322a lab quiz 2 fall 06 blank

322a lab quiz 2 fall 06 blank - CHEMISTRY 322aL/‘3'2-5-aL...

Info iconThis preview shows pages 1–5. Sign up to view the full content.

View Full Document Right Arrow Icon
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 2
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 4
Background image of page 5
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: CHEMISTRY 322aL/‘3'2-5-aL W6 W pro c'fc‘V-Q, 74hr SECOND LAB QUIz- 5"0 0 7 BY Rigs: l.(12) NAME Lab time This test comprises this sheet and four numbered pages. _A x. 1;! is Thu, Dec 7, 8:00 - 10:00 a.m., If told to use less than a certain number of words in an answer, no SO——deduction for excessive verbiage. Graded quizzes will be available frOm TAs Thu Vnd ri, NOV 30 and Dec 1, at (check-out)v los and offi - ours ("Special office hp 5" for TAs with no lab or r-n lar office '-rs Thu or Fri). If you initial her- “?-——> , your quiz will be put in. e Study -~-m, SGM 102, at 5 p m., Fri, Dev ‘. All quizzes will be rown out Jan 26. in the ~.me ooms as all previous exams. 1. (12) Dehydration of 4-methyl—2-pentanol, (CH3)CHCH2CH(OH)CH3, produces the following methylpentenes (MPs) without garbgn aggletgn rearrangement: 4—Me-1, 4—Me—2, 2—Me—l, and 2-Me-2. (a)(4) The 60% and 70% H2804 give different quantitative results. Circle one choice in each pair separated by / below: In the 60% acid the total MP yield is lower / higher, and one gets a larger / smaller fraction of rearranged products. This is because the 60% acid is more / less basic, and therefore average carbocation lifetime is longer / shorter. (b)(6) (i)(2) Only two of the MPs can come directly from the initial carbocation. Yet substantial amounts of other MP5 formed rapidly at < 100°. Tell what this implies about the magni- tude of £3 , the free energy of actiyatign,for the required carbocation rearrangements by circling one choice: large negative small negative large positive small positive (ii)(4) The only MP listed above which must form via a 3° carbocation is 2—Me-1. Draw the 3° ion and the main Product which forms from this ion. (c)(2) Methylpentene samples re—distilled on a steam bath generally left a small pot residue of material, not dimers and higher polymers from methylpentenes. Name or draw this material. -2- 2.(6) Note the gas chromatogram of the commercial solvent, "Skelly B". 'l J . , "_;r ; I I' l‘ ’ i '+‘ l- ' IIIIIIIII-IIIIIIIIII ’ h+ w-I... -:I‘ II. .- TII'IIIIIIIIIIIIIIIII !ll...III-IIIIEIINIIIIIIIIII l -' I..--. ': III-IIII-IIIIIIIIIII.I.......Ih“"flINIIIIIIIIIIIII Illll l; I III-II...-=:=....--.-.-.--. VJINIUI=HIIIIIIIIII -| 5 I. ===.. II... I v III-II...- A IILCII I......... .- LL: I"h 'III-IIIIIIIIIIHI-KZHIIIIIIIIIIIIIII II... ' I ..T' v I. II."..III-l.....==:=:.....' . i' v-n == 15‘; * II... :I-MIIIII-UIIIII il P . I n... E I ' . v n..- — i . ‘ ' =a===lllllli==: .75 III-Illlll=====EEEilIIH==IIIII===:======IIIIHw- - flyfiflflfiflfififififlfifififi %fifiE%fi%fififimfifififlfi =======E==============="=-- ==32§"""""""'""':llil'lllllll-I-Iul- V III-IIIIIIIIIaIIIIII III-IIIIIIIIIII-III III-IIIIIIIIII-IIIIIII III-IA-§IIIIU-IHIIIIIHIIII III-IIIIIIII-I-l"IIIIIIII .- EIErlflllilI-IEI:=:IFIn.“-IlIfllHI.IIIHIIIIII-“II-IIIIIIIIIIIZHIV-III... Ilflr Ill 71". mm All-III- ]. I! [III [IZJ'IIIIIIII—IIIIIII-IIIIll-lI-I'll 1' !I III...- IIu-Ifill-IIII-IIIIIII-IIIIIIHQEJ7‘IIU-PQIIIIIIIIII 'IIIIIIII=:IIIII'll-III..- MIIIIIII III-IIIIIIIIIIL‘CZIIII'E'IIll-"IIIIII I'll-III IIIIF‘K‘I-I-III [III-I-Il-I-l III-IIII=:IIIIIIIIIIIII-VIIIIIMIIIII-Ill. :=:===:III=!-‘EIIIII-l I.- A «III-IIIIIIIIIII-I-III III-P-IIII'ZIIIVII-IIIIIIIIII.“ I‘m-III..- F5'=:====:flfi:3.1::=====E=fiflfil§g§=====Eigfi========E=§u==""'==¥====i§fl=== .- . Ell-IIIIIIII IIIII.-I-IIIIii-Ill...III-IIIIIIIIIIIIII-II”... I-IL III-IIQNIIII =IIIIIIIIIII Ill-IIIIlllIIhIIIIIIIIIII-lIII-II-IIIIII-lllflllIIIIIZIIIIIIIIILNIII I...III--III-I...-III-.2...-I...-I...-III-II...-I...-I..:..l..5............f.. l..-I...-III-IIIIIIIIIIIIIIIIIIIIIIIII..l-I-I-Ill.‘IIIIIIIIII-fllll-l...--.-.§ 1Ill-I...-III-.I-l-IIIIIIIIIIIIIII-II...-III-l-Iflfllfl-I-I-l-IlIII-II...--_---- II...III-lIll-Ill-IIIIIIIII-IIIIIIIIII-IIIIIII-IIIll-IIIII-Ill-lI-IIIII .- III-i , Ill-IIIIIIIIII...-Ill-IIIIII‘JIIIIIIIIF5.....l-IlI........'.-....-..ifi--=...-. (HIE-II-IIII-IIIIIIII‘IIIIIII III.-IIII-I-IIIEIIIIIII-IIJIIIII=II III-III.- lI-IIII-III-IIII-III-I-Ill- fl 'vII.!.:!-III-IIIIIII-IIIIIII-Illl III-I'll...- ,‘ A _ (a)(2) Which peak is associated with the most volatile component? Number (b)(4) Recall that one estimates the areas of GC peaks by assuming they are triangular and calculating "Kb-h". Mark peak 4 to show how this is done: Draw a proper baseline, the height and Kb. Then calculate the area in "squarelets" (use h and Kb to the nearest % squarelet edge). Also state how one determines Kb using <10 words. 3. (9) (a)(4) (b)(2) (c)(3) -3- This ques and 4 on p 4 deal with the bromine-catalyzed isomer— ization of I, gis—X—CH=CH—X (X = COOMe) to the trans isomer, III. The Br-containing species, II-gauche and II-anti, dif— ferent conformations of X—gH-—$H-X, are intermediates. Br Consider a mechanism for I 4 III in which a propagation step is II-anti + I 9 III + II-gaughg, i.e., a mechanism in which II-anti does not expel Br- but "hands" it directly to I. (1) Write the other propagation step(s) of this mechanism and circle the chain;carry;ng species. (2) Sum the steps, showing that they do add up to the overall reaction. In this prep, liquid dimethyl maleate (I) is converted to solid dimethyl fumarate (III). Maximum conversion of I a III requires enough Br2 that about 12 % by mass solid Br—contain— ing by-products form. Tell why one might choose to use less Brz while leaving more unreacted I; use <15 words. Exactly 80 mg Br2 (MW = 160) reacts with 10.0 mmol I (MW = 144). The yield of III is 60% based on I; all the Br2 reacts by addILiQn to I. CaICulate the mass ratio of III to Br—containing product(s). -4- 4. (4) Consider I a III as occurring in two steps: (1) I in soln a ' III as supercooled liq (scl) with.AG = 0; then (2) IIIscl a IIIsolid,.i.e., some III crystallizes, with AGxtln < 0. For Simplic1ty, assume mmolsolute = 100-(Xsolute), e.g., XIII = 0.02 means 2 mmol III in soln. Calculate mmol solid III from 20 mmol I if rxn is run to equilibrium at the T—- (a) Where = 0.03. (b) where XIII = 0.06; after rxn is over, mixture is cooled to T at which XIII to 0.02. M (6) Note the data. M W AG .- (l) n—Bu—Br -—-—> Q—Bu+ + Br‘ + 1 (2) -Bu+ + I' -—--> Q—Bu-I - 71 Sum: n—Bu—Br I‘ -——-> n—Bu-I + Br~ +7 (a)(3) Set up the equilib 'um expression f-* the Sum, and calculate Reg (2 sig figs), = e tfii/RT); r%‘= 0.66 kcal/mole. (b)(3) In ace "e saturated in both NaI and NaBr, ake [I'] = 2.0 M and =r‘] = 3.010"7 M. Plug these values i o the above Ke * iression, calculate [n—BuI1/[n—BuBr], and stas- what this ratio means about the Keq value for the overall x~, 3-; —Br + NaI ———-> n-Bu-I + NaBr in acetone satd in both 5; ts. ...
View Full Document

Page1 / 5

322a lab quiz 2 fall 06 blank - CHEMISTRY 322aL/‘3'2-5-aL...

This preview shows document pages 1 - 5. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online