CEM 351, Fall 2010 Midterm Exam 1 Friday, October 1, :50 2:40 p.m. Room 138, Chemistry

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1 Name (print) Signature Student # EM 351, Fall 2010 Midterm Exam 1 Friday, ctober 1, :50 2:40 p.m. Room 138, hemistry Wright N. Swers 1.(20 2.( (20 4.(20 5.(20 6.(20 Section Number (2 pts extra credit) Grade this question? YES N! TTAL (100 Score Note: Answer any 5 questions for a total score of 100 pts. Be sure to look over all the questions first before beginning the exam, and indicate which five questions are to be graded by checking the corresponding box. Recitation Day and Time Instructors: Room 1. T 9:10-10:00 a.m. Jason Lam M 10:20-11:10 a.m. armin Burrell T 10:20-11:10 a.m. armin Burrell Th 10:20-11:10 a.m. Arvind Jaganathan T 10:20-11:10 a.m. Jason Lam T 11:30-12:20 p.m. Jason Lam Th 11:30-12:20 Arvind Jaganathan M 4:10-5:00 p.m. armin Burrell Th 4:10-5:00 p.m. Arvind Jaganathan 183

2 1. (20 pts) (a) (12 pts) Draw structures to represent the following IUPA names (i) 2-isopropoxypropane (ii) cis-4-methylcyclohexanol (iii) 2,2,3,3-tetramethylbutane or (b) (2 pts) Which of the compounds in (a) has no dipole? Please circle your answer (i) (ii) (iii) (c) (4 pts) Are any of the groups on the cyclohexane in (ii) axial? Please explain with a drawing and a few words; here is a model to work from but you must make your own drawing in the box provided. Explanation: The cis related groups are on the same face of the ring [imagine it flat, as in the answer to (ii)]. ere the 3 and groups point up ; the ʼs on carbons 1 and 4 then show the contrasting down (other face) position. 3 (d) (2 pts) ow many resonances would you expect in the 1 NMR of (i)? of (iii)? Explanation: (i) Two resonances: (a) a doublet [at ca. 1.1] and (b) a septet [at ca. 3.5 ppm], with the a:b integral ratio of 6:1 (= 12:2). The 6 equivalent ʼs in the isopropyl groupʼs 3 groups split the one on the between them into a septet, while they are split into doublets. (iii) ne singlet [at about 0.8 ppm]. All six of those 3 groups are equivalent. Note: though rough chemical shifts are given above, they werenʼt required for the answer

3 2. (20 pts) (a) (3 pts each) Form acceptable Lewis structures for the following uncharged compounds by adding the necessary nonbonding electrons and multiple bonds: N: :N N: ydrogen cyanide Urea (b) (3 pts each) Assign formal charges as necessary to the atoms in the following Lewis structures. 1-hloroethyl cation Nitromethane (a racing fuel) (c) (3 pts) Which of the following alcohols would have the lowest pka (greatest acid strength)? Explain your answer in terms of the reaction that defines Ka. 3 2 F pkaʼs: For dissociation of acid A (below), K is the ordinary equilibrium constant, but since the pka scale is defined in 2, Ka leaves out the essentially constant 2 water concentration [ 2 ]. The more dissociated the acid, the higher the 3 + and A concentrations. Since 3 + is common to all acids, we ask which A can best accept the negative charge left on it when A dissociates. F 3 2 has three atoms of F, the most electronegative element, that pull electrons negative charge toward themselves, leaving a more positive carbon next to. This in turn attracts the negative charge in F 3 2, a stabilization mechanism that 3 2 and 3 donʼt share. (d) (5 pts) Which of the following compounds has a dipole moment (circle Yes or No): 3 N 2 Yes No 2 Yes No 3 3 Yes No (remember, dimethyl ether is bent at ) ( 3 ) 2 =( 3 ) 2 Yes No BF 3 Yes No K A A K = [ 3 + ][A ] [ 2 ][A] but Ka = N [ 3 + ][A ] [A] (water omitted) 3

4 3. (20 pts) onsider the four compounds A-D: 1-chloroethyl A B D (a) (3 pts) Which one of A-D would you expect to show the 1-chloroethyl cation (seen in problem 2b, at left) as a major fragment in its mass spectrum (circle the letter): A B D (b) (6 pts) Fill in the three Newman projection blanks below to represent the three staggered conformations you would see viewing down the bond in compound A shown by the arrow as the sp 3 carbon at the other end underwent rotation. Be sure to pay close attention to the wedged/dashed bond notation as used above between 2 3 and (c) (1 pt) (1 pt) ircle your drawing of the highest energy conformation. (d) (8 pts) Name the four compounds according to IUPA conventions: A B D 3-chloropentane 2-chloro-3-methylbutane 1-chloro-3-methylbutane 2-chloro-2-methylpentane (e) (2 pts) ne of the compounds A-D has a different formula than the other three. Which is it, and what is its formula (circle the letter, fill in blank)? A B D Formula:

5 4. (20 pts) Structural analysis by spectroscopy (a) (3 pts each) For each pair of compounds in the grid below, propose a spectroscopic method to tell them apart, and explain what you would look for with the suggested method. ompounds Method & Explanation (Many possible) A IR: Detects - stretch and only one - stretch in the left hand compound, just two - stretches in right. 1 NMR: Left: three singlets in 9:2:1 ratio; Right: two (9:3). MS: Left: huge 31 peak ( 2 + ) from tbu loss; Right: mainly 3 losses. B IR: Left: nly 2 stretches and bends; Right: Strong = stretch 1 NMR: Left: A broad singlet; Right: two triplets in 1:1 ratio; 13 NMR: Left: 1 peak; Right: 3 peaks, one (=) downfield MS: Left: Strong molecular ion; others at M mol -(n x 14); Right: Weak molecular ion; large -28 peak IR: Left: Aldehyde - stretch doublet; Right: 3 bends 1 NMR: Left: doublet:triplet in 2:1 ratio; Right: sharp singlet ca. 2.5 ppm MS: Left: strong 29 peak ( + ); Right: weak molecular ion; huge 3 (43) loss. D 1 NMR: Left: four resonances: singlet, triplet, multiplet, triplet with integrals 6:2:2:3; Right: three resonances: doublet, septet, singlet (6:1:6). 13 NMR: Left: 5 resonances, one (quaternary) weak; Right: 4 resonances, one weak (b) (8 pts) Below are 1 NMR and IR spectra for ethyl acetate, a solvent in laquers and fingernail polish remover. Please identify key features that confirm this structure i.e., assign the 1 NMR peaks to sites in the molecule, explain their splitting patterns, and predict their respective integrals. Similarly, identify at least three features of the IR spectrum that confirm the structure. b ( 2 next to 3 ) c (2) c a b (3) ( 3 group, no -'s next door) a (3) 1 NMR ( 3 group next to 2 ) PPM IR stretches = stretch -- 3 stretches 5

6 5. (20 pts) More on structural language and analyses: consider the following compounds (i) (ii) (iii) (iv) (a) (6 pts) Identify the number of sp 3, sp 2, and sp carbons in compound (i): sp sp sp (careful!) (b) (2 pts) Identify the number of secondary carbons in compound (i): (c) (2 pts) What is the hybridization of the oxygens in (iii)? sp 3 sp 2 sp (d) (4 pts) Identify among the compounds above the one that should show only one resonance in both 13 and 1 NMR spectra. (i) (ii) (iii) (iv) (e) (6 pts) Write the chemical formula--e.g for (iv) for compounds (i)-(iii). (i): 6 12 (ii): 8 18 (iii):

7 6. (20 pts) (a) (4 pts) Draw the 3D (i.e. wedges and dashes) structure of a molecule of formula 2 4 Br 2 that has no dipole in its lowest energy conformation. Br Br (b) (6 pts) Now draw the 3D representation of the 2 4 Br 2 isomer that has two peaks in the 1 NMR spectrum a doublet (1:1) and a quartet (1:3:3:1) and in its Mass Spectrum, it has a 1:2:1 triplet for the parent ion, with the sub peaks separated by 2 mass units. Explain your choice of structure and how it gives rise to the above spectroscopic behavior. What will be the parent ion masses, and where will the MS show a 1:1 doublet? Br Br 1 NMR: 3 1:1 doublet split by single next door 1 NMR: 1:3:3:1 quartet split by 3 s next door MS: Mass is 2 4 = Br's, thus it M mol can be (Bromine isotopes) or 160 or 160 or 162 = 186, 188 (2 ways), 190 (c) (10 pts) In hapters 1 and 2 we saw that bonding comes from lowering electron pairs energies. But discussing mass spectrometry, we learned about knocking the least tightly held (i.e. highest energy) electrons out of molecules to make positive ions. Now let s look at the following 8-valence electron systems: Ne -F 2 N 3 4 (all have a full octet like Neon). The cost of pulling out an electron, known as the ionization energy (IE) or ionization potential (IP) seems to follow electronegativity trends until we get to 4! an you explain this seemingly surprising fact? Yes you can! [Units of IE are electron volts (ev); one ev = kcal/mol, so these numbers are big compared, say, to the - bond strength in 2 (104.2 kcal/mol = 4.52 ev) or the - bond strength in methane (105.1 kcal/mol = 4.55 ev).]. int: consider where each valence electron pair resides; how does the valence electron set of 4 differ from the others? Ne 21.6 ev F N (just like 2 s value!) All but 4 have lone pair (nonbonding) electrons. 4 has only bonding electron pairs, and since bonding lowers the electrons' energies by sharing them between atoms, those require higher energy to pull out. ompare the IEs of the atoms, where in each case we're pulling an electron from a p orbital: Ne: 21.6; F: 17.4; : 13.6; N: 14.5; : 11.3; B: 8.3 ere the "hiccup" is between N and ; Why? 7

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