Acetone \(\left[\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CO}\right]\) 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 mol-ecules? (\boldsymbol{d} 1 Propanol ~ ( C H ~ \(_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{OH}\) ) 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.

First and foremost, we will draw the Lewis structure for acetone (\((\mathrm{CH}_{3})_{2} \mathrm{CO}\)). The carbon atom in the middle is the central atom, double bonded to an oxygen and single bonded to two methyl groups. The hydrogen atoms are bonded to each of the carbon atoms in the methyl group. Here's the Lewis structure: H H H | | | H-C-C-C=O | | | H H H Now let's predict the geometry around each carbon atom in acetone.

Since the carbon atoms in the two methyl groups have three neighboring atoms and no lone pairs, they exhibit a trigonal planar geometry (120° bond angles). The central carbon atom is double bonded to an oxygen atom and single bonded to two methyl groups, so it also has three neighboring atoms with no lone pairs, and therefore, also has trigonal planar geometry (120° bond angles).

To determine if the acetone molecule is polar or nonpolar, we need to consider the electronegativities of the constituent atoms and the geometry of the molecule. Oxygen is more electronegative than carbon, so the carbon-oxygen double bond is polar. However, due to the symmetry of the acetone molecule resulting from its trigonal planar geometry, the dipole moments around the central carbon atom cancel out each other. Therefore, the overall acetone molecule is nonpolar.

Between acetone (\((\mathrm{CH}_{3})_{2} \mathrm{CO}\)) molecules, the main intermolecular attractive forces are London dispersion forces. These are relatively weak forces and arise due to temporary fluctuations in electron distribution which induce temporary dipoles.

Acetone and 1-propanol have similar molecular weights, but their boiling points are significantly different (acetone boils at \(56.5^{\circ} \mathrm{C}\), while 1-propanol boils at \(97.2^{\circ} \mathrm{C}\)). The main reason for this difference is the type of intermolecular attractive forces present in each molecule. 1-propanol (\(CH_{3} CH_{2} CH_{2} OH\)) has an -OH group, which can form hydrogen bonds with other molecules. Hydrogen bonding is a much stronger intermolecular force than London dispersion forces, which are present in acetone. As a result, 1-propanol has a higher boiling point since more energy is required to break these stronger hydrogen bonds. In summary, the difference in boiling points between acetone and 1-propanol is due to the difference in the types and strength of intermolecular attractive forces present in each molecule—with hydrogen bonding in 1-propanol being stronger than London dispersion forces in acetone.