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Advantage of Internet Essays Advantage of Internet Paper Advantage of Internet Paper 1 Organic Chemistry 261(270, 271) Midterm-I Examination September 29, 2008 Name (print) ID No. Time: 50 minutes Total number of pages: 6 Answer all questions in the space provided. Question I. (15. 0) II. (14. 0) III. (14. 0) IV. (15. 0) V. (10. 0) Total (68. 0) Percentage Mark GOOD LUCK 2 (9. 0) 1. a) Write a Lewis structure for each of the following compounds showing any unshared electron pairs. b) Calculate the formal charge on each atom other than hydrogen. Be sure to show your calculations. a) CH3OH2 EC(c) = 1/2(8) = 4 FC(c) = 4- 4 = 0 H H H C O H H EC (O) = 1/2 (6) + 2 = 5 FC (O) = 6-5 = +1 b) (CH3)3CO H C H H H C C O HH C H c) (CH3)2O-BF3 H HH C F H H C O B F H F (6. 0) 2. H H Formal charge calculations for (b, c) as of part (a) Which resonance form in the following pairs would contribute more to the Hybrid (more stable)? Explain the reason for your choice. a) CH3CH CH CH OH CH3CH CH CH OH Carbon of the other structure do not meet the octet rule O CH3 More stable because of more covalent bonds O c) CH2 C CH3 CH2 O C CH3 Negative charge resides on more electronegative atom. O C NH2 b) CH3 C NH2 3 (8. ) II a) Write a dash formula for each of the following compounds showing any unshared electron pairs. b) predict the hybridization of the indicated atom in each molecule? a) CH3 CH N CH3 Answers: sp3 a) CH3 sp3 O b) CH3 C C C H sp2 sp2 sp3 CH N CH3 sp sp O C C C H sp2 sp2 b) CH3 sp3 sp c) CH3BeCH3 c) CH3BeCH3 sp3 d) BH4 d) BH4 (6. 0) 2. Which compound in each of the following pairs would have the higher boiling point? Explain a reason for your answer. a) CH3CH2CH2OH or CH3CH2OCH3 Alcohol, because of hydrogen bonding b) or O Ketone, because of dipole-dipole intractions c) N H or N CH3 Primary amine, because of hydrogen bonding 4 (8. 0) III 1. a) Draw structures of three alkyl bromide with the formula C4H9Br b) Classify each as to whether it is primary, secondary, or tertiary alkyl bromide. CH3CH2CH2 CH2Br primary CH3 CH3CH2 CHBr CH3 CH3 C Br tertiary CH3 (6. 0) 2. Write a condensed structural formula for each of the following compound. O seconday a) O (CH3)2CHCOCH(CH3)2 or (CH3)2CH C CH(CH3)3 b) NH CH3 CH3CH2CH(CH3)CH2NHCH2CH3 or CH3CH2 H CHCH2NCH2CH3 OH OH CH2 CH2 CH CH CH CH CH3 c) 5 (15. 0) IV 1. Draw a structure for compounds that meet the following descriptions. a) Two amines with the formula C3H9N CH3CH2CH2NH2 and CH3NHCH2CH3 and Many other possibilities. b) Two ketones with the formula C5H10O O CH3CH2 and O CH2CH3 CH3 C CH2CH2CH3 C Many other possibilities. c) Two ethers with the formulas C4H8O O O and Many other possibilities. d) Draw bond-line structures of two cyclic compounds with molecular formula C4H8. e) Draw an isomer of CH3CH2CH2CH2C N CH3 CH3CHCH2C N with the same functional group. and one more possibility 6 (10. 0) V. What is the relationship between the members of the following pairs? That is, are they Stereoisomers, constitutional isomers, the same, or resonance structure. Explain the reason for your choice. CH3 CH a) CH2 CHCH2CH3 and H2C CH2 Answer: constitutional isomers same molecular formula, but different connectivity of atoms). b) NH2 and NH2 Answer: Resonance structures- (same connectivity of atoms, but different distribution of electrons. CH3 c) CH3 C CH3 CH3 Answer: different drawing of the same molecule , same Molecular formula or (CH3)3C CH3 d) H H3C C C CH3 H and H3C H C C CH3 H Answer: stereoisomers (cis-trans isomers) different location of atoms in space, but same molecular formula. 7 Periodic Table of the Elements 1 1 18 2 H 1. 00794 3 2 4 13 5 14 6 15 7 16 8 17 9 He 4. 002602 10 Li 6. 941 11 Be 9. 012182 12 B 10. 811 13 C 12. 0107 14 N 14. 0067 15 O S 32. 065 34 F Cl 35. 453 35 Ne Ar 39. 948 36 15. 9994 18. 9984032 20. 1797 16 17 18 Na 22. 989770 Mg 24. 3050 20 3 21 4 22 5 23 6 24 7 25 8 26 9 27 10 28 11 29 12 30 Al Sc 44. 95591 39 Si Ge 72. 64 50 P As 74. 92160 51 19 26. 981538 28. 0855 30. 973761 31 32 33 K 39. 0983 37 Ca 40. 078 38 Ti 47. 867 40 V 50. 9415 41 Cr Mo 95. 94 74 Mn Tc [97. 9072] 75 Fe 55. 845 44 Co 58. 9332 45 Ni 58. 6934 46 Cu 63. 546 47 Zn 65. 39 48 Ga 69. 723 49 Se 78. 96 52 Br 79. 904 53 Kr 83. 80 54 51. 9961 54. 938049 42 43 Rb 85. 678 55 Sr 87. 62 56 Y 88. 90585 57 Zr 91. 224 72 Nb 92. 90638 73 Ru 101. 07 76 Rh 102. 9055 77 Pd 106. 42 78 Ag 107. 8682 79 Cd 112. 411 80 In 114. 818 81 Sn 118. 71 82 Sb 121. 76 83 Te 127. 60 84 I At Xe Rn 126. 90447 131. 293 85 86 Cs Fr Ba Ra La * 138. 9055 89 Hf 178. 49 104 Ta 180. 9479 105 W 183. 84 106 Re 186. 207 107 Os 190. 23 108 Ir 192. 217 109 Pt Ds [281] 64 Au Rg [272] 65 Hg 200. 59 112 Tl 204. 3833 113 Pb 207. 2 114 Bi Po 132. 90545 137. 327 87 88 195. 078 196. 96655 110 111 208. 98038 [208. 9824] [209. 9871] [222. 0176] 115 116 117 118 Ac** Rf 58 Db 59 Sg 60 Bh 61 Hs [277] 62 Mt [268. 1388] 63 Uub [285] 66 Uut [285] 67 Uuq Uup Uuh [289] 68 [288] 69 [289] 70 71 [223. 0197] [226. 0254] [227. 0277] [261. 1088] [262. 1141] [266. 1219] [264. 12] * Ce ** Th Pr Pa Nd 144. 24 92 Pm [144. 9127] 93 Sm 150. 36 94 Eu 151. 964 95 Gd 157. 25 96 Tb 158. 92534 97 Dy 162. 50 98 Ho Es Er Fm Tm Md Yb 173. 04 102 Lu 174. 967 103 140. 116 140. 90765 90 91 164. 93032 167. 259 168. 93421 99 100 101 U Np Pu Am Cm Bk Cf No Lr 232. 0381 231. 03588 238. 02891 [237. 0482] [244. 0642] [243. 0614] [247. 0704] [247. 0703] [251. 0796] [252. 0830] [257. 0951] [258. 0984] [259. 1010] [262. 1097]