Problem 92 - Chemistry The Central Science In Si Units

Question

Ethyl propanoate, \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COOCH}_{2} \mathrm{CH}_{3},\) gives a fruity pineapple-like smell. (a) Draw the Lewis structure for the molecule, assuming that carbon always forms four bonds in its stable compounds. (b) How many \(\sigma\) and how many \(\pi\) bonds are in the molecule? (c) Which CO bond is shortest in the molecule? (d) What is the hybridization of atomic orbitals around the carbon atom associated with that short bond? (e) What are the approximate bond angles around each carbon atom in the molecule?

Step-by-Step Solution

Answer

(a) Draw the Lewis structure. (b) 11 \(\sigma\) and 1 \(\pi\). (c) C=O bond. (d) sp\(^2\) hybridization. (e) 109.5° around sp\(^3\), 120° around sp\(^2\).

Step 1: Lewis Structure

To draw the Lewis structure of ethyl propanoate, first identify the atoms and the connectivity. The structure is as follows: \( \mathrm{CH_3CH_2COOCH_2CH_3} \). Connect the atoms: start with the central \( \mathrm{COO} \) group. The first carbon (C) is bonded to two oxygens (O) and the other C is part of a \( \mathrm{CH_2} \) group. Each \( \mathrm{C} \) in the \( \mathrm{CH_3} \) and \( \mathrm{CH_2} \) groups forms four bonds with hydrogens (H) or carbons, ensuring that each carbon makes four bonds. Connect the remaining atoms accordingly to complete the structure with covalent bonds.

Step 2: Count Sigma and Pi Bonds

In the Lewis structure, count the number of \(\sigma\) and \(\pi\) bonds. Every single bond is a \(\sigma\) bond. In ethyl propanoate, all C-H and C-C single bonds are \(\sigma\) bonds. Additionally, the C=O double bond contains one \(\sigma\) and one \(\pi\) bond. In total, there are 11 \(\sigma\) bonds and 1 \(\pi\) bond.

Step 3: Shortest CO Bond

In the structure, a C=O bond typically represents a shorter bond as compared to C-O bonds. The double bond between carbon and oxygen in the \(\mathrm{COO} \) group is the shortest due to the additional \(\pi\) bonding which draws the atoms closer together, making the bond stronger and shorter.

Step 4: Hybridization of Carbon

The carbon atom involved in the C=O bond (shortest bond) is sp\(^2\) hybridized. This is because it is double-bonded to oxygen and single-bonded to another carbon, involving one s and two p orbitals to form three \(\sigma\) bonds and the remaining p orbital participates in the \(\pi\) bond.

Step 5: Bond Angles Around Each Carbon

The approximate bond angles depend on the hybridization. For sp\(^3\) hybridized carbon atoms (like those in \(\mathrm{CH_3} \) and \(\mathrm{CH_2} \) groups), angles are about 109.5°. For the sp\(^2\) hybridized carbon in the COO group, the bond angles are approximately 120°.

Key Concepts

Sigma Bonds

In ethyl propanoate, sigma (() bonds are a fundamental component. These are the single covalent bonds formed by the end-to-end overlap of atomic orbitals. In this molecule:

• All carbon-hydrogen (C-H) single bonds are sigma bonds.
• Carbon-carbon (C-C) single bonds are also sigma bonds.
• Even in the double bond, one of the bonds is a sigma bond while the other is a pi bond.

Altogether, the ethyl propanoate molecule has 11 sigma bonds.
This includes all single bonds and the sigma bond part of the double bond between carbon and oxygen in the central COO group.

Pi Bonds

Pi (() bonds occur from the side-to-side overlap of p orbitals and are found in double or triple bonds. In ethyl propanoate:

• Only one pi bond exists, which is part of the carbon-oxygen (C=O) double bond in the COO group.

Unlike sigma bonds, pi bonds provide additional electron density above and below the bond axis, contributing to the bond's strength but reducing flexibility. This pi bonding in the C=O double bond is why it is shorter and stronger than a typical C-O single bond.

Hybridization

Hybridization is the mixing of atomic orbitals to form new equivalent hybrid orbitals. These determine the geometry and bonding properties of a molecule. In ethyl propanoate:

• The carbon atom involved in the C=O bond is sp1 hybridized. This involves one s and two p orbitals mixing to form three sp1 hybrid orbitals. The unhybridized p orbital forms the pi bond.
• Other carbon atoms, like those in the methyl (CH1) and methylene (CH2) groups, are sp3 hybridized, using one s and three p orbitals to form four equivalent sp3 hybrid orbitals.

Sp1 hybridization results in planar geometry around the carbon, leading to larger bond angles compared to the tetrahedral arrangement from sp3 hybridization.

Bond Angles

Bond angles are influenced by the hybridization of the central atom. They dictate the overall shape of the molecule:

• In ethyl propanoate, carbon atoms forming single bonds in the CH1 and CH2 groups are sp3 hybridized, resulting in bond angles approximately 109.5°, consistent with tetrahedral geometry.
• The carbon in the COO group, bonded with a double bond, is sp1 hybridized. This leads to bond angles of around 120°, aligned with planar geometry.

This variances in bond angles define the molecule's stereochemistry, enabling its unique three-dimensional structure.