Chem 31.1 ATQ Experiment 1

Experiment 1 – Solubility of Organic Compounds Theoretical Data 1

Theoretical Data 2

Test compound

Hexane Aniline Benzamide Cellulose dichloromethane benzyl alcohol ethyl alcohol ethyl acetate Acetone Sucrose Toluene Glycine benzaldehyde benzoic acid

Classification:

Water

Ether

+ + + + + + + -

+

5% NaOH

Solvents 5% NaHCO3

5% HCl

Conc. H2SO4

-

+

-

-

+ +

-

-

-

-

-

+ +

-

+ + + -

Answers to Questions:

1. State what types of intermolecular forces are present in solutions formed due to intermolecular attractions between the solute and the solvent

For Water-Soluble Compounds: Acetone – Water: Hydrogen bonding and van der Waals forces Acetone – Diethyl Ether: Dipole – (induced) dipole and van der Waals forces Sucrose – Water: Hydrogen bonding and van der Waals forces Ethyl alcohol – Water: Hydrogen bonding and van der Waals forces Ethyl alcohol – Dietheyl Ether: Dipole – (induced) dipole and van der Waals forces For Acidic Compounds: Benzoic Acid – NaOH: Ion – (induced) dipole and van der Waals forces Benzoic Acid – NaHCO3: Ion – (induced) dipole and van der Waals forces Phenol – NaOH: Hydrogen Bonding, Ion – dipole, and van der Waals forces For Basic Compound(s): Aniline – HCl: Ion – dipole and van der Waals forces For Neutral Compounds: Benzyl alcohol – H2SO4: Hydrogen bonding and van der Waals forces Benzaldehyde – H2SO4: Ion – (induced) dipole and van der Waals forces  

Solutions with water: ethanol and water: H-bonding sucrose with water: H-bonding phenylmethanol with water: H-bonding (partial solubility) Solutions with ether: sucrose with ether: London dispersion phenylmethanol with ether: London dispersion

2. Write the balanced chemical equations for solute-solvent combinations that are formed due to chemical reactions.

.

benzoic acid: C6H5COOH + NaOH benzamide: C6H5CONH2 + H2SO4 phenylmethanal: C6H5CHO + H2SO4 phenol: C6H5OH + NaOH aniline: C6H5NH2 + HCl A2. For Acidic Compounds:

-> C6H5COO- Na+ + H2O -> C6H5COSH2 + HNO2 -> C6H5 -> C6H5O- Na+ + H2O -> C6H5NH3 + Cl-

Benzoic Acid – NaOH 〖C_6 H_5 COOH+NaOH→ H_2 O+ C_6 H_5 COO〗^-+Na^+ Benzoic Acid – NaHCO3 〖C_6 H_5 COOH+NaHCO_3→ H_2 O+ CO_2+C_6 H_5 COO〗^-+Na^+ Phenol – NaOH PhOH+NaOH→H_2 O+ PhO^-+Na^+

3. On the basis of solubility behaviour, show how each of the following pairs of compounds may be distinguished from each other: a. CH3COOH and CH3(CH2)5COOH - Diethyl ether b. CH3CH2OCH2CH3 and CH3CH2OH – NaOH c. toluene and benzaldehyde - concentrated H2SO4 d. tert-butyl chloride and tert-butanol - H2O Reference: http://webapps.utsc.utoronto.ca/chemistryonline/solubility.php