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 Step 3: KCN in ethanol, heat (under reflux) Step 4: LiAlH4 in dry ether OR H2, Ni, heat (c) Reagents and conditions Step 1: HCl(aq), heat (under reflux) Step 2: SOCl2 OR PCl5 OR PCl3 Step 3: LiAlH4 in dry ether

-10- (d) Reagents and conditions Step 1: LiAlH4 in dry ether Step 2: SOCl2 OR PCl5 OR PCl3 Step 3: excess conc NH3 in ethanol, heat in sealed tube Step 4: SOCl2 OR PCl5 OR PCl3 Step 5: LiAlH4 in dry ether

-11- Q8a Functional groups present: 3o aromatic amine, 2o aliphatic amine and 2o amide Q8b The three nitrogen-containing groups can be arranged in the following order of increasing pKb value: (2) < (1) < (3). The smaller the pKb value, the larger is the Kb value and hence the stronger is the basicity of the nitrogen-containing group. In this case, the basicity is determined by the availability of the lone pair of electrons on the N atom to form a dative covalent bond with a proton. The greater this availability, the greater the basicity. (2) is a secondary amine and is the strongest base here. The two electron-donating alkyl groups bonded to the N atom increases the electron density at the N atom and makes the lone pair of electrons at the N atom more available to form a dative covalent bond with a proton. (1) is an aromatic amine. It is a weaker base than (2) because the lone pair of electrons on the N atom delocalises into the benzene ring and is less available to form a dative covalent bond with a proton. (3) is an amide. It is the weakest base here because the orbital containing the lone pair of electrons on the N atom overlaps with the  electron cloud of the adjacent C=O group and the lone pair of electrons is delocalised, and hence not available to form a dative covalent bond with a proton. Q8c (i) N N N H O H H + H +Cl- Cl-  acid-base reaction, amide is not protonated (iv) O N N N O H  N(2) undergoes condensation to form amide (ii) N N N H O H H + H + HO H H+ +Cl- Cl- Cl-  acid-base reaction and amide undergoes acid hydrolysis (v) N N N H + + I- O I-  nucleophilic substitution occurs with N(1) and N(2) functioning as nucleophiles (iii) N N N H H H + CH3COO-Na+  amide undergoes base hydrolysis (vi) N N N H O HBr Br  benzene ring of aromatic amine undergoes electrophilic substitution (similar to phenylamine)

-12- Q8d Type of reaction: nucleophilic substitution X: O H Br Br NN Q9a Q9b Q10(a) – Q, R, S and T Information Deduction / Explanation Q has molecular formula, C8H9NO. Q is likely to contain a benzene ring due to C:H ≈ 1:1. Q, C8H9NO, is a neutral compound. Q may contain an amide or –CN group. Q on heating under reflux with NaOH(aq) yields a salt R and S. Alkaline hydrolysis of Q occurred. Q contains the amide functional group. (Note to tutors: Q cannot be a nitrile because while the basic hydrolysis of a nitrile will yield a carboxylate salt, NH3 (Mr = 17) will also be produced, which cannot be S. R is a carboxylate salt. S is an amine. S (Mr = 93) reacts with Br2(aq) to form T, C6H4Br3N. Electrophilic substitution of S occurred. S is an aromatic amine. S is which has Mr = 93. T is Since the sum of carbon atoms in R and S is the same as that in Q, then the number of carbon atoms in R = 8 – 6 = 2. Hence R is CH3COO−Na+. Hence Q is NH2 NH2Br Br Br N C CH3 H O

-13- Q10(b) – U (C8H9NO) and V U V Q10(b) – W (C8H9NO), X, Y and Z W C N CH3 O H X CH3NH2 Y Z Q11 Q12a Proteins can be hydrolysed in the laboratory by prolonged heating with aqueous HCl or H2SO4 of higher concentration (e.g. 6 mol dm−3 HCl). Q12b Overlapping regions are underlined. Chymotrypsin Trypsin Sequence of fragments in P asp-lys-gly-phe gly-phe-lys asp-lys-gly-phe-lys lys-val-arg val-arg lys-val-arg val-phe val-phe-asp-lys val-phe-asp-lys Based on sequences of fragments (last column), find overlapping regions again. val-phe-asp-lys asp-lys-gly-phe-lys lys-val-arg Sequence of amino acids in P: val-phe-asp-lys-gly-phe-lys-val-arg Q13a There are 6 ways of forming the tripeptide. Note: Each tripeptide is formed from 3 different amino acids. BCD CBD DBC BDC CDB DCB C O CH3 NH2 CH3 COO-Na+ COO-Na+

-14- Q13b or the zwitterionic form Q13c (i) BCD tripeptide in (b) + dilute aqueous sodium hydroxide (ii) BCD tripeptide in (b) + dilute hydrochloric acid Q13d A zwitterion is an electrically neutral molecule with oppositely charged ends. Zwitterionic form of B Zwitterionic form of D Note to tutors: Link back to Q11 to explain why the