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Chapter 14: Problem 123

Many primary amines, RNH \(_{2},\) where \(R\) is a carbon-containing fragment such as \(C H_{3}, C H_{3} C H_{2},\) and so on, undergo reactions where the transition state is tetrahedral. (a) Draw a hybrid orbital picture to visualize the bonding at the nitrogen in a primary amine (just use a \(C\) atom for \(^{4} \mathrm{R}^{\prime \prime}\) . (b) What kind of reactant with a primary amine can produce a tetrahedral intermediate?

Short Answer

Expert verified
(a) To draw a hybrid orbital picture of a primary amine with a simple carbon atom as the carbon-containing fragment, we depict nitrogen with four sp^3 hybrid orbitals forming a tetrahedron. The R group, in this case, a carbon atom, is bonded to one hybrid orbital, and two hydrogen atoms are bonded to the other two hybrid orbitals. The nitrogen atom carries an unshared electron pair in the remaining hybrid orbital. (b) A reactant that can produce a tetrahedral intermediate with a primary amine is an electrophilic carbonyl compound (R'CO). The nitrogen atom in the primary amine donates its electron pair to the electron-deficient carbonyl carbon, generating a tetrahedral intermediate.

Step by step solution

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(a) Drawing hybrid orbital picture

To draw a hybrid orbital picture of a primary amine with a simple carbon atom as the carbon-containing fragment, let's start by understanding the electronic configuration of the nitrogen atom. Nitrogen has 5 valence electrons, with an electronic configuration of \(1s^{2}2s^{2}2p^{3}\). When nitrogen forms a primary amine, it uses one of its 2s electrons and three 2p electrons to create four hybrid orbitals, typically referred to as sp^3 hybridization. These hybrid orbitals form a tetrahedron with angles of approximately 109.5 degrees between them. These sp^3 hybrid orbitals will be involved in forming bonds with the R group (carbon-containing fragment) and two hydrogen atoms. To draw this hybrid orbital picture: 1. Draw a nitrogen atom surrounded by 4 hybrid orbitals (sp^3), pointing towards the corners of a tetrahedron. 2. Draw the R group attached to one of the hybrid orbitals. 3. Draw two hydrogen atoms, each attached to another hybrid orbital. 4. Show the unshared electron pair of the nitrogen atom in the remaining hybrid orbital.
02

(b) Reactant producing a tetrahedral intermediate

In order to produce a tetrahedral intermediate in a reaction with a primary amine (RNH2), we need a reactant that is electron-deficient, which can accept an electron pair from the nitrogen atom of the primary amine. Such a reactant will engage with the nitrogen atom and share an electron pair, forming a new bond. A common type of reactant that can produce a tetrahedral intermediate with a primary amine is an electrophile, such as a carbonyl compound (R'CO). Here's how the reaction takes place: 1. The nitrogen atom in the primary amine RNH2 acts as a nucleophile, donating its electron pair to the electron-deficient carbonyl carbon (C=O) in the reactant R'CO. 2. The carbonyl bond (C=O) breaks, and one of the carbonyl bond electrons shift towards the oxygen atom, creating a tetrahedral intermediate. The oxygen atom is now negatively charged, while the nitrogen atom still retains a positive charge due to its shared electron pair with the carbonyl carbon. Therefore, a reactant that can produce a tetrahedral intermediate with a primary amine is an electrophilic carbonyl compound (R'CO).

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Key Concepts

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

Hybridization in Amines
Amines, such as primary amines, have a unique structure due to the way nitrogen bonds. Nitrogen is vital in organic compounds and has 5 valence electrons with the configuration of \(1s^2 2s^2 2p^3\). When forming primary amines, nitrogen undergoes \(sp^3\) hybridization.

This means it mixes its 2s electrons with its 2p electrons to produce four sp鲁 hybrid orbitals. These orbitals are arranged in a tetrahedral shape. This arrangement creates bond angles of about 109.5 degrees, allowing the nitrogen atom to bond with other atoms in the most stable way possible.

- **Bonding Details:** - In a primary amine, the nitrogen atom forms bonds using these sp鲁 hybrid orbitals. - It connects to a carbon-containing group (designated as \(R\)) and two hydrogen atoms. - One hybrid orbital houses an unshared electron pair, which plays a crucial role in reactivity.

This understanding of hybridization helps us visualize molecul鈥嬧�媋r geometry and predict how amines interact in chemical reactions.
Tetrahedral Intermediates
In organic chemistry, reactions often involve the formation of unstable structures known as tetrahedral intermediates. These forms are crucial in understanding how primary amines react with certain compounds.

A tetrahedral intermediate occurs when a molecule temporarily adopts a tetrahedral shape during a reaction. For amines, this usually happens when they react with an electrophile, like a carbonyl compound. The nitrogen atom in the amine provides a pair of electrons to form a bond with the electrophilic center, typically a carbon atom.

- **Formation Process:** - The nitrogen in a primary amine (RNH\(_2\)) acts as a nucleophile, attacking an electron-deficient carbon. - The carbon center, often part of a carbonyl group, accepts the electrons and breaks its double bond with oxygen momentarily. - This generates a temporary structure where the carbon is bonded to four different groups, fitting the definition of tetrahedral geometry.

These intermediates are key to understanding reaction mechanisms. They are pivotal in enabling the nitrogen atom to engage in further reactions, eventually leading to the final stable product.
Electrophilic Reactions
Electrophilic reactions play a significant role in the chemistry of amines. An electrophile is an electron-deficient molecule that seeks out regions of high electron density to obtain the electrons it lacks.

Primary amines, with their lone pair of electrons on nitrogen, are prime targets for these types of reactions. By donating this electron pair, amines react with electrophiles to form new compounds, often creating tetrahedral intermediates along the way.

- **How It Works:** - The lone electron pair on nitrogen allows it to serve as a nucleophile, initiating reactions with many electrophiles. - Carbonyl compounds (\(R'CO\)) are classic examples where amines can create new bonds. - As the amine nitrogen donates its electron pair to form a bond with an electrophilic carbon, the double bond in the carbonyl group often breaks temporarily.

Understanding electrophilic reactions with amines is critical in organic synthesis. These reactions help chemists design pathways to create a variety of compounds, each with valuable applications in materials, pharmaceuticals, and beyond. By mastering these interactions, students can gain deeper insights into the chemical behavior of amines.

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Most popular questions from this chapter

(a) What is meant by the term reaction rate? (b) Name three factors that can affect the rate of a chemical reaction. (c) Is the rate of disappearance of reactants always the same as the rate of appearance of products?

The enzyme carbonic anhydrase catalyzes the reaction \(\mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{HCO}_{3}^{-}(a q)+\mathrm{H}^{+}(a q) .\) In water, without the enzyme, the reaction proceeds with a rate constant of 0.039 \(\mathrm{s}^{-1}\) at \(25^{\circ} \mathrm{C}\) . In the presence of the enzyme in water, the reaction proceeds with a rate constant of \(1.0 \times 10^{6} \mathrm{s}^{-1}\) at \(25^{\circ} \mathrm{C}\) . Assuming the collision factor is the same for both situations, calculate the difference in activation energies for the uncatalyzed versus enzyme-catalyzed reaction.

What is the molecularity of each of the following elementary reactions? Write the rate law for each. \(\begin{array}{l}{\text { (a) } 2 \mathrm{NO}(g) \longrightarrow \mathrm{N}_{2} \mathrm{O}_{2}(g)} \\ {\mathrm{CH}_{2}} \\ {\text { (b) } \mathrm{H}_{2} \mathrm{C}-\mathrm{CH}_{2}(g) \longrightarrow \mathrm{CH}_{2}=\mathrm{CH}-\mathrm{CH}_{3}(g)} \\ {\text { (c) } \mathrm{SO}_{3}(g) \longrightarrow \mathrm{SO}_{2}(g)+\mathrm{O}(g)}\end{array}\)

Based on their activation energies and energy changes and assuming that all collision factors are the same, rank the following reactions from slowest to fastest.\( \begin{aligned} \text { (a) } E_{a} &=45 \mathrm{kJ} / \mathrm{mol} ; \Delta E=-25 \mathrm{kJ} / \mathrm{mol} \\ \text { (b) } E_{a} &=35 \mathrm{kJ} / \mathrm{mol} ; \Delta E=-10 \mathrm{kJ} / \mathrm{mol} \\ \text { (c) } E_{a} &=55 \mathrm{kJ} / \mathrm{mol} ; \Delta E=10 \mathrm{kJ} / \mathrm{mol} \end{aligned}\)

For the elementary process \(\mathrm{N}_{2} \mathrm{O}_{5}(g) \longrightarrow \mathrm{NO}_{2}(g)+\mathrm{NO}_{3}(g)\) the activation energy \(\left(E_{a}\right)\) and overall \(\Delta E\) are 154 \(\mathrm{kJ} / \mathrm{mol}\) and 136 \(\mathrm{kJ} / \mathrm{mol}\) , respectively. (a) Sketch the energy profile for this reaction, and label \(E_{a}\) and \(\Delta E\) . (b) What is the activation energy for the reverse reaction?
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