Lab+Lecture+9+-+Methyl+Diantilis_Formal+Lab+Report_Annotated_2023Oct24
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1
CHEM 12AL
Organic Chemistry Lab
2
Synthesis of Methyl Diantilis
1)
Two step synthesis
- Everyone performs the first step of the reaction
- Split into 4 groups of 4 and carry out second step to produce different products
2)
Combine all techniques up to this point
- Set up reactions
- Purify intermediates and final products
- Characterize products by 1
H NM and 13
C NMR
3)
Formal Lab report is written based on Methyl Diantilis Synthesis
- Detailed formal lab report guidelines available on bCourses
3
Synthesis of methyl diantilis
Two steps: i) aldehyde reduction, ii) etherification
NaBH
4
reduction mechanism
⊥
H-
0
-
H
-
N
Hy
-
0
-
Hz
I
is
oft
->
Ot
ON
.
not
a
1
,
'Yout
3
equiv
of
hydride
remaining
(H0)
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4
Amberlyst-15 (H
+
) catalyzed etherification mechanism
Synthesis of methyl diantilis
Two steps: i) aldehyde reduction, ii) etherification
Sheterogeneous
H
*
/Acid
Catalyst
Ju
T
⑰
i
,
it
"
In t
-
-
"
os
wo
,
an
I
->
dH
·
benzylic
cation
R
-OH
·
resonance
stabilized
↓
F
·
Sul
-
H
I
hi
It
en
1
-
I
·
similar
stability
tab"
catius
In
-Ott
·
compare
purity
5
•
Due on Monday, Dec 4 (30pts)
•
Typed: 12pt font, double spaced
•
Page count: 3-5 pages including figures
•
Use ChemDraw to draw molecules and reaction schemes
•
Formal Lab Report is a separate assignment to the experimental lab notebook.
*
Can not be used as your dropped lab score*
Synthesis of methyl diantilis –
Formal Lab Report
Formatting Guidelines
•
Use ChemDraw to draw molecules and reaction schemes
•
Formal Lab Report is a separate assignment to the experimental lab notebook.
6
Synthesis of methyl diantilis –
Formal Lab Report
Lab Report Outline
1) Introduction
2)
Results and Discussion
3) Conclusion
4) Experimental
5) References
6)
Supplemental Information
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7
Synthesis of methyl diantilis –
Formal Lab Report
Lab Report Outline
Introduction [5 pt]
•
Introduce the synthesis of methyl diantilis in general (why it’s important, etc). In this introduction, include at least two reference to a peer-reviewed journal article. Indicate the references using superscripted numbers which correspond to a numbered list of properly formatted citations at the end of the report. •
Draw a reaction scheme for the specific reactions that you carried out. •
Describe this reaction scheme in 2-3 sentences.
Results, Discussion, and Conclusions [10 pt]
•
Write a summary of your results.
•
Write a cohesive discussion regarding the experimental details for the reaction that you carried out. Your discussion should include:
An analysis of all NMR spectra of your group
A rational arrow-pushing mechanism for both reactions
A discussion of your yields with all four alcohols
Discuss the composition of each product. In this discussion, include information gained from TLC as well as 1
H NMR and 13
C NMR. Include the purity calculations and a qualitative comparison of your group’s products with the products from other groups in your lab section.
Describe the experimental factors that you believe had the biggest effect in causing any variability between different groups in your laboratory section. •
Write a concise conclusion that summarizes the results of your experiment.
8
Synthesis of methyl diantilis –
Formal Lab Report
Lab Report Outline
Experimental [6 pt]
•
Write a concise description of the steps that were taken to carry out this reaction in the lab. This should be written in the third person passive tense and should constitute an overview of the procedure without including all the details. •
Report the 1
H NMR and 13
C NMR data for your intermediate and products.
References [2 pt]
Formatted according to the American Chemical Society style
Supporting Information: [7 pt]
Attach copies of each NMR spectrum, annotated as typical for this course.
9
NMR Spectroscopy
1
H NMR of Aromatic Molecules
- Circulating pi electrons create ring currents
- Induced magnetic field increases B
effective
near H atoms
Increase in B
effective
deshields H atoms and shifts
downfield benzene
ethylene
Induced magnetic field
B
0
Induced magnetic field
= 6-8 ppm
= 5-6 ppm
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10
NMR Spectroscopy
1
H NMR of Aromatic Molecules
2-Isopropylphenol
7.00
6.75
7.25
7.50
1.0
1.0
1.0
1.0
7.35
7.21
7.01
6.80
Ch
-
He
Haz
CHf
H
!
Ad
·
4
sets
of
aromatic
'It's
f
·
complex
splitting
-
Of
He
11
NMR Spectroscopy
1
H NMR of Aromatic Molecules
7.00
6.75
7.25
7.50
1.0
1.0
1.0
1.0
7.35
7.21
7.01
6.80
How do we assign the aromatic protons given the information that we have available: i) Chemical shift
values
, ii) Peak Areas
, and iii) Peak Splitting
?
Origins of aromatic 1
H Chemical Shift Values
Consider
ring
substituent
resonance
forms
⊥
it
*
3
Elite
j
I
↑
-
=I
E
-
OH
is
pi-donating
I
I
-
I
->
lone
pairs
on
oxygen
donate
intoring's
shield
ortholpara
positions
-
-oH
is
withdrawing
->
oxygen
is
more
dectronegative
than
carbons
is
an
Eltria
the
o-csigma
bond
,
hart
to
predict
mary
:
substituents
affect
shielding/deshielding
of
aromatict's
in
sometimes
wats
.
12
NMR Spectroscopy
1
H NMR of Aromatic Molecules
7.00
6.75
7.25
7.50
1.0
1.0
1.0
1.0
7.35
7.21
7.01
6.80
Origins of aromatic 1
H Chemical Shift Values
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13
NMR Spectroscopy
1
H NMR of Aromatic Molecules
Origins of aromatic 1
H Peak
Splitting
3-bund
coupling
Hist
1X
I
I
ing
strengths
-H
1
Ortho
-
H
-
- H
or
tho
coupling
:
6-10
Hz
(normal
3-bond
coupling)
pare
#
meta
coupling
:
1-3Hz
(4-bond
coupling)
para
para
coupling
:
0-1H1
/S-bond
coupling
(
·
Conjugation
enables
long-range
coupling
·
S-bond
coupling
is
uncommon
in
nonvaromatic
systems
14
NMR Spectroscopy
1
H NMR of Aromatic Molecules
7.00
6.75
7.25
7.50
Origins of aromatic 1
H Peak
Splitting
fanity
t
3-bond
coupling
⊥
a
Stab
,
"Ybc
,
"Tod
Ab-lit
I
-
·
Ha
or
Hd
e
ef
4-bund
coupling
HborHe
-
"Fac
,
"
Abd
Hb:
He
have
more
coupling
Partiers
15
NMR Spectroscopy
Aromatic 1
H Splitting trees
3
J
a
b
= 8 Hz
4
J
a
c
= 2 Hz
3
J
d
c
= 8 Hz
4
J
d
b
= 1 Hz
1 box = 1 Hz
↑
Ha
Ald
--
--
N
-
Ha-Hb
/Ortho)
N
↑
Hd-H
,
(orthol
-
8
-
-
r
IT
Ha-He
(metc)
l
Hd-Hb
(meta)
1
:
1
:
I
:
1
1
:
1
iii)
I
I
us
it
te
-
doublets
of
doublets
doublets
of
doublets
(dd]
(dd]
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16
NMR Spectroscopy
Aromatic 1
H Splitting trees
3
J
b
a
= 8 Hz
3
J
b
c
= 7 Hz
4
J
b
d
= 1 Hz
3
J
c
d
= 8 Hz
3
J
c
b
= 7 Hz
4
J
c
a
= 2 Hz
1 box = 1 Hz
⑧
⑧
Hb
Ha
I
N
-
=
Hb-
Ha
(orthol
F
->
Ha-Hd
(ortha)
V
8
8
---
--
I
-
-
8
y
8
S
/Hb-Hc(orthol
-
-
TH
-
Hb
lot
#
T
ST
Ab-Hd
(metal
11
is
is
~
MHc-Ha
1
:
1
2
:
2
1
:
1
:
I
2
:
2
1
:
I
expected
Mee
le
triple
of
doublets
triplet
of
doublets
(
+
d)
(
+d)
17
NMR Spectroscopy
1
H NMR of Aromatic Molecules
7.00
6.75
7.25
7.50
1.0
1.0
1.0
1.0
7.35
7.21
7.01
6.80
C
-
He
n
CH
Ad
Hc
Hb
Ha
-
dd
dd
I
d
Id
HA
-
OH
He
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