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Many primary amines, \(\mathrm{RNH}_{2}\), where \(\mathrm{R}\) is a carboncontaining 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 \({ }^{\text {"R" }}\) ). (b) What kind of reactant with a primary amine can produce a tetrahedral intermediate?

Short Answer

Expert verified
(a) In a primary amine, nitrogen forms three sp3 hybrid orbitals, bonding with hydrogen atoms and the carbon-containing fragment (R). The structure is as follows: R | N - H | H The lone electron pair is in the 2p orbital of Nitrogen. (b) A molecule containing an electrophilic center, such as a carbonyl group (C=O), can interact with a primary amine to produce a tetrahedral intermediate. The nitrogen atom in the primary amine acts as a nucleophile, donating its electron pair to the carbonyl carbon.

Step by step solution

01

Identify the orbitals involved in bonding

In a primary amine, the nitrogen atom will have three bonding sites: one with the carbon-containing fragment (R) and two with hydrogen atoms. Nitrogen has one 2s and three 2p orbitals. Since three bonding sites need to be formed, Nitrogen uses its one 2s and two of its 2p orbitals to form three hybrid sp3 orbitals, which will be involved in the bond formation. The remaining 2p orbital hosts the lone electron pair.
02

Draw the hybrid orbital picture

In the hybrid orbital picture, depict the Nitrogen atom with its three sp3 hybrid orbitals bonding with hydrogen atoms and the carbon-containing fragment. Additionally, indicate the 2p orbital with the lone electron pair. The resulting structure is as follows: R | N - H | H And the unpaired electron is in the p-orbital of Nitrogen. #Step 2: Identifying the type of reactant that can form a tetrahedral intermediate with a primary amine#
03

Recognize the nature of the reaction

The primary amine has a lone pair of electrons on the nitrogen atom. Therefore, the nitrogen atom can act as a nucleophile (electron donor). To identify the type of reactant that can interact with primary amines to produce a tetrahedral intermediate, we need to look for a molecule that can accept the electron pair from nitrogen.
04

Determine the type of reactant molecule

The molecule needs to be an electrophile, as it will accept an electron pair from the nucleophilic nitrogen atom in the primary amine. A common electrophile is the carbonyl group (C=O), as the carbon atom is electron-deficient, making it a suitable candidate for accepting an electron pair. When the nucleophilic nitrogen in the primary amine interacts with the carbonyl carbon in the electrophile molecule, it forms a new bond, leading to the formation of a tetrahedral intermediate. In summary, the reactant that can form a tetrahedral intermediate with a primary amine is a molecule containing an electrophilic center, such as a carbonyl group.

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Key Concepts

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

Tetrahedral Intermediates
In the realm of organic chemistry, tetrahedral intermediates are central to understanding how molecules interact in many reactions. These intermediates are crucial components in processes involving species such as primary amines.

When a primary amine encounters an electrophile, the resulting interaction often leads to the formation of a tetrahedral configuration at the nitrogen atom. This happens because the nitrogen's lone pair of electrons are donated to bond with the electrophile, increasing the coordination number around the nitrogen from three to four.

Consider a simple reaction between a primary amine and a carbonyl compound. The nitrogen atom attacks the electrophilic carbon, and a new bond forms. As the bond is made, the electrons in the carbonyl bond shift towards the oxygen, creating a temporary tetrahedral intermediate at the nitrogen.

It's important to ensure we visualize the molecule correctly. The nitrogen, now bonded to four different groups, adopts a tetrahedral geometry rather than its original trigonal pyramidal shape, due to the extra bonding. These intermediates are unstable and quickly rearrange to form more stable products, but their brief existence is a pivotal moment in the reaction pathway.
sp3 Hybrid Orbitals
The concept of sp3 hybrid orbitals is fundamental to the structure of primary amines and their ability to form tetrahedral intermediates. The understanding of hybridization helps explain the behavior of molecules during chemical reactions.

In primary amines, the nitrogen atom has one s and three p orbitals. To form bonds with three different atoms or groups, the nitrogen hybridizes its orbitals into three sp3 hybrid orbitals. These orbitals have an equal amount of s and p character, which makes them ideal for forming sigma (σ) bonds.

Each sp3 hybrid orbital is directed towards one corner of an imaginary tetrahedron, with the lone pair of electrons usually residing in the fourth corner. This arrangement allows the primary amine to form stable bonds while maintaining an optimal distance and angle from its neighboring atoms.

When visualizing these orbitals, it's helpful for students to draw them as slightly lobe-shaped, radiating out from the central nitrogen atom, and directed towards the vertices of a tetrahedron. This depiction aids in understanding how the nitrogen can form bonds with other atoms while also accommodating its lone pair in a spatially efficient manner.
Nucleophilic Reactions
Diving into the world of nucleophilic reactions can provide deeper insight into the behavior of primary amines. A nucleophile is essentially a 'nucleus lover,' an atom or molecule that is rich in electrons and seeks a positive center to bond with – typically an electron-deficient (electrophilic) site.

Primary amines possess a nucleophilic nitrogen atom because of the lone pair of electrons it harbors. These lone electrons carry a negative charge and are attracted to positively charged or electron-deficient areas of other molecules. During a nucleophilic reaction, the nucleophile donates its electron pair to form a new chemical bond.

Reactants such as carbonyl compounds are ideal targets for nucleophilic attack because their carbon atoms are electron-poor, creating a perfect match for the electron-rich nitrogen. When the nitrogen of a primary amine attacks such a carbon, it forms a tetrahedral intermediate, as mentioned previously.

These reactions are not just a theoretical concept; they underpin many biological processes and synthetic pathways in chemistry. Understanding how nucleophiles behave is key to mastering a wide array of chemical reactions, including those that form the backbone of organic synthesis.

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Most popular questions from this chapter

Platinum nanoparticles of diameter \(\sim 2 \mathrm{nm}\) are important catalysts in carbon monoxide oxidation to carbon dioxide. Platinum crystallizes in a face-centered cubic arrangement with an edge length of \(3.924 \AA\). (a) Estimate how many platinum atoms would fit into a \(2.0-\mathrm{nm}\) sphere; the volume of a sphere is \((4 / 3) \pi r^{3}\). Recall that \(1 \AA=1 \times 10^{-10} \mathrm{~m}\) and \(1 \mathrm{nm}=1 \times 10^{-9} \mathrm{~m} .\) (b) Estimate how many platinum atoms are on the surface of a \(2.0-\mathrm{nm} \mathrm{Pt}\) sphere, using the surface area of a sphere \(\left(4 \pi r^{2}\right)\) and assuming that the "footprint" of one \(\mathrm{Pt}\) atom can be estimated from its atomic diameter of \(2.8 \AA\). (c) Using your results from (a) and (b), calculate the percentage of \(\mathrm{Pt}\) atoms that are on the surface of a \(2.0-\mathrm{nm}\) nanoparticle. (d) Repeat these calculations for a 5.0 -nm platinum nanoparticle. (e) Which size of nanoparticle would you expect to be more catalytically active and why?

The following data were collected for the rate of disappearance of \(\mathrm{NO}\) in the reaction \(2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{NO}_{2}(g)\) : $$ \begin{array}{llll} \hline \text { Experiment } & {[\mathrm{NO}](M)} & {\left[\mathrm{O}_{2}\right](M)} & \text { Initial Rate }(M / s) \\ \hline 1 & 0.0126 & 0.0125 & 1.41 \times 10^{-2} \\ 2 & 0.0252 & 0.0125 & 5.64 \times 10^{-2} \\ 3 & 0.0252 & 0.0250 & 1.13 \times 10^{-1} \end{array} $$ (a) What is the rate law for the reaction? (b) What are the units of the rate constant? (c) What is the average value of the rate constant calculated from the three data sets? (d) What is the rate of disappearance of \(\mathrm{NO}\) when \([\mathrm{NO}]=0.0750 \mathrm{M}\) and \(\left[\mathrm{O}_{2}\right]=0.0100 \mathrm{M} ?(\mathrm{e})\) What is the rate of disappearance of \(\mathrm{O}_{2}\) at the concentrations given in part ( \(\mathrm{d}\) )?

The reaction $2 \mathrm{ClO}_{2}(a q)+2 \mathrm{OH}^{-}(a q) \longrightarrow \mathrm{ClO}_{3}^{-}(a q)+\( \)\mathrm{ClO}_{2}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}(l)$ was studied with the following results:

Based on their activation energies and energy changes and assuming that all collision factors are the same, which of the following reactions would be fastest and which would be slowest? Explain your answer. (a) \(E_{a}=45 \mathrm{~kJ} / \mathrm{mol} ; \Delta E=-25 \mathrm{~kJ} / \mathrm{mol}\) (b) \(E_{a}=35 \mathrm{~kJ} / \mathrm{mol} ; \Delta E=-10 \mathrm{~kJ} / \mathrm{mol}\) (c) \(E_{a}=55 \mathrm{~kJ} / \mathrm{mol} ; \Delta E=10 \mathrm{~kJ} / \mathrm{mol}\)

Urea \(\left(\mathrm{NH}_{2} \mathrm{CONH}_{2}\right)\) is the end product in protein metabolism in animals. The decomposition of urea in \(0.1 \mathrm{M} \mathrm{HCl}\) occurs according to the reaction $$ \begin{aligned} \mathrm{NH}_{2} \mathrm{CONH}_{2}(a q)+\mathrm{H}^{+}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow & \mathrm{NH}_{4}^{+}(a q)+\mathrm{HCO}_{3}^{-}(a q) \end{aligned} $$ The reaction is first order in urea and first order overall. When \(\left[\mathrm{NH}_{2} \mathrm{CONH}_{2}\right]=0.200 \mathrm{M},\) the rate at \(61.05^{\circ} \mathrm{C}\) is \(8.56 \times 10^{-5} \mathrm{M} / \mathrm{s}\). (a) What is the rate constant, \(k ?\) (b) What is the concentration of urea in this solution after \(4.00 \times 10^{3} \mathrm{~s}\) if the starting concentration is \(0.500 \mathrm{M}\) ? (c) What is the half-life for this reaction at \(61.05^{\circ} \mathrm{C}\) ?

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