Problem 118 - Chemistry The Molecular Science

Question

Consider the compound \(\mathrm{C}_{3} \mathrm{H}_{7} \mathrm{NO}\) in which there is no nitrogen-to-oxygen bond and there is an ethyl group. (a) Write a Lewis structure for the compound. (b) Draw two resonance structures for this molecule. (c) Use VSEPR theory to predict the geometry around each carbon atom and the nitrogen atom in each resonance structure. (d) Experimental evidence indicates that the nitrogen atom is triangular planar with bond angles close to \(120^{\circ} .\) Which prediction in part (c) is correct? (e) Describe the hybridization of the nitrogen atom.

Step-by-Step Solution

Answer

(a) The structure is \( \text{CH}_3\text{CH}_2\text{C}(=\text{O})\text{NH}_2 \). (d) Correct: trigonal planar, \(120^{\circ}\). (e) Nitrogen is \( \text{sp}^2 \) hybridized.

Step 1: Determine the Structure

To write a Lewis structure for \( \text{C}_3\text{H}_7\text{NO} \) without nitrogen-oxygen bonds and with an ethyl group, identify that an ethyl group \( \text{C}_2\text{H}_5 \) is attached to an acyl \( \text{R}-\text{CO} \) group with nitrogen. This suggests a secondary amide as the structure: \( \text{CH}_3\text{CH}_2\text{C}(=\text{O})\text{NH}_2 \).

Step 2: Draw Lewis Structure

The Lewis structure is:- Ethyl group \( \text{CH}_3\text{CH}_2- \) attached to a carbonyl \( \text{C}(=\text{O})- \).- The nitrogen \( \text{N} \) is bonded to \( \text{H}_2 \). Thus, the structure is \( \text{CH}_3\text{CH}_2\text{C}(=\text{O})\text{NH}_2 \). No \( \text{N}-\text{O} \) bond exists.

Step 3: Draw Resonance Structures

The resonance structures involve electron delocalization between the carbonyl and amine nitrogen:1. The typical amide form, where nitrogen is single-bonded to the carbon and no double-bonded resonance structure.2. A minor resonance where the lone pair on nitrogen forms a double bond with carbon (\( \text{C}=\text{N} \)), and the \( \text{O} \) has a negative charge (\( \text{O}^- \)).

Step 4: Predict Geometries Using VSEPR

For the carbon atoms: - The first carbon (in ethyl) is tetrahedral (\(109.5^{\circ}\)).- The second carbon (in carbonyl) is trigonal planar (\(120^{\circ}\)) due to \( \text{sp}^2\) hybridization.For the nitrogen: - Under typical conditions, it would appear trigonal pyramidal, but due to resonance, it appears planar which aligns with \( \text{sp}^2 \) hybridization.

Step 5: Confirm Observation

The experimental observation indicates the nitrogen in amides is trigonal planar with bond angles close to \(120^{\circ}\), supporting the idea that resonance leads to \( \text{sp}^2 \) hybridization with a planar structure.

Step 6: Determine Nitrogen Hybridization

The nitrogen atom is typically \( \text{sp}^3 \) hybridized, but in this compound, resonance between nitrogen, carbonyl carbon, and oxygen results in \( \text{sp}^2 \) hybridization, leaving nitrogen planar with bond angles close to \(120^{\circ}\).

Key Concepts

Lewis Structure

Understanding the Lewis structure is vital for representing molecules' electronic configuration. For the compound \(\mathrm{C}_{3} \mathrm{H}_{7} \mathrm{NO}\), constructing the correct Lewis structure involves identifying how atoms like nitrogen, carbon, and oxygen coordinate without forming unwanted bonds such as nitrogen-to-oxygen. In this molecule, nitrogen is part of an amide group attached to an ethyl group (\(\mathrm{C}_{2}\mathrm{H}_{5}\)). This results in the nitrogen atom bonding with two hydrogen atoms and the carbonyl group (\(\mathrm{C}(=\mathrm{O})\)).
The Lewis structure is drawn by arranging electrons around the atoms satisfying the octet rule where applicable, showing single bonds between carbon and nitrogen and the configuration of ethyl group attached to it, as a secondary amide: \(\text{CH}_3\text{CH}_2\text{C}(=\text{O})\text{NH}_2\). It's important to remember that Lewis structures provide a basic blueprint of how atoms share or transfer electrons to form molecules.

VSEPR Theory

Valence Shell Electron Pair Repulsion (VSEPR) Theory helps us predict the shape of a molecule based on electron pair repulsions. For \(\mathrm{C}_{3} \mathrm{H}_{7} \mathrm{NO}\), understanding the spatial arrangement of atoms is key to functioning as intended. The theory posits that electron pairs around a central atom will arrange themselves to minimize repulsion, thereby determining molecular geometry.
In this molecule, examining carbon atoms reveals differing geometries. The first carbon in an ethyl group follows a tetrahedral shape, yielding bond angles of approximately \(109.5^{\circ}\). Conversely, the carbon in the carbonyl group is trigonal planar due to its planar \(\text{sp}^2\) hybridization, leading to bond angles of \(120^{\circ}\). This theory helps us visualize the likely arrangement in 3D space, ensuring accurate predictions and understanding of molecular behavior.

Molecular Geometry

Molecular Geometry offers a focused view on how molecules are shaped based on bonded atoms and electron pairs. By examining \(\mathrm{C}_{3} \mathrm{H}_{7} \mathrm{NO}\), we can dissect the arrangement of atoms within this compound. With the secondary amide structure, the nitrogen atom usually presents a trigonal pyramidal shape due to its lone pair and bonding with hydrogen and carbon atoms.
However, experimental data and resonance effects indicate a trigonal planar form with bond angles near \(120^{\circ}\). This flattening into a plane occurs because the nitrogen's lone pair delocalizes over the carbonyl group, altering its expected shape. Such geometric transformations allow the molecule to stabilize further, which is crucial for understanding real-world molecular interactions.

Hybridization

Hybridization is an essential concept when determining atomic bonding capabilities within molecules. In \(\mathrm{C}_{3} \mathrm{H}_{7} \mathrm{NO}\), nitrogen's hybridization is interesting due to the resonance effects seen in amides. Typically, nitrogen in amines features \(\text{sp}^3\) hybridization, leading to a non-planar geometry.
Yet, in this secondary amide, nitrogen exhibits \(\text{sp}^2\) hybridization as electron delocalization between the nitrogen, carbon, and oxygen alters the bond environment. This \(\text{sp}^2\) hybridization provides a planar formation around nitrogen, facilitating a bond angle nearing \(120^{\circ}\), characteristic of trigonal planar structures. Understanding hybridization clarifies how atomic orbitals mix in molecules, providing a basis for molecular geometry, stability, and reactivity predictions.