orgo postlab 2
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University of Illinois, Urbana Champaign *
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Course
440
Subject
Chemistry
Date
Jan 9, 2024
Type
Pages
4
Uploaded by PresidentHeatHare29
●
Q1
○
(a) Isopropyl alcohol: 83C, Acetone: 56C, Al(O
i
Pr)3: 125C
○
(b) Isopropyl alcohol forms an azeotrope with water; add water to this
mixture, use a Dean-Stark Trap, and boil away the water, letting the
isopropyl alcohol flow back into the reaction flask. It would be hard to
separate Al(O
i
Pr)3 and acetone using methods we’ve learned in class, like
a separatory funnel, since acetone is miscible in water and organic
solvents.
○
(c) A large excess of isopropyl alcohol and an azeotrope of acetone with a
compound like heptane in a Dean-Stark trap can drive this reaction
towards the products; in this set-up, a large excess of the reactant
(alcohol) is used, and acetone is continuously removed from the
environment, both of which favor product formation.
●
Q2
○
(1) lower temp (2) use less fluid bro idk
○
(a) One solution would be to use a large excess of isoamyl alcohol, which
would drive the reaction towards the products as per Le Chatelier’s
Principle. Another way to remove water from the reaction and drive it
continuously towards the products without using a Dean-Stark trap would
be to use MgSO4 to dry the water being formed.
●
Q3
○
(a) If you don’t fill up the trap with heptane, and particularly, if the amount
of heptane in the flask is below the amount of water that the reaction
generates, then you won’t know when your reaction is complete and
won’t be able to end it on time.
○
(b) adding brine to the extraction helps to transfer water from the organic
layer to the aqueous layer and make sure that the organic layer is
adequately dried out.
○
(c) Isoamyl alcohol is among the starting materials, and would have a
strong peak at the 3200-3500 mark because of the presence of a
hydrogen-bonded OH group; this wouldn’t be present in the pure isoamyl
acetate product that is desired. Monitoring the presence/absence of this
peak would indicate whether or not the reaction has gone to completion.
Follow the Lab Report Guidelines included with this syllabus. Report should include:
ü Balanced equation for esterification reaction
In notebook (???)
ü Mechanism of esterification reaction (It is advisable to look up esterification in
your Chemistry 236 text as well as through online resources)
In notebook
ü Calculate percent yield (experimental yield/theoretical yield * 100%)
1.996 / 8.957 * 100 = 22.28% yield of isoamyl acetate
ü Summary table with experimental yield, percent yield and boiling point range for
all fractions. Percent yield for non-product fractions can be omitted.
Experimental Yield
Percent yield
Boiling Point
Range
Fraction 1
Not measured (waste)
---
92-99C
Fraction 2
1.996 g
22.28% yield
129-133C
ü Summary table for IR spectrum data (include original printout with report). See
Sample lab report for IR summary table.
Wavelength (cm^-1)
Intensity
Functional Group
3430.12
m
O-H
2930.56 - 2959.93
m
Alkane C-H
1743.00
m
Acyclic ester C=O
DISCUSSION
ü Discuss LeChatelier’s principle and why a Dean-Stark trap is used
LeChatelier’s principle states that if you disrupt the environment for a reaction that’s in
dynamic equilibrium, then that equilibrium shifts to counteract the change. In this
specific example, one of the products of the esterification reaction is water; by using a
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Related Questions
G.204.
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Consider the reaction:
2 NH3(g)
N₂(g) + 3H₂(g).
At T = 740 K, the equilibrium constant Kc for this reaction is Kc = 10.
What is the correct form of the equilibrium constant Kc for this reaction?
(A) Kc = [N₂] [H₂]³
(B) Kc
(C) Kc =
(D) Kc
=
(E) Kc
-
2[NH3]
[N₂]+3 [H₂]
[N₂] +3 [H₂]
2[NH3]
2
[NH3]
[N₂] [H₂] ³
13
[N₂] [H₂] ³
[NH3]²
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Calculate the value of the equilibrium constant, Ke, for the reaction
Q(g) + X(g)
2M(g) + N(g)
given that
M(g) Z(g)
6R(g)
2N(g) + 4 Z(g)
3 X(g) + 3Q(g) =9R(g)
Kc =
Ke1 = 3.01
Ke2 = 0.504
K3 = 14.3
4
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Calculate the value of the equilibrium constant, Ke, for the reaction
Q(g) + X(g)
2 M(g) + N(g)
given that
Ke
=
M(g) = Z(g)
6R(g) = 2N(g) + 4 Z(g)
3 X(g) + 3 Q(g) = 9R(g)
.116
Kel = 3.57
Kc2 = 0.524
Kc3 = 14.6
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Consider the overall reaction for t-butyl bromide (2-bromo-2-methylpropane).
(CH3)3CBr(aq) + H2O(1) →(CH3)3COH(aq) + H*(aq) + Br "(ag)
k[(CH3)3CB1]. The accepted mechanism for the
The experimental rate law is rate =
%3D
reaction is shown below.
(1) (CH3)3C-Br(aq) →(CH3)3C*(aq) + Br "(aq)
[slow]
(2) (CH3)3C*(aq) + H2O(1) →(CH3)3C-OH2*(aq)
[fast]
(3) (CH3)3C-OH2*(aq) →H*(aq) + (CH3)3C-OH(aq) [fast]
a) Why doesn't H2O appear in the rate law?
b) Write rate laws for the elementary steps.
c) What intermediates appear in the mechanism?
d) Show that the mechanism is consistent with the experimental rate law.
ed to the
of consumption of N
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Ff.278.
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Consider the following equilibrium reaction: 2 CH3OH + CH2O ⇌CH2(OCH3)2+ H2O ΔH = + 46
indicate the direction that the equilibrium will shift when the following changes are made.(No states are provided because the solvent for the reaction is not water, so aq for aqueous would not be appropriate.)
(a)Concentration of CH3OH increases
(b)H2O is removed from the reaction
(c)Reaction is heated
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Construct the expression for Kc for the following reaction.
Question 8 of 35
[H0]
CH,COOH(aq) + H₂O (1) CH,COO (aq) + H₂O'(aq)
Drag the tiles into the numerator or denominator to form the expression. Each reaction participant
must be represented by one tile. Do not combine terms.
Kc =
2[H₂01
[H,0⁰
[H,01
2[H,0²
2[H,0]
[CH₂COO] 2(CH,COO] (CH,COOP 2[CH,COO [CH,COOH] 2(CH,COOH] [CH,COOH!" 2[CH,COOH!"
2[H,01¹
RESET
[H₂O]"
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Trimethylamine, (CH3)3N, is a common reagent. It interacts
readily with diborane gas, B₂H6. The latter dissociates to BH3
and this forms a complex with the amine, (CH3)3N→BH3.
The reaction between trimethylamine and borane is shown.
CH3
CH3 H
CH3-N : + B-H
CH3-N B-H
CH3 H
CH3 H
Is the BH3 fragment a Lewis acid or a Lewis base? To decide
this, answer the following questions:
H
a Which of the following is a definition of a Lewis acid?
proton donor
electron pair acceptor
proton acceptor
electron pair donor
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Give detailed Solution with explanation (no need Handwritten answer
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Calculate the value of the equilibrium constant, Kc, for the reaction
Q(g) + X(g)
2 M(g) + N(g)
given that
Kc =
M(g) = Z(g)
6R(g) — 2N(g) + 4 Z(g)
3 X(g) + 3 Q(g) = 9R(g)
Kc1 = 3.01
Kc2
= 0.433
Kc3
= 12.5
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Hw.29.
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3. In the following reaction, acetic acid and ethanol react together to form ethyl ethanoate
and water. Use the following concentrations under equilibrium to calculate the equilibrium
constant, K.
CH,COOH (I) + CH,CH,OH (1) = CH,COOC,H¿(1) + H,O (1)
Species
Concentration (mol dm)
CH;CH,OH
0.30
CH;COOH
0.13
H,0
0.39
CH,COOC,H;
0.39
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Please don't provide handwriting solution
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A,Hº
4,Sº
Compound
4,Gº
ΗΝΟΣ
HNO3
N₂O4
H₂O
AgH° (101)
mol
-76.73
-134.31
Use this table of values to calculate the values listed below for the reaction,
HNO₂ + HNO3 → N₂O4 + H₂O
9.08
-241.83
J
5° (mol-K)
249.41
266.39
304.38
188.84
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1
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mol
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O-H
H
CH,OH
(b)
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H--0=C
(b)
0-H--
(a)
CH,
CH,
CH,
CH
H-C CH,
(CH.)4–NH,
CH,
CH,
|(d)
(a)
CH,
(d)
(c)
CH,
CH,
HC
CH
CH,
CH,COOH
(b)
[The image above is from a different textbook. Ask me later if you would like to know the
reference.]
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represented by the letter (d) in the image?
O hydrogen bonding
3 of these 4 responses are correct.
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O Kb2
O Ka2
Ο Και
Kbl
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reaction
CO(g)+ H₂O(g) → CO₂(g)+ H₂(g)
Na₂CO3(s)+H3PO4(aq) → Na₂HPO4(aq) + CO₂(g) + H₂O(1)
highlighted atom is being...
oxidized reduced
O
neither
oxidized nor
reduced
圖∈
Ar
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Additional UO2 (s) is added to the system.
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3C2H2 (8) C6H6 (8).
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- Calculate the value of the equilibrium constant, Ke, for the reaction Q(g) + X(g) 2 M(g) + N(g) given that Ke = M(g) = Z(g) 6R(g) = 2N(g) + 4 Z(g) 3 X(g) + 3 Q(g) = 9R(g) .116 Kel = 3.57 Kc2 = 0.524 Kc3 = 14.6arrow_forwardPlease explain to me in detail!arrow_forwardKc = 0.193 at 45 degrees Celsius for the reaction: Cycloheptane <--> Methylcyclohexane. What must be the initial concentration of cycloheptane for it to have an equilibrium concentration of 0.10 M?arrow_forward
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ISBN:9781285199047
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Publisher:Cengage Learning