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Chapter 7: Problem 39

Propose a synthesis of each of the following compounds starting with 2-butanol and any other reagents: (a) 2-chlorobutane (b) 2 -methoxybutane (c) 2-butanamine (d) 2 -butanethiol

Short Answer

Expert verified
To obtain 2-chlorobutane, use HCl; for 2-methoxybutane, use methanol and H2SO4; for 2-butanamine, first oxidize to 2-butanone and then use NH3 and H2; for 2-butanethiol, convert to 2-bromobutane with HBr and then use thiourea.

Step by step solution

01

Synthesis of 2-Chlorobutane

To synthesize 2-chlorobutane from 2-butanol, use hydrochloric acid (HCl). The reaction proceeds via an S_N2 mechanism where the hydroxyl group of 2-butanol is replaced by a chlorine atom.
02

Reaction Equation

The chemical reaction is as follows: 2-Butanol + HCl → 2-Chlorobutane + H2O
03

Synthesis of 2-Methoxybutane

To synthesize 2-methoxybutane from 2-butanol, you need to use methanol and an acid catalyst (e.g., H_2SO_4). This is an example of a Williamson ether synthesis where the alcohol reacts with the methanol to form the corresponding ether.
04

Reaction Equation

The chemical reaction is as follows: 2-Butanol + MeOH → 2-Methoxybutane + H2O
05

Synthesis of 2-Butanamine

For the synthesis of 2-butanamine from 2-butanol, first convert 2-butanol into 2-butanone via oxidation (using PCC or another oxidizing agent). Then, apply reductive amination using NH_3 (ammonia) and H_2 (hydrogen) in the presence of a catalyst such as Raney Nickel.
06

Reaction Equation

The chemical reactions are as follows: 2-Butanol → 2-Butanone 2-Butanone + NH_3 + H_2 → 2-Butanamine
07

Synthesis of 2-Butanethiol

To synthesize 2-butanethiol from 2-butanol, convert 2-butanol to 2-bromobutane using HBr. Then, react 2-bromobutane with thiourea, followed by hydrolysis, which yields 2-butanethiol.
08

Reaction Equation

The chemical reactions are as follows: 2-Butanol + HBr → 2-Bromobutane + H2O 2-Bromobutane + NH2CSNH2 → 2-Butanethiol

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Key Concepts

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

S_N2 Mechanism
The S_N2 mechanism is a bimolecular nucleophilic substitution reaction. Here, 'S_N2' stands for Substitution Nucleophilic Bimolecular, indicating that two reactants are involved in the rate-determining step. This mechanism is particularly useful for synthesizing alkyl halides from alcohols. For instance, when synthesizing 2-chlorobutane from 2-butanol, an S_N2 mechanism is followed. The nucleophile (in this case, Cl^-) attacks the electrophilic carbon from the opposite side, leading to the inversion of configuration. Consequently, 2-butanol undergoes substitution, replacing the hydroxyl group with a chlorine atom to form 2-chlorobutane.
Williamson Ether Synthesis
The Williamson ether synthesis is a straightforward method to create ethers from alcohols and alkyl halides (or tosylates). This reaction involves the deprotonation of an alcohol to form an alkoxide ion, which then acts as a nucleophile. The alkoxide ion attacks an electrophilic carbon in an alkyl halide, leading to the formation of an ether. For example, synthesizing 2-methoxybutane from 2-butanol involves using methanol and an acid catalyst, typically sulfuric acid (H_2SO_4). The methanol is first protonated, and the 2-butanol alkoxide ion attacks the protonated methanol, resulting in the formation of 2-methoxybutane along with water as a byproduct.
Reductive Amination
Reductive amination is a method to convert carbonyl compounds into amines. The process involves the formation of an imine intermediate, which is then reduced to form the corresponding amine. For synthesizing 2-butanamine from 2-butanol, the alcohol is initially oxidized to 2-butanone using an oxidizing agent like PCC (Pyridinium chlorochromate). Subsequently, the 2-butanone undergoes reductive amination. In this step, ammonia (NH_3) reacts with the carbonyl group to form an imine. This imine is then reduced using hydrogen gas (H_2) and a catalyst, such as Raney Nickel, resulting in the formation of 2-butanamine.
Thiourea Hydrolysis
Thiourea hydrolysis is a process used to convert alkyl halides into thiols. The reaction involves the interaction between an alkyl halide and thiourea, forming an intermediate isothiourea compound. Hydrolysis of this intermediate yields the desired thiol. For instance, to synthesize 2-butanethiol from 2-butanol, the alcohol is first converted into 2-bromobutane using hydrobromic acid (HBr). The resulting 2-bromobutane then reacts with thiourea to form an isothiourea intermediate. Finally, hydrolyzing this intermediate gives 2-butanethiol as the final product. This multi-step process efficiently introduces a thiol group into the molecule.

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Most popular questions from this chapter

Write a mechanism for the formation of 1-bromopropane from the reaction of 1-propanol and \(\mathrm{PBr}_{3}\).

You are given a supply of ethyl iodide, tertbutyl iodide, sodium ethoxide, and sodium tert-butoxide. Your task is to use the \(\mathrm{S}_{\mathrm{N}} 2\) reaction to make as many different ethers as you can. In principle, how many are possible? In practice, how many can you make?

You may be familiar with the compound tert-butyl methyl ether. Its common acronym is MTBE. It has been an additive for gasoline since 1979 , although its use has declined in the United States since 2003 . How would you make MTBE if you have bottles of tert-butyl alcohol and methyl alcohol? You may use any other inorganic reagents needed. Your proposed route should avoid mixtures containing undesired products.

Wait! The argument just presented ignores angle strain in the starting material. Cyclopropane itself is strained. Won't that strain raise the energy of the starting material and offset the energy rise of the transition state? Comment. Hint: Consider angle strain in both starting material and transition state-in which will it be more important?

The \(\mathrm{S}_{\mathrm{N}} 2\) reaction that occurs between an alkyl halide and a nucleophile is a very important process in organic chemistry. For this reason, it is necessary to recognize the nucleophilicity of various reagents. In the following pairs of compounds, indicate the more nucleophilic reagent. Explain your reasoning. (a) \(\left(\mathrm{CH}_{3} \mathrm{CH}_{2}\right)_{3} \mathrm{~N} \quad\left(\mathrm{CH}_{3} \mathrm{CH}_{2}\right)_{3} \mathrm{P}\) (b) \(\left(\mathrm{CH}_{3} \mathrm{CH}_{2}\right)_{3} \mathrm{~N} \quad\left(\mathrm{CH}_{3} \mathrm{CH}_{2}\right)_{2} \mathrm{~N}^{-}\) (c) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{SNa} \quad \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OK}\) There is much detail in this chapter, but that is not usually a problem for most students. Good nucleophiles can be distinguished from good bases, good leaving groups can be identified, and so on. What is hard is to learn to adjust reaction conditions (reagents, temperature, and solvent) so as to achieve the desired selectivity to favor the product of just one of these four reactions. That's hard in practice as well as in answering a question on an exam or problem set. To a certain extent, we all have to learn to live with this problem!
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