Experiment 3
Determining the Structures of Unknown Molecules Using
NMR, IR and Mass Spectroscopy
Xiaolei Gao, Ph.D.
Background
The theory of infrared spectroscopy, mass spectroscopy and NMR spectroscopy will be covered
in the first two lab periods. Please read chapters 12 and 13 in McMurry for a complete
discussion of these techniques.
Procedure
Part I
The lab consists of three parts. The first part is a thought experiment. Consider the following
four compounds:
Using the knowledge of spectroscopy, you gained in the prior two classes, predict what the IR,
NMR and mass spectrum of each would look like and fill in the following table (I completed part
of table for you as an example):
Compound
# Carbon
# Proton
Approximate Chemical Shift
#
Signals
Signals
of Protons
Main IR Stretch
Expected
Molecular Ion
in M.S.
One Possible Fragment in M.S.
1
5
5
1) The proton in
OH appears
3200-3600
108
cm-1
randomly on
1
HNMR
2) The protons in
CH2 appears
@ ~4-5 ppm
(singlet)
3) Protons on
the benzene
ring appear
@~7 ppm
(multiplet)
2
3
4
100
43
Part II
In person students will come to the lab and carry out the IR experiment of the four unknown
compounds.
Remote students will click the link below or copy it to your browser, then scroll down the page
to find "Virtual Lab 2: Infrared Spectroscopy". Follow the instruction there to complete the
virtual lab (remember you can use your cursor to see different view of the lab). You need to
carry out the virtual lab four times to obtain IR data for all four compounds. The credit of this
virtual lab goes to North Carolina State University.
https://sites.google.com/ncsu.edu/ncstatevrorganicchemistrylabs/home
After you complete the in-person or virtual lab, report here which compound is which and
explain why (Compound 1 was used as an example here to show you how this is done).
Based on the IR result, I conclude that compound A is pentan-2-ol because IR of compound A
show a strong and broad peak at 3200-3600 cm-1, which is an indication of the presence of
alcohol functional group. I also observed a medium peak at 2850-3000 cm-1,but no peaks at
3000-3100 cm-1 .This indicates the presence of sp3 C-H, and absence to aromatic sp2 C-H. The
only aliphatic alcohol within the four compounds is pentan-2-ol, therefore I conclude
compound A is pentan-2-ol.
Based on the IR result, I conclude that compound B is … because…
Based on the IR result, I conclude that compound C is … because…
Based on the IR result, I conclude that compound D is … because…
Part III
1. Which of the four compounds has a MS as shown below? Explain all the major
fragments,
2. The following 1HNMR belong to which of the four compounds in this experiment (s:
singlet; d: doublet; t: triplet; m: multiplet)? Explain.
3. Which of the four compounds has the following 13CNMR? Explain.
2
5
6
1) Proton on Carbon
3200-3600 cm-1
88
#1 1 ppm
(doulet)
OH
2) Proton on Carbon
H
#2 4 ppm
(multiplet)
+
3) Proton on Carbon
CH3
#3 1 ppm
(quartet)
4) Proton on Carbon
MW = 45
#4 1 ppm
(multiplet)
5) Proton on Carbon
#5 1 ppm
(triplet)
4) OH proton
appears randomly
on 1HNMR
3
6
4
1) Proton on Carbon
1700 cm-1
120
#1 2 ppm
O
(singlet)
2) Proton on Carbon
#4 7 ppm
+
(doublet)
CH3
3) Proton on Carbon
#5 7 ppm
(triplet)
MW = 43
4) Proton on Carbon
#6 7 ppm
(triplet)
4
5
4
1) Proton on Carbon
#1 2 ppm
(singlet)
1715 cm-1
100
O
O
2) Proton on Carbon
43
#3 1 ppm
(doublet)
3) Proton on Carbon
#4 1 ppm
(multiplet)
4) Proton on Carbon
#5 1 ppm
(doublet)
Part II
In person students will come to the lab and carry out the IR experiment of the four unknown compounds.
Remote students will click the link below or copy it to your browser, then scroll down the page
to find "Virtual Lab 2: Infrared Spectroscopy". Follow the instruction there to complete the
virtual lab (remember you can use your cursor to see different view of the lab). You need to
carry out the virtual lab four times to obtain IR data for all four compounds. The credit of this
virtual lab goes to North Carolina State University.
https://sites.google.com/ncsu.edu/ncstatevrorganicchemistrylabs/home
After you complete the in-person or virtual lab, report here which compound is which and
explain why (Compound 1 was used as an example here to show you how this is done).
Based on the IR result, I conclude that compound A is pentan-2-ol because IR of compound A
show a strong and broad peak at 3200-3600 cm-1, which is an indication of the presence of
alcohol functional group. I also observed a medium peak at 2850-3000 cm-1, but no peaks at
3000-3100 cm-1 .This indicates the presence of sp3 C-H, and absence to aromatic sp2 C-H. The
only aliphatic alcohol within the four compounds is pentan-2-ol, therefore I conclude
compound A is pentan-2-ol.
Based on the IR result, I conclude that compound B is benzyl alcohol because the IR of
compound B shows a broad peak at 3200-3600 cm-1, which indicates the presence of an alcohol
functional group. There was also a peak between 3100-3000 cm-1 which indicates the presence
of aromatic sp2 hybridized carbons and a peak at 3000-2850 cm-1 which indicates the presence
of alkane sp3 hybridized carbons. Since both carbons were shown on the IR of compound B, I
can conclude that compound B is benzyl alcohol.
Based on the IR result, I conclude that compound C is 4-methyl-2-pentanone because the IR of
compound C shows a strong peak at 1700-1800 cm-1 which means there is a carbonyl group.
The peak was shown to be a ketone saturated (C=O) at 1725-1705 cm-1. Therefore, this leads to
compound C being either 4-methyl-2-pentanone or acetophenone. However, for the compound
to be acetophenone, there would have been a peak at 3100-3000 cm-1 for the aromatic sp2
hybridized carbons. Since there was a peak at 2850-3000 cm-1 , this indicates the presence of
alkane sp3 hybridized carbons, I can conclude that compound C is 4-methyl-2-pentanone.
Based on the IR result, I conclude that compound D is acetophenone because the IR of
compound D shows a strong peak at 1700-1800 cm-1 which means there is a carbonyl group.
This peak was shown to be an aryl (C=O) at 1700-1680 cm-1. There was also a peak at 3100-3000
cm-1 which indicates the presence of aromatic sp2 carbons. Therefore, I can conclude that
compound D is acetophenone.
Part III
1. Which of the four compounds has a MS as shown below? Explain all the major
fragments.
This MS shown above is for benzyl alcohol because the base peak at 43 indicates alpha cleavage
on carbon #2 to produce a fragment with a molecular weight of 43. Also, the parent peak at 100
matches the full molecular weight of benzyl alcohol as well as there are multiplets to match the
aromatic ring in the compound.
2. The following 1HNMR belong to which of the four compounds in this experiment (s:
singlet; d: doublet; t: triplet; m: multiplet)? Explain.
in.
The following 1HNMR belongs to 2-pentanol because the highest ppm is 4 ppm which
would represent carbon #2 that has one hydrogen. The doublet represents carbon #1 with
three hydrogens that is a CH3 near the OH group at about 1 ppm. The triplet represents carbon
#5 with three hydrogens that appears at about 1 ppm. The multiplet with 4 hydrogens
represents carbon #4 that lies between 1 ppm and 2 ppm. Lastly, the OH proton can appear
randomly on the HNMR. Therefore, this OH proton is located between 2.5 ppm and 3 ppm as a
singlet.
3. Which of the four compounds has the following 13CNMR? Explain.
The following CNMR is shown for acetophenone for two main distinct peaks. The first
one lies between 190 ppm to 200 ppm which indicates a C=O ketone. By knowing there
is a C=O in the compound, this limits our options to acetophenone and 4-methyl-2pentanone. The second peak(s) lie between 120 ppm and 140 ppm which indicates the
C=C of an aromatic benzene ring. Therefore, between our two options, the only one that
has a benzene ring and a C=O is acetophenone.
Experiment 3
Determining the Structures of Unknown Molecules Using
NMR, IR and Mass Spectroscopy
Xiaolei Gao, Ph.D.
Background
The theory of infrared spectroscopy, mass spectroscopy and NMR spectroscopy will be covered
in the first two lab periods. Please read chapters 12 and 13 in McMurry for a complete
discussion of these techniques.
Procedure
Part I
The lab consists of three parts. The first part is a thought experiment. Consider the following
four compounds:
O
O
OH
OH
2
Pentan-2-ol
1
Benzyl Alcohol
3
Acetophenone
4
4-Methyl-pentan-2-one
Using the knowledge of spectroscopy, you gained in the prior two classes, predict what the IR,
NMR and mass spectrum of each would look like and fill in the following table (I completed part
of table for you as an example):
Compound
# Carbon
# Proton
Approximate Chemical ShiQ
#
Signals
Signals
of Protons
Main IR Stretch
Expected
Molecular
Ion in M.S.
One Possible Fragment in M.S.
1
5
5
1) The proton in
OH appears
randomly on
1HNMR
2) The protons
in CH2
appears @
~4-5 ppm
(singlet)
3) Protons on
the benzene
ring appear
@~7 ppm
(mul]plet)
3200-3600
cm-1
108
H 2C
OH
31
2
5
6
1) The proton in
OH appears
randomly on
1HNMR
2) The proton in
CH3 appears
@ 1 ppm
(3H, triplet)
3) Protons on COH appears
@~3-4 ppm
(1H, sextet)
4) CH3 aaached
to C-OH
appears @
~1ppm
(duplet)
5) Protons on
CH2 aaached
to C-OH
appears @~1
ppm (2H,
quartet)
6) Protons on
CH2 appears
@~0-1 ppm
(2H, sextet)
3400-3600
cm-1
88
45
3
6
4
1) The protons
in CH3
appears @ ~2
ppm (3H,
singlet)
2) The protons
in the
benzene ring
appears @~ 7
ppm (5H,
mul]plet)
1660-1780
cm-1
120
43
4
5
4
1) The protons
in CH3
1670-1780
100
O
O
cm-1
appears @ ~ 1
ppm (6H,
duplet)
2) The protons
43
aaached to
the CH2
appears @~2
ppm (2H,
duplet)
3) The protons
aaached to
CH3 right next
to the C==O
appears @~ 2
ppm (3H,
singlet)
4) The proton
aaached to
the CH
appears @
~0-1 ppm
(1H,
mul]plet)
Part II
Remote students will click the link below or copy it to your browser, then scroll down the page
to find "Virtual Lab 2: Infrared Spectroscopy". Follow the instruction there to complete the
virtual lab (remember you can use your cursor to see different view of the lab). You need to carry
out the virtual lab four times to obtain IR data for all four compounds. The credit of this virtual
lab goes to North Carolina State University.
https://sites.google.com/ncsu.edu/ncstatevrorganicchemistrylabs/home
After you complete the in-person or virtual lab, report here which compound is which and
explain why (Compound 1 was used as an example here to show you how this is done)
- Based on the IR result, I conclude that compound A is pentan-2-ol because IR of compound A
show a strong and broad peak at 3200-3600 cm-1, which is an indication of the presence of
alcohol functional group. I also observed a medium peak at 2850-3000 cm-1, but no peaks at
3000-3100 cm-1 .This indicates the presence of sp3 C-H, and absence to aromatic sp2 C-H. The
only aliphatic alcohol within the four compounds is pentan-2-ol, therefore I conclude compound
A is pentan-2-ol.
- Based on the IR result, I conclude that compound B is benzyl alcohol because we are able to
determine the presence of a hydroxide functional group on the compound. This is indicated by a
strong broad peak at 3200-3600 cm-1. After observing a medium peak at 3000–3100 cm-1, we can
determine the presence of an aromatic sp2 C-H and a sp2 C-H concluding that compound B is
benzyl alcohol.
- Based on the IR result, I conclude that compound C is 4-methyl-pentan-2-one because we are
able to see a carbonyl group through a distinct peak at 1700-1800 cm -1. Since there is no peak at
3000-3100 cm-1, but a medium peak at 2850-3000 cm-1, this shows the presence of sp3 C-H as
well as the absence to an aromatic sp2 C-H.
- Based on the IR result, I conclude that compound D is acetophenone because an aryl group is
shown through a distinct peak at 1700-1800 cm-1. An indication of an sp2 C-H and the aromatic
sp2 C-H is observed through a medium peak at 3000-3100 cm-1.
Part III
1. Which of the four compounds has a MS as shown below? Explain all the major
fragments.
-
- From the chart below, the major peaks of fragmentation emerge from the alpha cleavage
of the corresponding C-C bonds to the carbonyl group. The 4-methyl-pentan-2-one,
(CH3) 2CHCH2COCH3 molecular formula should be fragmented as:
(CH₃)₂CHCH₂-COCH₃ → C₄H₉ + C₂H₃O which gives m/z = 57 and 43.
(CH₃)₂CHCH₂CO-CH₃ → C₅H₉O + CH₃ which gives m/z = 85 and 15. We can also observe the
highest peak in the spectrum, which is the base peak at m/z = 43 and at m/z = 15, 57, and 85 all
other peaks. Thus, the MS mentioned below belongs to 4-Methyl-pentan-2-one as the four
compounds.
2. The following 1HNMR belong to which of the four compounds in this experiment (s:
singlet; d: doublet; t: triplet; m: multiplet)? Explain.
- The 1HNMR belongs to the Pentan-2-ol because the CH proton attachment to the hydroxyl
oxygen group has a peak at ~3-4 ppm with (1H, multiplet) There is also a peak of (3H, triplet) at
~1ppm due to the coupling of the methyl group's neighboring two protons and at ~1.5ppm there's
another peak with (3H, duplet).
3. Which of the four compounds has the following 13CNMR? Explain.
- The compound that has 13CNMR is Acetophenone from the four compounds because it has a
double-bond, carbon-carbon aromatic group and a carbonyl group. It also has six distinct atoms
of carbon, which the 13CNMR has six peaks in total.
Name
Exp #4
Date: 2/17/2021
Lab Partner: n/a
Title: Nucleophilic Substitution Reactions of Organic Halides
1. Abstract and Purpose: (3 point).
When studying organic chemistry, there are many different types of reactions that
occur. This lab will focus on nucleophilic substitution reactions known as SN2 and SN1,
specifically examining the factors that affect the relative rates of SN2 and SN1 reactions
of alkyl halides. These factors include alkyl halide structure, nature of the leaving group,
properties of the nucleophile, and steric hindrance.
The SN2 reaction is a one-step, concerted substitution process meaning it will
make new bonds and break old bonds at the same time. Therefore, SN2 reactions do not
proceed by an intermediate. Since there is no intermediate, the rate-determining step
involves two components: the alkyl halide and the nucleophile. The SN2 reaction also
gives inversion of stereochemistry at the reaction center.
There are four main factors that can affect an SN2 reaction. However, the most
important are the steric effects which means the more congested and hindered the alkyl
halide, the slower the reaction will occur. The leaving group is another factor that can
affect this type of reaction, where the weakest bases are the best leaving group for SN2
reactions as they can best stabilize the anion. The choice of solvent can affect the rate of
an SN2 reaction such that polar aprotic solvents are ideal as they solvate the metal
counterion of the nucleophile, making it more reactive. Lastly, the nucleophile has an
effect as SN2 reactions would rather have larger atoms since they are more polarizable.
The SN1 reaction is a two-step unimolecular process with one intermediate and
two transition states. The rate depends upon only the concentration of the alkyl halide,
therefore making the rate of this reaction independent of the nucleophile concentration.
Name
Lab Partner: n/a
Exp #4
Date: 2/17/2021
Once again, many factors can affect an SN1 reaction. The alkyl halide must form
a stable carbocation, in which the more stable carbocation, the faster the reaction will
happen. This is due to lowering the activation barrier. Along with a stable carbocation,
resonance structures can occur such that they provide a tremendous amount of
stabilization to the carbocation. The leaving group can also have the same effect on an
SN1 reaction as an SN2 reaction such that the weakest bases are the best leaving group as
they can best stabilize the anion. Unlike SN2 reactions and as stated above, the
nucleophile will not affect the rate of an SN1 reaction because it is not part of the ratedetermining step. Lastly, the choice of solvent can affect the rate of an SN1 reaction such
that polar protic solvents are ideal as it stabilizes the carbocation intermediate as it forms,
thus increasing the rate of the reaction.
Overall, this lab will allow for the examination on all of the following factors that
affect the relative rates of an SN2 or SN1 reaction. For the experiment on SN2 reactions,
a solution of sodium iodide in acetone will be used. Sodium iodide is an excellent
nucleophile that is soluble in acetone which is a polar aprotic solvent. For the experiment
on SN1 reactions, a solution of silver nitrate in ethanol will be used. Ethanol is a polar
protic solvent that will favor a stable carbocation. Silver halide salts are very insoluble
and will precipitate from the solution, indicating if a reaction has occurred.
Name
Exp #4
Date: 2/17/2021
Lab Partner: n/a
2. Balanced equation: (2 point)
SN2 Reaction
SN1 Reaction
3. Reagent Table: (Add more rows when needed) (2 points)
Name
Silver Nitrate
M.W.
(g/mol)
169.87
Density
*
(g/mL)
M.P.*
(°C)
4.35
212
B.P.*
(°C)
440
Amount
(grams or
mL)
n/a
Moles
n/a
Hazards/Precautions
Oxidizing Solids,
Corrosive to Metals,
Skin Corrosion,
Serious Eye Damage,
Acute Aquatic Hazard,
Chronic Aquatic
Hazard
Name
Exp #4
Date: 2/17/2021
Lab Partner: n/a
Neopentyl
Bromide
151.04
1.199
n/a
105-106
n/a
n/a
Flammable Liquids,
Skin Irritation, Eye
Irritation, Specific
Target Organ Toxicity
Ethanol
46.07
1.60
-144
78.29
n/a
n/a
Flammable Liquids,
Eye Irritation
Benzyl Chloride
126.58
4.36
-43
177-181
n/a
n/a
Flammable Liquids,
Acute Oral Toxicity,
Acute Inhalation
Toxicity, Skin
Irritation, Serious Eye
Damage, Skin
Sensitisation, Germ
Cell Mutagenicity,
Carcinogenicity,
Specific Target Organ
Toxicity, Acute
Aquatic Hazard
Acetone
58.08
0.79
-94
56
n/a
n/a
Flammable Liquids,
Eye Irritation, Specific
Target Organ Toxicity
2-Iodobutane
184.02
1.598
-104
119-120
n/a
n/a
Flammable Liquids,
Acute Aquatic Hazard
2-Bromobutane
137.02
1.255
n/a
91
n/a
n/a
Flammable Liquids,
Acute Aquatic Hazard,
Chronic Aquatic
Hazard
2-Bromo-2methylpropane
137.02
1.22
-20
71-73
n/a
n/a
Flammable Liquids
Name
Exp #4
Date: 2/17/2021
Lab Partner: n/a
1-Chlorobutane
1-Bromobutane
92.57
0.886
-123
77-78
n/a
n/a
Flammable Liquids,
Aspiration Hazard,
Acute Aquatic Hazard,
Chronic Aquatic
Hazard
137.02
1.276
-112
100-104
n/a
n/a
Flammable Liquids,
Skin Irritation, Eye
Irritation, Specific
Target Organ Toxicity,
Acute Aquatic
Toxicity, Chronic
Aquatic Toxicity
5. Procedure, Observations and Data
Procedure (3 point)
Observations and Lab Data (4 point)
Name
Lab Partner: n/a
Exp #4
Date: 2/17/2021
A summary of the procedure done with bullet points)
Report all observations and all data that you collect in the
(2 points)
lab here
Part I – Reactivity with NaI in acetone (SN2)
Part A: Effect of Structure of the Alkyl Halide on
Notes from Virtual Lab on Sn2 Reaction.
•
the Relative Rates of SN2
•
3 key features of an SN2 reaction is the
position of the nucleophile attack, the
Add approximately 1-inch depth of hot tap
transition state, and the stereochemistry of
water to a 100 mL beaker and place on the
the product.
hotplate at a setting of 2 to set up a water bath
and stabilize the water temperature at 45°C.
•
3 factors that affect the rate of an SN2
reaction is the structure of the alkyl halide,
Clamp a thermometer in the water so that the
temperature can be monitored. Make sure the
the effect of sterics, and the identity of
thermometer is not touching the bottom of the
leaving group.
beaker or else the temperature being recorded
•
•
•
The alkyl halides used were 1-
will be of the glass beaker and not the water!
bromobutane, 2-bromobutane, and 2-
Measure 2 mL of 15% sodium iodide in
methyl-2-bromopropane.
acetone into each of three clean, dry 10-cm
•
test tubes. Make sure to mark the test tubes
For the structure of alkyl halides, it was
predicted that 1-bromobutane will react the
respectively with number 1, 2, and 3.
•
fastest while 2-methyl-2-bromopropane
Add 2 drops of 1-bromobutane to test tube 1
will react the slowest.
and mix thoroughly by swirling. Be sure to
note the start time.
•
•
white and cloudy.
Add 2 drops of 2-bromobutane to test tube 2
and mix thoroughly by swirling. Be sure to
note the start time.
The test tube with 1-bromobutane was
•
The test tube with 2-bromobutane was
clear, therefore showing no reaction.
Name
Lab Partner: n/a
•
•
Exp #4
Date: 2/17/2021
Add 2 drops of 2-methyl-2-bromopropane to
•
The test tube with 2-methyl-2-
test tube 3 and mix thoroughly by swirling.
bromopropane was also clear, therefore
Be sure to note the start time.
showing no reaction.
Watch each test tube carefully and note the
•
time at which a precipitate appears and its
Since 1-bromobutane produced white
precipitate, also the only test tube to
color.
•
produce a precipitate.
After 5 minutes, place any test tubes that have
no precipitate into the 45°C water bath.
•
fastest under the structure of alkyl halides.
Again, watch for the formation of
precipitates.
•
In conclusion, 1-bromobutane was the
•
For the steric effects, it was predicted that
Heat for up to 6 minutes (11 minutes total)
1-bromobutane will react the fastest while
then remove the test tubes and allow to cool
1-bromo-2,2-dimethylpropane will react
to room temperature. As some acetone may
the slowest.
have evaporated during heating, there is the
•
possibility of a precipitate of unreacted
The test tube with 1-bromobutane was
white and cloudy.
starting material, NaI, forming on the walls of
the test tube. To guard against this “false
•
dimethylpropane was clear, therefore
positive” result, gently shake these tubes for a
showing no reaction.
few minutes which should redissolve any NaI
but leave other precipitates unchanged. Note
•
The test tube with 1-bromo-2,2-
•
Since 1-bromobutane produced white
if precipitate disappeared upon shaking.
precipitate, also the only test tube to
Record all results.
produce a precipitate.
Empty the test tubes into the halide waste
container and dispose the test tube to the
white cardboard box for broken glasses.
•
In conclusion, 1-bromobutane was the
fastest under the steric effects.
Name
Lab Partner: n/a
Exp #4
Date: 2/17/2021
Part B: Steric Effects and the Relative Rates of
•
SN2 Reactions
•
For the effect of leaving group, it was
predicted that 1-bromobutane will react the
Measure 2 mL of 15% sodium iodide in
fastest while 1-chlorobutane will react the
acetone into each of two new clean, dry 10-
slowest.
cm test tubes. Make sure to mark the test
•
The test tube with 1-bromobutane was
tubes respectively with number 4 and 5.
•
white and cloudy.
Add 2 drops of 1-bromobutane to test tube 4
and mix thoroughly by swirling. Be sure to
•
clear, therefore showing no reaction.
note the start time.
•
Add 2 drops of 1-bromo-2,2-dimethylpropane
•
Since 1-bromobutane produced white
(neopentyl bromide) to test tube 5 and mix
precipitate, also the only test tube to
thoroughly by swirling. Be sure to note the
produce a precipitate.
start time.
•
The test tube with 1-chlorobutane was
•
Repeat the process as described in Part A.
In conclusion, 1-bromobutane was the
fastest under the steric effects.
Part C: Effect on the Leaving Group on the
Relative Rates of SN2 Reactions
•
Predictions of SN1 Reaction.
Measure 2 mL of 15% sodium iodide in
acetone into each of two new clean, dry test
•
•
For Part 1 of the reactivity with AgNO3 in
tubes. Make sure to mark the test tubes
Ethanol, the 2-methyl-2-bromopropane will
respectively with number 6 and 7.
react the fastest. This is due to SN1
Add 2 drops of 1-bromobutane to test tube 6
reactions preferring a tertiary alkane.
and mix thoroughly by swirling. Be sure to
note the start time.
•
This reaction should produce a pale yellow
precipitate.
Name
Lab Partner: n/a
•
•
Exp #4
Date: 2/17/2021
Add 2 drops of 1-chlorobutane to test tube 7
•
For Part 2 of the reactivity with AgNO3 in
and mix thoroughly by swirling. Be sure to
Ethanol, the 2-iodobutane will react the
note the start time.
fastest. This is due to having the best
Repeat the process as described in Part A.
leaving group.
Part II – Reactivity with AgNO3 in Ethanol (SN1)
•
This reaction should produce a very bright
Part A
•
yellow precipitate.
Keep the hot water bath from part I and
maintain at 45°C.
•
•
Measure 2 mL of 1% silver nitrate in ethanol
undergo since SN1 reactions do not react
into each of three clean, dry 10-cm test tubes.
with primary alkanes.
Make sure to mark the test tubes respectively
•
Add 2 drops of 1-bromobutane to test tube 8
and mix thoroughly by swirling. Be sure to
note the start time.
•
However, if this reaction were to go, it
would produce a white precipitate.
with number 8, 9, and 10.
•
The reaction with 1-chlorobutane will not
•
The reaction with benzyl chloride will be
faster than 1-chlorobutane because the
benzene ring helps stabilize due to the
Add 2 drops of 2-bromobutane to test tube 9
resonance structures making better
and mix thoroughly by swirling. Be sure to
note the start time.
•
Add 2 drops of 2-methyl-2-bromopropane to
test tube 10 and mix thoroughly by swirling.
Be sure to note the start time.
•
Watch each test tube carefully and note the
time at which a precipitate appears and its
color.
intermediates.
Name
Lab Partner: n/a
•
Exp #4
Date: 2/17/2021
After 5 minutes, place any test tubes that have
no precipitate into the 45°C water bath.
Again, watch for the formation of
precipitates.
•
Heat for up to 6 minutes (11 minutes total)
then remove the test tubes and allow to cool
to room temperature.
•
Empty the test tubes into the halide waste
container and dispose the test tube to the
white cardboard box for broken glasses.
Part B
•
Measure 2 mL of 1% silver nitrate in ethanol
into each of three clean, dry 10-cm test tubes.
Make sure to mark the test tubes respectively
with number 11, 12, 13, and 14.
•
Add 2 drops of 2-bromobutane to test tube 11
and mix thoroughly by swirling. Be sure to
note the start time.
•
Add 2 drops of 2-iodobutane to test tube 12
and mix thoroughly by swirling. Be sure to
note the start time.
•
Add 2 drops of 1-chlorobutane to test tube 13
and mix thoroughly by swirling. Be sure to
note the start time.
Name
Lab Partner: n/a
•
Exp #4
Date: 2/17/2021
Add 2 drops of benzyl chloride to test tube 14
and mix thoroughly by swirling. Be sure to
note the start time.
•
Repeat the process as described in Part A.
Name
Lab Partner: n/a
Exp #4
Date: 2/17/2021
6. Conclusions and discussions (4 points)
Sn2 Reaction Conclusion
The alkyl bromide that reacted the fastest with sodium iodide in acetone was 1bromobutane. The alkyl bromide that reacted the slowest with sodium iodide in acetone
was 2-methyl-2-bromopropane. By looking at the structure of the alkyl bromides, it can
be seen that 1-bromobutane is a primary alkyl bromide. Therefore, the SN2 reaction
prefers primary structures. The 2-methyl-2-bromopropane was the slowest because the
more congested and sterically hindered the alkyl halide, the slower the reaction will
undergo SN2.
Under the steric effects, the alkyl bromide that reacted the fastest with sodium
iodide was 1-bromobutane compared to the slowest that was 1-bromo-2,2dimethylpropane. Even though both structure are primary, there was still a difference in
reactivity due to steric effects. For 1-bromobutane, there are two hydrogens attached to
the neighboring carbon. Unlike 1-bromo-2,2-dimethylpropane that has two methyl groups
attached to the neighboring carbon. The two methyl groups are very bulking for iodide to
go through, therefore it slows the rate of the reaction. The iodide has a much easier time
to go through the two hydrogens in 1-bromobutane.
For the effect of the leaving group, it was seen that the alkyl halide that reacted
the fastest with sodium iodide in acetone was 1-bromobutane. The larger the leaving
Name
Lab Partner: n/a
Exp #4
Date: 2/17/2021
group, the less electronegative the leaving group is, therefore it is the better leaving group
under SN2 reactions. It is also known that the weakest bases are the best leaving group
for SN2 reactions as they can best stabilize the anion. Therefore, the bromine is a better
leaving group than the chlorine. However, if 1-iodobutane was used then it would react
the fastest over the 1-bromobutane due to it being larger than bromine and chlorine. The
reaction would take place as it is a primary structure, there are no bulky groups, and it has
the best leaving group.
SN1 Reaction Conclusion
Since this lab was not completed in person or virtually, there can not be any
definite conclusions. However based on the predictions made, it can be estimated what
the conclusions would be. The alkyl bromide that would react the fastest with silver
nitrate in ethanol would be 2-bromo-2-methylpropane. It is estimated that this would
react the fastest due to it being a tertiary alkyl halide. For SN1 reaction, it prefers the
tertiary structures compared to secondary or primary. Looking under the affects of the
leaving group, it can be estimated that the alkyl halide that would react the fastest is 2iodobutane. This is because iodide is the largest compared to bromine, therefore it is less
electronegative. Lastly, for SN1 reactions, benzyl chloride would react faster than 1chlorobutane. Even though they are both primary halides, the benzyl chloride allows for
better intermediates as it produces resonance structures. The reaction of 1-chlorobutane
would not undergo an SN1 reaction.
Nucleophilic Substitution Reaction of Organic Halides
Abstract
Organic chemistry involves a variety of chemical reactions, which proceed at different
rates. Nucleophilic substitution reactions are some of the organic reactions that proceed at
different rates and compete with each other. There are two types of nucleophilic substitution
reactions namerly SN1 and SN2 reactions. This lab experiment therefore focuses on the two types
of the nucleophilic substitution reactions of the alkyl halides, deeply analysing factors that
influence their relative rates of reactions. Factors that have been shown to affect SN1 and SN2
reactions include; chemical properties of the nucleophile, the structure of the alkyl halide, steric
hindrance, and the leaving group.
Background information.
SN2 reactions organic reactions are one-step process reactions that proceed without
intermediate. SN2 involves simultaneous bond breaking and bond formation. Given that SN2
reactions do not proceed by an intermediate, nucleophile and alkyl halide are the two
components that influence the rate determining step of the SN2 reactions. The process of the SN2
reactions results in the formation of the stereochemistry at the center of the reaction. Steric
hindrance is the most important factor that affects the rate of the SN2 reaction. The higher the
concentration of the alkyl halide, the slower the rate of the reaction. Other factors such as the
properties of the nucleophile ,which were mentioned at the beginning, may also influence the
rate of the SN2 reaction. The polarization and the size of the nucleophilic atom can either
increase or reduce the rate SN2 reactions. Larger atoms increases the rate of the SN2 reactions
because they can be easily polarized. Nevertheless, the PH strength of the leaving group greatly
determines the SN2 reaction. Strong bases are poor leaving groups and do not increase the rate of
the SN2 reaction. Weak acids on the other hand, are good leaving groups and can increase the
rate of SN2 reaction greatly. Thus an anion solution can stabilize the rate of the SN2. Finally,
solvents can also lower or increase the rate of the SN2 reaction. The role of solvent in dissolving
the metal counter ions are such that they can either make the reaction more reactive or lower its
reactivity. For instance, polar aprotic solvents makes the nucleophile more reactive by dissolving
the metal counter ions, making the SN2 more reactive..
As opposed to SN2 reaction, SN1 is a double step unimolecular reaction that proceeds in
two distinct states: intermediate and two transition states. The rate of the SN1 reaction majorly
depends on the concentration of the alkyl halide but not nucleophile species. Higher
concentration of the alkyl halide increases the rate of SN1 ration while lower concentration
reduces the rate of reaction. Additionally, the stability of the carbocation formed by alkyl halide
determines the rate of the SN1 reaction. Stable carbocation increases the rate of the reaction
while unstable carbocation reduces the rate of the reaction. Nevertheless, the occurrence of the
resonance structure during the reaction may improve the rate of the reaction. This is because
resonance structures improve the stability of the carbocation and lowers the activation barrier
thus increasing the rate of the SN1 reaction. As aforementioned, nucleophiles do not have any
effect on the reaction rate of the SN1 reaction. This is quite different from the SN2 reaction
where the size and polarization of the nucleophile increases the rate of reaction. Along with the
noted factors, the HP leaving group may stabilize or weaken the rate of SN1 reactions. As
mentioned in SN2 reaction weak bases are good leaving groups that increase the rate of SN2
reactions. The same is applicable to SN1 reactions. Weak bases increase the stability of the ions
thus increasing the rate of the SN1 reactions. The solvent used in the reaction process can either
increase or reduce the rate of SN1 reaction. Just like SN2 reaction, protic solvents improve the
stabilization of the carbocation intermediate, increasing the rate of the SN1 reaction.
As mentioned in the introduction this lab experiment focuses on the above discussed
factors that affect the rate of nucleophilic substitution reactions. Reagents used for the SN2
reaction are acetone and sodium iodide solution. Sodium iodide which is nucleophile is dissolved
in acetone which is a good polar and aprotic solvent. For the SN1 reaction experiment, a solution
of the silver nitrate dissolved in ethanol, which is a polar protic solvent, will be used. The protic
nature of the ethanol will favor the formation of stable carbocation, resulting in a complete
reaction of the reagents. The appearance of silver halide at the end of the reaction is an indication
of the complete reaction of the process. Below are the balanced chemical equations for both SN2
and SN1 reactions respectively.
Sn2 reaction
Sn1 reaction.
Name
M.W.
Densit
M.P.
y*
*
(g/mol)
B.P.*
Amoun
Moles
t
Hazards/Precauti
ons
(°C)
(grams
(g/mL)
(°C)
or mL)
Silver Nitrate
169.87
4.35
212
440
n/a
n/a
Oxidizing Solids,
Corrosive to
Metals, Skin
Corrosion, Serious
Eye Damage,
Acute Aquatic
Hazard, Chronic
Aquatic Hazard
Neopentyl
Bromide
151.04
1.199
n/a
105-106 n/a
n/a
Flammable
Liquids, Skin
Irritation, Eye
Irritation, Specific
Target Organ
Toxicity
Ethanol
46.07
1.60
-144
78.29
n/a
n/a
Flammable
Liquids, Eye
Irritation
Benzyl
126.58
4.36
-43
177-181 n/a
n/a
Chloride
Flammable
Liquids, Acute
Oral Toxicity,
Acute Inhalation
Toxicity, Skin
Irritation, Serious
Eye Damage, Skin
Sensitisation,
Germ Cell
Mutagenicity,
Carcinogenicity,
Specific Target
Organ Toxicity,
Acute Aquatic
Hazard
Acetone
58.08
0.79
-94
56
n/a
n/a
Flammable
Liquids, Eye
Irritation, Specific
Target Organ
Toxicity
2-Iodobutane
184.02
1.598
-104
119-120 n/a
n/a
Flammable
Liquids, Acute
Aquatic Hazard
2-
137.02
1.255
n/a
91
n/a
n/a
Bromobutane
Flammable
Liquids, Acute
Aquatic Hazard,
Chronic Aquatic
Hazard
2-Bromo-2-
137.02
1.22
-20
71-73
n/a
n/a
methylpropane
1Chlorobutane
Flammable
Liquids
92.57
0.886
-123
77-78
n/a
n/a
Flammable
Liquids,
Aspiration Hazard,
Acute Aquatic
Hazard, Chronic
Aquatic Hazard
1-
137.02
1.276
Bromobutane
-112
100-104 n/a
n/a
Flammable
Liquids, Skin
Irritation, Eye
Irritation, Specific
Target Organ
Toxicity, Acute
Aquatic Toxicity,
Chronic Aquatic
Toxicity
Procedure (3 point)
Observations and Lab Data (4 point)
Summary of the procedure (2 points)
Part I – Reactivity with NaI in acetone
Observation
Notes from Virtual Lab on Sn2 Reaction.
(SN2)
1. Features of an SN2 reaction include; the
Part A: Effect of Structure of the Alkyl
position of the nucleophile attack, the
Halide on the Relative Rates of SN2
stereochemistry of the product, and the
transition state.
1. 1-inch depth of hot tap water was
2. The alkyl halides used were 1added to a 100 mL beaker and
bromobutane, 2-bromobutane, and 2placed on the hotplate at a setting of
methyl-2-bromopropane.
2 to set up a water bath and stabilize
3. The key 3 factors affecting the rate of an
the water temperature at 45°C.
SN2 reaction are the effect of sterics, the
2. Thermometer was clamped in the
structure of the alkyl halide, and the
water so that the temperature can be
identity of the leaving group.
monitored.
4. The assumption was that the reactions of
3. 2 mL of 15% sodium iodide was
the alkyl halide would differ; it was
measured in acetone into each of
predicted that 1-bromobutane will react
three clean, dry 10-cm test tubes.
the fastest while 2-methyl-2Make sure to mark the test tubes
bromopropane will react the slowest.
respectively with number 1, 2, and
5. There was no reaction in the test tube
3.
with 2-bromobutane. Thus it was clear.
4. 1-bromobutane was added to the
test tube 1 and mixed thoroughly by
swirling. Be sure to note the start
time.
5. Add 2 drops of 2-bromobutane to
6. The test tube that had 1-bromobutane
appeared white and cloudy.
7. Since 1-bromobutane produced white
test tube 2 and mix thoroughly by
precipitate, also the only test tube to
swirling. Be sure to note the start
produce a precipitate.
time.
6. Add 2 drops of 2-methyl-2bromopropane to test tube 3 and
mix thoroughly by swirling. Be sure
to note the start time.
7. Watch each test tube carefully and
note the time at which a precipitate
appears and its color.
8. After 5 minutes, place any test tubes
that have no precipitate into the
45°C water bath. Again, watch for
the formation of precipitates.
9. Heat for up to 6 minutes (11
minutes total) then remove the test
tubes and allow to cool to room
temperature. As some acetone may
have evaporated during heating,
there is the possibility of a
8. The test tube with 2-methyl-2bromopropane was also clear, therefore
showing no reaction.
9. In conclusion, 1-bromobutane was the
fastest under the structure of alkyl
halides.
10. For the steric effects, it was predicted
that 1-bromobutane will react the fastest
while 1-bromo-2,2-dimethylpropane will
react the slowest.
11. The test tube with 1-bromo-2,2dimethylpropane was clear, therefore
showing no reaction.
12. The test tube with 1-bromobutane was
white and cloudy.
13. The test tube with 1-bromobutane was
white and cloudy.
precipitate of unreacted starting
14. Since 1-bromobutane produced white
material, NaI, forming on the walls
precipitate, also the only test tube to
of the test tube. To guard against
produce a precipitate.
this “false positive” result, gently
shake these tubes for a few minutes
which should redissolve any NaI but
15. 1-bromobutane was the fastest under the
steric effects.
16. Since 1-bromobutane produced white
leave other precipitates unchanged.
precipitate, also the only test tube to
Note if precipitate disappeared upon
produce a precipitate
shaking. Record all results.
10. Empty the test tubes into the halide
17. For the effect of leaving the group, it was
predicted that 1-bromobutane will react
waste container and dispose of the
the fastest while 1-chlorobutane will
test tube to the white cardboard box
react the slowest.
for broken glasses.
18. The test tube with 1-chlorobutane was
clear, therefore showing no reaction.
Part B: Steric Effects and the Relative
19. In conclusion, 1-bromobutane reacted
Rates of SN2 Reactions
faster under steric effect.
1. Measure 2 mL of 15% sodium
Predictions of SN1 Reaction.
iodide in acetone into each of two
new clean, dry 10-cm test tubes.
1. For Part 1 of the reactivity with AgNO3
Make sure to mark the test tubes
in Ethanol, the 2-methyl-2-
respectively with number 4 and 5.
bromopropane will react the fastest. This
2. Add 2 drops of 1-bromobutane to
test tube 4 and mix thoroughly by
is due to SN1 reactions preferring a
tertiary alkane.
swirling. Be sure to note the start
time.
3. Add 2 drops of 1-bromo-2,2-
2. This reaction should produce a pale
yellow precipitate.
3. For Part 2 of the reactivity with AgNO3
dimethylpropane (neopentyl
in Ethanol, the 2-iodobutane will react
bromide) to test tube 5 and mix
the fastest. This is due to having the best
thoroughly by swirling. Be sure to
leaving group.
note the start time.
4. Repeat the process as described in
Part A.
4. This reaction should produce a very
bright yellow precipitate.
5. The reaction with 1-chlorobutane will
not undergo since SN1 reactions do not
Effect on the Leaving Group on the
react with primary alkanes.
Relative Rates of SN2 Reactions Part C
6. However, if this reaction were to go, it
1. Measure 2 mL of 15% sodium
iodide in acetone into each of two
would produce a white precipitate.
7. The reaction with benzyl chloride will be
new clean, dry test tubes. Make sure
faster than 1-chlorobutane because the
to mark the test tubes respectively
benzene ring helps stabilize due to the
with number 6 and 7.
resonance structures making better
2. Add 2 drops of 1-bromobutane to
test tube 6 and mix thoroughly by
swirling. Be sure to note the start
time.
3. Add 2 drops of 1-chlorobutane to
test tube 7 and mix thoroughly by
intermediates.
swirling. Be sure to note the start
time.
4. Repeat the process as described in
Part A.
Reactivity with AgNO3 in Ethanol (SN1)
Part A
1. Measure 2 mL of 1% silver nitrate
in ethanol into each of three clean,
dry 10-cm test tubes. Make sure to
mark the test tubes respectively with
number 8, 9, and 10.
2. Keep the hot water bath from part I
and maintain at 45°C.
3. Add 2 drops of 1-bromobutane to
test tube 8 and mix thoroughly by
swirling. Be sure to note the start
time.
4. Add 2 drops of 2-bromobutane to
test tube 9 and mix thoroughly by
swirling. Be sure to note the start
time.
5. Add 2 drops of 2-methyl-2bromopropane to test tube 10 and
mix thoroughly by swirling. Be sure
to note the start time.
6. Watch each test tube carefully and
note the time at which a precipitate
appears and its color.
7. After 5 minutes, place any test tubes
that have no precipitate into the
45°C water bath. Again, watch for
the formation of precipitates.
8. Heat for up to 6 minutes (11
minutes total) then remove the test
tubes and allow to cool to room
temperature.
9. Empty the test tubes into the halide
waste container and dispose of the
test tube to the white cardboard box
for broken glasses.
Part B
1. Measure 2 mL of 1% silver nitrate
in ethanol into each of three clean,
dry 10-cm test tubes. Make sure to
mark the test tubes respectively with
number 11, 12, 13, and 14.
2. Add 2 drops of 2-bromobutane to
test tube 11 and mix thoroughly by
swirling. Be sure to note the start
time.
3. Add 2 drops of 2-iodobutane to test
tube 12 and mix thoroughly by
swirling. Be sure to note the start
time.
4. Add 2 drops of 1-chlorobutane to
test tube 13 and mix thoroughly by
swirling. Be sure to note the start
time.
5. Add 2 drops of benzyl chloride to
test tube 14 and mix thoroughly by
swirling. Be sure to note the start
time.
● Repeat the process as described in
Part A.
Conclusion
SN2 reaction
The reaction of the alkyl bromide with sodium iodide in acetone proceeded to
form two organic compounds: 1-bromobutane and 2-methyl-2 bromopropane. From the reaction,
the fastest reaction of the alkyl bromide and sodium iodide resulted in the formation of the 1bromobutane while the lowest reaction formed the 2-methyl-2 bromopropane. From the analysis
of the reaction it is clear that SN2 reaction proceeds in the primary alkyl halide. The slowest
reaction could have been hindered by the statistically congested alkyl halide. This is in line with
initial discussion that highlighted that steric alkyl halide and congestion hinders the rate of
reaction of the SN2 reaction. The steric effect on the alkyl halide caused the difference in the
reaction with the acetone. From the structural difference of the halides, two hydrogen atoms were
attached to the neighbouring carbon atoms 1-bromobutane which was a result of faster reaction
between sodium iodine and alkyl bromide. In the bromo-2-2-dimethylpropane, the two methyl
groups were attached to the neighbouring atoms, hindering the reaction thus slowing down the
rate of reaction. The two methyl groups prevented the passing of iodine because they are bulky. 1
bromobutane on the other hand had no bulky methyl groups making it easy for the iodine to go
through. Hence the faster rate of reaction for the 1-bromobutane for SN2 reaction.
The reaction alkyl halide with sodium iodide in acetone in terms of the leaving group was
faster for the 1-bromobutane than 2-methyl-2-bromopropane. This was due to the fact the 1bromobutane was the largest leaving group. Larger leaving group increase the rate of reaction
because larger leaving groups are less electronegative making them better leaving groups in Sn2
reaction. Additionally leaving groups are the weakest base and can therefore stabilize anion,
resulting in a faster Sn2 reaction. From the result, bromine can be described as a better leaving
group compared to chlorine.
SN1 reaction
This conclusion of the Sn1 reaction is based on assumption of what would have happened
if the experiment was to be physically performed and analyzed. First, this experiment was never
concluded virtually or physically due to technical issues. Back to the assumption, if the
experiment were to be conducted and analyzed, it is hypothesized that 2-bromo-2-methylpropane
would have reacted faster with the silver nitrate than the other alkyl bromide. This could have
been due to the fact that 2-bromo-2-methylpropane is a tertiary alkyl halide. As mentioned
before, Sn1 reaction proceeds better in secondary or tertiary halide compared to primary halide.
Considering the effect of the leaving group, the reaction would have been faster in 2-iodobutane
because it is the largest alkyl halide compared to bromine. This would have been the case as the
larger the alkyl halide, the faster the rate of the reaction. The size of the alkyl halide is inversely
proportional to electronegativity of the halide. Hence, less electronegative halide increases the
rate of the Sn1 reaction. The reaction of benzyl chloride would be faster because it intermediates
the further produce resonance structures which increase the rate of reaction. Chlorobutane on the
other hand would not undergo Sn1 reaction due to bulkiness and much concentration of the
electronegative.
Х
Experiment 3 Lab Report (...
(doulet)
2) Proton on Carbon
OH
H
KH
CH3
#2 : 4 ppm
(multiplet)
3) Proton on Carbon
#321 ppm
(quartet)
4) Proton on Carbon
#41 ppm
(multiplet)
5) Proton on Carbon
MW = 45
#521 ppm
(triplet)
4) OH proton
appears randomly
on 'HNMR
3
6
4
1) Proton on Carbon
1700 cm
120
#1 22 ppm
(singlet)
2) Proton on Carbon
#47 ppm
(doublet)
3) Proton on Carbon
CH3
#5:7 ppm
(triplet)
4) Proton on Carbon
MW = 43
#6 27 ppm
(triplet)
1) Proton on Carbon
4
5
4
1715 cm
100
#12 ppm
f
(singlet)
2) Proton on Carbon
43
#31 ppm
(doublet)
3) Proton on Carbon
#4 = 1 ppm
(multiplet)
4) Proton on Carbon
#5 a 1 ppm
(doublet)
Part 11
In person students will come to the lab and carry out the IR experiment of the four unknown compounds.
Remote students will click the link below or copy it to your browser, then scroll down the page
to find "Virtual Lab 2: Infrared Spectroscopy". Follow the instruction there to complete the
virtual lab (remember you can use your cursor to see different view of the lab). You need to
carry out the virtual lab four times to obtain IR data for all four compounds. The credit of this
virtual lab goes to North Carolina State University.
https://sites.google.com/ncsu.edu/ncstatevrorganicchemistrylabs/home
After you complete the in-person or virtual lab, report here which compound is which and
explain why (Compound 1 was used as an example here to show you how this is done).
Experiment 3- Elyse Rebelo
Determining the Structures of Unknown Molecules Using
NMR, IR and Mass Spectroscopy
Xiaolei Gao, Ph.D.
Background
The theory of infrared spectroscopy, mass spectroscopy and NMR spectroscopy will be covered in the first two lab
periods. Please read chapters 12 and 13 in McMurry for a complete discussion of these techniques.
Procedure
Part 1
The lab consists of three parts. The first part is a thought experiment. Consider the following four compounds:
OH
OH
2
1
Pentan-2-ol
Acetophenone
4-Methyl-pentan-2-one
Benzyl Alcohol
Using the knowledge of spectroscopy, you gained in the prior two classes, predict what the IR, NMR and mass
spectrum of each would look like and fill in the following table ( completed part of table for you as an example):
Compound #
# Carbon
# Proton
Approximate Chemical
Main IR Stretch
Expected
Molecular
One Possible Fragment in
M.S.
Signals
Signals
Shift of Protons
lon in M.S.
1
5
5
3200-3600 cm
108
1) The proton in OH
appears randomly
HC
on 'HNMR
OH
31
2) The protons in
CH2 appears @
--4-5 ppm (singlet)
3) Protons on the
benzene ring
appear @ 7 ppm
(multiplet)
2
5
6
1) Proton on Carbon
3200-3600 cm
88
#1: 1 ppm
(doulet)
2) Proton on Carbon
OH
#24 ppm
KA
(multiplet)
3) Proton on Carbon
CH3
#31 ppm
MW = 45
(quartet)
4) Proton on Carbon
#41 ppm
(multiplet)
5) Proton on Carbon
#51 ppm
(triplet)
4) OH proton
appears randomly
on 'HNMR
3
6
4
1) Proton on Carbon
1700 cm
120
#12 ppm
(singlet)
2) Proton on Carbon
#47 ppm
(doublet)
3) Proton on Carbon
CH3
#5 7 ppm
(triplet)
MW = 43
4) Proton on Carbon
#6 27 ppm
Х
Experiment 3 Lab Report (...
Part III
1. Which of the four compounds has a MS as shown below? Explain all the major
fragments.
LOO
80
60
40
20
0.0
0.0
20
40
60
80
100
120
This MS shown above is for benzyl alcohol because the base peak at 43 indicates alpha cleavage
on carbon #2 to produce a fragment with a molecular weight of 43. Also, the parent peak at 100
matches the full molecular weight of benzyl alcohol as well as there are multiplets to match the
aromatic ring in the compound.
2. The following 'HNMR belong to which of the four compounds in this experiment (s:
singlet; d: doublet;t: triplet; m: multiplet)? Explain.
3H, d
3H, t
4H, m
1H, m 1H, s
o
in.
The following 'HNMR belongs to 2-pentanol because the highest ppm is 4 ppm which
would represent carbon #2 that has one hydrogen. The doublet represents carbon #1 with
three hydrogens that is done Pupptiplet represents carbon
Х
Experiment 3 Lab Report (...
2. The following HNMR belong to which of the four compounds in this experiment (s:
singlet; d: doublet;t: triplet; m: multiplet)? Explain.
3H, d
3H, t
4H, m
1H, m 11, s
5
4
in.
The following "HNMR belongs to 2-pentanol because the highest ppm is 4 ppm which
would represent carbon #2 that has one hydrogen. The doublet represents carbon #1 with
three hydrogens that is a CH3 near the OH group at about 1 ppm. The triplet represents carbon
#5 with three hydrogens that appears at about 1 ppm. The multiplet with 4 hydrogens
represents carbon #4 that lies between 1 ppm and 2 ppm. Lastly, the OH proton can appear
randomly on the HNMR. Therefore, this OH proton is located between 2.5 ppm and 3 ppm as a
singlet.
3. Which of the four compounds has the following 13CNMR? Explain.
ppm
Int.
标记读
197.85
137.23
133.04
128.56
128.29
26.47
174
229
463
866
1
2
3
4
5
6
1000
259
Х
Experiment 3 Lab Report (...
in.
The following HNMR belongs to 2-pentanol because the highest ppm is 4 ppm which
would represent carbon #2 that has one hydrogen. The doublet represents carbon #1 with
three hydrogens that is a CH3 near the OH group at about 1 ppm. The triplet represents carbon
#5 with three hydrogens that appears at about 1 ppm. The multiplet with 4 hydrogens
represents carbon #4 that lies between 1 ppm and 2 ppm. Lastly, the OH proton can appear
randomly on the HNMR. Therefore, this OH proton is located between 2.5 ppm and 3 ppm as a
singlet.
3. Which of the four compounds has the following 13CNMR? Explain.
ppm
Int.
标记歲
174
229
197.85
137.23
133.04 463
128.56 866
128.29 1000
26.47 259
1
2
3
4 +
5 +
6
200
180
160
140
120
100
80
60
40
20
0
CDS-00-674
pom
The following CNMR is shown for acetophenone for two main distinct peaks. The first
one lies between 190 ppm to 200 ppm which indicates a C=0 ketone. By knowing there
is a C=0 in the compound, this limits our options to acetophenone and 4-methyl-2-
pentanone. The second peak(s) lie between 120 ppm and 140 ppm which indicates the
C=C of an aromatic benzene ring. Therefore, between our two options, the only one that
has a benzene ring and a C=0 is acetophenone.
5
un
6
e
1) Th e around
proton at 3400-3200
C2 is a
cm-1
sextet and
appears at
around 4
ppm.
2) Th
proton of
the OH
group
appears
randomly.
3) The
proton at
C3 is a
quintet
an d
appears at
around 1
ppm.
4) Th
proton at
C4 is a
sextet and
appears at
around 1
ppm.
5) Th
proton at
C1 is a
doublet
e
e
2
5
6
88
OH
v
m/z=73
1) The around
proton at 3400-3200
C2 is a
cm 1
sextet and
appears at
around 4
ppm.
2) Th e
proton of
the OH
group
appears
randomly.
3) The
proton at
C3 is a
quintet
a n d
appears at
around 1
ppm.
4) Th e
proton at
C4 is a
sextet and
appears at
around 1
ppm.
5) Th е
proton at
C1 is a
doublet
and
appears at
around 1
ppm.
6) Th e
3
6
4
120
CH2
1) One of the 1680 cm 1
aromatic
protons is
a doublet
n d
appears at
around 8
a
m/z=91
ppm.
2) Another
aromatic
proton is a
triplet and
appears at
around 7-8
ppm.
3) The third
aromatic
proton
appears as
a triplet at
around 7
ppm.
4) The
protons of
the methyl
group
appear at
a singlet at
around 2-3
5
4
e
1700 cm
100
IUO
m/z=85
1) Th
proton at
с 3
appears as
a doublet
at around
2-3 ppm.
2) The
proton at
с 1
appears as
a singlet at
around
2-2.5 ppm.
3) The
proton at
C4 is a
multiplet
an d
appears at
around 2
ppm.
4) The
protons at
C5 and the
methy!
group are
equivalent
and
appear as
a doublet
at around
1 ppm.
Part II
88
nd
200
ОН
73
2
5
6
3200-3600 cm
88
OH
I
KH
+
CH3
MW = 45
1) Proton on Carbon
#11 ppm
(doulet)
2) Proton on Carbon
#24 ppm
(multiplet)
3) Proton on Carbon
#31 ppm
(quartet)
4) Proton on Carbon
#41 ppm
(multiplet)
5) Proton on Carbon
#521 ppm
(triplet)
4) OH proton
appears randomly
on 'HNMR
1) Proton on Carbon
#12 ppm
(singlet)
2) Proton on Carbon
#47 ppm
(doublet)
3) Proton on Carbon
#57 ppm
(triplet)
4) Proton on Carbon
#67 ppm
3
6
4
1700 cm 1
120
O
+
CH3
MW = 43
N
5
6
3400-3600
88
45
cm-1
1) The proton in
OH appears
randomly on
HNMR
2) The proton in
CH3 appears
@ 1 ppm
(3H, triplet)
3) Protons on C
OH appears
@3-4 ppm
(1H, sextet)
4) CH3 attached
to C-OH
appears @
1ppm
(duplet)
5) Protons on
CH2 attached
to C-OH
OH
+ CH
`CH₃
3
appears @ 1
ppm (2H,
quartet)
6) Protons on
CH2 appears
@ 0-1 ppm
(2H, sextet)
3
6 4
1660-1780
120
43
1) The protons
in CH3
cm-
H
-C-H
appears @ 2
ppm (3H
Singlet)
2) The protons
in the
benzene ring
appears @7
ppm (SH
multiplet)
Lab Partner: n/a
Title: Nucleophilic Substitution Reactions of Organic
Halides
1. Abstract and Purpose: (3 point).
When studying organic chemistry, there are many
different types of reactions that occur. This lab will focus on
nucleophilic substitution reactions known as SN2 and SN1,
specifically examining the factors that affect the relative rates
of SN2 and SN1 reactions of alkyl halides. These factors
include alkyl halide structure, nature of the leaving group,
properties of the nucleophile, and steric hindrance.
The SN2 reaction is a one-step, concerted substitution
process meaning it will make new bonds and break old bonds
at the same time. Therefore, SN2 reactions do not proceed by
an intermediate. Since there is no intermediate, the rate-
determining step involves two components: the alkyl halide
and the nucleophile. The SN2 reaction also gives inversion of
stereochemistry at the reaction center.
There are four main factors that can affect an SN2
reaction. However, the most important are the steric effects
which means the more congested and hindered the alkyl halide,
the slower the reaction will occur. The leaving group is another
factor that can affect this type of reaction, where the weakest
bases are the best leaving group for SN2 reactions as they can
best stabilize the anion. The choice of solvent can affect the
rate of an SN2 reaction such that polar aprotic solvents are
ideal as they solvate the metal counterion of the nucleophile,
making it more reactive. Lastly, the nucleophile has an effect as
SN2 reactions would rather have larger atoms since they are
more polarizable.
The SN1 reaction is a two-step unimolecular process
with one intermediate and two transition states. The rate
depends upon only the concentration of the alkyl halide,
therefore making the rate of this reaction independent of the
nucleophile concentration.
Once again, many factors can affect an SN1 reaction.
The alkyl halide must form a stable carbocation, in which the
more stable carbocation, the faster the reaction will happen.
This is due to lowering the activation barrier. Along with a
stable carbocation, resonance structures can occur such that
they provide a tremendous amount of stabilization to the
carbocation. The leaving group can also have the same effect
on an SN1 reaction as an SN2 reaction such that the weakest
bases are the best leaving group as they can best stabilize the
anion. Unlike SN2 reactions and as stated above, the
nucleophile will not affect the rate of an SN1 reaction because
it is not part of the rate-determining step. Lastly, the choice of
solvent can affect the rate of an SN1 reaction such that polar
protic solvents are ideal as it stabilizes the carbocation
intermediate as it forms, thus increasing the rate of the reaction.
Overall, this lab will allow for the examination on all
of the following factors that affect the relative rates of an SN2
or SN1 reaction. For the experiment on SN2 reactions, a
solution of sodium iodide in acetone will be used. Sodium
iodide is an excellent nucleophile that is soluble in acetone
which is a polar aprotic solvent. For the experiment on SN1
reactions, a solution of silver nitrate in ethanol will be used.
Ethanol 15 polar proti solvent that will favor stable
Х
Lab 4 Nucleophillic Substi...
rate of an SN2 reaction such that polar aprotic solvents are
ideal as they solvate the metal counterion of the nucleophile,
making it more reactive. Lastly, the nucleophile has an effect as
SN2 reactions would rather have larger atoms since they are
more polarizable.
The SN1 reaction is a two-step unimolecular process
with one intermediate and two transition states. The rate
depends upon only the concentration of the alkyl halide,
therefore making the rate of this reaction independent of the
nucleophile concentration.
Once again, many factors can affect an SN1 reaction.
The alkyl halide must form a stable carbocation, in which the
more stable carbocation, the faster the reaction will happen.
This is due to lowering the activation barrier. Along with a
stable carbocation, resonance structures can occur such that
they provide a tremendous amount of stabilization to the
carbocation. The leaving group can also have the same effect
on an SN1 reaction as an SN2 reaction such that the weakest
bases are the best leaving group as they can best stabilize the
anion. Unlike SN2 reactions and as stated above, the
nucleophile will not affect the rate of an SN1 reaction because
it is not part of the rate-determining step. Lastly, the choice of
solvent can affect the rate of an SN1 reaction such that polar
protic solvents are ideal as it stabilizes the carbocation
intermediate as it forms, thus increasing the rate of the reaction.
Overall, this lab will allow for the examination on all
of the following factors that affect the relative rates of an SN2
or SN1 reaction. For the experiment on SN2 reactions, a
solution of sodium iodide in acetone will be used. Sodium
iodide is an excellent nucleophile that is soluble in acetone
which is a polar aprotic solvent. For the experiment on SN1
reactions, a solution of silver nitrate in ethanol will be used.
Ethanol is a polar protic solvent that will favor a stable
carbocation. Silver halide salts are very insoluble and will
precipitate from the solution, indicating if a reaction has
occurred.
2. Balanced equation: (2 point)
А
A
slow
Nu-
의
B
Nu-
B
+ i
Nu-
B
Substrate
Transition State
Product (Inverted)
SN2 Reaction
Nu
slow
fast
SN1 Reaction
Х
Lab 4 Nucleophillic Substi...
Name
Silver Nitrate
Neopentyl
Bromide
Ethanol
Benzyl Chloride
M.W. Density* M.P.* B.P.* Amount Moles Hazards/Precautions
(g/mol) (g/mL) (°C) (°C) (grams or
mL)
169.87 4.35 212 440 n/a n/a Oxidizing Solids,
Corrosive to Metals,
Skin Corrosion,
Serious Eye Damage,
Acute Aquatic
Hazard, Chronic
Aquatic Hazard
151.04 1.199 n/a 105- n/a n/a Flammable Liquids,
106
Skin Irritation, Eye
Irritation, Specific
Target Organ Toxicity
46.07 1.60 -144 78.29 n/a n/a Flammable Liquids,
Eye Irritation
126.58 4.36 -43 177- n/a n/a
Flammable Liquids,
181
Acute Oral Toxicity,
Acute Inhalation
Toxicity, Skin
Irritation, Serious Eye
Damage, Skin
Sensitisation, Germ
Cell Mutagenicity,
Carcinogenicity,
Specific Target Organ
Toxicity, Acute
Aquatic Hazard
58.08 0.79 -94 56 n/a n/a Flammable Liquids,
Eye Irritation,
Specific Target Organ
Toxicity
184.02 1.598 -104 119- n/a n/a Flammable Liquids,
120
Acute Aquatic Hazard
137.02 1.255 n/a 91 n/a n/a Flammable Liquids,
Acute Aquatic
Hazard, Chronic
Aquatic Hazard
137.02 1.22 -20 71-73 n/a n/a Flammable Liquids
Acetone
a
2-Iodobutane
2-Bromobutane
2-Bromo-2-
methylpropane
1-Chlorobutane
92.57 0.886
-123 77-78 n/a
n/a
1-Bromobutane 137.02 1.276
n/a
n/a
-112 100-
104
Flammable Liquids,
Aspiration Hazard,
Acute Aquatic
Hazard, Chronic
Aquatic Hazard
Flammable Liquids,
Skin Irritation, Eye
Irritation, Specific
Target Organ
Toxicity, Acute
Aquatic Toxicity,
Chronic Aquatic
Toxicity
4. Calculations: Shown each calculation for moles of reagents,
limiting reagent, theoretical yield and percent yield. (3 points)
Calculating moles of reagents: n/a
Determining limiting reagent: n/a
The limiting reagent is n/a.
Calculating theoretical yield: n/a
The theoretical yield is n/a.
The actual yield is n/a.
Calculating Percent yield: n/a
Reaction percent yieldiena
solution of sodium iodide in acetone will be used. Sodium
iodide is an excellent nucleophile that is soluble in acetone
which is a polar aprotic solvent. For the experiment on SN1
reactions, a solution of silver nitrate in ethanol will be used.
Ethanol is a polar protic solvent that will favor a stable
carbocation. Silver halide salts are very insoluble and will
precipitate from the solution, indicating if a reaction has
occurred.
2. Balanced equation: (2 point)
A
slow
1
Nu
B
+
1
Nu-
D
B
Substrate
Transition State
Product (Inverted)
SN2 Reaction
Nu
Nu
-CY
slow
Z
como
fast
х
yic
-Nu
SN1 Reaction
3. Reagent Table: (Add more rows when needed) (2 points)
Name M.W. Density* M.P.* B.P.* Amount Moles Hazards/Precautions
(g/mol) (g/mL) (°C) (°C) (grams or
mL)
Silver Nitrate 169.87 4.35 212 440 n/a n/a Oxidizing Solids,
Corrosive to Metals,
Skin Corrosion,
Serious Eye Damage,
Acute Aquatic
Hazard, Chronic
Aquatic Hazard
Neopentyl 151.04 1.199 n/a 105- n/a n/a Flammable Liquids,
Bromide
106
Skin Irritation, Eye
Irritation, Specific
Target Organ Toxicity
Ethanol
46.07 1.60 -144 78.29 n/a n/a Flammable Liquids,
Eye Irritation
Benzyl Chloride 126.58 4.36 -43 177- n/a n/a Flammable Liquids,
181
Acute Oral Toxicity,
Acute Inhalation
Toxicity, Skin
Irritation, Serious Eye
Damage, Skin
Sensitisation, Germ
Cell Mutagenicity,
Carcinogenicity,
Specific Target Organ
Toxicity, Acute
Aquatic Hazard
Acetone
58.08 0.79 -94 56 n/a n/a Flammable Liquids,
Eye Irritation,
Specific Target Organ
Toxicity
2-Iodobutane 184.02 1.598 -104 119- n/a n/a Flammable Liquids,
120
Acute Aquatic Hazard
2-Bromobutane 137.02 1.255 n/a 91 n/a n/a Flammable Liquids,
Acute Aquatic
Hazard, Chronic
Aquatic Hazard
2-Bromo-2- 137.02 1.22 -20 71-73 n/a n/a Flammable Liquids
methylpropane
1-Chlorobutane 92.57 0.886 -123 77-78 n/a n/a
Flammable Liquids,
Aspiration Hazard,
Acute Aquatic
Hazard, Chronic
Aquatic Hazard
1-Bromobutane 137.02 1.276 -112 100- n/a n/a Flammable Liquids,
104
Skin Irritation, Eye
Irritation, Specific
Target Organ
n
CON
box (98) - sdhani.
OH
ning appear
Send
Feedback to Lea
3/5/21 1.29 PM
Part 1 5/5
218
Part II 6/6
Part III
@-7 ppm
(multiplet)
1) The proton on
the hydroxyl group
will appear
randomly in the
HNMR.
OA
it
1.2/3 The key peaks need to be explained are 100, 85. 58 ar
compound 4 a ketone
2.1.53 Explain all peaks (ppm, splitting patterns) to supp
conclusion
2) The methyl
3.253
group near the OH
appears at
approximately
1ppm (3H, d)
3) the proton
Count number of carbon peaks in NMR and make
accountable
6
CH3
6
attached to
5
COH appear
4ppm (1H, m
0
4) The CH2 group
closest to the COH
S.
shows up at
1 ppm (2H, al. The
other CH2 group
shows up at
"1ppm/2H, m)
woryromoureur
esdh...
?content_id
wy
Amazon.com:
che other
3
mpty group
.nows up at
1ppm (3H, t)
1)The protons on
the aromatic rings
show up at 7ppm
(SH,m)
2)The methyl
4
group bonded to
C=O shows up at
wood
2 ppm (3H, )
B B
1695
120
- 43
1) The ethyl
group bonded to
CHO shows up at 2
CH
1715 100
0
ppm (3H)
2) The CH2 shows
up as at2 ppm
فال
(2H.d)
4) the single
proton that is
ros
attached to the
two methyl groups
shows up at
Part
43
1.213 The
compound 4
2.1.5/3 Explain all
conclusion
3.25/3
Count number of carbon peat
accountable
los
2
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