Why does diethyl ether have a lower dielectric constant than diethyl sulfide? Since O is more electronegative than S, I expected that the dielectric constant of diethyl ether would be higher. However, it is known that dielectric constant of diethyl ether is lower than that of diethyl sulfide.
Please let me know
The dielectric constant is not directly proportional to the electronegativity of the solvent atom. Other factors, such as the size and shape of the solvent molecule, also play a role in determining its dielectric constant.
The dielectric constant of diethyl ether (ETH) is lower than that of diethyl sulfide (DES) because the structure of ETH molecules is more symmetrical, with two symmetrical ether oxygen atoms.
This symmetrical structure results in a lower dipole moment and a weaker ability to separate charged species compared to DES, which has a more polar sulfur atom.
Two factors that contribute directly to a higher dipole moment for diethyl sulfide vs. diethyl ether:
1. C-S bonds are longer than C-O bonds; dipole moments are proportional to the distance between the charges.
2. C-S-C angle is close to 90 degrees, while C-O-C angle is close to 108 degrees. The sharper angle leads to a larger portion of the two bond moments being additive to each other, hence an overall larger dipole moment for the entire molecule. (In diethyl ether relatively larger portions of the two bond moments cancel each other out.)
You can add to this
3. Higher polarizability of the sulfur atom compared to the oxygen atom.
On the other hand, the C-S bond has a lower charge separation than the C-O bond due to the smaller electronegativity difference. But overall the two or three first factors "win". How do we know? Not because of majority rule but because the data says so.
- Badges
- Science topic