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Show how the following transformations might be achieved with the aid of azide derivatives. (a) \(\mathrm{CH}_{2}=\mathrm{CHCH}_{2} \mathrm{CH}_{2} \mathrm{OH} \rightarrow \mathrm{CH}_{2}=\mathrm{CHCH}_{2} \mathrm{CH}_{2} \mathrm{NH}_{2}\)

Short Answer

Expert verified
To convert pent-2-en-4-ol to pent-2-en-4-amine using azide derivatives, follow these steps: 1. Convert the alcohol group (OH) to an alkyl halide (Cl) using thionyl chloride (SOCl2). 2. Replace the alkyl halide (Cl) with an azide group (N3) using sodium azide (NaN3) in an SN2 reaction. 3. Reduce the azide group (N3) to an amine group (NH2) using either hydrogen gas (H2) with a metal catalyst (Pd-C or Pt) or lithium aluminum hydride (LiAlH4).

Step by step solution

01

Convert the alcohol group to an alkyl halide

To convert the alcohol group (OH) to an alkyl halide, we can use thionyl chloride (SOCl2) as a reagent. This will convert the hydroxyl group (OH) into a chloro group (Cl). Reactant: \(\mathrm{CH}_{2}=\mathrm{CHCH}_{2} \mathrm{CH}_{2} \mathrm{OH}\) Reagent: \(\mathrm{SOCl}_{2}\) Product: \(\mathrm{CH}_{2}=\mathrm{CHCH}_{2} \mathrm{CH}_{2} \mathrm{Cl}\)
02

Replace the alkyl halide with an azide group

Next, use sodium azide (NaN3) as a nucleophile to replace the chloro group (Cl) with an azide group (N3). This is an SN2 reaction. Reactant: \(\mathrm{CH}_{2}=\mathrm{CHCH}_{2} \mathrm{CH}_{2} \mathrm{Cl}\) Reagent: \(\mathrm{NaN}_{3}\) Product: \(\mathrm{CH}_{2}=\mathrm{CHCH}_{2} \mathrm{CH}_{2} \mathrm{N}_{3}\)
03

Reduce the azide group to an amine

For the final step, we will reduce the azide group (N3) to an amine group (NH2) by using either hydrogen gas (H2) with a metal catalyst (such as palladium on carbon or platinum) or using LiAlH\(_{4}\) (lithium aluminum hydride) as the reducing agent. Reactant: \(\mathrm{CH}_{2}=\mathrm{CHCH}_{2} \mathrm{CH}_{2} \mathrm{N}_{3}\) Reagent (option 1): \(\mathrm{H}_{2} / \mathrm{Pd-C}\) (or \(\mathrm{H}_{2} /\mathrm{Pt}\)) Reagent (option 2): \(\mathrm{LiAlH}_{4}\) Product: \(\mathrm{CH}_{2}=\mathrm{CHCH}_{2} \mathrm{CH}_{2} \mathrm{NH}_{2}\) Through these three steps, we have successfully converted pent-2-en-4-ol to pent-2-en-4-amine using azide derivatives.

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Key Concepts

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

Alkyl Halide
Alkyl halides, also known as haloalkanes, are organic compounds that contain one or more halogen atoms (such as chlorine, bromine, or iodine) bonded to an alkyl group. They serve as fundamental building blocks in various chemical reactions. In our transformation process, the alkyl halide plays a crucial intermediary role.

When converting an alcohol to an alkyl halide, reagents like thionyl chloride (\( \mathrm{SOCl}_{2} \)) are commonly used. This process involves replacing the hydroxyl group of the alcohol with a halogen atom, resulting in a chloroalkane.
  • This step is critical because it activates the molecule for subsequent nucleophilic substitution reactions.
  • Alkyl halides are more reactive compared to alcohols, facilitating further transformation steps.
By converting the alcohol into an alkyl halide, the molecule becomes primed for the nucleophilic attack needed in the upcoming SN2 reaction.
SN2 Reaction
The SN2 (Substitution Nucleophilic Bimolecular) reaction is a type of nucleophilic substitution where the nucleophile attacks the electrophile directly, resulting in a one-step reaction mechanism. It is characterized by its simultaneous bond making and bond breaking process, leading to an inversion of stereochemistry at the carbon center.

In this transformation, we used sodium azide (\( \mathrm{NaN_3} \)) to perform an SN2 reaction on the alkyl halide. This forms an azide compound, preparing it for the final conversion to an amine.
  • The SN2 reaction requires a strong nucleophile and a suitable electrophile, in this case, a primary alkyl halide.
  • Since the reaction proceeds with inversion, it’s important to start with a configuration that leads to the desired final product.
  • The reaction occurs in one concerted step, making it highly efficient for transformations like the one highlighted here.
This reaction is ideal for synthesizing compounds where controlling the stereochemical outcome is crucial.
Amine Synthesis
Amine synthesis in the context of our transformation involves converting the azide compound produced from the SN2 reaction into an amine. Amines are organic compounds that contain nitrogen and are widely used in a vast array of chemical processes. The transformation from azide to amine typically involves a reduction step.

There are multiple ways to achieve this reduction, including using catalysts and reagents like hydrogen gas with palladium on carbon (Pd-C) or lithium aluminum hydride (\( \mathrm{LiAlH_4} \)).
  • Each approach has its own advantages. Hydrazine-palladium provides high selectivity and yields, whereas \( \mathrm{LiAlH_4} \) is versatile and commonly used in different contexts.
  • Choosing the right method depends on the reaction conditions and the desired purity of the final amine product.
  • The overall transformation to amine is notable because it provides a method to access compound frameworks that play critical roles in pharmaceuticals and natural products.
By understanding these options, chemists can effectively synthesize a variety of amines using starting materials similar to the alkyl halides and azides in this exercise.

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Most popular questions from this chapter

Choline, a constituent of phospholipids (fat-like phosphate esters of great physiological importance), has the formula \(\mathrm{C}_{5} \mathrm{H}_{15} \mathrm{O}_{2} \mathrm{~N}\). It dissolves readily in water to form a strongly basic solution. It can be prepared by the reaction of ethylene oxide with trimethylamine in the presence of water. (a) What is a likely structure for choline? (b) What is a likely structure for its acety1 derivative, acetylcholine, \(\mathrm{C}_{7} \mathrm{H}_{17} \mathrm{O}_{3} \mathrm{~N}\), important in nerve action?

Novocaine, a local anesthetic, is a compound of formula \(\mathrm{C}_{13} \mathrm{H}_{20} \mathrm{O}_{2} \mathrm{~N}_{2}\). It is insoluble in water and dilute \(\mathrm{NaOH}\), but soluble in dilute HC1. Upon treatment with \(\mathrm{NaNO}_{2}\) and \(\mathrm{HCl}\) and then with \(\beta\) -naphthol, a highly colored solid is formed. When Novocaine is boiled with aqueous \(\mathrm{NaOH}\), it slowly dissolves. The alkaline solution is shaken with ether and the layers are separated. Acidification of the aqueous layer causes the precipitation of a white solid A; continued addition of acid causes \(A\) to redissolve. Upon isolation \(A\) is found to have a melting point of \(185-6^{\circ}\) and the formula \(\mathrm{C}_{7} \mathrm{H}_{7} \mathrm{O}_{2} \mathrm{~N}\). Evaporation of the ether layer leaves a liquid \(\mathrm{B}\) of formula \(\mathrm{C}_{6} \mathrm{H}_{15} \mathrm{ON}\). B dissolves in water to give a solution that turns litmus blue. Treatment of \(B\) with acetic anhydride gives C, \(\mathrm{C}_{8} \mathrm{H}_{17} \mathrm{O}_{2} \mathrm{~N}\), which is insoluble in water and dilute base, but soluble in dilute HCl. \(\mathrm{B}\) is found to be identical with the compound formed by the action of diethylamine on ethylene oxide. (a) What is the structure of Novocaine? (b) Outline all steps in a complete synthesis of Novocaine from toluene and readily available aliphatic and inorganic reagents. $$ \begin{array}{|l|l|l|} \hline {\text { CARBOXYLIC ACIDS }} \\ \hline \text { Name } & \text { Formula } & \begin{array}{c} \text { M.p. } \\ { }^{\circ} \mathrm{C} \end{array} \\ \hline \text { o-Nitrobenzoic } & \circ-\mathrm{O}_{2} \mathrm{NC}_{6} \mathrm{H}_{4} \mathrm{COOH} & 147 \\ \hline \text { m-Nitrobenzoic } & \mathrm{m}-\mathrm{O}_{2} \mathrm{NC}_{6} \mathrm{H}_{4} \mathrm{COOH} & 141 \\ \hline \text { p-Nitrobenzoic } & \mathrm{p}-\mathrm{O}_{2} \mathrm{NC}_{6} \mathrm{H}_{4} \mathrm{COOH} & 242 \\ \hline \text { Phthalic } & \mathrm{o}-\mathrm{C}_{6} \mathrm{H}_{4}(\mathrm{COOH})_{2} & 231 \\ \hline \text { Isophthalic } & \mathrm{m}-\mathrm{C}_{6} \mathrm{H}_{4}(\mathrm{COOH})_{2} & 348 \\ \hline \text { Terephthalic } & \mathrm{p}-\mathrm{C}_{6} \mathrm{H}_{4}(\mathrm{COOH})_{2} & 300 \text { sub1. } \\ \hline \text { Salicylic } & \mathrm{o}-\mathrm{HOC}_{6} \mathrm{H}_{4} \mathrm{COOH} & 159 \\ \hline \text { p-Hydroxybenzoic } & \mathrm{p}-\mathrm{HOC}_{6} \mathrm{H}_{4} \mathrm{COOH} & 213 \\ \hline \text { Anthranilic } & \mathrm{o}-\mathrm{H}_{2} \mathrm{NC}_{6} \mathrm{H}_{4} \mathrm{COOH} & 146 \\ \hline \text { m-Aminobenzoic } & \mathrm{m}-\mathrm{H}_{2} \mathrm{NC}_{6} \mathrm{H}_{4} \mathrm{COOH} & 179 \\ \hline \text { p-Aminobenzoic } & \mathrm{p}-\mathrm{H}_{2} \mathrm{NC}_{6} \mathrm{H}_{4} \mathrm{COOH} & 187 \\ \hline \text { o-Methoxybenzoic } & \mathrm{o}-\mathrm{CH}_{3} \mathrm{OC}_{6} \mathrm{H}_{4} \mathrm{COOH} & 101 \\ \hline \text { m-Methoxybenzoic } & \mathrm{m}-\mathrm{CH}_{3} \mathrm{OC}_{6} \mathrm{H}_{4} \mathrm{COOH} & 110 \\ \hline \text { p-Methoxybenzoic(Anisic) } & \mathrm{p}-\mathrm{CH}_{3} \mathrm{OC}_{6} \mathrm{H}_{4} \mathrm{COOH} & 184 \\ \hline \end{array} $$

Synthesize m-chloroaniline from benzene. Any inorganic reagents may be used.

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Consider possible ways of formulating the electronic structures of nitroso dimers with the knowledge that X-ray diffraction studies indicate the presence of nitrogen-nitrogen bonds.

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