulfuric acid, is used to enhance the reaction rate.
  • The reaction produces acetylsalicylic acid and acetic acid as a byproduct.
Understanding this esterification step is crucial, as it connects the pain-relieving salicylic acid with its acetyl group to form aspirin.
Methyl Salicylate
Methyl salicylate, also known as oil of wintergreen, is another ester derived from salicylic acid. It possesses a distinctive minty aroma and is often used in muscle rubs for its soothing properties. The chemical formula is \( ext{CH}_3 ext{OC}_6 ext{H}_4 ext{COOH}\).

To produce methyl salicylate:
  • Salicylic acid undergoes esterification with methanol.
  • This reaction is often acid-catalyzed, using sulfuric or hydrochloric acid to speed up the process.
  • The product formed is methyl salicylate along with water.
This synthesis highlights the versatility of salicylic acid, showcasing how its conversion into different esters results in products with varying applications like in pain relief or as flavor and fragrance agents.
Kolbe-Schmitt Reaction
The Kolbe-Schmitt reaction is an important step in synthesizing salicylic acid from phenol. This reaction capitalizes on the interaction of phenol with carbon dioxide under basic conditions to introduce a carboxylate group onto the phenol molecule.

Here's how the Kolbe-Schmitt reaction unfolds:
  • Phenol reacts with sodium hydroxide to form sodium phenoxide, preparing it for carboxylation.
  • The sodium phenoxide then encounters carbon dioxide, forming sodium salicylate.
  • An acidification step, typically using hydrochloric acid, converts sodium salicylate into salicylic acid.
Understanding the Kolbe-Schmitt reaction is essential because it lays the foundational chemistry needed for further conversion of salicylic acid into valuable derivatives like acetylsalicylic acid and methyl salicylate. It demonstrates a clever use of carbon dioxide as a reactant to transform simple phenol into a key aromatic compound.

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Most popular questions from this chapter

Give structures of all compounds below: (a) \(p\) -nitrophenol \(+\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Br}+\mathrm{NaOH}(\mathrm{aq}) \rightarrow \mathrm{A}\left(\mathrm{C}_{8} \mathrm{H}_{9} \mathrm{O}_{3} \mathrm{~N}\right)\) \(\mathrm{A}+\mathrm{Sn}+\mathrm{HCL} \rightarrow \mathrm{B}\left(\mathrm{C}_{8} \mathrm{~B}_{11} \mathrm{ON}\right)\) \(\mathrm{B}+\mathrm{NaNO}_{2}+\mathrm{HCL}\), then phenol \(\rightarrow \mathrm{C}\left(\mathrm{C}_{14} \mathrm{H}_{14} \mathrm{O}_{2} \mathrm{~N}_{2}\right)\) \(\mathrm{C}+\) ethyl sulfate \(+\mathrm{NaOH}\) (aq) \(\rightarrow \mathrm{D}\left(\mathrm{C}_{16} \mathrm{H}_{18} \mathrm{O}_{2} \mathrm{~N}_{2}\right)\) \(\mathrm{D}+\mathrm{SnCL}_{2} \rightarrow \mathrm{E}\left(\mathrm{C}_{8} \mathrm{H}_{11} \mathrm{ON}\right)\) \(E+\) acetyl chloride phenacetin \(\left(\mathrm{C}_{10} \mathrm{H}_{13} 2 \mathrm{~N}\right)\), an analgesic ("pain-killer") and antypyretic ("fever- killer") (b) \(\beta-(-0-\) hy droxyphenyl \()\) ethyl alcohol \(+\mathrm{HBr} \rightarrow \mathrm{F}\left(\mathrm{C}_{8} \mathrm{H}_{9} \mathrm{OBr}\right)\) \(\mathrm{F}+\mathrm{KOH} \rightarrow\) coumarane \(\left(\mathrm{C}_{8} \mathrm{H}_{8} \mathrm{O}\right)\), insoluble in \(\mathrm{NaOH}\).

Reduction of the carbony1 group of quercitin gives a substance that loses water readily to give a brilliant-violet compound, which, when treated with hydrochloric acid, is converted reversibly to a red salt. Consider possible ways that the reduction product might dehydrate to give a violet substance and show how addition of a proton to this substance could occur in such a way as to give a substantial color change:

Give for each of the following pairs of compounds a chemical test, preferably a test-tube reaction, that will distinguish between the two compounds. Write a structural formula for each compound and equations for the reactions involved. (a) phenol and cyclohexanol. (b) methy1 \(p\) -hydroxybenzoate and \(p\) -methoxybenzoic acid. (c) 9,10 -anthraquinone and 1,4 -anthraquinone.

Outline all steps in the synthesis from toluene of: (a) p-cresol via diazotizatiori; (b) p-cresol via thallation; (c) m-cresol via diazotizatiori; (d) m-cresol via thallation.

Write resonance structures that account for the stability of the cation of Wurster's salts. Explain why octamethyl- p-phenylenediamine does not form a similarly stable cation radical.

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