Identifying Unknown Organic Compounds By Solubility Properties

Summary
In this experiment we learned to identify an unknown organic compound based on its solubility properties. 
We were provided with an unknown solid compound and five liquid solvents. The unknown was one of three possible organic compounds which we were able to identify based on its solubility in each of the provided solvents. The following report explores some of the molecular properties of the given solutes and solvents and explains the effects of these properties on solubility.

Note: For the purposes of this experiment, we consider the salute soluble if 0.01 g of it can be dissolved in 1ml of the solvent at room temperature.


How intermolecular forces effect solubility
There are three main steps involved in dissolution of a solid solute in a liquid solvent.
  1. Breakdown of the bonds between solute particles. This step requires energy and is therefore endothermic.
  2. Breakdown of the bonds between solvent particles which is also an endothermic step.
  3. Formation of new bonds between solute and solvent particles. This step is exothermic since it results in energy release.
The more exothermic the entire process is, the more soluble the solute. In order for the dissolution to proceed in the standard laboratory conditions, the process has to be exothermic or require very little energy (as much as can be provided by room temperature). In other words, the total energy required to break the bonds between solute particles and solvent particles has to be less than, equal to or only slightly greater than the energy release during formation of the bonds between the salute and solvent particles. Therefore, salute-solvent intermolecular forces has to be greater than or similar to the salute-salute and solvent-solvent intermolecular forces. 

The strength of the intermolecular forces of a compound depends on the size, structure, polarity and hydrogen-bonding capabilities of its particles. Therefore, compounds with similar characteristics to the solvent are more likely to be soluble. 

The solvents and their properties
Water molecules are highly polar with a relatively low mass. In addition, they are capable of forming hydrogen bonds. Each water molecule is capable of forming four hydrogen bonds with its surrounding molecules. 
A hydrogen bond is a strong form of dipole–dipole interaction but it also has some qualities that is similar to covalent bonds. Hydrogen bonds are stronger than van der Waals and dipole–dipole interactions but weaker than ion-dipole, ionic and covalent bonds. A compound is soluble in water if its particles are capable of forming bonds with water molecules that are similar to or greater than hydrogen bonds in force. 
Many organic compounds contain a non-polar backbone but also have a polar functional group. Water solubility of such compounds depends on their size, structure and the functional group's bonding capabilities. We will see some examples of these compounds in this experiment.


Hydrochloric Acid is a very strong aqueous acid and reacts with compounds with basic functional groups to form salts. If an organic  compound is insoluble in water but dissolves in hydrochloric acid, we can conclude that the compound has a basic functional group and reacts with hydrochloric acid to form a water soluble salt. The water insolubility of such compound probably means that it contains more than 5 carbons since longer carbon chains decreases water solubility.  The reactions are as follows:
Formation of Hydrochloric Acid
Hydrogen chloride (HCl) + Water (H2O) → Hydronium ion (H3O+) +  Chloride ion (Cl−)
Reaction of basic water-insoluble organic compound with hydrochloric acid
Amine (R-NH2) + Hydrochloric acid (H3O+, Cl-) → R-ammonium chloride (R-NH3+ Cl-) +  Water (H2O) 

The reaction produces R-NH3+ Cl- which is a water soluble salt.




Aqueous Sodium Hydroxide (NaOH(aq)) is a strong basic solution and can help dissolve organic water-insoluble compounds that have an acidic functional group. NaOH(aq) reacts with the acidic functional group and forms a water soluble salt. For example, sodium hydroxide can be used to de-protonate weak acidic organic compounds such as phenols to create water soluble salts:

C6H5OH + Na+ + OH- + H2O → C6H6ONA+ + OH- + H3O+
Phenoxide anion (C6H6O-) is highly soluble in water.
Aqueous Sodium Bicarbonate (NaHCO3(aq)) is a weak basic solution that can also help dissolve water-insoluble organic compounds with an acidic functional group by salt formation. However, the acidic functional group needs to be stronger than the previous example since sodium bicarbonate is a much weaker base than sodium hydroxide. The following is an example of such reaction:

Reaction of Sodium Bicarbonate in Water
NaHCO3 + H2O \rightarrow H2CO3 + OH- + Na+
H2CO3 + H2O  \rightleftarrows  HCO3- + H3O+
HCO3- + H2O \rightleftarrows CO32-  + H3O+

Reaction of Aqueous Sodium Bicarbonate with a Carboxylic Acid

R-COOH + Na+ + OH- + H2O → R-COONA+ + OH- + H3O+

The above reaction produces R-COO- NA+ which is a water soluble salt. 

Diethyl ether (C2H5)2O is a colourless organic solvent suitable for dissolving compounds with weaker intermolecular forces. Ether molecules are slightly polar due to the electronegative oxygen. They are however unable to form hydrogen bonds with each other and are only connected by dipole-dipole interactions which is much weaker than hydrogen bonds and ion-dipole interactions that existed in the previously discussed solvents. In other words, breaking the bonds between ether molecules takes less energy and so ether is able to dissolve even non-polar compounds. Keep in mind that because of its weak intermolecular forces, ether is not a good solvent for more polar solvents.





The solutes and their properties
The unknown compound that we had to identify was one of the following three compounds:

Glucose C6H12O6

In Water: Glucose has five hydroxyl groups and they are all capable of forming hydrogen bonds with water. This makes glucose highly soluble in water.

In Hydrochloric Acid: The high polarity of glucose makes is soluble in hydrochloric acid. The hydroxyl groups are likely to form ion-dipole bonds with the H+ ion. In addition, as before, many hydrogen bonds with water molecules will help dissolve glucose in hydrochloric acid.

In Aqueous Sodium Hydroxide: Glucose is soluble in the solution of sodium hydroxide in water. Hydrogen bonds with water molecules will help the solubility of glucose in this solvent. Also, glucose is likely to react with the hydroxide ions, losing some of the hydrogens attached to the hydroxyl groups to them. This might result in some ion-dipole interactions which also help the the solubility of glucose.

In Aqueous Sodium Bicarbonate: The high polarity of glucose in addition to the hydroxyl groups will again make glucose soluble in aqueous sodium bicarbonate solution.  

In Diethyl ether (C2H5)2O: The high polarity of glucose has the reverse effect this time. As mentioned before, ether has a low polarity and is not good at dissolving highly polar compounds. The ether molecules are unable to break the intermolecular forces of glucose and this makes glucose insoluble in ether.



P-toluidine C7H9N 


In Water: P-toluidne is slightly basic due to the amino functional group but it is insoluble in water because of the benzene ring. The polarity of the amine group is simply not enough to make a molecule of p-toluidine's shape and size water soluble.

In Hydrochloric Acid: P-toluidine reacts with HCl.H2O to form a water soluble ammonium salt. It is therefore soluble in hydrochloric acid. This is an example of water solubility by salt formation.

In Aqueous Sodium Hydroxide: Sodium Hydroxide is a strong base and so it is unable to protonate p-toluidine to form a water soluble salt. P-toluidine is therefore insoluble in aqueous sodium hydroxide.

In Aqueous Sodium Bicarbonate: Unlike hydrochloric acid, sodium bicarbonate is not a strong enough acid to protonate the amine group. As a result, P-toluidine remains insoluble in aqueous sodium bicarbonate solution.

In Diethyl ether (C2H5)2O: P-toluidine is soluble in ether due to the low polarity of its molecules and weak intermolecular forces.


Benzoic Acid C6H5COOH


In Water: Benzoic acid is a kind of carboxylic acid (R-COOH). Water-solubility of carboxylic acids decrease as the number of carbons increase. In case of benzoic acid, the molecule contains seven carbons and six of them form a benzene ring. Benzoic acid is capable of forming hydrogen bonds with water molecules and if you increase the temperature some benzoic acid molecules even loose their hydrogen to water molecules and form ions. Solubility of benzoic acid increases as more molecules lose a proton to form ions. This causes benzoic acid to be insoluble in cold water but soluble in hot water. 

In Hydrochloric Acid: Hydrochloric acid is a very strong acid. Hydrogen chloride molecules like to lose a proton and become ions. In order for benzoic acid to become soluble in an aqueous solution the benzoic acid molecules need to also lose protons and this is harder to accomplish in the presence of acidic compounds such HCL. This makes benzoic acid insoluble in hydrochloric acid.

In Aqueous Sodium Hydroxide: Sodium Hydroxide is a strong base and helps deprotonate benzoic acid  molecules. As a result, more benzoic acids turn to water soluble ions and this makes benzoic acid soluble in the aqueous sodium hydroxide solution

In Aqueous Sodium Bicarbonate: Although the aqueous sodium bicarbonate solution is slightly basic, it is not strong enough to deprotonate enough benzoic acid molecules to dissolve it. As a result, benzoic acid is insoluble in aqueous sodium bicarbonate solution.

In Diethyl ether (C2H5)2O: Benzoic acid is soluble in ether due to the low polarity of its molecules and weak intermolecular forces.

The Experiment

  1. Add 0.01 g of the unknown compound to five different test tube.
  2. Add 1 ml of each of the available solvents to each test tube. 
  3. Record the solubility of the unknown compound in each solvent. 
  4. If the compound is insoluble in water, try heating the test tube. Record your observation.
  5. Identify the unknown compound based on its solubility in each solvent.
My sample displayed similar solubility properties to benzoic acid. I was therefore able to conclude that my sample is in fact benzoic acid.



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