CH251Midterm1AU2010KEY

CH251Midterm1AU2010KEY - Callam Chemistry 251 Autumn 2010...

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Unformatted text preview: Callam Chemistry 251 Autumn 2010 Midterm 1 200 points HI 131 — Tuesday, October 19, 2010 — 7:00 pm — 9:00 pm Name (PRINT) I have neither given or received aid on this exam (SIGN) ~————-—————.—_—_.—._—_—__—_ Place an X in the column next to your recitation section. Failure to mark your correct recitation section will result in —5 points from your exam total. I Time Room 3 ' Teaching Assietant ‘ I _ 1:30 pm — 2:18 pm MP1005 Kamala Kunchithapatham — - _ g Thursday 1:30 pm -— 2:18 pm MP1008 Ryan Yoder - 2:30 pm — 3:18 pm MP1005 Kamala Kunchithapatham _ _ $2186 — _ Thursday 3:30 pm — 4:18 pm 5M2186 Kamala Kunchithapatham Thursday 3:30 pm — 4:18 pm MP1045 Ryan Yoder Please read all of the directions for each section of the exam carefully. There are 11 pages including this cover page (Page 1) and a scoring page (Page 2) for Midterm #1, please check to make sure you have all of the pages. Make sure your answers are clear. Any structures or answers that are unclear or ambiguous are considered wrong. 1/11 Points/Page = Callam Chemistry 251 Autumn 2010 Part 1. Functional Groups and Nomenclature. (20 points) (a) For each of the molecules below write the name of the major functional group present on the line below each structure. (1 point each for a total of 10 points) QC]: ‘ EEO 0Y0 I \/Lo/\/ O 0 OJ LAC'IDNE Kenna esnn Acerm. eTHeK SULFONE , OH OH 0’0 A S I:>_< \l/ S/ \/ O \N,H PettOXIDt SULFIBE ALtoH'OL. CARBON”; IMIME, _— m — TH) (b) For each of the following structures provide the lUPAC name. (2 points each for a total of6 points) H H CH2CHZCH3 CHsCHZCH(CH3)CH20H(CH3)CHZCH3 ‘ HgCH CH3 3 cis- l-¢+M1l-%-wl-M4l . i r _ cyclolquw, 3'5 "A‘M‘H'fl’ hep-law. ‘5 - m+k1‘ “M.” (c) For each of the following IUPAC names draw the skeletal structure in the space provided below.‘(2 points each for a total of 4 points) 1—cyclopentyl—2,3-dimethy|cyc|ohexane 3,4-diethyl-3-methyI-octane 3/11 ‘ Points/ Page = Callam Chemistry 251 Autumn 2010 Part 2. Resonance. (40 points) For each of the following structures : (i) Redraw the molecule in complete skeletal form and draw the indicated number of MAJOR resonance forms (ii) Label the best resonance form. (iii) Draw the resonance hybrid. (iv) Show electron movement using curved arrow notation between resonance forms. (a) 4 additional resonance structures (20 points) \ / N02 N 4/11 Points/Page = Callam Chemistry 251 Autumn 2010 Part 3. Acids and Bases. (50 points] (a) Determine which molecule in the set is more basic and provide an explanation using pictures and a no more than 5 words. (5 points each for a total of 10 points) Li Li é Cf vs. MOM: MSW . 5,1 vs 5P hybridile F F 2 HC F>|\NH2 VS. 3 \NH2 . mam. BASK- F is tuecmouea, IU DU C-TION (b) Write a stepwise reaction sequence using your knowledge of acid base chemistry to show how the following reaction occurs. Your answer should include curved arrows to show which bonds are breaking and which bonds are formed. Two separate steps will be necessary to arrive at the product. One step will be a Bronsted-Lowry acid base reaction and one step will be a Lewis acid base reaction. (10 points) o O / [)ZS + NaOH + CH3! [)—8 + Nal + H20 N N H Step1. , o (9 a if >175 m?“ :2f (7:5 + Hvo-H n5 : N :4 V @ Nae Step2. /° 0 If H\ $35K}. 6" l(?—:“e c1143 : [0}.5/ Nata a + Nail: 5/11 Points/Page = Callam Chemistry 251 Autumn 2010 (c) For each of the following reactions: (10 points each for a total of 40 points) (i) Predict the product of the acid—base reaction (clearly showing important lone pairs and bonds) (4 points) (ii) Redraw the reactants (clearly showing important lone pairs and bonds) and use arrow-pushing to show the electron movement and changes in bonding (2 points) (iii) Determine which side of the equilibrium is favored. (2 points) (iv) Provide a factor to account for the side of the equilibrium that is favored. (2 points) F F F F FxNxF + HactNm cm H Li F c, /CF3 V 5 SN \ «t/ FBCK "/CFS \ / 5' + ' ..._._x N: N LK/e ""‘ e t i ‘ Li ==$ 1.33 H A Ac”) 355‘ tum:ch C- Base " "D O O C. 6 “7‘ fl 0 c. Ami: _ 0 Base . Base one reactant H; one product 12; So nawu, 2% A ‘ e u «‘93 e 9 ° 2.. NH2 6/11 e LW Wm L , Points/Page: Callam Chemistry 251 Autumn 2010 Part 4. Hybridization and Bonding. (30 points) (a) Draw three unique constitutional isomers in skeletal form for the molecular formula C6le with only alkane functional groups. (6 points) Cl Q—fi (b) Answer the following questions based on the structure shown below. (2 points each for a total of 12 points) A @ ? Acibic Q HYDRO MW ‘ Basic NxN LN"? PM "- CF3 (i) What is the molecular formula of the above molecule? C ‘ ?_ th F3 N3 0 (ii) How many un-hybridized p-orbitals are in the structure? 1 ? (iii) How many electron lone pairs are in sp2 hybridized orbitals? I (iv) What orbitals are overlapping to make the bond labeled A? Csfa ‘NSP a (v) Circle the most acidic hydrogen in the molecule. (vi) Circle the most basic lone pair in the molecule. (c) Draw a clear molecular orbital picture for the molecule shown below. You should draw all of the p orbitals in pen and all of the hybridized orbitals in pencil. Make sure that your drawing is clear. Your drawing should take into “account bond angle and dimensions of the orbitals (use descriptions such as parallel or perpendicular to the paper if needed). (12 points v ) H c' 3 >=CZO % f H c 7/11 Points/Page = Callam Part S. Newman Projections. (20 points) Chemistry 251 Autumn 2010 (a) Using Newman projections, draw all of the staggered and eclipsed conformers for the molecule shown below down the indicated bond. Assign the relative energies of the confomations. W éfii Energy = + 5.1! (Cch [Hal 8/!“ What is the barrier to rotation for this bond? 48.0 -LL: 5. 5 land, . Energy: +2.2 (aniline! Energy Values: CH3, CH2CH3 (gauche) = 1.0 kcal/mol CH3, CH3 (gauche) = 0.9 kcal/mol OH, CH3 (gauche) = 1.2 kcal/mol OH, CH2CH3 (gauche) = 1.4 kcal/mol H, H (eclipse) = 1.0 kcal/mol H, CH3 (eclipse) = 1.4 kcal/mol H,OH (eclipse) = 1.3 .gcal/mol H, CH2CH3 (eclipse) = 1.5 kcal/mol CH3, CH3 (eclipse) = 4.0 kcal/mol CH3, CHZCHg (eclipse) = 4.5 kcal/mol 0H, CH3 (eclipse) = 2.5 kcal/mol OH, CHZCHS (eclipse) 2 3.0 kcal/mol Energy: + 7.‘L( \ccal Points/Page = Callam Chemistry 251 Autumn 2010 Part 6. Cycloalkanes, (30 points) (a) Draw the chair flip for the cyclohexane chair shown below. (b) At 298 K, the equilibrium percentages between the two conformations is 83% :17%. Using the equation below, calculate the AG" between the two conformations and assign the value to the reaction indicating the correctsign. (10 points) 06.. = _2‘3°3 (O‘Domsqj (20.8) . ‘ca AG° = —2.303 RT log Keq R = 0.001987 kcaI/(K mol) T = Temperature Kelvin (K) OH 83? 0 We 043 AM 3 *’ 0’” kcql‘lMol (c) For each of the following, draw both possible chair conformations and circle the one that is preferred. Calculate the relative energy difference between the two conformations using the following information. A values : (CH3 = 1.7, C(CH3)3 = 5.0 kcal/mol, CH(CH3)2 = 1.85 kcal/mol, CH3, CH(CH3)2 gauche = 1.5 kcal/mol) (20 points) (0 CH3 ES * kw —-= w *— Cl-l CH3 Fevonw- 3 AQ ‘7 ‘l' 3.3 Karl/ml (ii) m :‘ \ +1.? teal/ml r 3.55 / 9/11 Act :7 +2.05- Kcallw‘ Wing/Page: ...
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This note was uploaded on 03/22/2011 for the course CHEM 251 taught by Professor Stambuli during the Spring '08 term at Ohio State.

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CH251Midterm1AU2010KEY - Callam Chemistry 251 Autumn 2010...

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