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322af02_e1_blank - CHEMISTRY 322aL EXAM NO 1 Please Print...

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Unformatted text preview: CHEMISTRY 322aL EXAM NO. 1 SEPTEMBER 18, 2002 Please ___________________ Print Last Name ____________________ First Name SSN ________________ name of TA (1) (20) _____ (2) (10) _____ (3) (20) _____ (4) (20) _____ (5) (15) _____ (6) (15) _____ (100) Grader ______ ______ ______ ______ ______ ______ first letter of last name I will observe all the rules of Academic Integrity while taking this exam. Signature Chemistry 322a -2Exam No. 1 (1) (20) Answer the following questions. Name______________ (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 electron pairs. Each pair of bonding electrons must be shown in the dash or line formulation, while nonbonding electron pairs must be shown in the dot convention. Place your answers in the boxes HOCH2CH(CH3 )CH2 CH(CH3)2 (CH3 )2CHNHCH2 CH=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. H H C N: H H C H H H Cl: : H (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. O NHCCH3 O (CH3 )2CHCH2 CCH2CH 3 CH3 CH3 -C-CO2 CH3 O CH3 Choices: alkane, alkene, alkyne, alcohol, aldehyde, amide, amine, aromatic, carboxylic acid, ester, ether, ketone, nitrile. : C C O HH C H H H :Cl : : : Al Cl: Cl: : : : : : = = Chemistry 322a -3Name______________ Exam No. 1 (2) (10) According to hybrid orbital theory, the bonding in ethyne, C2H2, 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 2s 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 energy 2px 2py 2s 2pz Show the electronic configuration for ground state atomic C in the valence level by placing electrons as dots ( ) in the above diagram. (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 2s and 2px atomic orbitals in your picture. Draw and label all the atomic orbitals according to the x,y,z coordinates shown below. z x C y energy Draw and label the set of atomic orbitals for sp-hybridized carbon, and add the valence level electrons as dots. Chemistry 322a Exam No. 1 (3) (20) Answer the questions below. -4- Name______________ (A) (4) In the boxes below, draw structural formulas for two constitutional isomers of C3H8O 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 C3 H8O with Different Functional Groups (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. CH3 CH2 CH2 CH2 Cl (CH3 )2CHCH2 OH ClCH2 CH2 CH2 CH3 OH CH3 CH2 CHCH3 Br Br CH3 CH2 CH2 CHCH2CH 3 CH3 CH2 NHCH2 CH(CH3)2 CH3 CH2 CH3 CH3 -C-CH2-N-H H CH3 CH3 -CH2 -CH2 -C -Cl H CH3 CH2 CH(CH3)CH 2Cl (C) (6) Circle the compounds below that have nonzero molecular dipole moments based on geometries predicted by VSEPR theory. CH3 Cl HC CH H2C=CCl 2 NH3 BH3 BeCl2 (monomer) Chemistry 322a Exam No. 1 -5- Name______________ (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. 24. 25. (B) The molecular formula of the structure below is HO (circle correct answer below) C 6H12O C6H14O C7 H12O C7H14O (C) The formal charges on the following carbon species are (circle correct answer below) : : H C H I H H C H II H H C H III I (0), II (-), III (0) I (+), II (-), III (0) I (+), II (-), III (-) I (0), II (-), III (-) (D) Nitrous oxide, N2O, has 16 valence electrons. Two correct Lewis structures with proper formal charges that are resonance structures for nitrous oxide are: (circle correct answer below) + - + + + : N N O: : N N O: I and IV : N=N=O: : N=N=O: : : : : : : : : 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): H2C=CH 2 I H3C-CH3 II HC CH III (circle correct answer below) I>II>III III>II>I II>III>I II>I>III (F) Noting that the benzene ring is planar, the aromatic compound with a molecular dipole moment of zero is (circle correct answer below) Cl Cl Cl Cl Cl Cl Cl Cl Chemistry 322a Exam No. 1 (4) Contd. -6- Name______________ (G) Identify all the tertiary alcohols among the structures below. OH CH3 I only I CH3 CH3 CHCH2OH II I and III I and IV OH (CH3 )2CHCHCH 3 III I, III and IV OH CH3 CH2 C(CH3)2 IV (circle correct answer below) (H) The important carbonyl functional group is found in -aldehydes, carboxylic acids, ketones, esters and ethers. -aldehydes, amides, ketones, carboxylic acids, and esters. -amines, carboxylic acids, esters, ethers and ketones. -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. MW MP BP O CH3 CCH3 acetone 58 -95oC 56oC OH CH3 CHCH3 isopropyl alcohol 60 -89oC 82oC 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 difference in van der Waals attractive forces in the two compounds. -intermolecular hydrogen bonding that is present only in isopropyl alcohol (J) Short range van der Waals or London attractive forces increase in magnitude with -increasing size of the molecule. -the presence of polarizable atoms. -the presence of n and p electrons. -all of the above factors. = Chemistry 322a -7Exam No. 1 (5) (15) Answer the following questions. Name______________ (A) (6) Diazomethane, CH2N2, is a highly reactive and toxic organic compound. The bond connectivities 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, determine the number of valence electrons in diazomethane. number of valence electrons: ________ H C H N N bond connectivities in diazomethane a Lewis 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 carbon atoms that are sp2 hybridized. CH2 OH HO CH C H O C O C C * OH HO ascorbic acid (vitamin C) 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 ascorbate anion is formed, which is relatively stable. The hybrid of the ascorbate anion has contributions from three resonance structures. Draw two resonance structures below that make major contributions to the hybrid, where a negative charge is on an oxygen atom. Show all nonbonding electron pairs and formal charges in your resonance structures. Chemistry 322a Exam No. 1 (6) (15) -8- Name______________ (A) (9) Provide condensed or line-dash structural formulas inside the boxes for the following compounds. isopropyl alcohol methyl ethyl ether a tertiary amine with formula C4 H11N (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 100oC. Circle the structure with the higher mp. Offer a short explanation for the higher mp. CH3 CH3 -C-CH3 CH3 neopentane CH3 CH2 CH2 CH2 CH3 pentane The boiling points of the two constitutional isomers below differ by more than 40oC. Circle the structure with the higher bp. Offer a short explanation for the higher bp. CH3 CH3 -N-CH3 trimethylamine CH3 CH2 CH2 NH2 propylamine ...
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