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Exam 1 Answers - CHEMISTRY 322aL/325aL Please g g 3 EXAM NO...

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Unformatted text preview: CHEMISTRY 322aL/325aL Please g g 3/ EXAM NO. 1 Print Last Name SEPTEMBER 14, 2005 First Name USC ID name of TA Because of the lost 1 PM lecture on Monday, some material that was to be covered for this exam was not discussed. Those sections have been deleted from this exam and the maximum point total for this exam is 90 instead of the usual 100. Grader / (1) (20) __ (2) (10) __ (3) (20) __ (4) (16) __.__ (5) (10) _ (6) (10) __ (7) (04) __ (90) t ' first letter of last name I will observe all the rules of Academic Integrity While taking this exam. Signature Chemistry 322a / 325a -2— Name Exam No. 1 (1) (20) Answer the following questions. (A) (4) Provide detailed dash-line structural formulas for the two organic molecules below written as condensed formulas. Your detailed structures must show all the atoms and all the bonding and nonbonding electron pairs. Show each pair of bonding electrons as a dash or line and show each nonbonding electron pair as dots. Draw the structures inside the boxes. (CH3)3CCH2CHC1CH3 CH3OCH2CH=C(CH3)2 (B) (2) Inside the box to the right, draw a bond-line (abbreviated) structure of the condensed formula written below the box. / OH CH2=C(CH3)CH(OH)CH2CH3 (C) (6) Add any missing formal charges in each of the following structures. All valence~level electrons are shown. Place the charges immediately next to the atoms on which they reside. Not all the structures may have missing charges. If there is no formal charge, write "none" below the structure. ~3r® ' Elli hill .. Iii©_© ./O\ / \ ' 1—1_C-H H-EI—Nzc. H—(IZ—Cl—H H—g—C —H HH NONE. \ (D)(8) Select the defining functional group or family name for each structure below from the list of choices. Write your answer on the line below the structure. CH 8H CH3CH2NHCH3 HCJOCH3 CH38NHCH3 oldz gig .__C~M_\'Y\S QYl'lV‘ (Ania-Lg Choices: alkane, alkene, alkyne, alcohol, aldehyde, amide, amine, aromatic, carboxylic acid, ester, ether, ketone, nitrile. Chemistry 322a/ 325a -3— Name Exam No. 1 (2) (10) Answer the following questions about the structure and bonding in borane, 8H3. (A) (2) What is the total number of valence electrons in borane (3H3)? SQ _ (B) (2) Draw inside the box below a geometrically correct Lewis structure for borane, where bonding electron pairs are shown as lines. Use VSEPR theory to determine the correct geometry. (C) (4) The molecular structure (shape) of borane is explained by having the boron atom use a set of hybrid atomic orbitals. Provide the requested information inside the boxes below. Note your answers will be in the form of electronic energy level diagrams, not pictures of atomic orbitals. >5 00 H a; I: cu .2' I: 0 LI a u a) _ a) Show the electron configuration of ground Show the electron configuration of atomic state atomic. boron in Schrodinger notation boron after electron promotion and as an energy diagram Label an the hybridization produces a set of atomic orbitals in the first and second energy orbitals used in the molecule borane. Label levels, both filled and unfilled: Use up and all the atomic orbitals in the first and down arrows for the electrons in showing second energy levels and Show their occupancy. occupancy with arrows for electrons. (E) (2) In the box below, draw a picture of the hybrid orbitals projected from the boron atom that are used in the bonding in borane. Describe the geometry of the projected orbitals. > ' Q Q Describe the geometry of the projected d : atomic orbitals. T [13% Q 330. X g S (Lama~ r~ l I ,r I, l “(A nob, WL qfoy- Mu ,‘JCF {nun Chemistry 322a / 325a -4- Exam No. 1 Name (3) (20) Answer the following questions. (A) (9) Provide the condensed structural formulas of the following compounds inside the boxes, using the information provided. On the line below each b0x, give the alkyl group name of the compound. C4HIOO C3H80 a structurally C4H11N a secondary alcohol symmetrical ether a tertiar amine Cele) CH3QH c143 CH3CHLOQHICH )1 \SOEWE¥)Q\COLIQS i -\- L j ¢.\ ‘ L «it m L¥\:f°;¥\3\ OLNU'n K (B) (6) Indicate whether each pair of structural formulas below represents the same compound, constitutional isomers, or different compounds that are not isomers. (CH3)2CHCHC1CH2CH3 CH3CHC1CHZQHCH3 1 CH3 I O canal—Ho hm) 350M! WM V . l c CH3CH2§HCH2CH2CH3 CHgCHZCHzcmcHZCHa)2 W CHZCI-I3 (C) (5) In each pair of compounds below, circle the one with the larger dipole moment (more polar). Assume geometries predicted by VSEPR theory. I HO _ graerrrriugtéyl322a/325a Q at A ’ a fl Name S43) 7 I. W M O '4 . (4) Circle the correct answer for each question or statement below. (A) The molecular formula of the structure below is (circle correct answer below) D4) C5H4O CSHSO C5H100 (B) The carbon—carbon six-electron bond ("triple bond") in an alkyne is composed of (circle correct statement below) —three sigma bonds. -tw i ma and one i bonds. (C) The hybridization states of the central carbon atoms in the three structures below are (circle correct answer below) (i) I sp II sp3 III sp3 H2C=O H3COH H2C(OH)2 (ii) I sp2 11 sp3 111 sp3 I 11 III (iii)Isp3 113p3 111sp3 (iv)Isp2 II sp3 III sp2 (D) Among the bolded C-H bonds in the structure below, (circle correct statement below) H c IV H—cE c-cH-CH=CH . ) a. A II; (E) What is the relationship of the structures in each pair below? + .- :O-H + 393 :Q: .-_ z“: CH2=CH-&-H CHs-CH-C-H CH3-CHzC-H CH3—CH—C-H A B (circle correct statement below. -A and B both have structures that are constitutional isomers. -A and B both have structures that are resonance structures. -A has resonance structure w ‘ nstitutional isomers. -A has constitutlonal isomers while B has resonance structures. A is the shortest an h -. . B is the lon est and the weakest. - is the shortest and the strongest. A is the longest and the weakest. Chemistry 322a / 325a -6- Name Exam No. 1 _ A 2m (4) Contd. M (F) The VSEPR model predicts the geometries of the compounds below are (circle answer below) C02 502 I linear, 11 linear Ilinear, II bent I . H I bent, 11 linear I bent, II bent (G) The functional groups present in the addictive analgesic drug demerol are (circle correct statement below) COCH2CH3 ‘ —aromatic, amide, ester. ’ -aromatic amine ketone, ether. IN. . o u . ' amine, ester. CH3 -aroma.tic, amide, ketone, ether. demerol (H) Classify the following alcohols as primary (1°), secondary (2°) or tertiary (3°). 3 ‘ (circle correct answer below) (CH3)2CHOH / OOH (CH3l3CCHZOH I and III are 1° 11 is 2° I II x! I and II are 2°, III is 10 > 111 I and II are 2°, III is 3° . I is 1°, 11 192°, III is 3° ‘U‘ Chemistry 322a / 325a -7- Name Exam No. 1 (5) (10) The organic compound, nitrobenzene (C6H5NOZ) has the atom connectivities shown below. Note, this structure does not show all the valence—level electrons. Atom connectivities (but not all the covalent bonds) in nitrobenzene. (A) (2) What is the number of valence-level electrons in nitrobenzene? HG - (B) (2) Provide a Lewis structure for nitrobenzene by adding all the missing valence-level electrons to the above structure. Use the dash-line convention for covalent bonds and show nonbonding electron pairs as dots. Your Lewis structure must conform to the Theory of Valence and the Octet Rule, and show any formal charges. (C) (4) It is known that the two nitrogen—oxygen bonds in nitrobenzene are equivalent and all the C-C bonds in the ring are very nearly the same length. In the box below, draw four resonance structures that explain these bOnding features. (D) (2) Draw the hybrid of the above resonance structures in the box to the right. The hybrid structure must shOw the proper magnitudes of all formal charges. (0K 1‘3 {W era 'lT’ 6" 50f cw ~ " Chemistry 322a/ 325a -8- Name Exam No. 1 (6) (10) Answer the questions below. (A) (6) Provide short definitions or descriptions of the following terms. 2’ “C Constitutional Isomers \Sowv—x—x ‘l‘\-\«.* dt‘c‘e‘v l-y\ (o:*0M\ (OM’KU—‘hbrh-LS The Octet Rule 0— 0. LC _ Electronegativity o M Mo doc’ftdo Tkk Lfikkm'l— +0 \Oll\x‘Q\/\ 6:7\ c\'—N\°_~Cl" «AA-ru‘fs Q..\Lc.-\T~o’V\3- (rm Q co\10u\<z—~N\~ looch Mt sq m‘ a (B) (4) Two pairs of structural formulas are shown below: A and B. Briefly explain why the structures in A are representations of the same compound while the structures in B represent different compounds. Your explanation must be based on the nature of chemical bonds. ‘ ° \1\ At 'ltots‘x- VD QTOUkfl/gk‘l‘lnk C—C., \n—mi OQLur-g \;a_e m$k Sigma Bamks‘ 04w SYMMLJfbK‘c-ak\ Eomk G_‘(\t§ - lac-31‘1be AOCS (KO-T “Mcou p\& +\I_ WVKQ‘QY’A? cu‘l'DMt; oc—‘m‘hb, Ir“ 3’ C. \Qx—jL \OCLFFl‘Lr‘ —\-o Vo+1H°M aroww‘i ’f‘lAL C~Q qfis L~ét$+8. Ro+a+t94\ wacoqxflfi omr\€gpfi>f’vx V3 Oink/fig erbf‘bis +\~°:\’ “MIX-K Chemistry 322a/ 325a -9- Name Exam No. 1 (7) (O4) Compounds A and B below have the same molecular formula, C4H100. While they have similar solubilities in water, the boiling point of B is more than 80°C higher than A. Explain the different boiling points in terms of intermolecular forces. Use pictures in ' your explanation. CH3CH20CH2CH3 A CH3CH2CH2CHZOH B b 345°C 117.2°C P Glubflity in 7.5 g/lOO mL 8-0 g/ 100 mL water, 25°C Ti" ‘DMM‘MZ “Powfi” 0’? ‘3 C—‘tlnv—aekcolwq \$ “Nd” k‘fikl“ \qua‘cfiL 0‘? Myth-032.“ bewil’vxj "Y‘arksK \V'tu‘: cLPMCQQ Tktj s‘hw'fi "K’Rt'tmolublcsh "cahci 5-90.51 lot 1a; 'V'l/‘K “EVIL 3% MON. m“ —\—\~L Artful; --Q"t-E)o\Q {lorezs VA TlmL \m-‘(zx-Nlncukcgr Runs CXR mo'k‘ \MVOr’V‘ur-rel— \’Y\ +\’\'~ \ICh~ For S-t-vJ‘L. MVV‘L +\’\Q“M"\ 1"erle \g xamclgi ‘Ngr +0 OVLV’LOML “mg S'l’nkbfltikfiam in B é 0" *‘0 H“ bO’KAC’VA? , EC— ,./\AJ‘—L S», CH3<H7§+2QHZSC§< "10}le CHgffl/y 6" S‘HZ 6km; Ha cuCHK C qéuflgf’ “:0 CS“; Var V3 i 45; (“25% Ur—‘Bmdimg \’l’\ ‘ d\?a(;——thPo\9\ hflUC& 34434—— Cd-ltkuAi-ao'k \‘flq‘ \\IU:& ...
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Exam 1 Answers - CHEMISTRY 322aL/325aL Please g g 3 EXAM NO...

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