Experiment 1 – Solubility of Organic Compounds

Theoretical Data 1

Theoretical Data 2

Test compound SolventsWater Ether 5%

NaOH5%

NaHCO3

5% HCl Conc. H2SO4

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

Classification:

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 forcesAcetone – Diethyl Ether: Dipole – (induced) dipole and van der Waals forcesSucrose – Water: Hydrogen bonding and van der Waals forcesEthyl alcohol – Water: Hydrogen bonding and van der Waals forcesEthyl alcohol – Dietheyl Ether: Dipole – (induced) dipole and van der Waals

forces For Acidic Compounds:Benzoic Acid – NaOH: Ion – (induced) dipole and van der Waals forcesBenzoic Acid – NaHCO3: Ion – (induced) dipole and van der Waals forcesPhenol – 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 forcesBenzaldehyde – H2SO4: Ion – (induced) dipole and van der Waals forces  

Solutions with water:ethanol and water: H-bonding sucrose with water: H-bondingphenylmethanol with water: H-bonding (partial solubility)

Solutions with ether:sucrose with ether: London dispersionphenylmethanol 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 -> C6H5COO- Na+ + H2Obenzamide: C6H5CONH2 + H2SO4 -> C6H5COSH2 + HNO2

phenylmethanal: C6H5CHO + H2SO4 -> C6H5

phenol: C6H5OH + NaOH -> C6H5O- Na+ + H2Oaniline: C6H5NH2 + HCl -> C6H5NH3 + Cl-

A2.For Acidic Compounds: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 – NaOHPhOH+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