Question

Isopropylamine and n-pentylamine are to be synthesized. In the laboratory synthesis, outline all the steps from alcohols of four carbons or less, using any needed inorganic reagents.

Short answer

The synthesis of isopropylamine initiates from isopropanol and that of n-pentylamine starts from n-butanol. Firstly, these alcohols are converted into alkyl halides, specifically isopropyl chloride and n-butyl chloride, through a reaction with hydrochloric acid. Subsequently, these alkyl halides are converted to azides; isopropyl azide and n-butyl azide, via a nucleophilic substitution reaction with sodium azide. Lastly, reduction of these azides with a reducing agent like lithium aluminium hydride, yields the desired amines, isopropylamine, and n-pentylamine. The relevant chemical reactions can be represented as follows: For Isopropylamine: Conversion to alkyl halide: \(CH_{3}CH(OH)CH_{3} + HCl \rightarrow CH_{3}CH(Cl)CH_{3} + H_{2}O\) Conversion to azide: \(CH_{3}CH(Cl)CH_{3} + NaN_{3} \rightarrow CH_{3}CH(N_{3})CH_{3} + NaCl\) Reduction to amine: \(CH_{3}CH(N_{3})CH_{3} + LiAlH_{4} \rightarrow CH_{3}CH(NH_{2})CH_{3} + LiAlH_{4}O\) For n-pentylamine: Conversion to alkyl halide: \(CH_{3}(CH_{2})_{2}CH_{2}OH + HCl \rightarrow CH_{3}(CH_{2})_{2}CH_{2}Cl + H_{2}O\) Conversion to azide: \(CH_{3}(CH_{2})_{2}CH_{2}Cl + NaN_{3} \rightarrow CH_{3}(CH_{2})_{2}CH_{2}N_{3} + NaCl\) Reduction to amine: \(CH_{3}(CH_{2})_{2}CH_{2}N_{3} + LiAlH_{4} \rightarrow CH_{3}(CH_{2})_{2}CH_{2}NH_{2} + LiAlH_{4}O\)

Step-by-step solution

Choose the correct Alcohol starter

For isopropylamine, the synthesis will start from isopropanol, an alcohol with three carbons. For n-pentylamine, the synthesis will start from n-butanol, which is an alcohol with four carbons.

Convert Alcohols to Alkyl Halides

The first step in the synthesis of amines from alcohols is to convert the alcohols to alkyl halides. For Isopropylamine: Isopropanol reacts with hydrochloric acid to produce isopropyl chloride. \( CH_{3}CH(OH)CH_{3} + HCl \rightarrow CH_{3}CH(Cl)CH_{3} + H_{2}O \) For n-pentylamine: n-butanol reacts with hydrochloric acid to produce n-butyl chloride. \( CH_{3}(CH_{2})_{2}CH_{2}OH + HCl \rightarrow CH_{3}(CH_{2})_{2}CH_{2}Cl + H_{2}O \)

Convert Alkyl Halides to Azides

The alkyl halides can then be converted to azides by a nucleophilic substitution reaction with sodium azide. For Isopropylamine: Isopropyl chloride reacts with sodium azide to form isopropyl azide. \( CH_{3}CH(Cl)CH_{3} + NaN_{3} \rightarrow CH_{3}CH(N_{3})CH_{3} + NaCl \) For n-pentylamine: n-Butyl chloride reacts with sodium azide to form n-butyl azide. \( CH_{3}(CH_{2})_{2}CH_{2}Cl + NaN_{3} \rightarrow CH_{3}(CH_{2})_{2}CH_{2}N_{3} + NaCl \)

Convert Azides to Amines

The azides can then be reduced to produce the amines. This is performed using a strong reducing agent such as lithium aluminium hydride. For Isopropylamine: Isopropyl azide is reduced by lithium aluminium hydride to form isopropylamine. \( CH_{3}CH(N_{3})CH_{3} + LiAlH_{4} \rightarrow CH_{3}CH(NH_{2})CH_{3} + LiAlH_{4}O{2} \) For n-pentylamine: n-Butyl azide is reduced by lithium aluminium hydride to form n-pentylamine. \( CH_{3}(CH_{2})_{2}CH_{2}N_{3} + LiAlH_{4} \rightarrow CH_{3}(CH_{2})_{2}CH_{2}NH_{2} + LiAlH_{4}O{2} \)

Key concepts

Alkyl Halide Formation

Understanding how to create alkyl halides from alcohols is a pivotal skill in synthetic organic chemistry. The process involves the reaction of an alcohol with a hydrogen halide, such as hydrochloric acid (HCl), yielding an alkyl halide and water as the byproduct. For instance, when isopropanol reacts with HCl, isopropyl chloride is formed through a mechanism known as 'nucleophilic substitution'.

Radicals are not typically involved in this transformation; what's important is the role of the hydrogen halide. It acts as both an acid, donating a proton to the hydroxyl group (-OH) of the alcohol, and as a source of halide ions. This converts the friendly hydroxyl group in alcohols into a good leaving group. Subsequently, the halide ion (Cl−, in our example) attaches to the alkyl group, giving rise to the alkyl halide product.

Nucleophilic Substitution

Nucleophilic substitution is a core mechanism in synthetic organic chemistry where a nucleophile, often negatively charged or neutral with a lone pair of electrons, replaces a leaving group in a molecule. There are two main types: SN1 and SN2 mechanisms.

The exact mechanism that occurs—SN1 or SN2—depends on the structure of the alkyl halide and the conditions of the reaction. In the textbook exercise, sodium azide (NaN3) acts as the nucleophile, carrying an azide group that is ready to displace the chloride in our alkyl halide to form an alkyl azide. The SN2 mechanism is likely in play here, where the nucleophilic attack and the expulsion of the leaving group occur simultaneously. This single-step process is favored by primary alkyl halides like what we see in the creation of n-pentylamine from n-butyl chloride.

Azide Reduction

After forming the alkyl azides, the last step of our amine synthesis journey is the azide reduction. This is the process of transforming the -N3 group into an -NH2 group, ultimately converting an azide into an amine. It's usually achieved through the use of a strong reducing agent.

In our exercise, lithium aluminium hydride (LiAlH4) is the reducing agent of choice. This powerful reagent donates hydride ions (H−)that attack the nitrogen of the azide group, triggering a series of bond rearrangements and ultimately leading to the formation of the free amine. The reduction of an azide to an amine is a great illustration of how specific reagents can help us achieve the transformation of functional groups, a common theme in synthetic organic chemistry.

Synthetic Organic Chemistry

The synthesis of amines from alcohols showcases the power and versatility of synthetic organic chemistry. It involves several steps, each reliant on fundamental chemical reactions and principles, such as nucleophilic substitution and reduction reactions. Mastering synthetic organic chemistry means understanding not just individual reactions, but how to string them together to build complex molecules from simple starting materials.

The approach used here is a classic synthetic strategy: transform functional groups in a stepwise fashion while contemplating factors like reactivity and selectivity. Throughout the process, chemists must consider the influence of reactant structure, steric hindrance, electronic effects, and reaction conditions. For students and chemists alike, the ability to devise and execute multi-step syntheses is essential, and exercises like these are vital in honing that skill.