Midterm Exam III. CHEM 181: Introduction to Chemical Principles Solutions C HC N H 3 C. pka = 9.7

Transcription

1 Midterm Exam III EM 181: Introduction to hemical Principles Solutions 1. For reference, here are the pk a values for four weak acids (not all resonance structures shown): N 3 N N pka = 3.5 pka = 9.7 pka = 4.8 pka = 16 Now consider this list of compounds (not all resonance structures shown): N N N 3 2 N N 1

2 (a) An unknown acid has a pk a of 9.9 and the following IR spectrum: i. Identify this compound from the list on page 2 (redraw the Lewis structure below). ii. ircle the most acidic proton of this compound. iii. Draw any resonance structures that are important in determining the acidity of this proton. iv. Label any peaks on the IR spectrum that you can assign (there will likely only be a few.) Just as important is that there are not peaks for either = or N stretches. 2

3 (b) An unknown acid has a pk a of 4.6 and the following IR spectrum: i. Identify this compound from the list on page 2 (redraw the Lewis structure below). ii. ircle the most acidic proton of this compound. iii. Draw any resonance structures that are important in determining the acidity of this proton. N N N Two minor structures (single/triple-bonded, with a 2 charge on one nitrogen) could be included, too. iv. Label any peaks on the IR spectrum that you can assign (try to label most.) The large peak at 2200 cm 1 can t be a =. Given how close it is to a N, it s a sensible guess that the N N stretch would be around here, too. (Note that bond orders are probably a little under 3 and a little over 1 for N N bonds.) 3

4 2. For each of the following, rank the compounds by most acidic (basic) to least acidic (basic). (a) Write strongest next to the strongest acid and weakest next to the weakest. (b) Write strongest next to the strongest acid and weakest next to the weakest. (c) Write strongest next to the strongest BASE and weakest next to the weakest BASE. (It is the N 2 group that acts as the base.) (d) Write strongest next to the strongest BASE and weakest next to the weakest BASE. (It is the N (or N) that acts as the base.) 4

5 3. Match the compounds on the following page to the 1 NMR spectra on this page. (All peaks are shown. Ignore very small spikes in the baseline, as well as the calibration peak at 0 ppm.) The data here are plotted with an integration trace (red) which lets you measure the area of each peak and makes it much easier to tell which peaks are real and which are blips of noise. Because you didn t have this information on the exam, we accepted two correct answers for #4. 5

6 4. Succinic acid, 4 6 4, is a diprotic acid: K a ( ) = K a ( ) = Aniline, 6 5 N 2, is a weak base: K b ( 6 5 N 2 ) = (a) If you start with a 0.1 M solution of aniline in water, and add succinic acid until 50% of the aniline is converted to its conjugate acid ( 6 5 N + 3 ), what is the p of the resulting solution? The important thing here is that there are equal quantities of aniline and its conjugate base (aniline + ) at equilibrium. So K b ( 6 5 N 2 ) = [ 6 5 N + 3 ][ ] [ 6 5 N 2 ] = [ 6 5 N + 3 ] [ 6 5 N 2 ] [ ] [ ] = At equilibrium, it also must be true that so K w = = [ + ][ ] [ + ][ ] = [ + ] = [ ] = = M p = log 10 [ + ] =

7 (b) At this point, what are the concentrations of the succinic acid and its two conjugate bases ( 4 6 4, 4 5 4, and )? We know the + concentration at equilibrium, and we can use that to figure out where the two acid equilibria lie for succinic acid: K a ( ) = [Succ ][ + ] [ 2 Succ] = [Succ ] [ 2 Succ] [Succ ] = 4.67 [ 2 Succ] K a ( ) = [Succ2 ][ + ] [Succ ] = [Succ2 ] 1.35 [Succ 10 5 ] [Succ 2 ] = 0.19 [Succ ] In other words, because we know the p, we also know the proportion of 2 Succ, Succ, and Succ 2 in solution at equilibrium. For every Succ we create, we must also create one aniline +, while we create two aniline + with every Succ 2. So the stoichiometry is [aniline + ] = [Succ ] + 2[Succ 2 ] and plugging in the above proportions lets us solve 0.05 = [Succ ] + 2[Succ 2 ] 0.05 = [Succ ] [Succ ] [Succ ] = M [Succ 2 ] = M [ 2 Succ] = M 7

8 5. Decanoic acid ( ) has a K a of It is a white solid that is only slightly soluble in water K sp ( ) = M If 85 g (0.5 mole) of decanoic acid is added to 1 L of water, what is the lowest p at which all of it will dissolve? Assume that p is changed by adding strong base (e.g. Na), the solution volume does not change, either when adding base or adding decanoic acid, and The reactions are is completely soluble in water. Dec(s) Dec(aq) Dec(aq) Dec (aq) + + (aq) The equilibrium expressions are Dec(aq) = M = [Dec (aq)][ + (aq)] [Dec(aq)] Stoichiometry is So Dec(aq) + Dec (aq) = 0.5 M = [Dec (aq)][ + (aq)] [Dec(aq)] = [ + (aq)] [ + (aq)] = M p =