A) Phenol Red is a weakly acidic dye which is yellow in the HA form and red in the A form. What concentration of Phenol Red is required to obtain an absorbance of 0.50 at 550 nm at pH 7.8? Note: ε550 of A = 2.5 x 104 M-1 cm-1; ε550 of HA = 0; pKa = 7.8. B) What would the absorbance of the solution in part (a) be if the pH was 8.8?

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Which set of H's in the phenacetin analog shown below correspond to HA/HA' in the 1H NMR spectrum shown below? Phenacetin1HNMR Chem235.12.fid 7.9 7.8 7.7 7.6 HA/HA' HB/HB' N A/A' 7.5 11 Group 1 one H3C 10 B 7.2 7.1 7.0 F1 (ppm) C/C' ZI 6.8 6.7 6.6 6.5 6 f1 (ppm) 5 [Choose ] [Choose ] CH3 Group 2 900 -800 700 -600 -500 400 -300 -200 100 -0 > -3800 3600 3400 3200 -3000 2800 2600 2400 -2200 -2000 1800 -1600 1400 - 1200 -1000 -800 600 -400 -200 -200

Show all atoms in the structure most consistent with each NMR spectrum (and additional spectroscopic data)

The NMR spectra for compound 1 were acquired in a 7.5 mg / 0.6 mL solution ofCDCl3 . The 1H and 13C peaks are also listedbelow. Provide a full analysis of the NMR spectra for compound 1. correct assignment of NMR spectra of both 13C spectra.  correct rationalisation of 13C spectrum1H NMR (400 MHz, CDCl3) δ 7.73 (d, J = 9.5 Hz, 1H), 7.56 (ddd, J = 8.5, 7.5, 1.6 Hz, 1H),7.51 (dd, J = 7.5, 1.6 Hz, 1H), 7.36 (d, J = 8.5 Hz, 1H), 7.30 (dd, J = 8.5, 7.5 Hz, 1H), 6.45(d, J = 9.5 Hz, 1H).13C NMR (101 MHz, CDCl3) δ 160.79, 154.09, 143.43, 131.85, 127.87, 124.44, 118.86,116.94, 116.74.Note: There are two carbon peaks in the 13C spectrum that are so close together that they are not differentiable at the resolution in this experiment.  you should be able to assign these peaks to one of two carbon atoms in 1.

2. (Chapter 13 - Q58b) The compound whose H NMR spectrum is shown has the molecular formula C7H7B1. Follow the following questions to predict the unknown structure. Chem. shift Rel. area 2.31 1.50 701 1.00 7.35 1.00 TMS 10 7 6 O ppm 4 3 2 Chemical shift (8) e20s Cenge leaming 2(a) Degree of the unsaturation of this compound is= 2(b) The two distinct peaks in the aromatic region of the H NMR indicate that compound is .disubstituted = 2(c) The splitting pattern of the peak at 2.31 ő is = 2(d) The group that corresponds to the splitting pattern in 1(c) is = 2(e) This compound has the plane of symmetry Yes or No = 2(f) The name of the unknown compound = Intensity

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