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Many primary amines, \(\mathrm{RNH}_{2}\), where \(\mathrm{R}\) is a carbon- containing fragment such as \(\mathrm{CH}_{3}, \mathrm{CH}_{3} \mathrm{CH}_{2}\), 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 \(\mathrm{C}\) atom for " \(\mathrm{R}^{w}\) ). (b) What kind of reactant with a primary amine can produce a tetrahedral intermediate?

Short Answer

Expert verified
In a primary amine, the nitrogen atom is sp^3-hybridized, with the R group (carbon atom) and the two hydrogen atoms occupying three sp^3 orbitals, and the lone pair of electrons occupying the remaining sp^3 orbital. The reactant for a tetrahedral intermediate would be an electrophile, like a carbonyl compound, which can accept the lone pair of electrons from the nitrogen atom in the primary amine.

Step by step solution

01

Draw a hybrid orbital picture

For a primary amine, the nitrogen atom is bonded to a carbon-containing fragment, R, and two hydrogen atoms, and it also contains a lone pair of electrons. We can visualize the bonding using hybrid orbitals that are derived from the atomic orbitals of nitrogen. In a primary amine, the nitrogen atom is sp^3-hybridized. The sp^3 orbitals are formed by the hybridization of one s and three p orbitals. To visualize the bonding in a primary amine, we can place each sp^3 orbital along the corners of a tetrahedron, with the nitrogen atom at the center. The R group (carbon atom) and the two hydrogen atoms will occupy three of the sp^3 orbitals, while the lone pair of electrons occupies the remaining sp^3 orbital.
02

Identify the reactant for a tetrahedral intermediate

A tetrahedral intermediate can be formed when a primary amine reacts with an electrophile, a species that can accept a lone pair of electrons. The electrophile should have an atom with a partial positive charge or a positively charged atom that can attract the lone pair of electrons on the nitrogen atom of the primary amine. An example of such a reactant is a carbonyl compound (R'CO, where R' is a carbon-containing group). The carbonyl carbon atom has a partial positive charge because of the electronegativity difference between the carbon and oxygen atoms. When a primary amine reacts with a carbonyl compound, the lone pair of electrons on the nitrogen attacks the carbonyl carbon, forming a tetrahedral intermediate. So, the type of reactant that can react with a primary amine to produce a tetrahedral intermediate is an electrophile, like a carbonyl compound.

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Key Concepts

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

sp3 Hybridization
Understanding sp3 hybridization is fundamental when exploring organic chemistry, as it explains the 3D shape and bonding in many organic compounds. In a primary amine, the nitrogen atom is centrally located with its electron configuration initially in the form of one s and three p orbitals. Through the process of hybridization, these orbitals merge to form four equivalent sp3 hybrid orbitals. Imagine taking a set of differently shaped balloons and inflating them to blend into one unified shape; this is akin to orbital hybridization.

When these new hybrid orbitals form, they arrange themselves in a way that maximizes the distance between them—resulting in a tetrahedral structure. Three of these orbitals will form sigma bonds with a hydrogen atom and an R group, which is simply a placeholder for any carbon-containing group. The fourth orbital houses the lone pair of electrons, essential for the amine's reactivity as a nucleophile. This arrangement dictates the molecule's geometry and its ability to engage in downstream reactions.
Tetrahedral Intermediate
Let's dive into the concept of a tetrahedral intermediate, which often arises in organic reaction mechanisms. Following the idea of sp3 hybridization, when a primary amine encounters a suitable electrophile, the lone pair on the nitrogen can form a new bond. This action results in a temporary species known as a tetrahedral intermediate, named for its resemblance to the geometric shape of a tetrahedron.

This intermediate is a pivotal point in numerous reactions, serving as a stepping stone to the final products. For instance, when a primary amine reacts with a carbonyl compound, it forms a new bond with the carbonyl carbon, temporarily creating a tetrahedral species. The generation of this intermediate can be seen as the 'giving a handshake' moment in a reaction, where the nucleophile (amine) reaches out and forms a new bond, setting the stage for further transformation.
Nucleophile-Electrophile Reactions
In the molecular dance of nucleophile-electrophile reactions, the participants—the nucleophile and the electrophile—play well-defined roles. The nucleophile, in this case, a primary amine, can be envisioned as an entity with a pair of electron 'gloves', ready to donate to anyone willing to accept them. When a nucleophile encounters an electrophile, a substance that is seeking electrons, a chemical bond is formed, initiating a reaction.

An example highlight is the interaction between a primary amine's lone pair and the partially positive carbon on a carbonyl group. The amine, as the nucleophile, 'attacks' the electrophile, paving the way for the formation of a tetrahedral intermediate. These interactions are central to building complex organic molecules and are a cornerstone of organic synthesis concepts.
Carbonyl Compound Reactions
Delving into carbonyl compound reactions, we focus on the behavior of the carbonyl group which consists of a carbon atom double-bonded to an oxygen atom. This distinctive configuration places a partial positive charge on the carbon, making it a prime target for attack by nucleophiles.

In the context of primary amines, the lone pair on the nitrogen atom of the amine is attracted to the electron-deficient carbonyl carbon, initiating a reaction. Upon this nucleophilic attack, the double bond of the carbonyl shifts to form a single bond and the oxygen becomes negatively charged, resulting in the previously mentioned tetrahedral intermediate. The inherent polarization of carbonyl compounds and their propensity to undergo such reactions make them highly valuable in the synthesis of a myriad of organic substances, from simple esters to complex natural products.

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Most popular questions from this chapter

For each of the following gas-phase reactions, write the rate expression in terms of the appearance of each product and disappearance of each reactant: (a) \(2 \mathrm{H}_{2} \mathrm{O}(g) \longrightarrow 2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g)\) (b) \(2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{SO}_{3}(g)\) (c) \(2 \mathrm{NO}(g)+2 \mathrm{H}_{2}(g) \longrightarrow \mathrm{N}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g)\) (d) \(\mathrm{N}_{2}(g)+2 \mathrm{H}_{2}(g) \longrightarrow \mathrm{N}_{2} \mathrm{H}_{4}(g)\)

(a) What is meant by the term molecularity? (b) Why are termolecular elementary reactions so rare? (c) What is an intermediate in a mechanism? (d) What are the differences between an intermediate and a transition state?

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

The following mechanism has been proposed for the gasphase reaction of \(\mathrm{H}_{2}\) with \(\mathrm{ICl}\) : $$ \begin{aligned} &\mathrm{H}_{2}(g)+\mathrm{ICl}(g) \longrightarrow \mathrm{HI}(g)+\mathrm{HCl}(g) \\ &\mathrm{HI}(g)+\mathrm{ICl}(g) \longrightarrow \mathrm{I}_{2}(g)+\mathrm{HCl}(g) \end{aligned} $$ (a) Write the balanced equation for the overall reaction. (b) Identify any intermediates in the mechanism. (c) If the first step is slow and the second one is fast, which rate law do you expect to be observed for the overall reaction?

The oxidation of \(\mathrm{SO}_{2}\) to \(\mathrm{SO}_{3}\) is accelerated by \(\mathrm{NO}_{2}\). The reaction proceeds according to: $$ \begin{aligned} &\mathrm{NO}_{2}(g)+\mathrm{SO}_{2}(g) \longrightarrow \mathrm{NO}(g)+\mathrm{SO}_{3}(g) \\ &2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{NO}_{2}(g) \end{aligned} $$ (a) Show that, with appropriate coefficients, the two reactions can be summed to give the overall oxidation of \(\mathrm{SO}_{2}\) by \(\mathrm{O}_{2}\) to give \(\mathrm{SO}_{3}\). (b) \(\mathrm{Do}\) we consider \(\mathrm{NO}_{2}\) a catalyst or an intermediate in this reaction? (c) Is this an example of homogeneous catalysis or heterogeneous catalysis?

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