Question

(a) Suggest how the reaction of \(\mathrm{NH}_{3}\) with 1-bromopropane might proceed. (b) Suggest a method of preparing 1-hexylamine in high yield from 1-bromohexane.

Short answer

(a) Ammonia attacks 1-bromopropane to form propylamine via an S\(N^2\) mechanism. (b) Use phthalimide and hydrazine to convert 1-bromohexane to 1-hexylamine.

Step-by-step solution

Understand the Reaction Context for (a)

The reaction between ammonia (\(\mathrm{NH}_3\)) and 1-bromopropane is an example of an organic substitution reaction, where the nucleophilic ammonia will replace the bromine atom.

Identify the Nucleophilic Substitution Mechanism for (a)

1-bromopropane (\(\mathrm{C_3H_7Br}\)) is a primary alkyl halide and is likely to undergo an S\(\mathrm{N}^2\) reaction. Here, the lone pair of electrons on the nitrogen atom in ammonia attacks the carbon in 1-bromopropane, displacing the bromide ion.

Write the Reaction Equation for (a)

The balanced chemical equation for this process is: \[\mathrm{NH}_3 + \mathrm{C_3H_7Br} \rightarrow \mathrm{C_3H_7NH_2} + \mathrm{HBr}\]This reaction forms propylamine and hydrobromic acid (HBr).

Plan the Reaction Sequence for (b)

To prepare 1-hexylamine from 1-bromohexane, a similar S\(\mathrm{N}^2\) reaction path is used, but a better nucleophile than ammonia is often employed to achieve high yields.

Use a Stronger Nucleophile for (b)

React 1-bromohexane with sodium or potassium phthalimide to create a phthalimide-substituted derivative. This is then followed by a hydrazine treatment to yield 1-hexylamine.

Write the Reaction Equations for (b)

The reaction steps can be written as:1. Formation of a phthalimide intermediate:\[\mathrm{C_6H_{13}Br + C_8H_5NO_2Na} \rightarrow \mathrm{C_8H_5NO_2C_6H_{13}} + \mathrm{NaBr}\]2. Treatment with hydrazine:\[\mathrm{C_8H_5NO_2C_6H_{13}} + \mathrm{N_2H_4} \rightarrow \mathrm{C_6H_{13}NH_2} + \mathrm{C_8H_8N_2O_2}\]

Key concepts

Nucleophilic Substitution

Nucleophilic substitution reactions are a cornerstone of organic chemistry and involve the replacement of an atom or group within a molecule by a nucleophile. A nucleophile is an atom or molecule that donates an electron pair to form a new chemical bond. In these reactions:

• The nucleophile attacks a positive or partially positive atom in the substrate molecule.
• This attack can lead to the ejection of a leaving group.

In the case of 1-bromopropane reacting with ammonia (\(\mathrm{NH}_3\)), ammonia acts as the nucleophile because it has a lone pair of electrons. The bromine in 1-bromopropane leaves, allowing ammonia to bond with the carbon.
Different mechanisms can govern these reactions, such as the S\(\mathrm{N}^1\) and S\(\mathrm{N}^2\) mechanisms, which will be further discussed. These reactions are fundamental in the synthesis of many different organic compounds.

Primary Alkyl Halides

Primary alkyl halides are compounds where a halogen atom, like bromine, is attached to a primary carbon. A primary carbon is directly bonded to only one other carbon atom. These halides typically undergo nucleophilic substitution reactions following an S\(\mathrm{N}^2\) mechanism.
An S\(\mathrm{N}^2\) mechanism is characterized by:

• A single, concerted step where the nucleophile attacks the electrophilic carbon from the side opposite to the leaving group.
• The leaving group, such as bromide, departs as a new bond forms.
• This kind of reaction leads to inversion of configuration at the carbon center.

In the specific case of the reaction between 1-bromopropane and ammonia, this means that ammonia's lone pair attacks the carbon bonded to bromine, displacing bromide and forming propylamine.

Chemical Reaction Mechanisms

Understanding the detailed steps—or mechanisms—through which chemical reactions occur is crucial for mastering organic chemistry. Reaction mechanisms can include a variety of elementary steps and intermediate formations, which together describe the overall transformation.
For nucleophilic substitution reactions, the S\(\mathrm{N}^1\) and S\(\mathrm{N}^2\) mechanisms are key.

The S\(\mathrm{N}^2\) Mechanism

This mechanism is typical for reactions involving primary alkyl halides:

• The nucleophile attacks from the side opposite to the leaving group, leading to a concerted process.
• No intermediate is formed, and the inversion of stereochemistry can occur.

The S\(\mathrm{N}^1\) Mechanism

In contrast, the S\(\mathrm{N}^1\) mechanism occurs in two steps and typically involves secondary or tertiary alkyl halides:
• The leaving group first departs, forming a carbocation intermediate.
• The nucleophile then attacks the planar carbocation, which may allow for racemization.
By understanding these mechanisms, chemists can predict and control the outcomes of nucleophilic substitution reactions, crucial for effective synthesis of organic compounds.