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Exam 1 Answer - CHEMISTRY 322aL Please K Q EXAM NO 1 Print...

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Unformatted text preview: CHEMISTRY 322aL Please K Q\, EXAM NO. 1 Print Last ame SEPTEMBER 18, 2002 First Name SSN name of TA Grader (1) (20) __ (2) (10) __ ___. (3) (20) __ __ (4) (20) _ _._____._ (5) (15) __ __ (6) (15) __ __ (100) first letter of last name I will observe all the rules of Academic Integrity while taking this exam. Signature Chemistry 322a -2- Name Exam No. 1 (1) (20) Answer the following questions. (A) (6) Provide detailed dash-line structural formulas for the two organic compounds below that are written as condensed formulas. Your structures must show all the atoms and all the bonding and nonbonding electro - irs. Each - .. ‘ of bonding electrons must be shown in the dash or line formulation, while onbon- mg electron pairs .~, ust be shown in the dot convention. Place your answers in the boxes C. HOCH2CH(CH3)CH2CH(CH3)2 (CH3)2CHNHCH2CH=CH2 (B) (6) Show the formal charges in each of the following structures. Place the charges immediately next to the atoms on which they reside. (C) (8) Identify the defining functional group or family name for each structure below from the list of choices. Write your answers on the lines below the structures. 2 (Ac/t f? (CH3)2CHCH2CCH2CH3 CHg-fiZ-COZCH3 CH3 Es Choices: alkane, alkene, alkyne, alcohol, aldehyde, amide, carboxylic acid, ester, ether, ketone, nitrile. _-. mine, aromatic, ofilflm" Chemistry 322a -3- Name Exam No. 1 (2) (10) According to hybrid orbital theory, the bonding in ethyne, Csz, is understandable by invoking an sp hybridization state for each carbon. Complete the diagrams and pictures below that illustrate the generation and spatial arrangement of this set of hybrid orbitals. Assume a mixing of the valence-level 25 and 2px atomic orbitals. (A) (5) Generation of valence-level atomic orbitals for sp-hybridized carbon: hybrid orbitals for sp-hybridized electronic ground state carbon after electron promotion for atomic carbon and hybridization Show the electronic Draw and label the set of configuration for ground state atomic orbitals for atomic C in the valence level sp-hybridized carbon, and by placing electrons as dots add the valence level (.) in the above diagram. electrons as dots. (B) (5) Draw the spatial projections of all the valence-level atomic orbitals for an sp-hybridized carbon by completing the diagram below. Assume mixing of the 25 and 2pX atomic orbitals in your picture. Draw and label all the atomic orbitals according to the x,y,z coordinates shown below. Chemistry 322a -4- Name Exam No. 1 (3) (20) Answer the questions below. (A) (4) In the boxes below, draw structural formulas for two constitutional isomers of C3H30 that have different functional groups. Use the line-dash formulation for all covalent bonds and the dot formulation for all nonbonding electrons. Constitutional Isomers of C3H80 with Different Functional Groups fl: W1; 0 My... boxflh (B) (10) Indicate whether the structural formulas in each pair below are the same compound, constitutional isomers of each other, or different compounds that are not isomers. CH3CH2CiH2 CICHZCHZCHZCH3 CHZCI QH (CH3)2CHCHZOH CH3CH2CHCH3 ?r W CH3CH2CH2CHCH2CH3 Br CH3 CHZCH3 S CH3CH2NHCH2CH(CH3)2 CH3-q-CH2-N-H wmd H 9H3 . CH3CH2CH(CH3)CH2C1 CH3-CH2-CH2-(i—Cl Whom - - H (C) (6) Circle the compounds below that have nonzero molecular dipole moments based on geometries predicted by VSEPR theory. . HC:CHI [email protected] BeClz (monomer) +1 “54 60va (9; 4 Lack WJE%@ Chemistry 322a -5- Name Exam No. 1 led/Ch (4) (20) Circle the correct answer for each question or statement below. (A) The number of valence electrons in the nitrate ion, NO3', is 22. 23. 25. (B) The molecular formula of the structure below is H (circle correct answer below) Celeo C6H14O C71'1120 (C) The formal charges on the following carbon species are (circle correct answer below) H—(;-H H—éli—H H—§ 1(0), 11 (-), III (0) H H H I 11 III I (+), 11 <-), 111 (-) I (0), II (—), III (-) (D) Nitrous oxide, N20, has 16 valence electrons. Two correct Lewis structures with proper formal charges that are resonance structures for nitrous oxide are: - + .. + _ +_ .'. _ + :1 (circle correct answer below) =N=H—H : =N—H I II III IV I and III II and III II and IV (E) The carbon-hydrogen bond lengths in the hydrocarbons below are in the order (longest to shortest): (circle correct answer below) H2C=CH2 Hsc-CHa HC=CH I>II>III II>III>I I 111 H 111>11>1 (F) Noting that the benzene ring is planar, the aromatic compound with a molecular dipole moment of zero is (circle correct answer below) 1 1 1 1 r i :C: l (3 c1 \Cl Chemistry 322a -6— Name Exam No. 1 (4) Contd. (2‘ t (G) Identify all the tertiary alcohols among the structures below. H cH3 pH H CH3 CH3CHCH20H (CH3)2CHCHCH3 CH3CH2C(CH3)2 1 11 111 IV (circle correct answer below) only 1 I and 111 I, 111 and 1v (H) The important carbonyl functional group is found in —a1dehy_des, _cgr_bo\>g&acids, ketones, esters and ethers. (aFdEl’tydes, amides, ketones, carboxylic acids, and esters. 5, car oxy c acids, esters, e ers an e one. -aldehydes, amines, carboxylic acids, esters and ethers. (I) The melting points and boiling points of acetone and isopropyl alcohol are shown in the table below. (PH CH3CCH3 CH3CHCH3 acetone i50pr0pyl alcohol 58 60 —95°C -89°C 56°C 82°C The differences in melting and boiling points for the two compounds are due to (circle correct answer below) - the difference in the molecular weights. -the difference in the dipole moments of the two compounds. —the diff nce in van der Waals attractive forces in the two com ounds. -1ntermolecular hy rogen on g a 15 present only in isopropyl alcohol (I) Short range van der Waals or London attractive forces increase in magnitude with -increasing size of the molecule. -the presence of polarizable atoms. -the o - - n - of n and 1t electrons. a l of the above factors. ‘ _ Chemistry 322a -7- Name Exam No. 1 (5) (15) Answer the following questions. (A) (6) Diazomethane, CH2N2, is a highly reactive and toxic organic compound. The bond connect in diazomethane are shown in the box to the left below. In the box to the right, complete a Lewis structure for diazomethane. Your Lewis structure must follow the octet rule and include all formal charges. As a first step, d he the number of valence electrons in diazomethane. flu number of valence electrons: ”0 bond connectivities . . in diazomethane a Lems structure for diazomethane (B) (9) The structure of ascorbic acid (vitamin C) is shown below. In the structure below, identify with an arrow all the Watoms that are sp2 hybridized. ascorbic acid (vitamin C) Ma r (£4 Ascorbic acid is a weak acid. The site of the acidity is the hydrogen noted with an asterisk. When. this proton (H+) is removed, the relatively stable ascorbate anion is formed. The hybrid of the ascorbate anion has contributions from several resonance structures. Two resonance structures, where there is a negative charge on an oxygen atom, make major contributions to the hybrid. Draw these two resonance structure below. Show all nonbonding electron pairs and formal charges in your resonance structures. Chemistry 322a -8- Exam No. 1 (6) (15) Name (A) (9) Provide condensed or line—dash structural formulas inside the boxes for the following compounds. 0 v- 0&0“ ”(if OH gl’ formula. CH chH . CH—O-Cl—l cu CH CH rhea ’46 “‘"M— 1’ s 3 3 z 3 s .1 s /\ W .. isopropyl alcohol methyl ethyl ether a tertiary amine with formula C4H11N «MI, W (B) (6) Predict the difference in physical properties in each pair of constitutional isomers below, as requested, and offer a short explanation for the difference. The melting points of the two constitutional isomers below differ by more than 100°C. Circle the structure with the higher mp. Offer a short explanation for the higher mp. CH3CH2CH2CH2CH3 neopentane pentane TL‘k more CoQOg—V Ol‘ SPMQV\¢a—\ Swg‘l'vw‘K F‘KQ—‘S ‘omepr ex fit QDKICL Sing uxcL lugs. cs \A‘Skf‘ W‘L\‘\-\'Y\3 ?O\M’T. The boiling points of the two constitutional isomers below differ by more than 40°C. Circle the structure with the higher bp. Offer a short explanation for the higher bp. ‘FH3 CH3CH2CH2NH2 CHs-N-CHs propylamine trimethylamine l‘YV’W-waolt ikKou— \(x alt—o 1% \DDMQli/n \&c~.i$ +0 q Vast“ \sq—v. 21+?“ «wvs7 ‘5 Wju‘smlg-to WK‘QMP +\,\Qg9\ o.-l—+ro~c.'l‘\vk l’K+¢ro..9——"l—K\OW\5 Cl-LLF‘}’P\8 1le, lxciuli‘l—i) \lOc'F—mr- flwt—Hok' ...
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