2009 – 14077 Ma’am Rea Abuan
Experiment # 1
SOLUBILITY OF ORGANIC COMPOUNDS
State what types of intermolecular forces are present in solutions formed due to intermolecular attractions between the solute and the solvent.
Compound 1
Compound 2
Intermolecular Forces
Class S (Water-soluble) Compounds
Acetone
Water
Hydrogen Bonding & London Dispersion Forces
Diethyl Ether
Dipole – Induced Dipole & London Dispersion Forces
Ethyl Alcohol
(Ethanol)
Water
Hydrogen Bonding & London Dispersion Forces
Diethyl Ether
Dipole – Induced Dipole & London Dispersion Forces
Sucrose
Water
Hydrogen Bonding & London Dispersion Forces
Class N (Neutral) Compounds
Benzyl Alcohol
Sulfuric Acid
Hydrogen Bonding & London Dispersion Forces
Benzaldehyde
Sulfuric Acid
Ion – Induced Dipole & London Dispersion Forces
Class B (Basic) Compounds
Aniline
Hydrochloric Acid
Ion – Dipole & London Dispersion Forces
Class A (Acidic) Compounds
Benzoic Acid
NaOH
Ion – Induced Dipole & London Dispersion Forces
NaHCO3
Ion – Induced Dipole & London Dispersion Forces
Phenol
NaOH
Hydrogen Bonding, Ion – Dipole & London Dispersion Forces
Write the balanced chemical equations for solute-solvent combinations that are formed due to chemical reactions.
Compound 1
Compound 2
Chemical Reactions
Class N (Neutral) Compounds
Benzyl Alcohol
Sulfuric Acid
C6H5CH2OH + H2SO4 → C6H5COOH + 2H2O + SO
Benzaldehyde
Sulfuric Acid
C6H5CHO + H2SO4 → H2O + CH3CH2CH=CH
Class B (Basic) Compounds
Aniline
Hydrochloric Acid
C6H5NH2 + HCl → C6H5NH3+ + Cl-
Class A (Acidic) Compounds
Benzoic Acid
NaOH
C6H5COOH + NaOH → H2O + C6H5COO- + Na+
NaHCO3
C6H5COOH + NaHCO3 → H2O + CO2 + C6H5COO-+ Na+
Phenol
NaOH
PhOH + NaOH →PhO- + Na+ + H2O
On the basis of solubility behavior, show how each of the following pairs of compounds may be distinguished from each other:
CASE 1: CH3NH2 and CH3(CH2)5CH2NH2
The two compounds each possesses an amine group (N-containing) and their only difference lies in their alkyl group. The amine group is determined to be the polar part of both compounds, making them both uniform in that respect. However, the length of the non-polar alkyl region will determine the overall polarity of each organic compound. As the non-polar region lengthens and the polar region remains the same, the compound becomes increasingly water-insoluble [Organic Chemistry Laboratory Manual]. This means that CH3NH2 is more polar and will more readily dissolve in water than CH3(CH2)5CH2NH2.
CASE 2: CH3CHO and HOCH2CHO
The second case presents an opposite scenario of the first case. This time, the number of polar groups is increased while the non-polar region is kept constant. These functional groups represent potential hydrogen bonding sites for the molecules with water. Upon dissolution, it can then be concluded that HOCH2CHO is more polar and will more readily dissolve than CH3CHO.
CASE 3: Benzylamine and Benzyl Alcohol
Both of these organic compounds, upon dissolution in water, are determined to be non-polar substances. Following the flowchart further on, however, reveals that upon reaching the step where the addition of 5% HCl is required, the two separate ways. In the case of benzylamine, it is known to be strongly alkaline and thus reacts with HCl, classifying it as a B (basic) compound [Chemical Land 21]. Benzyl alcohol is insoluble in 5% HCl and goes on to react with concentrated H2SO4, resulting in a class N (neutral) category verdict.
Sources: http://www.chemicalland21.com/industrialchem/organic/BENZYLAMINE.htm Organic Chemistry Laboratory Manual 2008 Edition
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