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1 Quizzes and Results from an Organic Chemistry Molecular Modeling Study American Chemical Society New York City September 11, 2003 Jim Hermanson & Julie.

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Presentation on theme: "1 Quizzes and Results from an Organic Chemistry Molecular Modeling Study American Chemical Society New York City September 11, 2003 Jim Hermanson & Julie."— Presentation transcript:

1 1 Quizzes and Results from an Organic Chemistry Molecular Modeling Study American Chemical Society New York City September 11, 2003 Jim Hermanson & Julie B. Ealy Penn State University Berks-Lehigh Valley

2 2 Quiz #1 On structure A the arrow is pointing to a nitrogen On structure B, the arrow is pointing to a nitrogen On structure C, the arrow is pointing to an oxygen 1. Which of the structures are aromatic? A. B, C B. A, B C. A, B, C D. A, C 2. Fill in the missing electron pairs on the nitrogens or oxygen on the three structures making sure to position them so that they support your choice in Question 1. 3. Explain your choice of electron pairs on Structure C.

3 3 Results from Quiz #1 – Aromaticity (1 & 2) Question 1: Which of the structures are aromatic? 31.6% - correct answered68.4% - incorrect answer Answer ABCD # of students3565 %15.826.331.626.3 Question 2: Placement of electrons on N or O to support choice in Question 1. % Correct# of studentsComments 26.35 Choice correct; electron placement correct 15.83 Choice incorrect; electron placement supported choice 31.66 Choice correct; some electron placement correct 26.35 Choice incorrect; electron placement incorrect In addition to incorrect placement of electrons, the following incorrect answers were seen supplied by students: 1. no electrons 2. (+) sign on the oxygen or nitrogen 3. Presence of a p-orbital with or without electrons 4. Placement of hydrogen on nitrogen or oxygen and no electrons

4 4 Results from Quiz #1 – Aromaticity (3) Question 3: Explanation of pairs on structure C. 36.8% (7) Correct63.2% Incorrect Example responses Correct Explanation: 36.8%(7 students) 1) …2 [oxygen electrons]are in the system for aromatic and 2 are on the oxygen only. 2) Structure C has an electron pair with a set outside of the ring and inside the ring, therefore making it aromatic. Not a correct explanation for the question asked or incorrect: 10.5% (2 students) Valence electron explanation: Oxygen has 6 valance electrons and two are used. So 4 others have to be placed. 31.6% (6 students) Octet explanation: 1) …2 electrons are placed on the oxygen to complete the octet. 2) Oxygen should have 4 pairs of electrons. Octet rule. 10.6% (2 students) sp 3 explanation: The molecule will not have sp 3 bonding at the site if the electrons are positioned in this manner (2 electrons inside and 2 electrons outside). 10.6% (2 students) Miscellaneous explanations: The electrons on the top and bottom. The electrons need to move easily around the ring. (Both electrons outside on the oxygen).

5 5 Quiz #2 Black-carbon Red-oxygen Green-hydrogen Yellow-fluorine Blue-nitrogen 1. Rank the molecules in order from the most shielded methyl protons to the least shielded methyl protons. Most shielded Least shielded 2. Rank the molecules from farthest downfield to farthest upfield. Farthest downfield Farthest upfield 3. Explain the rankings in question 1. 4. Explain the rankings in question 2. A B C D

6 6 Results from Quiz #2 – Shield/Deshield (1) Question # 1: Rank the molecules in order from the most shielded methyl protons to the least shielded methyl protons. Most shielded Least shielded Correct answer was ADBC 73.7% Correct 26.3% Incorrect Though a large percentage of students answered this question correctly it would be more revealing if the molecules consisted of different atoms as follows: 1) As a series of halogens + the methyl group following a family trend), or 2) Atoms bonded to a methyl group where the atoms used move across a row. Answer ADBCACBDABDCDBCAACDB # of students141121 %73.7*5.26 10.55.26

7 7 Results from Quiz #2 – Shield/Deshield (2) Question #2: Rank the molecules from farthest downfield to farthest upfield. Farthest downfield Farthest upfield Correct answer CBDA. 63.2% Correct 36.8% Incorrect Answer CBDADBCAACBDCABCBDCAACBACDBA # of students12111121 %63.2*5.26 10.55.26 *It’s notable that a number of students (about 10%) got the first question correct, but failed to see that the answer to the second question is just the reverse of the first answer-- as predicted by theory taught in lecture. It might be instructive to have asked the second question designed to keep the answer in the same order as that of question # 1, ranking from farthest upfield to farthest downfield.

8 8 Results from Quiz #2 – Shield/Deshield (3) Question 3: Explain the rankings in Question 1 – most shielded methyl protons to least shielded methyl protons. I. 52.6% (10 students) Correct multiple choice order – Satisfactory explanation: Correct explanation required a proper usage of electronegativity theory as applied to the deshielding phenomenon exhibited on the molecule. Minimum explanation; Electronegative elements deshield nearby (vicinal) protons. Best explanation: Fluorine is the most electronegative atom in this series, hence it pulls electron density toward itself and away from nearby (vicinal) protons; in the process, deshielding these same protons. Oxygen- -being the second most electronegative element--generates the second highest level of deshioelding on nearby protons. Nitrogen is the least electronegative and exhibits the least amount of deshielding effect on nearby protons. In ethane, the electronegativities of hydrogen and carbon are roughly the same, therefore, no--or very little-- deshielding of vicinal protons occurs. II. 21.1% (4 students) Correct multiple choice order – Unsatisfactory explanations: Explanations included: 1) Simply stating“electronegativity” 2) Stating that the more polar the molecule, the greater the deshielding 3) Stating that it depended on how much the surrounding molecules tended to deshield hydrogens around them (wrong use of chemistry language) 4) Unpaired electrons from the oxygen, nitrogen and fluorine cause the surrounding protons to be more deshielded. The more unshared pairs, the more deshielded (also drew lone pairs on O, N, and F)

9 9 Results from Quiz #2 – Shield/Deshield (3) III. 26.3% (5 students) Incorrect multiple choice order – Unsatisfactory explanation. Example explanations: 1) In molecule A electrons are being pulled towards hydrogen, making it the most shielded whereas in the other molecules they are being pulled towards the heteroatoms. 2) Because the oxygen will absorb or withdraw the electrons from methyl groups. 3) The more protons are pulled from hydrogens, the higher the shielding

10 10 Results from Quiz #2 – Shield/Deshield (4) Question 4: Explain rankings in Question 2 – farthest downfield to farthest upfield 57.9% (11 students) Question #2 ranking correct; satisfactory explanation: Explanations: 1) Deshielding gives a downfield shift; shieldeding gives an upfield shift 2) Least shielded or deshielded/downfield 3) C-H molecule A farthest upfield, then N, O, and F 5.26% (1) Question #2 ranking correct; incomplete explanation: Explanation: Electronegativity/backwards 10.5% (2) Question #2 ranking incorrect; unsatisfactory explanation: Explanation: 1) The more downfield, the more deshielded – bulkier groups are more shielded, which should also be taken into consideration. 2) O and N will appear about the same region on a spectro plate 21.1% (4) Question #2 ranking incorrect, though, reverse of Question 1; satisfactory explanation: Explanation: Deshield/downfield 5.26% (1) Question #2 ranking incorrect, though, reverse of Question 1; unsatisfactory explanation: Explanation: Question 2 is just asking the opposition of Question 1.

11 11 Quiz #3 Structure A Structure B 1. On Structure B there are how many unique carbons? A. 2 B. 4 C. 6 D. 8 2. On Structure A there are how many equivalent sets of hydrogens? A. 1 B. 2 C. 3 D. 4 3. Describe the equivalent sets of hydrogens on Structure A. Also identify the equivalent sets of hydrogens by number.

12 12 Results from Quiz #3 – Unique C/H (1 & 2) Question 1. On Structure B there are how many unique carbons? Correct answer was choice D (8) 26.3% Correct 73.7% Incorrect # of unique carbons 2468 # of students1855 % of students5.2642.126.3 Though only 26.3% of the students answered the question correctly it is recommended changing the choices to 4, 5, 6, and 8 carbons instead of 2, 4, 6, and 8 carbons Question 2. On Structure A there are how many equivalent sets of hydrogens? Correct answer was choice C (3) 73.7% Correct 26.3% Incorrect # of equivalent sets 1234 # of students04141 % of students021.173.75.26 We recommend changing the choices to 2, 3, and 4, not 1, 2, 3, and 4.

13 13 Results from Quiz #3 – Unique C/H (3) Question 3. Describe the equivalent sets of hydrogens on Structure A. Also identify the equivalent sets of hydrogens by number. 68.4% Correct 26.3% Incorrect 5.26% No response Correct responses: A. Students either drew an axis of symmetry or there was a statement about an axis. Example responses: 1) Axis of rotation and 3 sets of hydrogens 2) Hydrogens on each side of the double bond can be rotated into each other 3) The equivalent hydrogens fall on equivalent carbons. By symmetry across the axis, those are equivalent. B. Students either listed the equivalent sets of hydrogens by number or circled them. Incorrect responses: 21.1% (4) of the students had a similar incorrect response. Example response: Equivalence set one is composed of hydrogen attached to carbons 4, 3, 2, and 1 and set two is the two hydrogens attached to carbons 5 and 6. 5.26% (1) of the students had the response. 2 of the 3 sets were correct - 1, 4 and 2, 3 Instead of 5, 6 as one equivalent set, they were separated into two different sets.

14 14 Quiz #4 A B C D 1. The structure that fits with the attached spectroscopic information is (circle your choice) ABC D 2. Explain how your choice was made.

15 15 Results from Quiz #4 - Spectroscopy Structure ABCD # of students01511 % of students088.25.88 Question 1. The structure that fits with the attached spectroscopic information is (circle your choice) ABC D Correct answer was choice B (15) 88.2% correct11.8% incorrect Question 2. Explain how your choice was made. (The percentages do not total 100%. A student’s responses were included in more than one category.) 47.1% (8 students) IR – carbonyl or C=O 41.4% (7 students) 2 equivalent or 2 unique or 2 identical or proper # of equivalent sets of hydrogens 35.3% (6 students) H-NMR 29.4% (5 students) Ratio of: - 6:4 - one set of two hydrogens and one set of three hydrogens - 4:6 ratio Group of 6 with 4 vicinals and group of 4 with 6 vicinals 2 sets of hydrogen – one with 3 hydrogens (4 vicinal) and one with 2 hydrogens (6 vicinal)

16 16 Results from Quiz #4 – Spectroscopy (2) 29.4% (5 students) Process of elimination - C-13 NMR – one set of carbons with no hydrogens – eliminate C - eliminate D and A – no carbonyl - IR – no evidence of OH – eliminate A - IR shows C=O. Can’t be D. - Peak of alcohol not there – cannot be A - C has more than 3 unique sets of hydrogens and would have CH 3, CH 2, and CH peak – not equivalent to C-13 NMR 23.5% (4 students) IR – no evidence of OH 23.5% (4 students) 1 CH 3, 1 CH 2, and possibly mentioned 1 C (no protons attached) 11.8% (2 students) Mass spec - parent peak of 57 – corresponds to CH 3 CH 2 C=O – student was correct - molecular weight of molecule is 86 g/mole which can be seen on the mass spec – student did NOT choose the correct structure Miscellaneous - C 13 – 195-220 peak is for ketones - 3 peaks on C-NMR – 3 sets of equivalent carbons - C-NMR backs up H-NMR

17 17 Results from Quiz #4 - Spectroscopy 11.8% (2 students) Incorrect responses - It cannot be A because the peak for an alcohol is not there. The compound cannot be B because the hydrogen sets are symmetric and therefore there wouldn’t be two sets of 2 and 4 peaks showing up. The compound D cannot be chosen because of the same reason for B. Even if it were to have two sets it would create 2 triplets. The answer has to be C because the hydrogen on the #1 carbon creates the doublet and the set of the hydrogens on carbons 2-5 create the triplet. - I looked at the NMR and saw the peak at 4, meaning a benzene ring. Also, one C-O bond can be seen on the IR spectra. Also, the molecular weight of the molecule is 86g/mol, which can be seen on the mass spec.


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