 The most acidic species will be A, which has the fluorine to the carboxylic acid (maximum stabilization by F) and has no alkyl sidechains (minimum destabilization by methyl groups).  The least acidic of the optically-active species is H, which has the fluorine to the COOH group (minimum inductive donation from F to COO–) and which also has  a methyl group to the COOH (maximum inductive withdrawing from COO–). d). This reaction produces an ester: R–COOH + R'–OH R–COO–R' + H2O e). The labelled ester is: The reaction occurs as nucleophilic attack of the carboxyl carbon: f). i). The expected product is ii) When ethanoic acid is added to the NaOH solution, an acid-base equilibrium is established: CH3CO*OH + OH– CH3(COO*)– + H2O The ethanoate anion has two equivalent oxygens. When the solution is the label is scrambled: CH3(COO*)– + D+ with the two isotopomers formed in equal amounts (provided that the ethanoic acid / hydroxide system had sufficient time to attain equilibrium). The former isotopomer will yield the same product as i); the latter will give unlabelled propyl ethanoate and labelled water (or, to be more precise, labelled HOD). PROBLEM 17.  a). The volume of a sphere is V = (4/3) r3.  r = 3.5 x 10-10 m, so V = 1.796 10-28 m3. Now for the pressure: PV = nRT, so P = nRT / V.   n = 1 atom = (6.02205 1023)-1 = 1.661 10-24 moles.  so P = 2.2907 104 kPa (i.e. 229 atmospheres)
The most acidic species will be A, which has the fluorine  to the carboxylic acid (maximum stabilization by F) and has no alkyl sidechains (minimum destabilization by methyl groups). The least acidic of the optically-active species is H, which has the fluorine  to the COOH group (minimum inductive donation from F to COO–) and which also has a methyl group  to the COOH (maximum inductive withdrawing from COO–). d). This reaction produces an ester: R–COOH + R'–OH R–COO–R' + H2O e). The labelled ester is: The reaction occurs as nucleophilic attack of the carboxyl carbon: f). i). The expected product is ii) When ethanoic acid is added to the NaOH solution, an acid-base equilibrium is established: CH3CO*OH + OH– CH3(COO*)– + H2O The ethanoate anion has two equivalent oxygens. When the solution is the label is scrambled: CH3(COO*)– + D+ with the two isotopomers formed in equal amounts (provided that the ethanoic acid / hydroxide system had sufficient time to attain equilibrium). The former isotopomer will yield the same product as i); the latter will give unlabelled propyl ethanoate and labelled water (or, to be more precise, labelled HOD). PROBLEM 17. a). The volume of a sphere is V = (4/3)  r3. r = 3.5 x 10-10 m, so V = 1.796  10-28 m3. Now for the pressure: PV = nRT, so P = nRT / V. n = 1 atom = (6.02205  1023)-1 = 1.661  10-24 moles. so P = 2.2907  104 kPa (i.e. 229 atmospheres)
IChO30AU1998PP.docx