Question 21.4: Dimethyl sulfide, CH3SCH3, is one of the products of the sul...

Collect and Organize We need to draw Lewis structures for two sulfur compounds, determine their geometry, and compare their boiling points. Guidelines for drawing Lewis structures were discussed in Chapter 8. We are also asked to describe the hybrid orbitals of S in these compounds. Hybrid orbitals were described in Chapter 9. The effect of structure on boiling points was discussed in Chapter 10.

Analyze The Lewis structure for a molecule depends on the total number of valence electrons available, distributed over the atoms so that each atom has a complete octet (except H, which has a duet). Atoms with Z \geq 13 may have an expanded octet if such an arrangement leads to lower formal charges on the atoms. The arrangement of the bonding and lone pairs on the central atom allow us to predict intermolecular forces such as dipole–dipole interactions, hydrogen bonds, and dispersion forces. We can account for observed molecular geometries by combining s, p, and sometimes d orbitals to form hybrid atomic orbitals such as sp,  sp^{2},  sp^{3},  sp^{3}d, and sp^{3}d^{2}.

Solve (a) Dimethyl sulfide has a total of 20 valence electrons: 6 from the S atom, 6 from the H atoms, and 8 from the C atoms. These electrons can be distributed in six C—H single bonds and two S—C single bonds, with four electrons remaining as two nonbonding pairs on S. Dimethyl sulfoxide has an oxygen atom bonded to the sulfur in dimethyl sulfide. This oxygen brings an additional six valence electrons to the molecule, giving (CH_{3})_{2}SO a total of 26 valence electrons. Sharing one of the S lone pairs with O will complete the octets of both S and O but will leave O with a formal charge of –1 and sulfur with a formal charge of +1. Forming a S=O double bond makes the formal charges on both S and O equal to zero but requires that sulfur have an expanded octet. The two structures for (CH_{3})_{2}SO represent resonance forms:

(b) Methanethiol has 14 valence electrons: 6 from the S atom, 4 from the H atoms, and 4 from the C atom. The electrons are distributed in three C—H single bonds, one C—S single bond, and one S—H single bond. As in dimethyl sulfide, there are two nonbonding pairs left on S. Both dimethyl sulfide and methanethiol contain a sulfur atom surrounded by two bonding pairs and two nonbonding pairs of electrons for a total of four electron pairs. The electron-pair geometry is tetrahedral, and the molecular geometry is bent:

Both methanethiol and dimethyl sulfide are polar as a result of their molecular geometry and experience dipole–dipole interactions. Both molecules interact through dispersion forces as well. One might expect that the stronger dipole dipole forces in CH_{3}SH would lead to a higher boiling point than for CH_{3}SCH_{3}, but we observe the opposite. We conclude that the dispersion forces in CH_{3}SCH_{3} have a greater effect than the dipole–dipole interactions in CH_{3}SH, leading to dimethyl sulfide boiling at a temperature about 32°C higher than methanethiol.
(c) The similar geometry for CH_{3}SCH_{3} and CH_{3}SH, with each molecule’s central sulfur atom having two bonded atoms and two lone pairs, is consistent with sp^{3} hybrid orbitals on the sulfur atom in both cases.

Think About It Methanethiol and dimethyl sulfide differ only in that the latter has a CH_{3} group in place of a hydrogen atom. (This relationship between thiols and sulfides corresponds to the relationship between alcohols and ethers.) Sulfur does not expand its octet in these compounds because using the lone pairs on S to form multiple bonds is not needed. The boiling points of CH_{3}SCH_{3} and CH_{3}SH reveal an important observation: many weaker bonds (dispersion forces) in CH_{3}SCH_{3} can outweigh a few stronger forces (dipole–dipole forces) in CH_{3}SH.