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Unformatted text preview: CHEMISTRY 322bL anmmtﬁmké77‘2306 —PA££—2096 1H
Pracfm gr SECONDLABQUIZ‘ 07 m KEY Lab time BY Eigﬁ This test comprises this page and four numbered pagesJ plus junta“. will be available at lab checkout a
hours; Tu a.m. lab meets
will be available at SGM
your quiz wil 9 Fri Dec 1, quizzes
initial here
2,
l unclaimed tests will be thrown out Jan 1 I Fri a.m.,
, 2007. put out in the study room, (1 FINAL EXAM is Mon, Dec 8  10 a.m. in the lecture
room, SGM 124, as were all previous exams. (a)(2) To get clean PhNHAc from aq PhNH}? (6) Recall that aniline, Ph—NHZ, was converte 1 £:/0 to pnitroaniline
(PNA) via acetanilide, PhNHAc, and pOZNC6H4—NH—Ac. I0 one added Ac20,
swirled, and then added aq NaOAc. Reversing the order of
addition leaves some unreacted Ph—NHZ. Explain in <20 words.
Answer THIS question onlypossible NET negative otherwise. fqu/ NaO/itc {(1rny (I'M 6? (Cato [eh]. {Wk
Gyro gigIce? ﬂL‘Nﬁ/L’OM 2‘4)“ Mtg mf‘ Mcf. (b)(4) Give two reasons why the hydrolysis of pnitroacetanilide was
done using refluxing 20% aq HCl, bp ~ 108°. One reason re—
lates to ease of handling; for the other tell why the rate is
g9; slower than with "conc"(38%) HCl. Use <40 words total. 2 ,9 mm a w» M «WW
\
6 kt “Xi gag} ,4“ mm ,o/ 3 1 TM 4/1 /x “W 4“
cant, MJ (4) 2Naphthol (HA)'s pKa = 10. Assume its anion, A", reacts 105x
faster than HA with Ar—Néa. From pH — pKa = loglo([H1/[HA]),
the rxn rate falls 10X per pH unit from 9 a 6. (a) Calculate
the pH (pH=) at which amt prod from #3,: amt prod from HA. (b) Explain why from (pH: —1) to pH = O, the rxn rate stays
nearly constant; use <15 words.  2 — {{Jf f: :LI’CD 3. (10) This question deals with the formation of glucosazone (GPO)
and its cyclization to a 1,2,3—triazole. (a)(4) Balance the chemical equation below. Include correct coeffi—
cients for all species, including three gins: Products. c=o C=NNHC6H5 H» , #11011 J (:I=NNHC6H5 f“ [9 A N [71 1/ +' N 3 f. 14’; ($HOH)H + QHSNHNH2—> ((:H0H)n .6 M W
t‘ pilsggzzm a) Wu; (b)(6) The overall scheme for GPO —————’ triazole is shown below: ' Dissolution Step 1 Step 2
k  k k
d(issoln) 1 2
GPO  —————> GPO ——" m ____., p
solid in aq i_PrOH in soln d—k 1 (1) (3) Given the usual observations, namely that solid GPO dissolves
rapidly with simultaneous formation of the deep red IM, tell
what this means about the ratedetermining step and the rela—
tive values of kd, k1, and k2. Use <20 words. £05 2: 57‘ L CIM
[447‘ #Je (7m! jg .L,/, (2) (3) A few students got slower reaction with slower GPO dissolu
tion. All had heated more strongly, boiling away much of the
iPrOH. Explain these results in <25 words, referring to
which step(s) are slowed and why. CJW‘ law ("Iota/1‘ 6'00“; I W4 1w “ML, m. 14 #6 W2 3 4. (5) The three crystalline forms of glucose and the temperature
range over which each is in stable equilibrium with a saturat—
ed aqueous solution are as follows: (1) anhydrous a (55° 
98°), (2) a—hydrate (< 55°, MW = 198, i.e , 10 9 glucose
corresponds to 11 g hydrate), and (3) (anhydrous) B (> 98°). An equilibrated solution contains 48 g B and 27 g a—anomer.
The solution is slowly evaporated at 30° until 25 9 total
glucose remains in solution. Explaining your answers, give
(a) the comp sition and amount of each solid that crystal
lizes, and ( ) the amount of each anomer in solution. 5.(6)(a)(3) A 10—mL filtered portion of the GPA reaction mixture
stands 8 cm high in the polarimeter tube; it has an observed
rotation, aobsd = +13°. .After adding 2 mL conc H280 , aobsd
rises to +45°. The speCific rotation of aGPA = +10 °; take
that of BGPA = 0°. Assume the final GPA mixture is 90% a and
10% 8. Calculate the BGPA fraction before adding the acid.
Full credit only for minimum path calculation. :2 +130 :7 %=~0.26 (26%)=i7‘f%
oV" W030 W,” 8 /
(b)(3) The literature mp's for the GPAs are 134° for 3; 112° fOr a. A student gets 130°  132° for his/her 3, 103°  111° for a,
and a mixt mp of equal amounts = 116°  120°. Assuming the B
is >95% pure, interpret and explain these data; use <15 words. Sue“... «Aaf 6%.; “#va
up, gamma A :2. «c 7 '“ /7€=7 6. (9) The mutarotation of glucose is a firstorder reaction, i e.,
the reaction rate is proportional to the conc of the starting material: —d[A]/dt = k[A], where t is the elapsed time and k
is the rate constant. The solution to this differential
equation is ln([A]/[A]O) = kt; note that the  sign means k is positive since [A] is decreasing. In a firstorder proc ess, the rate of relative change is constant, e.g., if 20%
reacts in one minute the 20% of what remains reacts in the next minute. Specifically, the half—time (or halflife), t”,
the amount of time for [A]now a %[A]now, is often reported. (a)(3) Refer to, but do not detach, the following page, which is a
semi—log plot of optical rotation data vs time for a—glucose.
Determine tx, including units, from the time interval between two values on the vertical axis. Mark the plot and/or state
your reasoning. 15y» = /3’7’0 YMQ .A‘M (b)(2) From the basic equation, ln(%) = kt%. Use this and your
answer to (a) to get the rate constant, k; include its units. 1,4 AU/ﬂaaé‘l? €01) =3
wig. :o.1 _ 0‘37}
a<EvIQTM¢CVQ ‘i’;:E:;’~ N
(c)(2) The rate constant gotten from the plot is k for a—anomer » equilibrium mixture. Given that the eq mixt anomeric ratio is
B:a = 5:3, calculate kaag from (b). ICMLQ e7 lM/(‘Lf v3 '72 ﬂ £$_.g> gjﬁgvml.Kf: (d)(2) Note that the first data point, the one at 7 min/18°, does not
"fit". It was in fact ignored in drawing the line. Rational—
ize its position, given that time was counted from the begin;
ﬂing of the dissolving of a—glucose in water; use <15 words. ﬂof 4,6? Mom/L 6% Moi: 6k
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 Spring '07
 Jung
 Organic chemistry

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