WWU -- Chemistry Experiment 56: Synthesis of an Ester from an Alcohol and Carboxylic Acid Structure determination by Infrared Spectroscopy.

WWU -- ChemistryWWU -- Chemistry

Experiment 56: Synthesis of an Ester from an

Alcohol and Carboxylic Acid

Structure determination by Infrared Spectroscopy


Infrared Spectroscopy

• Read Chapter 25, pp 833 to 867 • Work problems on page 866-867


Typical Infrared AbsorptionRegions


O-H stretch

• The OH peak is very strong and broad!

• It usually appears near 3400 cm-1


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


C=C stretch

• C=C stretch occurs near 1650 cm-1

• It is often weak unless it is conjugated to a C=O


Normal Base Values for C=O Stretching

Functional Group Frequency (cm-1)

Ester Aldehyde Ketone Carboxylic acid Amide

1735 - 1740 1725 1715 1710 1690


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


C-H bendings

• Bendings aren’t as important as stretches.

• See page 848-850

• CH2 bending: 1450 cm-1

• CH3 bending: 1375 cm-1


Cyclohexanol

OHO-Hstretch

bending

C-Ostretch

sp3 C-H stretch


Ethyl Butanoate

CO

O

CH2 CH2 CH3CH2CH3

sp3 C-H

C=O stretch

C-O stretch


4-Methyl-2-pentanoneC-H < 3000, C=O @ 1715 cm-1

CH3 CH CH2 C CH3

OCH3

C-H stretch C=O stretch


4-Methyl-3-penten-2-one

C C

CH3

CH3

C CH3

O

H

C-H stretch

C=O stretch

C=Cstretch


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)


Spearmint and Caraway Oils

O

H

(R) - carvoneSpearmint oil

O

(S) + carvoneCaraway oil

H

H

(S) - Limonene

Limonene is presentIn both oils


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


Nucleophilic Substitution: Competing Nucleophiles

Experiment 21


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)


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


The SN2 Mechanism

.. _

:..

+ Br

_..

..: :

..

..

BrH O C Br

H

HH

H O

H O C

H

HH

C

H

HH


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


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


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


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


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


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%


Do the results of this experiment with 1-pentanolfit with what you expected?


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


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%


Why are we obtaining mixtures of halides in this reaction?

Time for chalk!!


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


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


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!


Important notice about next week’s lab lecture

Friday, May 19th

Bond Hall 1092:00 to 2:50


Important notice: Final Exam

The exam will be held in SL 110 rather than SL 130 on

Thursday, June 8th 3:30-5:30

WWU -- Chemistry Experiment 56: Synthesis of an Ester from an Alcohol and Carboxylic Acid Structure determination by Infrared Spectroscopy.

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