Chem 216 H, W12 Exam 1 Name Key Page 1chem215/216 H W12 E1-key.pdf · Chem 216 H, W12 Exam 1...

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Chem 216 H, W12 Exam 1 Name___________Key___________ Page 1 I. (16 points) Silica gel thin layer chromatography (TLC) is often used to monitor the progress of an organic reaction. For each of the following two reaction steps, a solvent system is selected to given the starting material an R f value of about 0.5. (1) Complete the reaction scheme by providing in the boxes the structures of the expected reaction products, and (2) fill in the TLC spots that would be expected for each of the two reactions when the reaction is 50% complete and 100% complete, each after acidic workup. Consider only the compounds that can be visualized as a spot on a TLC plate upon exposure of each solvent-dried plate to a 254 nm-UV lamp. II. (6 points) There are two types of filtration methods used in organic experiments. These are: (i) gravity filtration using a short-stemmed, glass-filter funnel and fluted filter paper and (ii) vacuum (or suction) filtration using, e.g., a Hirsch funnel and round filter paper to cover the inside bottom of the funnel. For each of the following experimental operations, suggest which of the two filtration methods should be used. Write, in the boxes provided, (i) or (ii) for gravity filtration or vacuum filtration, respectively.

Transcript of Chem 216 H, W12 Exam 1 Name Key Page 1chem215/216 H W12 E1-key.pdf · Chem 216 H, W12 Exam 1...

Page 1: Chem 216 H, W12 Exam 1 Name Key Page 1chem215/216 H W12 E1-key.pdf · Chem 216 H, W12 Exam 1 Name_____Key_____ Page 1 I. (16 points) Silica gel thin layer chromatography (TLC) is

Chem 216 H, W12 Exam 1 Name___________Key___________ Page 1

I. (16 points) Silica gel thin layer chromatography (TLC) is often used to monitor the progress of an organic reaction. For each of the following two reaction steps, a solvent system is selected to given the starting material an Rf value of about 0.5. (1) Complete the reaction scheme by providing in the boxes the structures of the expected reaction products, and (2) fill in the TLC spots that would be expected for each of the two reactions when the reaction is 50% complete and 100% complete, each after acidic workup. Consider only the compounds that can be visualized as a spot on a TLC plate upon exposure of each solvent-dried plate to a 254 nm-UV lamp.

II. (6 points) There are two types of filtration methods used in organic experiments. These are: (i) gravity filtration using a short-stemmed, glass-filter funnel and fluted filter paper and (ii) vacuum (or suction) filtration using, e.g., a Hirsch funnel and round filter paper to cover the inside bottom of the funnel. For each of the following experimental operations, suggest which of the two filtration methods should be used. Write, in the boxes provided, (i) or (ii) for gravity filtration or vacuum filtration, respectively.

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Chem 216 H, W12 Exam 1 Name___________Key___________ Page 2 III. (10 points) Draw the structure of the expected product from the reaction shown and provide a step-by-step mechanism for its formation by the use of the curved-arrow convention.

IV. (12 points) For the oxidation reaction shown below, the following experimental results were obtained.

Cyclohexanol (20 g) in water was treated with 10 g of CrO3 in the presence of H2SO4. The reaction resulted in the formation of 7.35 g of cyclohexanone. Note: Atomic weights for H, C, O, and Cr, are, respectively, 1, 12, 16, and 52.

(1) What is the limiting reagent in this experiment? Explain your answer.

Mw of cyclohexanol: 100 g/mol; CrO3: 100 g/mol. So, 0.2 moles of cyclohexanol and 0.1 mole of CrO3 are used in the experiment. Since cyclohexanol and CrO3 react in a 3 : 2 mol ratio, CrO3 is the limiting reagent in this expt.

(2) What is the theoretical yield (by weight) of cyclohexanone? Show your work.

Answer: theoretical yield:14.7 g

As 0.15 mol of cyclohexanol is going to undergo reaction, 0.15 mol of cyclohexanone (Mw 98 g/mol) should be produced. Therefore, the theoretical yield is: 0.15 mol x 98 g/mol = 14.7 g.

(3) What is the percent yield of the cyclohexanone product, in this experiment? Show your work.

Answer: 50%

(7.35 g/14.7 g) x 100 = 50 %

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Chem 216 H, W12 Exam 1 Name___________Key___________ Page 3 V. (14 points)

The reaction of glycolic acid and acetone in the presence of P2O5 produces a new compound, C5H8O3. The product has the following spectroscopic properties.

IR (neat): 2990-2900 (a few weak peaks) and 1790 (strong) cm-1. No peaks in the 4,000-3,100 cm-1 region.

1H NMR (CDCl3): δ 1.50 (s, 6H) and 4.35 (s, 2H) ppm. 13C NMR (CDCl3): δ 25.86 (q), 63.67 (t), 112.66 (s), and 171.53 (s) ppm [q, t, and s designate the peak splitting in the proton-coupled 13C NMR spectrum].

(1) (1 point) What is (are) the unit(s) of unsaturation of the compound? ______Two_________

(2) (1 point) Does this compound have a C=C bond? ____Yes or No (circle one that applies)____

(3) (2 points) Is the compound a ketone? ____Yes or No (circle one that applies)___ (4) (10 points) On the basis of the spectroscopic data shown above, propose the structure of the compound in the box below and assign all 1H and 13C chemical shifts by writing each of the chemical shifts next to the corresponding nuclei.

VI. (8 points) The 1H NMR data for H2 and H3 of triol 4 are shown below.

(1) On the basis of the NMR data above, show the stereochemistry of alcohol 4 at C-2 and C-3 by writing in the boxes above Hs and OHs (4 points).

(2) Indicate above the coupling constants for each of the following (4 points): 3J2,3 = __3.2__ Hz; 3J3,4 = __9.2__ Hz

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Chem 216 H, W12 Exam 1 Name___________Key___________ Page 4 VII. (16 points) An unknown compound, C6H12O2, has the following spectroscopic properties.

IR (neat): 2990-2900 (several weak peaks) and 1739 (strong) cm-1. No peaks in the 4,000-3,100 cm-1 region.

1H NMR (CDCl3): δ 0.96 (3H, t, J = 7.0 Hz), 1.26 (3H, t, J = 7.0 Hz), 1.65 (2H, sextet, J = 7.0 Hz), 2.27 (2H, t, J = 7.0 Hz), and 4.13 (2H, quartet, J = 7.0 Hz) ppm.

13C NMR (CDCl3): δ 13.70 (q), 14.32 (q), 18.57 (t), 36.32 (t), 60.13 (t), and 173.61 (s) ppm [q, t, and s designate the peak splitting in the proton-coupled 13C NMR spectrum].

(1) (1 point) What is (are) the unit(s) of unsaturation of this compound? ____One______

(2) (1 points) Does the compound have a C=C bond? ____Yes or No (circle one that applies)___ (3) (2 points) Is the compound a ketone or ester? _ketone or ester (circle one that applies)___ (4) (12 points) On the basis of the spectroscopic data shown above, propose the structure of the compound in the box below and assign all 1H chemical shifts and 13C peaks at 60.13 and 173.61 ppm by writing each of the chemical shifts next to the corresponding nuclei.

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Chem 216 H, W12 Exam 1 Name___________Key___________ Page 5 VIII. (18 points) The IR spectrum of an unknown compound, C12H16O, does not show peaks in the 4,000-3300 and 1800 – 1640 cm-1 regions. Its 1H NMR spectrum is shown below. 13C NMR spectrum of the compound shows the following peaks at: 22.6 (q), 34.0 (d), 55.8 (q), 114.2 (d), 127.7 (d), 128.7 (s), 130.2 (d), 142.4 (d), and 159.8 (s) ppm [q, t, d, and s designate the peak splitting in the proton-coupled 13C NMR spectrum].

(1) (2 points) Does the compound have an alkenic C=C bond? _Yes or No (circle one that applies)__

(2) (2 points) From the 1H and 13C NMR data above, what would you suggest about the substitution pattern of a benzene ring? _1,4- (or para-) disubstituted benzene_ (3) (2 points) Peaks a and b are separated by 16 Hz. What does this mean in terms of the stereo-chemistry of the compound? ____trans-(di-substituted) alkene________ (4) (12 points) On the basis of the spectroscopic data shown above, propose the structure of the compound in the box below and assign all 1H chemical shifts by writing each of the chemical shifts next to the corresponding nuclei.