225Sp08 Test 1 Key - "u- E“ CHEMISTRY 225 TEST...

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Unformatted text preview: "u- E“ CHEMISTRY 225 TEST #1 MAY 16,2008 1. (16) l. 2. (12) NAME égg g 3. __ (21) PERIODIC TABLE OF THE ELEMENTS 4. (12) m E 5' —~— “4) E ii. E ii: ii. ' ‘ ' ' ' ' TOTAL 05) 56 5? : 72 1'3 74 1'5 76 71' 78 1'9 8|] 3} 8?. 33 84 35 86 Ba La : HI 1'- W R: 09 [r Pl All [is 11 Ph Bi Po At Rn ms an I nu um um M2 mu 191: ml WM} m0. m4 201.2 299.0 mm m rm: 101 105 [06 ID? 108 110 111 [12 H4 Db ‘% Bl: Ba rm: 1 W) --- F--- (159'; (112] WT) GM 58 59 I51 62 63 64 65 66 61' 68 69 T} C: Pr Pm Sm Eu Gd Th Dz. Bo Er Tm Lu 1-01 am u :m 1510 um 158.9 15 164.9 1513 MM :15» 90 91 92 93 94 95 96 9’? 93 99 100 “33 Th P: [I g}: Pu An Cm Bk CF E9 Fm Lt mo 23m m n n- (m: cm) at: asn um: .~. .- M l. (16 Points) ORGANIC FUNCTIONAL GROUPS. GOOD LUCK! a. ( f 7) Chemists in Paris have developed a “green”, solvent—free method for the synthesis of certain carboxylic acid derivatives that utilizes microwave heating (MW) [Tetrahedron Lett. 2008, 49, 3004]. An example of this methodology is shown below. 0 w H CH3(CH2)1,NH2 + CH3COZH M—r CH3CNH(CH2)11CH3 - (1)1der1tify the functional group in the product. QM) (\é, o (3) Use the functional group approach to write a structure for the product of the analogous reaction below. compound A - (3) In other news from the world of organic chemistry research, Indian chemists have utilized compound A above as a reactant in an organic reaction [Tetrahedron Lett. 2008, 49, 2527]. - Determine the molecular formula of compound A. C'no H 2 5 N — in the common nomenclature system a primary amine is named as alkylamz'ne where the prefix refers to the number of carbons in the molecule. Provide the common name for compound A. clécxfi 4W; W‘L 2 b. ( f 6) Write condensed structural formulas for compounds satisfying each of the following descriptions. Each compound must have at least five carbons - an aldehyde 0 an ether 0 an ester l l a? O . u I one C2010 C” C”; a”; c out; Up, cos i ester c. ( f 3) In the presence of a strong acid certain alkenes will accept a proton to form a G) cationic species. Such an electron-deficient i-r CH3_C_CH3 . . . . - —fi-— 51360185 is identified as a carboeation. An example (EH 3 is shown to the right. :- (2) Deduee the structure of the alkene precursor to the carboeation. HINT: the carbon skeleton is unchanged as a result of the reaction. I (1) Describe the geometry at the central carbon of the carboeation product. __ \i "'13 0 mal . (12 Points) BONDING IN ORGANIC COMPOUNDS. a. ( _ f 8) 4-Hydr0xyornithine is a nonproteinogenic amino acid found abundantly in marine organisms and plants. Chemists in India have recently deveIOped a laboratory synthesis of 4-hydroxyomithine [Tetrahedron Left. 2008, 49, 3297]. The structure of 4-hydroxyornithine is shown below. Determine the geometry and hybridization at each of the indicated (arrows) atoms of this compound. NOTE: for the sake of clarity lone pairs of electrons have been omitted from N and O. Atom Geometry Hybridization a l3 O ' Na F7vrfimf€£tfitl if b / OH NH2 Cc “lfrll‘vaLf (L, a ( (Pfi 4-hydroxyomithine Cd I 13 9 ._... a l 4,9 3 b. ( f 4) Dimethylamine, (CH3)2NH, is the simplest secondary amine. Using the atomic framework to the right sketch a picture of the covalent bonding in dimethylamine. Be sure to show the geometry at each atom clearly. Identify the orbitals used to overlap. . (2] Points) STRUCTURAL ISOMERISM- a. ( __ f 9) Glycolic acid (structure to the right) is the simplest hydroxy acid and the precursor to a bond 3 bond [3 potentially useful polymer. As noted in the April 28”" issue of Chemical & Engineering News, a Japanese firm J 3 ' ' J is investing $100 million to build a plant to produce H P9 so - glycolic acid in West Virginia. H 0 CH2 - H i o (2) Complete the structure of glycolic acid by adding lone airs to the a ro riatc atoms. . . p pp p glycohc and I (2) Identify the orbitals that overlap to form the pi bond _ _ . __. f l 1' id. ° mm C? and Om t” I o (3) Which of the indicated carbon-oxygen bonds of glycolic acid is the longer of the two? E l' ‘ b ' fl . a xp am ['16 y r; O “(l CL :4..- iC~~J€K ‘ I‘ll f’wcri L o-vrwim. ‘3"! C911!” curl C3905. gmwri Lg .-.9‘L]£E.M,A7L7 “all? “(f {Cg/)2. “.1 152-; T“ m Cfirt éifi I“? C 06%)», ‘LD “lie murL’uc— 1% use“ Lad . a vtflfi‘av—GHCC. wd-i/LCA log:ch L Lac. 996%; £116 VLLLL 0 (2) Write a structural formula showin all bonds for an isomer of glycolic acid. I” “A at (,La L A (1, , o (2, BONUS QUESTION!) Write a structural formula for a second isomer of glycolic acid. O O .F".. I! H ’4 Cde—Ohfi- CW5C OOH k'élycolic acid isomers 4 b. ( I 8) Write condensed struc a] formulas satisfying each of the following molecular fOHHUlaS- W10»; aw Gufl myypr Lama—we! 0 (341130 (mp isomers) o Cqu, with two tertiary carbons o C5H9Cl l l on; on (frag. Ct; a; Cer on; as C/t-LC/HO at s: on car out all: = at cat, can OH L C4HSO (two isomers) C5HQC| c. ( hm / 4) Write structures for the complete set of alkyl halides with the molecular formula C4H9Bl’ W"? _l i l . only? at cut, oil, 5“ 9‘" CA4; (“J—tux, 8., CI"; Gib», are, U'fC/té War?” 3v i C4HQBr isomers “é . (12 Points) THE RESONANCE METHOD 3. ( _#_fl__ f 4) Alkyl azides are useful intermediates for the synthesis of primary amines. The structure of methyl azide is shown below. Using the atomic framework provided write two resonance structures for this compound. methyl azide b. ( I 8) When treated with a particularly strong acid, acetic acid can serve as a base, i.e. as a proton acceptor. This property is depicted below via condensed structural formulas. CH3C02H ——H—)“' CH3CO2H2+ The product of this reaction can be represented by three resonance structures, one of which is shown below. 5 o (3) Circle the least stable resonance structure and briefly justify your choice. CI-ivcl-«EA {Shwc \L-AvL lac“; “[‘fwé’s-‘l' CLuaiF «Cl L7?- imtlé- okfirl or gesa‘le'l‘ {I {'[ngLWfi-AC— a‘l- CGLLB‘a-m. 5. (l 4 Points) MULTIPLE CHOICE. Write the letter of the correct response in the space provided. A Which molecular formula could represent a compound with only sp2 hybridized carbons? A. C3H4O B. C3l150 C. C3H602 D. C3H302 . . . . l Consrder the structure to the 11 ht. Estlmate the C-N-C bond an le 111 L -- C. g g CH3C H:N—CH3 this compound A. 90° B. 109.50 C. 120° D. 180° D_d___ When atomic orbitals on different atoms overlap they combine to form a pair of molecular orbitals. Which statement concerning these molecular orbitals is TRUE? A. The molecular orbitals are formed as a bonding and antihonding pair. B. Antibonding molecular orbitals are always higher in energy than bonding molecular orbitals. C. Each molecular orbital can accommodate two electrons. D. A, B, and C are all true & Identify the shortest carbon—carbon single bend in CH3—-CH=CH—CH=CH~—CH2—CH3 the molecule to the right. I) J j j a- b c d - A. Bond a B. Bondb C. Bond c D. Bond d Q Which two structures below represent the same compound? (ill-h Elli-Ia (I3He ‘ Cl-bCl-bCI—bCIZI-ICHCHQCHa % Cl-bCHzCHCHzCHQCHCHg CH3 I II III IV A. I and II B. ll and III C. H1311le D. I and IV A Deduce the identity of the molecular orbital to the right. A- OhicspS ' Csp3l 3- 5*lcsp3 ‘ Csp3l l C. ‘Itb[C2p - C2D] D. n* [C213 - C213] Which of the following would be a legitimate IUPAC name for a branched alkane? A. 2,4,4—nimethylpentane B. 3-dirnethylhexane C. 2-ethylhexane D. 2,2-dirnethylhexane ...
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225Sp08 Test 1 Key - "u- E“ CHEMISTRY 225 TEST...

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