Solubility of Organic Compounds

Experiment 1: Solubility of Organic Compounds
(Answers to Questions)

Q1. State what types of inetmolecular forces are present in solutions formed due to intermolecular attractions between the solute and the solvent.

A1.
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

Q2. Write the balanced chemical equations for solute-solvent combinations that are formed due to chemical reactions.

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 – NaOH
PhOH+NaOH→H_2 O+ PhO^-+Na^+
For Basic Compound(s): Aniline – HCl
〖C_6 H_5 NH_2+ HCl→ C_6 H_5 NH〗^(3+)+ 〖Cl〗^-
For Neutral Compounds: Benzyl Alcohol – H2SO4
C_6 H_5 CH_2 OH+H_2 SO_4→C_6 H_5 COOH+2H_2 O+SO Benzaldehyde – H2SO4
C_6 H_5 CHO+H_2 SO_4→ H_2 O+ 〖CH〗_3 CH_2 CH=CH Q3. On the basis of solubility behavior, show how each of the following pairs of compounds may be distinguished from each other.

A3. CH3NH2 and CH3(CH2)5CH2NH2
Both are Nitrogen-containing (amine) compounds but differ in the attached alkyl group. Immediately, we can conclude that CH3NH2 > CH3(CH2)5CH2NH2 in terms of solubility. This conclusion is based on the statement that as the length of the non-polar part of the compound increases (e.g. hydrocarbon, R), with the polar parts kept constant, the polarity then decreases which consequentially decreases its solubility in water [1]. CH3CHO and HOCH2CHO
Both compounds can hydrogen bond with water through O in the carbonyl carbon site. However, HOCH2CHO, aside from bonding with O in the carbonyl site, can also bond with water through the hydroxide site. That being said, we can conclude that HOCH2CHO is more polar than CH3CHO and therefore, in terms of solubility, HOCH2CHO > CH3CHO. Benzylamine and Benzyl Alcohol
Large, non-polar, and benzene ring-containing compounds tend to be quite insoluble in water [2], so both compounds mentioned are insoluble in water. However, benzylamine has an amine group capable of forming/becoming salts when exposed in an acidic solvent (e.g. HCl). Benzyl alcohol, on the other hand, is an alcohol which has stronger H-bonding. Benzyl alcohol is not soluble in HCl (as observed in the results of Experiment 1).
So in terms of solubility in water, both are on par. But in an acidic solvent, Benzylamine > Benzyl alcohol.

References:
[1] Organic Chemistry Laboratory Manual, 2008 Edition, Institute of Chemistry, University of the Philiipines Diliman.
[2] Graziano, G., “Benzene solubility in water: A reassessment,” (2006) Chem. Phys. Letter, Volume 429, Issues 1-3, p 114-118.

References: [1] Organic Chemistry Laboratory Manual, 2008 Edition, Institute of Chemistry, University of the Philiipines Diliman. [2] Graziano, G., “Benzene solubility in water: A reassessment,” (2006) Chem. Phys. Letter, Volume 429, Issues 1-3, p 114-118.