Exp. 13*:
CALCULATION,
CHROMATOGRAPHIC,
AND
SPECTRAL
APPLICATIONS
Objectives:

To review common laboratory calculations,
chromatographic, and IR spectral
techniques.

To learn to calculate efficiency of synthetic
methods, and determine overall
marketability of synthetic products.

To learn to use mass spectrometry and NMR
spectroscopy for future use in structure
elucidation.
PROPOSED CHEMICAL EQUATION
OH
H 2 SO 4 (cat)
2 -m e th yl-2 -p e n ta n o l
M W : 1 0 2 .1 7 g /m o l
d : 0 .8 3 5 g /m L
c o s t: $ 6 7 .2 0 /1 2 m L
Am o u n t u s e d : 2 .0 m L
REACTANT
s u lfu ric a c id
c o s t: $ 2 5 .0 0 /5 0 0 m L
A m o u n t u s e d : 0 .5 m L
CATALYST
2 -m e th yl-1 -p e n te n e
M W : 8 4 .1 6 g /m o l
d : 0 .6 8 2 g /m L
c o s t: $ 1 2 9 .5 0 /2 5 m L
P ro d u c t m a s s : 1 .2 0 g
PRODUCT
Dehydration of 2-methyl-2-pentanol
LIMITING REAGENT &
THEORETICAL YIELD






Mass RCT  Moles RCT (using MWrct)
Moles RCT  Moles PROD (using stoichiometry)
Moles PROD  Mass PROD (using MWprod)
Determine limiting reagent.
Theoretical yield is always in the units of GRAMS of PRODUCT!
Theoretical yield =
# g REACTANT
1 mol of REACTANT
#g
Amount
you
started
with
Molecular
weight of
starting
material
1 mol PRODUCT
#g
1 mol REACTANT 1 mol PRODUCT
Stoichiometric ratio
Molecular
weight of
product
PERCENT YIELD
Percent yield =
ACTUAL PRODUCT MASS (g)
THEORETICAL YIELD (g)
X 100
Theoretical yield and percent yield
CALCULATIONS
Theoretical Yield (g) based on alcohol
Show these calculations in your lab
notebook!
Complete this table
electronically and
copy/paste into your
final lab report!
Theoretical Yield (g)
Actual Yield (g)
Percent Yield
Percent Yield
GREEN CHEMISTRY CALCULATIONS

Green chemistry calculations are used to
determine how “environmentally friendly”
your choice of reagents, solvents, and
conditions were.

This includes ATOM ECONOMY,
EXPERIMENTAL ATOM ECONOMY, AND
“Eproduct”.
ATOM ECONOMY
Atom economy : based on the efficiency of
reactant atoms converted to product atoms.
Q: Were ALL of the reactant atoms converted to
product atoms?




Any atoms of the reactants that did NOT appear in the
product structure were converted to side products or
waste.
Sometimes the side products and waste generated is
harmful to the environment.
An experiment should be designed to minimize
the generation of waste and unnecessary side
products.
ATOM ECONOMY
Atom economy = MW desired product
S MW reactants
* 100
• Atom economy is based on which reactants
were selected to make the product.
• It assumes that the reactants were used in
equivalent amounts, meaning that no excesses of
any reactant were used.
• The closer the atom economy is to 100%, the
better!
EXPERIMENTAL ATOM ECONOMY
Experimental atom economy = theoretical yield of product (g) * 100
S mass reactants
Q: Did we use ONLY the amount necessary to
generate the product?
• Sometimes an excess of one reactant is used in order to
drive the reaction to completion. For this reason, the
experimental atom economy is calculated.
• Experimental atom economy is a more precise
measure of efficiency than the atom economy—it
takes into account the mass of each reactant used.
“EPRODUCT “
“Eproduct” =
•
(% yield X % experimental atom economy)
100
“Eproduct” is the ultimate measure of efficiency, since
both the conditions used and the amount of
product that resulted under those conditions is
taken into account.
• “Eproduct” is a number, not a percentage!
• The higher “Eproduct” is, the better!
IMPROVING EFFICIENCY…

An efficient reaction would have 100% Atom
Economy and 100% Experimental Atom
Economy.

One could improve the efficiency of a reaction
by adjusting the conditions of a reaction in an
effort to improve either of these values, such
as:


Use different reactants to form the product (AE)
Use different amounts of reactants to form the
product (EAE).
COST ANALYSIS…
Cost per synthesis

In order to calculate the cost of your synthesis,
you must first determine the cost of the
amount of each chemical used, including
reactants, solvents, and catalysts!
Cost of solid ($) = Mass (g) of SOLID “A” used
*This
calculation is performed for
each solid used during course of
synthesis!
Cost of liquid ($) = Volume (mL) of LIQUID “B” used
Cost per synthesis ($) = Cost of “A” + Cost of “B”…etc.
*This calculation is performed for
each liquid used during course of
synthesis!
COST ANALYSIS…
Cost per gram

Now that the cost per synthesis has been
determined, based on the amount of
product generated…
Cost per gram ($/g) =
cost per synthesis ($)
actual yield (g)
COST ANALYSIS…
Cost per bottle

Now that the cost per gram has been
determined, when compared to the
manufacturers cost, how marketable is
your product?
Cost per bottle ($/g) = cost per gram x desired bottle size
(*Can also be calculated in mL)
Green Chemistry Calculations
CALCULATIONS
Atom Economy (%)
Experimental Atom Economy (%)
“E” product
Cost per Synthesis ($)
Cost per Gram ($/g)
Cost per 25 mL bottle
Atom Economy (%)
Experimental Atom Economy (%)
“E” product
Cost per Synthesis ($)
Cost per Gram ($/g)
Cost per 25 mL bottle
HPLC and TLC Chromatography

Introduced in Experiments 4 and 5.


Difference between ANALYTE POLARITY and
SOLVENT POLARITY.
UV detector is used in both. In order to be detected,
compounds must be UV active.

Most solvents used in TLC and HPLC are not UV active,
therefore do not appear on the TLC plate or in the HPLC
chromatogram!
GC Chromatography

Introduced in Experiment 2.

ADJUSTED AREA % must be calculated to eliminate solvent quantity!
All chromatograms
and spectra are
available in folder
on course website!
IR SPECTROSCOPY

Introduced in Experiment 10.


Base values are given in correlation tables.
Actual values are reported from actual spectrum

Don’t ever mention sp3 CH when using IR spectroscopy
to differentiate between reactants and products! They
are too common!
EXAMPLE OF IR SPECTROSCOPY
THINGS TO CONSIDER…
•What kinds of bonds are
present?
OH
CH 3
• If they appeared in the IR
spectrum, where would they
be?
CH 3
• Now, look at the spectrum.
Are they there?
Actual spectra are
available in folder
on course website!
IR SPECTROSCOPY BASE VALUES
Base values for Absorptions of Bonds (cm-1)
O-H
3200-3600
C-O
1000-1300
***(Esters have two!)***
C-H (sp2)
3000-3100
C-H (sp3)
2800-3000
Aldehyde C-H
2700 & 2800
***(there are two!)***
C=O
1650-1740
***(location depends on functional group!)***
C-X
500-700
MELTING POINT ANALYSIS

Introduced in Experiment 7.


Detects all impurities!
Recorded as Ti-Tf range.
Pure = matches literature mp EXACTLY!
 Impure = lower Ti = higher DT!

Before coming to the next lab…




Go to the website: www.ochem.com
From the left menu, select TUTORIALS.
From the right column, PRELECTURES, scroll
¾ of the way down the page.
Watch the following:



MASS SPECTROMETRY
SPECTROSCOPY (Part 3 of 4)
SPECTROSCOPY (Part 4 of 4)
(YOU’LL BE GLAD YOU DID! )
CALCULATING
DEGREE OF UNSATURATION
CcHhNnOoXx
DU = (2c + 2) – (h – n + x)
2
• 1o unsaturation = 1 C=C or 1 ring
• 2o unsaturation = 2 C=C, 2 rings, or CΞC, or combination of C=C & rings
• 3o unsaturation = combination of double bonds, triple bonds, rings
• 4o unsaturation = typically indicates an aromatic ring
13C-NMR


SPECTROSCOPY
d (ppm) = tells what type of carbon it is.
# signals = tells whether or not there is
symmetry within the molecule.
13C
NMR CHEMICAL SHIFT
CORRELATION CHART
R
R
O
C
O
C
H
R
R
R
210
O
C
OR
OH
R
R
O
C
O
C
Fn
NR 2
C sp3
sp3 C
o
C
X
C
C
o
C sp3
o
4 --3 --2 --1
C
o
0-50d
5 0 -1 1 0 d
1 1 0 -1 6 0 d
1 9 0 -2 2 0 d
220
O
C
1 6 0-1 90 d
200
180
160
140
120
p. 118 in lab manual
100
80
60
40
20
0
13C
NMR Spectral Analysis
14.68 d
17.67 d
29.20 d
70.98 d
46.46 d
1a
OH
4
5
2
4
1a
3
5
2
3
1b
C#
1a
1b
3
13.79d
20.91 d
22.34d
40.16 d
109.87 d
145.95 d
2
4
5
d
(ppm)
C#
1a
1b
2
3
4
5
All chromatograms
and spectra are
available in folder
on course website!
d
(ppm)
22.34
20.91
1b
1H-NMR

SPECTROSCOPY
d (ppm) = tells what type of proton it is.

# signals = tells whether or not there is symmetry
within the molecule.

Integration = tells # protons of each type.
Multiplicity = tells # neighboring protons, using
n + 1 rule.

1H-NMR
R
O
O
C
C
OH
10.0-12.0 d
R
SPECTROSCOPY
H
H
Fn
C C
H
C H
H
9.0-10.0 d
6.5-8.5 d
sp3C
H
C C H
5.0-6.5 d
2.0-4.5 d
0.0-2.0 d
(3o > 2o > 1o)
12
11
10
9
8
7
6
5
4
3
2
1
0
1H-NMR
Spectral Analysis
1.20 d
6H, s
1.38 d
2H, hex
1.44 d
2H, t
1.71 d
3H, s
4.66 d
1H, s
4.70 d
1H, s
0.93 d
3H, t
1.46 d
2H, pent
2.04 d
1H, s
1.99 d
2H, t
1a
OH
4
5
2
4
5
1b
H
#
1a
1b
2
3
4
5
d
(ppm)
2.04
2
1a
3
Int.
1
0.90 d
3H, t
Mult.
s
H#
1a
1b
2
3
4
5
1b
3
d
(ppm)
4.66
4.70
X
Int.
1
Mult.
1
X
s
s
X
All chromatograms
and spectra are
available in folder
on course website!
Notice some
signals are already
assigned for you in
the tables!
IR Spectral Analysis
OH
3616
Base
Values
(cm-1)
OH stretch
C-O stretch
sp3 CH
stretch
sp2 CH
stretch
C=C stretch
3200-3600
1000-1300
2800-3000
1162
2965
3076
1661
2962
Functional
Group
2-methyl2-pentanol
Frequency
(cm-1)
3000-3100
X
1600-1680
X
2-methyl-1pentene
Frequency
(cm-1)
X
X
MASS SPECTROMETRY—
How it works

Small amount of sample is vaporized into the
ionization source, then bombarded with high energy
electrons.

When a high energy electron hits the organic
molecule, it dislodges a valence electron, to produced
a radical cation, called the molecular ion.
M
+
e
-
M
Io n iza tio n
m o le cu la r io n
(= ra d ic a l ca tio n )
+
2 e
-
MASS SPECTROMETRY—
How it works


Electron bombardment transfers so much energy
that the bonds in the cation fragments begin to
break.
Some pieces retain the positive charge, some are
neutral.
m1
M
F ra g m e n ta tio n to :
+
c a tio n
+
m2
ra d ic a l
(= n e u tra l lo s s )
MASS SPECTROMETRY—
How it works



Fragments then flow through a strong magnetic field,
where the charged fragments are sorted onto a detector
based on their mass to charge ratio (m/z).
Since the charge number on each ion is usually +1, the
value of m/z for each ion simply = mass.
The output is a plot of the m/z value of each fragment
based on based on its relative abundance.
Relative
abundance of
fragment
Mass of
fragment
MASS SPECTROMETRY—
Interpretation

The way molecular ions break down can produce
characteristic fragments that help in
identification


Serves as a “fingerprint” for comparison with known
materials in analysis (used in forensics)
Positive charge goes to fragments that best can
stabilize it
MASS SPECTROMETRY—
Interpretation
Positive charge goes to fragments that best can
stabilize it!
 Carbocation stability is discussed in McMurry text,
p. 377.

H
H
H
H
H 3C
C
C
H
C
H
H
M eth yl
C
<
P rim ary
H
C
H
H 3C
C
H
<
H
H
~
B en zylic
In creasin g carb o catio n stab ility
C
H 3C
C
CH 3
H
Allylic
CH 3
~
S eco n d ary
CH 3
<
T ertiary
MASS SPECTROMETRY—
Alcohols


Functional groups cause common patterns of
cleavage in their vicinity
Alcohols undergo -cleavage (at the bond next
to the C-OH) as well as loss of H-OH to give C=C
MASS SPECTROMETRY—
Alkenes

R
Important fragment in terminal alkenes is the
allyl carbocation at m/z = 41 due to the following
cleavage:
CH 2
CH
CH 2
R
CH 2
CH
CH 2
H 2C
CH CH 2
resonance stabilized allyl carbocation
MASS SPECTROMETRY—
Carbonyl compounds


A C-H that is three atoms away leads to an internal
transfer of a proton to the C=O, called the McLafferty
rearrangement
Carbonyl compounds can also undergo  cleavage
Mass Spectral Analysis
59
69
84
102
OH
m/z
102
(M+)
59
(base)
Cation
formula
C6H14O
Structure
m/z
Molecular Ion
84
(M+)
C3H7O
Cationic Fragment
69
(base)
Cation
formula
C6H12
Structure
Molecular Ion
C5H9
Cationic Fragment
Mass Spectral Analysis
OH
m/z
102
(M+)
Cation
formula
C6H14O
Structure
m/z
Molecular Ion
84
(M+)
Cation
formula
C6H12
Structure
Molecular Ion
C5H9
Cationic Fragment
OH
59
(base)
C3H7O
Cationic Fragment
OH
69
(base)
Final Lab Report

In Lab…


Each student will perform all calculations in their own
laboratory notebooks and submit yellow copies to
instructor for grading.
Post Lab…



Each lab group will submit ONE copy of a typewritten,
paragraph style report addressing all points listed for
REACTION #2.
All data tables for REACTION #1 and REACTION #2
must be completed and copied into the document.
Original copies of tables provided from the course
website are UNACCEPTABLE.