Checking Hydrogen Charges and Correcting Hydrogen Numbers in a Chemical Structure - Prof. , Exams of Organic Chemistry

Download Checking Hydrogen Charges and Correcting Hydrogen Numbers in a Chemical Structure - Prof. and more Exams Organic Chemistry in PDF only on Docsity! First Three Letters of Last Name NAME a Network ID CHEMISTRY 332 SPRING 07 EXAMI Wednesday, February 14, 2007 The exam will be graded on a 100-point scale; there are 5 “ponus points”. Thus, it may be helpful to look over the entire exam before you begin. Select those problems that you know best to work on first. If you have enough time, answer all of the questions (i.e., we will grade every problem). Good luck! Average Bond Energies (kcal mole") H Cc N oO F Si $s cl Br 1 — 104 99 93 111 135 76 83 103 87 7 73° ser | 116? | 72 65 81 68 52 53° 65 46 45 | 108 52 48 56 37 83° 91 74 56 61 52, 58 *C=C 146, C=C 200 wnApaloy2zjal °C=N 147, CaN 213 ‘In CF, °C=O 176 (aldehydes) *Tn nitrites and nitrates 179 (ketones) Knowing an answer to a question and knowing how to answer a question are two entirely different propositions. Prof. Bob Grossman University of Kentucky CHEMISTRY 332 1) [15 pts.] Zyprexa is a drug manufactured by Eli Lilly for the treatment of bipolar disorder and schizophrenia. The chemical structure is shown at the Exam I, Spring 07 8.56 (conjugate acid) oN on right along with the calculated pK, values of the molecule’s acidic and basic sites. The plot of ionized N state vs. pH gives the percent of different ionized , 6.02 — (conjugate acid) -15.30 (conjugate acid) CHy @) In the boxes below, finish drawing the structures of the ionized species A — E corresponding to each curve. You will need to add or take away protons from Zyprexa and add formal charges. Note that species E is only present in very low concentration and therefore a second plot with an expanded y-scale is provided. species as a function of pH. N N; oC - Dea stl \ ‘a 6.09 ‘conjugate acid; r questions (2) and (ii) Answe: 1 ructures provided be \ Ionization State vs. pH / 100 ml NN ma . 80 * 0 te 2 fo —___.__. ¢ 60 N 2 50 3 a 40 1A 20 |+-—_—— - ——" 19 -— —— ° o 4 2 3 4 5 6 9 10 41 12 43 14 7 8 pH | Ionization State vs. pH |  Cremistry 332 Exam I, Spring 07 8) {10 pts.] As the above mechanism shows, the reagent N-chlorosuccinimide (NCS) is much more active in its protonated form compared to its neutral form. The goal of this problem is to explain this behavior with a qualitative molecular orbital (MO) diagram. Draw molecular orbital energy diagrams for the N-Cl bonds of the two structures shown below. Each diagram must show: - the energy level of the atomic orbital on chlorine used to make the N-Cl bond 4&4] - the energy level of the atomic orbital on nitrogen used to mak - d [rte +1] - the resulting MO energy levels iated with the N-Cl bond - labels and electrons for all lvls For reference, note that the 2s and 2p energy Tevel of neutral nitrogen are| energy to the 3s and 3p energy levels of neutral chiorine (you can assume they are equal). cl 1 ® N jerde of o— 9) [4 pts.] Sketch the shape of N.Cl o* for NCS and compare it to that of N*-Cl o* for NCS:H’. lobes about same si lavger Joke on Cl @--<O El ox NCS Bl on NCS-H* 10) (3 pts.] Explain why NCS-H’ is more reactive than NCS. NC S- H+ has a lower Iying Che fer eyes tan MES a, Pull credit CHEMISTRY 332 Exam I, Spring 07 oy" ge cgh, v0 Boge ie 8 wo og apm 8, Baet g wg ae he WS chee vp Hel Definition of angle ° ° 400, 200 0 Enea pan ° 30 60 ao ¢ Number of hydrogen bonds as a function of angle 11) [14 pts.] The purpose of this problem is to analyze data from the Cambridge Structural Database (CSD) and interpret it in the context of hydrogen bonding geometry. The CSD is the database you used for assignment #2. The stick figure at the left represents a carbonyl group; the shaded circle is the carbonyl oxygen. Each small circle is a “hit” from a search that looked for hydrogen-bonded interactions between the carbonyl oxygen and hydrogen bond donors (i.e., H-D where D is an electronegative atom). The position of each circle shows the location of the hydrogen atom, relative to the carbonyl, for that particular hit. The histogram at the right gives the angular distribution of hydrogen atoms (the angle @ is defined in the upper right). Use this data to answer the questions below. Reference J. Am. Chem. Soc. (1983) 105, 5761. (i) The midpoint of the tallest bar in histogram corresponds to an angle, 9, of 3S degrees. Use this midpoint value and the definition of ¢ to calculate the most frequently observed C=O-H “valence” angle: l2s fia] degrees. (38 +4 o) (ii) Check all the statements that are true: U1 The hydrogen atoms are distributed in a purely random way about the angle ¢. The data show a tendency for hydrogen atoms to reside in the directions associated with oxygen sp’ lone pairs. (The data show that all hydrogen atoms are found at the ideal angle associated with oxygen sp? lone pairs. The data provide no information about the orientation of the H-D bond. (The data show that H-D sigma* tends to be oriented for maximal orbital overlap with the oxygen sp” lone pairs. O = The electrostatic contribution to hydrogen bonding (i.e., charge attraction) is the same for each of the “hits”. The electrostatic contribution to hydrogen bonding depends on the HeeeO distance. OQ The “directional influence” of the lone pairs is independent of He**O distance. The “directional influence” of the lone pairs becomes less pronounced as the HeeeO distance decreases. The “directional influence” of the lone pairs becomes more pronounced as the HeeeO distance decreases. CHEMISTRY 332 Exam I, Spring 07 12) [6 pts.] Finish drawing the pictures below to illustrate why n—x* orbital overlap is zero for the tee-shaped geometry of H,O with nitrogen (N,), but nonzero for H,O with carbon monoxide (CO). Reference: Chem. Rev. (1988), 88, 899. an wt Te Uneguel te gual . Sige lobes: lobes ; C ‘ ne ner , is net overlap overlap between 13) [6 pts.} Draw pd hybrid atomic orbitals as plus and minus combinations of p, and d,,. The shapes of p, and d,, are provided for you. Hint: there are four lobes in each hybrid pd AO. Reference: Chem. Rev. (1994) 94, 1339. Zz Zz 4 . @.. . Pz de pd hybrid AO = pz + dz pd hybrid AO = pz — diz 14) [4 pts.] Qualitatively, for the interaction of a Is overlap is to be represented as positive, while destructive overlap is to be represented as negative. Zz On: sketch a plot showing how orbital overlap varies as a function of angle, 6, orbital with the d,, orbital (see the illustration below). Constructive plot Het Mar Miges Boab ase aad Miniaies at 135° @rhs PU cradee J eg .