Experiment 56: Synthesis of an Ester from an Alcohol and Carboxylic Acid
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Transcript of Experiment 56: Synthesis of an Ester from an Alcohol and Carboxylic Acid
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Experiment 56: Synthesis of an Ester from an
Alcohol and Carboxylic Acid
Structure determination by Infrared Spectroscopy
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Infrared Spectroscopy
• Read Chapter 25, pp 833 to 867 • Work problems on page 866-867
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Typical Infrared AbsorptionRegions
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O-H stretch
• The OH peak is very strong and broad!
• It usually appears near 3400 cm-1
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C-H stretch
• sp3 C-H stretch comes at 2950 cm-1
• sp2 C-H stretch comes at 3050 cm-1
• sp C-H stretch comes at 3300 cm-1
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C=C stretch
• C=C stretch occurs near 1650 cm-1
• It is often weak unless it is conjugated to a C=O
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Normal Base Values for C=O Stretching
Functional Group Frequency (cm-1)
Ester Aldehyde Ketone Carboxylic acid Amide
1735 - 1740 1725 1715 1710 1690
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C-O stretch
• The C-O band appears in the range of 1300 to 1000 cm-1
• Look for one or more strong bands appearing in this range!
• Ethers, alcohols, esters and carboxylic acids have C-O bands
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C-H bendings
• Bendings aren’t as important as stretches.
• See page 848-850
• CH2 bending: 1450 cm-1
• CH3 bending: 1375 cm-1
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Cyclohexanol
OHO-Hstretch
bending
C-Ostretch
sp3 C-H stretch
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Ethyl Butanoate
CO
O
CH2 CH2 CH3CH2CH3
sp3 C-H
C=O stretch
C-O stretch
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4-Methyl-2-pentanoneC-H < 3000, C=O @ 1715 cm-1
CH3 CH CH2 C CH3
OCH3
C-H stretch C=O stretch
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4-Methyl-3-penten-2-one
C C
CH3
CH3
C CH3
O
H
C-H stretch
C=O stretch
C=Cstretch
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Experiment 15Spearmint and Caraway
• Experiment 15, pp124-131• Essay: pp119-123• Technique 25 (Infrared spectroscopy)• Technique 22 (Gas Chromatography,
especially, Section 22.8 on page 806 and Fig 22.7)
• Technique 23 (Polarimetry)
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Spearmint and Caraway Oils
O
H
(R) - carvoneSpearmint oil
O
(S) + carvoneCaraway oil
H
H
(S) - Limonene
Limonene is presentIn both oils
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The experiment
• Run the IR spectrum on the two enantiomers
• Run a sample of each enantiomer through the gas chromatograph
• Determine the optical rotations for each enantiomer
• Calculate the specific rotation for each enantiomer
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Nucleophilic Substitution: Competing Nucleophiles
Experiment 21
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Reading Assignment
• pp. 180-190 (Exp 21) and 510-514 (exp 58)
• Chapter 10 in your lecture textbook• Review Techniques 1 through 6• Technique 7 (Sections 7.2, 7.4, 7.5, 7.8)• Technique 12 (Sections 12.5, 12.9, 12.11)• Technique 22• Technique 21 (omit this quarter)
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Changes to Exp 19 laboratory
• We will be using 1-pentanol, 2-pentanol, 3-pentanol and 2-methyl-
2-butanol instead of what is shown in the book.
• The procedure stays the same; just change the alcohols.
• Start 21A first, then do 21B, and then come back to finish 21A.
• Part 21C: gas chromatography only• Signup for 1, 2, and 3-pentanol
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The SN2 Mechanism
.. _
:..
+ Br
_..
..: :
..
..
BrH O C Br
H
HH
H O
H O C
H
HH
C
H
HH
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The SN1 Mechanism1)
2)
3)
: :..
slow
++ : Br :
..
..
_
++ : :
fast
:+
:+
fast
:..
+ H+
CH3 C CH3
CH3
Br
CH3 C CH3
CH3
CH3 C CH3
CH3
CH3 C CH3
CH3
O H
H
O
H
H
CH3 C CH3
CH3
O H
H
CH3 C CH3
CH3
O H
1935: Hughes & Ingold
carbocation
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Assisted SN2: Mechanism
1)
2)
R CH2 O + HH R CH2 O H
H
R CH2 O H
H
X +slow
X CH2 R + O
H
H
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Assisted SN1: Mechanism
CH3 C CH2
CH3
O
CH3
H
CH3 C CH2
CH3
O
CH3
H
H
1)
2)
3)
+ H
CH3 C CH2
CH3
O
CH3
H
H
CH3 C CH2
CH3
OCH3
H
H+
CH3 C CH2
CH3
CH3+ X CH3 C CH2
CH3
CH3
X
slow
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RBr/RClROHH2SO4NH4Br/NH4ClH2O
pentane
pentaneRBr/RClROHH2SO4 (trace)H2O (trace)
H2OH2SO4NH4Br/NH4ClROH
organicphase
aqueousphase
pentaneRBr/RClROH (trace)H2SO4 (trace)H2O (trace)H2Opentane
RBr/RClROHH2SO4 (trace)H2O (trace)
H2OH2SO4ROH
organicphase
aqueousphase
NaHCO3
organicphase
aqueousphase
pentaneRBr/RClROH (trace)H2O (trace)
H2OH2SO4ROH
pentaneRBr/RClROH (trace)H2O (trace)
Na2SO4
pentaneRBr/RClROH (trace)
H2O
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Gas Chromatography Results: 1-Chloro pentane and 1-Bromopentane
3.0 3.5 4.0 4.5 5.0 5.5
Minutes
-1.5
0.0
2.5
5.0
7.5
10.0
12.5
15.0
mVolts
4.57
9
5.33
2
s:\iln\chemistry\fall2003-summer2004\spring 2004\chem 354\wandler\competingnucs\np1p_8;45;53 pm.run
X:Y:
3.4135 Minutes0.0590 mVolts
WI:2SR-
R-Cl
R-Br
solvent
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Ret. Time Width Peak Peak Result Time Offset Area Sep. 1/2 Status No. Name () (min) (min) (counts) Code (sec) Codes ---- ------------ ---------- ------- ------- ---------- ---- ----- ------ 1 9.9306 4.579 0.000 2113 BB 1.5 2 90.0694 5.332 0.000 19166 BB 1.4 ---- ------------ ========== ------- ======= ========== ---- ----- ------ Totals: 100.0000 0.000 21279 Total Identified Counts associated with halides: 21279 counts
Gas Chromatography Results: 1-Chloropentane and 1-Bromopentane
from 1-pentanol
Assume that all response factors = 1.000
Round off the values to:9.9 % 1-chloropentane90.1 % 1-bromopentane
Important! Other components may appear on the printout. If so, be sure to recalculate for only 1-chloro and 1-bromopentane! Make sure that the total equals 100%
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Do the results of this experiment with 1-pentanolfit with what you expected?
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Gas Chromatography Results: 2-Pentanol in the solvent nucleophile
mixture
3.0 3.5 4.0 4.5 5.0
Minutes
-1
0
1
2
3
4
5
6
mVolts
4.15
1
4.21
6
4.85
2
4.93
2
s:\iln\chemistry\fall2003-summer2004\spring 2004\chem 354\wandler\competingnucs\hak2p_9;15;27 pm.run
X:Y:
3.7785 Minutes0.0240 mVolts
SR+WI:2SR-
solvent
2-chloropentane3-chloropentane
2-bromopentane3-bromopentane
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Gas Chromatography Results: 2-Pentanol in the solvent nucleophile
mixture Ret. Time Width Peak Peak Result Time Offset Area Sep. 1/2 Status No. Name () (min) (min) (counts) Code (sec) Codes ---- ------------ ---------- ------- ------- ---------- ---- ----- ------ 1 21.7171 4.151 0.000 4100 BV 1.4 2 11.6505 4.216 0.000 2200 VB 1.5 3 46.7971 4.852 0.000 8835 BV 1.4 4 19.8353 4.932 0.000 3745 VB 1.5 ---- ------------ ========== ------- ======= ========== ---- ----- ------ Totals: 100.0000 0.000 18880 Total Identified Counts associated with halides: 18880 counts
Round off the values to:21.7 % 2-chloropentane11.7 % 3-chloropentane46.8 % 2-bromopentane19.8 % 3-bromopentane
Important! Other components may appear on the printout. If so, be sure to recalculate for only the four halides! Make sure that the total equals 100%
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Why are we obtaining mixtures of halides in this reaction?
Time for chalk!!
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Gas Chromatography Results: 2-Methyl-2-butanol in the solvent
nucleophile mixture
3.0 3.5 4.0 4.5
Minutes
-0
0
1
2
3
4
5
mVolts
3.84
4
4.49
1
s:\iln\chemistry\fall2003-summer2004\spring 2004\chem 354\wandler\competingnucs\ncdtp_5;28;39 pm.run
X:Y:
3.1883 Minutes108 mVolts
SR+WI:2SR-
solvent2-chloro-2-methylbutane
2-bromo-2-methylbutane
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Gas Chromatography Results: 2-Methyl-2-butanol in the solvent
nucleophile mixture
Ret. Time Width Peak Peak Result Time Offset Area Sep. 1/2 Status No. Name () (min) (min) (counts) Code (sec) Codes ---- ------------ ---------- ------- ------- ---------- ---- ----- ------ 1 44.0704 3.844 0.000 5181 BB 1.4 2 55.9296 4.491 0.000 6575 BB 1.5 ---- ------------ ========== ------- ======= ========== ---- ----- ------ Totals: 100.0000 0.000 11756 Total Identified Counts for the two halides: 11756 counts
Assume that all response factors = 1.000
Important! Other components may appear on the printout. If so, be sure to recalculate for only the two halides! Make sure that the total equals 100%
Round off the values 44.1 % 2-chloro-2-methylbutane55.9 % 2-bromo-2-methylbutane
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We expect that the reaction of 2-methyl-2-butanolwith the solvent nucleophile mixture to be SN1. Why didn’t it come out as a 50-50 mixture?
Life is never straightforward!
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Important notice about next week’s lab lecture
Friday, May 19th
Bond Hall 1092:00 to 2:50
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Important notice: Final Exam
The exam will be held in SL 110 rather than SL 130 on
Thursday, June 8th 3:30-5:30