RI Nitrogen Compounds Tutorial - Answers to Self-check Questions

-6- Suggested Answers to Self-check Questions Q1 1o amines 2 o amines 3 o amines Q2 (a) NH2H2N (b) NH2 CHO (c) (d) (e) Q3 Order of increasing boiling point: CH3CH2CH2CH3 < (CH3)3N < CH3CH2CH2NH2 < CH3CH2CH2OH (b) (d) (a) (c) The 4 compounds similar electron cloud size. Boiling each compound involves overcoming the intermolecular forces within each compound. The stronger the intermolecular forces are, the greater is the energy needed to overcome these forces and the higher is the boiling point of the compound. (b) has the lowest boiling point as only weak instantaneous dipole-induced dipole interactions exist between these non-polar molecules. (d) has a higher boiling point than (b) because the molecules in (d) are polar and the permanent dipole-permanent dipole forces that exist between them are stronger than the instantaneous dipole- induced dipole forces between the molecules in (b). (a) has a higher boiling point than (d) because the hydrogen bonds that exist between the molecules of (a) are stronger than the permanent dipole-permanent dipole forces that exist between the molecules of (d). (c) has the highest boiling point because the hydrogen bonds between molecules of (c) are stronger than that between molecules of (a) as the O–H bond is more polar due to the greater electronegativity

-7- difference between O and H than between N and H. Thus the largest amount of energy is needed to overcome the intermolecular forces in (c) than in the other compounds. Q4a Test Add neutral FeCl3(aq) to each sample in a test tube. Note: Br2 cannot be used as both phenol and phenylamine will react to result in decolourisation. Observations For HO CH 2NH2 a violet colouration would be observed. For H2N CH 2OH there would be no violet colouration. Q4b Test Add 2,4-dinitrophenylhydrazine to each sample in a test tube. Add I2(aq) and NaOH(aq) to each sample in a test tube and heat each mixture in a hot water bath. Observations For an orange precipitate would be observed. For there would be no orange precipitate. For a yellow precipitate of CHI3 would be observed. For there would be no yellow precipitate. Note: Br2 cannot be used as both compounds will react to result in decolourisation.