RI Nitrogen Compounds Tutorial - Answers to Discussion Questions

-8- ANSWERS TO PRACTICE QUESTIONS Q5 Order of increasing basicity: (d) < (a) < (e) < (b) < (c). The basicity of each given compound depends on the availability of the lone pair of electrons on the nitrogen atom to form a dative covalent bond with a proton. The greater this availability, the greater the basicity of the compound. (d) is an amide and is the least basic (effectively neutral) because the orbital containing the lone pair of electrons on the nitrogen atom overlaps with the  electron cloud of both the benzene ring and the C=O bond. As such, the lone pair is delocalised and is the least available to form a dative covalent bond with a proton. The aromatic amines, (a) and (e), are less basic than the aliphatic amines, (b) and (c), because of the delocalisation of the lone pair of electrons on the nitrogen atom into the benzene ring. (a) is less basic than (e) because the electron-withdrawing –NO2 group further decreases the electron density at the nitrogen atom and hence further reduces the availability of the nitrogen lone pair to form a dative covalent bond with a proton. The aliphatic amines, (b) and (c), are stronger bases because the alkyl groups bonded to the nitrogen atom are electron-donating and this increases the electron density at the nitrogen atom, making the lone pair of electrons on the nitrogen atom much more readily available to form a dative covalent bond with a proton. (c) is a stronger base than (b) since there are two electron-donating alkyl groups bonded to the nitrogen atom in (c) compared to only one in (b). Q6a phenylamine and (phenylmethyl)amine Test Add a few drops of Br2(aq) to each sample in a test tube at room temperature. Observations For phenylamine, there would be decolourisation of orange Br2(aq) and the formation of a white precipitate. For (phenylmethyl)amine, there would be no decolourisation of orange Br2(aq) nor formation of a white precipitate. Q6b CH3CH2CH2NH3+Cl- and (CH3)4N+Cl- Test Add NaOH(aq) to each sample in a test tube and warm/heat each mixture gently. Observations For CH3CH2CH2NH3+Cl-, a pungent gas (CH3CH2CH2NH2) which turns moist red litmus paper blue will be evolved on warming. For (CH3)4N+Cl-, there will not be any gas evolved upon warming. Q6c CH3CONH2 and CH3COO–NH4+ Test Add NaOH(aq) to each sample in a test tube and warm each mixture. Observations For CH3COO-NH4+, NH3 gas which turns moist red litmus paper blue, would be evolved almost immediately upon warming. For CH3CONH2, there would not be any NH3 gas evolved upon gentle heating. The NH3 gas would only evolve after strong heating for some time.

-9- Q7 (a) Reagents and conditions Step 1: conc HNO3, conc H2SO4, heat (or 55 °C) Step 2: Sn, conc HCl, heat (under reflux), followed by NaOH(aq) Step 3: Br2(aq) (b) Reagents and conditions Step 1: LiAlH4 in dry ether OR NaBH4 OR H2, Ni, heat Step 2: SOCl2 OR PCl5 OR PCl3 St