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Ch. 19 - Amines
Wade - Organic Chemistry 9th Edition
Wade9th EditionOrganic ChemistryISBN: 9780135213728Not the one you use?Change textbook
All textbooksWade 9th EditionCh. 19 - AminesProblem 28
Chapter 19, Problem 28

Addition of one equivalent of ammonia to 1-bromoheptane gives a mixture of heptan-1-amine, some dialkylamine, some trialkylamine, and even some tetraalkylammonium bromide.
(a) Give a mechanism to show how this reaction takes place, as far as the dialkylamine. 
(b) How would you modify the procedure to get an acceptable yield of heptan-1-amine?

Verified step by step guidance
1
Step 1: Understand the reaction mechanism for part (a). The reaction involves the nucleophilic substitution of 1-bromoheptane with ammonia (NH₃). Ammonia acts as a nucleophile and attacks the electrophilic carbon attached to the bromine atom in 1-bromoheptane, displacing the bromide ion (Br⁻) in an SN2 reaction. This forms heptan-1-amine (primary amine) as the initial product.
Step 2: Explain the formation of dialkylamine. The primary amine (heptan-1-amine) formed in the first step can act as a nucleophile and react with another molecule of 1-bromoheptane in a similar SN2 mechanism. This results in the formation of a secondary amine (dialkylamine) and the release of another bromide ion.
Step 3: Address the formation of trialkylamine and tetraalkylammonium bromide. The secondary amine can further react with 1-bromoheptane to form a tertiary amine (trialkylamine). Additionally, the tertiary amine can react with yet another molecule of 1-bromoheptane to form a quaternary ammonium salt (tetraalkylammonium bromide). These side reactions reduce the yield of the desired primary amine.
Step 4: For part (b), to improve the yield of heptan-1-amine, use an excess of ammonia. By increasing the concentration of ammonia relative to 1-bromoheptane, the likelihood of the primary amine reacting further to form secondary, tertiary, or quaternary products is minimized. This is because the excess ammonia competes more effectively with the primary amine for the 1-bromoheptane.
Step 5: Additionally, perform the reaction under controlled conditions, such as maintaining a low temperature and using a solvent like ethanol to dissolve both reactants. These conditions help favor the formation of the primary amine by reducing the rate of secondary reactions. After the reaction, the product can be purified using techniques like distillation or extraction to isolate heptan-1-amine.

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Key Concepts

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

Nucleophilic Substitution Mechanism

The reaction of 1-bromoheptane with ammonia involves a nucleophilic substitution mechanism, specifically the SN2 pathway. In this process, the nucleophile (ammonia) attacks the electrophilic carbon atom bonded to the bromine, leading to the displacement of the bromine atom. This results in the formation of heptan-1-amine. Understanding this mechanism is crucial for illustrating how dialkylamines can also form through further reactions.
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Amine Formation and Reactivity

Amine formation from alkyl halides is a key concept in organic chemistry. When ammonia reacts with 1-bromoheptane, it can lead to the formation of primary amines, such as heptan-1-amine. However, the reactivity of amines allows them to further react with alkyl halides, resulting in the formation of secondary and tertiary amines. This reactivity is important for understanding the mixture of products generated in the reaction.
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Yield Optimization Techniques

To optimize the yield of heptan-1-amine, one can modify the reaction conditions, such as using excess ammonia or employing a controlled reaction temperature. Additionally, using a less reactive alkyl halide or removing by-products can help favor the formation of the desired primary amine. Understanding these techniques is essential for improving product yields in organic synthesis.
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Related Practice
Textbook Question

Show how to synthesize the following amines from the indicated starting materials by reductive amination.

(a) benzylmethylamine from benzaldehyde

(b) N-benzylpiperidine from piperidine

(c) N-cyclohexylaniline from cyclohexanone

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Textbook Question

Show how Gabriel syntheses are used to prepare the following amines.

(a) benzylamine

(b) hexan-1-amine

(c) γ-aminobutyric acid

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Textbook Question

Show how to synthesize the following amines from the indicated starting materials by reductive amination.

(e)

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Textbook Question

Show how you would accomplish the following synthetic conversions.

(b) 1-bromo-2-phenylethane → 3-phenylpropan-1-amine

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Textbook Question

Show how to synthesize the following amines from the indicated starting materials by acylation–reduction.

(a) N-butylpiperidine from piperidine

(b) N-benzylaniline from aniline

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Textbook Question

Show how you would accomplish the following synthetic conversions.

(a) benzyl bromide → benzylamine

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