Determining whether ethane, a simple hydrocarbon, can form a hydrogen bond with water requires examining the intermolecular forces that govern their interaction. The answer depends on the specific, stringent criteria required for this powerful attraction. Understanding these chemical principles is key to predicting the behavior and low solubility of ethane in water.
The Requirements for Hydrogen Bonding
Hydrogen bonding is a particularly strong type of intermolecular force, distinct from a true covalent bond. For this interaction to occur, a molecule must contain a hydrogen atom covalently bonded to one of three highly electronegative atoms: Fluorine (F), Oxygen (O), or Nitrogen (N). This requirement is often referred to as the F-O-N rule.
The strong pull of the electronegative atom creates a highly polar bond, leaving the hydrogen atom with a significant partial positive charge (\(\delta+\)). Because the hydrogen atom is small, its proton is left nearly “bare,” resulting in a high charge density. This partially positive hydrogen is then strongly attracted to a lone pair of electrons on a neighboring F, O, or N atom. This strong attraction defines the hydrogen bond, making it substantially stronger than standard dipole-dipole interactions. Water is a classic hydrogen-bonding molecule because its hydrogen atoms are bonded directly to oxygen, giving it unique properties like a high boiling point.
Chemical Structure and Properties of Ethane
Ethane has the chemical formula \(\text{C}_2\text{H}_6\), consisting of two carbon atoms bonded to each other and six hydrogen atoms. Its structure is characterized by a single bond between the two carbon atoms, with three hydrogen atoms attached to each carbon in a tetrahedral arrangement. Every bond in the ethane molecule is a Carbon-Hydrogen (\(\text{C}-\text{H}\)) bond.
The polarity of a bond is determined by the difference in electronegativity between the bonded atoms. Carbon (2.5) and hydrogen (2.1) have a difference of only 0.4. Because this difference is very small, the \(\text{C}-\text{H}\) bond is classified as non-polar or only very weakly polar.
Since ethane is composed entirely of \(\text{C}-\text{C}\) and \(\text{C}-\text{H}\) bonds, it does not contain a hydrogen atom attached to Fluorine, Oxygen, or Nitrogen. This structural reality means ethane cannot act as a hydrogen bond donor, as it lacks the necessary highly polarized hydrogen atom. Therefore, ethane is chemically incapable of forming true hydrogen bonds with water.
Interacting Forces and Solubility in Water
Because ethane cannot form hydrogen bonds, the primary intermolecular forces governing its interaction with water are the much weaker London Dispersion Forces (LDFs). These forces, also known as Van der Waals forces, are transient attractions arising from temporary fluctuations in electron distribution, creating momentary dipoles. Since ethane is a symmetrical, non-polar molecule, LDFs are the only significant force between ethane and water molecules.
The “like dissolves like” principle contrasts the non-polar ethane with the highly polar water. Water molecules are strongly attracted to each other through their extensive network of hydrogen bonds, which are much more powerful than the weak LDFs. For ethane to dissolve, it must break the strong water-water hydrogen bonds and replace them with the very weak ethane-water LDFs.
The energy required to break water’s strong internal attractions is not compensated by the formation of the weak attractions with ethane. This energy imbalance makes the mixing process energetically unfavorable, classifying ethane as a hydrophobic substance. Consequently, ethane is virtually insoluble in water; only minute traces of the gas will dissolve.