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Chapter 11: Problem 89

Acetone, \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CO},\) is widely used as an industrial solvent. (a) Draw the Lewis structure for the acetone molecule and predict the geometry around each carbon atom. (b) Is the acetone molecule polar or nonpolar? (c) What kinds of intermolecular attractive forces exist between acetone molecules? (d) 1-Propanol, \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{OH},\) has a molecular weight that is very similar to that of acetone, yet acetone boils at \(56.5^{\circ} \mathrm{C}\) and 1-propanol boils at \(97.2^{\circ} \mathrm{C}\). Explain the difference.

Short Answer

Expert verified
The Lewis structure of acetone shows trigonal planar geometry around each carbon atom. The molecule is polar due to C=O bond polarity and exhibits dipole-dipole interactions and London dispersion forces as intermolecular forces. The higher boiling point of 1-propanol compared to acetone can be attributed to the presence of hydrogen bonding in 1-propanol, a stronger intermolecular force.

Step by step solution

01

Drawing the Lewis Structure for Acetone

To draw the Lewis structure for acetone, first, count the total number of valence electrons. Carbon (C) has 4 valence electrons, Oxygen (O) has 6 valence electrons, and Hydrogen (H) has 1 valence electron. In acetone, there are 2 CH3 groups (or 2 sets of Carbon and 3 Hydrogens), 1 Carbon, and 1 Oxygen. Total valence electrons = (2 x (1C + 3H)) + 1C + 1O = 2 x (4 + 3) + 4 + 6 = 24 Now, draw the skeletal structure of the molecule, with carbon in the center, one oxygen bonded to the central carbon, and two CH3 groups bonded to the central carbon. Next, place the remaining electrons around the atoms to complete their octets, keeping in mind the molecule has 24 valence electrons in total.
02

Predicting the Geometry around Carbon Atoms

To determine the geometry around each carbon atom, we will use the VSEPR (Valence Shell Electron Pair Repulsion) theory. The central carbon atom is bonded to two other carbon atoms and one oxygen atom, making a total of 3 bonding domains around the central carbon. There are no lone pairs in this case. According to the VSEPR theory, with three bonding domains and no lone pairs, the geometry around the carbon atom is trigonal planar. For the other two carbon atoms (in CH3 groups), each carbon atom is bonded to three hydrogen atoms, with a total of 3 bonding domains, and no lone pairs. This also results in a trigonal planar geometry around these carbon atoms.
03

Determining Polarity of Acetone Molecule

To check the polarity of the molecule, we need to identify the polar bonds and the molecular geometry. There is a difference in electronegativity between Carbon and Oxygen (C-O bond) and Carbon and Hydrogen (C-H bond). The C=O bond is considered polar, while the C-H bonds are less polar. Due to the trigonal planar geometry around the oxygen atom, the C=O bond vectors do not cancel each other, leading to a net molecular dipole moment. Therefore, the acetone molecule is polar.
04

Identifying Intermolecular Attractive Forces

The types of intermolecular attractive forces that exist between acetone molecules are: 1. Dipole-dipole interactions: These occur between the polar molecules of acetone due to a difference in electronegativity between the carbon and oxygen atoms. 2. London dispersion forces: These are induced-dipole/induced-dipole interactions that occur in all molecules, including polar molecules like acetone.
05

Comparing Boiling Points of Acetone and 1-Propanol

We are given that acetone boils at \(56.5^{\circ} \mathrm{C}\), and 1-propanol boils at \(97.2^{\circ} \mathrm{C}\). The higher boiling point of 1-propanol can be attributed to the presence of hydrogen bonding between the molecules, due to the presence of an OH group in its structure. In acetone, dipole-dipole interactions and London dispersion forces exist, but hydrogen bonding is absent. Hydrogen bonding is a stronger intermolecular force compared to dipole-dipole interactions and London dispersion forces, which explains the higher boiling point of 1-propanol compared to acetone despite similar molecular weights.

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Most popular questions from this chapter

The generic structural formula for a 1 -alkyl-3-methylimidazolium cation is where \(\mathrm{R}\) is a \(-\mathrm{CH}_{2}\left(\mathrm{CH}_{2}\right)_{n} \mathrm{CH}_{3}\) alkyl group. The melting points of the salts that form between the 1 -alkyl-3-methylimidazolium cation and the \(\mathrm{PF}_{6}^{-}\) anion are as follows: \(\mathrm{R}=\mathrm{CH}_{2} \mathrm{CH}_{3}\left(\mathrm{~m} . \mathrm{p} .=60^{\circ} \mathrm{C}\right), \mathrm{R}=\mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3}\left(\mathrm{m.p.}=40^{\circ} \mathrm{C}\right)\) \(\mathrm{R}=\mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3}\left(\mathrm{~m} . \mathrm{p} .=10^{\circ} \mathrm{C}\right)\) and \(\mathrm{R}=\mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3}\left(\mathrm{~m} . \mathrm{p} .=-61^{\circ} \mathrm{C}\right) . \mathrm{Why}\) does the melting point decrease as the length of alkyl group increases?

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