Ch. 14 - Ethers, Epoxides, and Thioethers

Wade9th EditionOrganic ChemistryISBN: 9780135213728Not the one you use?Change textbook

Wade 9th Edition

Ch. 14 - Ethers, Epoxides, and Thioethers

Problem 17

Chapter 14, Problem 17

Show how you would synthesize butyl isopropyl sulfide using butan-1-ol, propan-2-ol, and any solvents and reagents you need.

Verified step by step guidance

Step 1: Convert butan-1-ol to butyl bromide. This can be achieved by reacting butan-1-ol with phosphorus tribromide (PBr₃) or hydrobromic acid (HBr) to replace the hydroxyl group (-OH) with a bromine atom, forming butyl bromide (C₄H₉Br).

Step 2: Convert propan-2-ol to isopropyl thiol. React propan-2-ol with phosphorus pentasulfide (P₂S₅) or Lawesson's reagent to replace the hydroxyl group (-OH) with a thiol group (-SH), forming isopropyl thiol (C₃H₇SH).

Step 3: Perform a nucleophilic substitution reaction. React butyl bromide (C₄H₉Br) with isopropyl thiol (C₃H₇SH) in the presence of a strong base such as sodium hydroxide (NaOH) or potassium hydroxide (KOH). The thiol group (-SH) will act as a nucleophile, displacing the bromine atom and forming butyl isopropyl sulfide (C₄H₉-S-C₃H₇).

Step 4: Purify the product. Use techniques such as distillation or recrystallization to isolate and purify the butyl isopropyl sulfide from the reaction mixture.

Step 5: Confirm the structure of the product. Use spectroscopic methods such as NMR (nuclear magnetic resonance) or IR (infrared spectroscopy) to verify the formation of butyl isopropyl sulfide and ensure the reaction was successful.

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Key Concepts

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

Nucleophilic Substitution

Nucleophilic substitution is a fundamental reaction in organic chemistry where a nucleophile attacks an electrophile, resulting in the replacement of a leaving group. In the synthesis of butyl isopropyl sulfide, the nucleophilic attack of the alcohols on a suitable electrophile (like a sulfonyl chloride) is crucial for forming the desired sulfide bond.

Alcohols as Reactants

Butan-1-ol and propan-2-ol are alcohols that can act as nucleophiles in organic reactions. Their hydroxyl (-OH) groups can be converted into better leaving groups, such as tosylates or mesylates, facilitating the nucleophilic substitution process. Understanding how to activate these alcohols is essential for the successful synthesis of butyl isopropyl sulfide.

Solvent Effects

The choice of solvent can significantly influence the rate and outcome of organic reactions. Polar protic solvents can stabilize ions and facilitate nucleophilic attacks, while polar aprotic solvents can enhance nucleophilicity. Selecting the appropriate solvent for the reaction involving butan-1-ol and propan-2-ol is vital for optimizing the synthesis of butyl isopropyl sulfide.

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Related Practice

Textbook Question

Show how you would accomplish the following transformations. Some of these examples require more than one step.

(a) 2-methylpropene → 2,2-dimethyloxirane

(b) 1-phenylethanol → 2-phenyloxirane

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Textbook Question

Predict the products of the following reactions. An excess of acid is available in each case.

(d)

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Textbook Question

Boron tribromide (BBr3) cleaves ethers to give alkyl halides and alcohols.

The reaction is thought to involve attack by a bromide ion on the Lewis acid–base adduct of the ether with BBr₃ (a strong Lewis acid). Propose a mechanism for the reaction of butyl methyl ether with BBr₃ to give (after hydrolysis) butan-1-ol and bromomethane.

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Textbook Question

Predict the products of the following reactions. An excess of acid is available in each case.

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Textbook Question

Show how you would use a protecting group to convert 4-bromobutan-1-ol to hept-5-yn-1-ol.

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Textbook Question

Mustard gas, Cl–CH₂CH₂–S–CH₂CH₂–Cl, was used as a poisonous chemical agent in World War I. Mustard gas is much more toxic than a typical primary alkyl chloride. Its toxicity stems from its ability to alkylate amino groups on important metabolic enzymes, rendering the enzymes inactive.

a. Propose a mechanism to explain why mustard gas is an exceptionally potent alkylating agent.

b. Bleach (sodium hypochlorite, NaOCl, a strong oxidizing agent) neutralizes and inactivates mustard gas. Bleach is also effective on organic stains because it oxidizes colored compounds to colorless compounds. Propose products that might be formed by the reaction of mustard gas with bleach.

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