Name:
Student No:
Page 1 of 13
Chemistry 2.222 Organic Chemistry II: Reactivity and Synthesis
FINAL EXAM – Winter 2003
Paper Number 546
Thursday April 24, 2003 6:00 – 9:00 pm
Frank Kennedy Brown Gym
Students are permitted to bring into the exam room ONE SHEET of 8½ x 11 paper, with any
HANDWRITTEN notes they wish (both sides). Molecular model kits are also permitted but no other
aids may be used.
The exam is in two parts. Answers to PART I are to be entered in the indicated spaces on the exam
paper itself.
An answer to ONE of the “Challenge Questions” in PART II may be written in the exam booklet for
EXTRA CREDIT.
PART I:
Question 1
(20 Marks)
Question 2
(10 Marks)
Question 3
(15 Marks)
Question 4
(15 Marks)
Question 5
(10 Marks)
SUB-TOTAL:
(70 Marks)
PART II (EXTRA CREDIT)
(7 Marks)
TOTAL:
Page 2 of 13
PART I: DO ALL QUESTIONS – There is choice in question 1 only.
1. (20 MARKS) Provide the missing product, starting compound or reagent/solvent/conditions
to correctly complete TEN of the following reactions. All reactions do in fact lead to
products. Clearly indicate which TEN responses you want marked!
1) Br2, AlBr3
CH2Cl2
2) Sn, HCl
Br
(a)
H
H
H
H
(b)
O
LDA, THF, -78 oC
then
Br
(c)
O
O
O
(d)
Hg(OAc)2 (cat.)
H2SO4 (aq.)
(e)
Page 3 of 13
O
Br2, AlBr3
O
CH2Cl2
(f)
O
NH
LiAlH4, THF
(H3O+ workup)
(g)
NaOMe
MeOH
O
O
OMe
MeO
(h)
O
O
OMe
reagent, solvent, workup
O
O
OH
O
(i)
HBr, CH2Cl2
MeO
(j)
OH
Page 4 of 13
O
O
(k)
C6H5NH2, Ether
O
N
H
(l)
O
O
OH
(m)
NH2
1) NaNO2,
HCl (aq.)
2) H2SO4 (aq.)
heat
(n)
LDA, THF
-78 oC
O
O
(o)
then heat
(H3O+ workup)
COOH
Page 5 of 13
2. (10 MARKS) Propose a synthetic route to prepare 5-methylheptane-4-one (A) starting from
ethyl acetoacetate (B). You may use any additional organic starting materials having three
or fewer carbons, as well as any reagents or solvents you require. This transformation can
be accomplished in fewer than 5 steps. Remember that although a retrosynthetic
analysis may be useful to you in solving this problem, the answer I want is the
“forward synthesis” complete with reagents and products.
O
Make
O
O
starting from
A
OEt
B
Page 6 of 13
3. (7 MARKS) Provide a detailed stepwise mechanism for the following transformation.
O
O
NaOCH3
CH3OH
O
O
O
+
O
OH
O
Page 7 of 13
(b) (8 MARKS) The Wieland-Miescher Ketone is an important starting material in steroid
synthesis. It can be prepared by the Robinson Annulation strategy shown here. Provide
a detailed stepwise mechanism for this reaction.
O
CH3
CH3
O
NaOH
O
O
+
H2O, 95 oC
O
Page 8 of 13
4. (a) (5 MARKS) Compound A below is suggested to be a trace impurity formed when 4,4'di-(tert-butyl)biphenyl is prepared from biphenyl and t-butylchloride in CH2Cl2, using FeCl3
as a catalyst. Explain how A could be formed under these conditions.
+
Cl
major product
FeCl3
H
CH2Cl2
+
H
C
A - trace impurity
(b) (5 MARKS) The aldehydes A and B below both undergo base-catalyzed condensation
reactions with acetone (C). The reaction of B under these conditions is much faster than
that of A. Briefly explain why this is so, taking into account electronic effects.
O
H3CO
H
A
O
O
O2N
H
B
H3C
CH3
C
Page 9 of 13
(c) (5 MARKS) Draw and label a diagram showing a reflux apparatus set-up. Give 2
advantages of using reflux conditions in a chemical reaction.
Page 10 of 13
5. (10 MARKS) The IR, 13C NMR and 1H NMR spectra of an organic compound having the
molecular formula C7H15N are shown below. NB: In the 1H NMR spectrum, m=multiplet,
d=doublet, tr=triplet, q=quartet.
4000
3000
2000
m
1500
m
Draw the structure of the compound in the box below.
1000
d
q
500
tr
Page 11 of 13
PART II: CHALLENGE PROBLEM (EXTRA CREDIT).
DO ONE PROBLEM ONLY. Part II is worth up to XX additional marks.
Write your answer in the exam booklet provided. Be sure your name and student number are on
the booklet. Insert the booklet into Part I when you hand it in.
A. SYNTHESIS
There is something seriously wrong with the following proposed synthesis of the drug
terfenadine. Identify the problem, and propose specific steps needed to make this route
workable.
H
N
Br
Mg
O
N
Ether
OH
HO
HO
terfenadine
B. MECHANISM
When carvone (A) is boiled in acidic water, it is converted to carvacrol (B). Give a mechanism
to explain this transformation.
O
OH
H3O+
100 oC
A
B
C. SPECTROSCOPY
The unusual compound diketene has the molecular formula C4H4O2. Its 1H and 13C NMR
spectra are shown on the next page. Heating diketene in the presence of cyclopentadiene leads
to the formation of bicyclo[2.2.1]hept-5-ene-2-one. What is the structure of diketene?
Diketene
+
heat
O
Page 12 of 13
Diketene 13C NMR
Four signals, δ 165.18, 147.66,
87.06 and 42.40 ppm.
PPM
150.0
140.0
130.0
120.0
110.0
100.0
90.0
80.0
70.0
Diketene 1H NMR
Three signals, δ 4.877,
4.494 and 3.897 ppm.
Double triplet
PPM
PPM
4.900
4.896
4.892
4.90
4.888
4.884
4.880
4.876
4.872
4.80
4.868
4.864
4.860
4.856
4.70
50.0
Double doublet
Double triplet
PPM
60.0
4.504
4.502
4.500
4.498
4.496
4.494
4.492
4.490
4.488
PPM
4.486
3.906
3.904
3.902
3.900
3.898
3.896
3.894
3.892
3.890
3.888
4.484
4.852
4.60
4.50
4.40
4.30
4.20
4.10
4.00
3.90
3.80
3.886
Spectroscopy “Crib Sheet” for 2.222 – Introductory Organic Chemistry II
1
H NMR – Typical Chemical Shift Ranges
Type of Proton
C
Chemical Shift (δ)
CH3
Type of Proton
0.7 – 1.3
C
C
Chemical Shift (δ)
2.5 – 3.1
H
O
C
CH2
C
1.2 – 1.4
9.5 – 10.0
H
C
C
O
C
10.0 – 12.0
(solvent dependent)
1.4 – 1.7
H
OH
C
C
H
1.5 – 2.5
C
OH
1.0 – 6.0
(solvent dependent)
O
H
Aryl
C
H
2.1 – 2.6
O
C
H
3.3 – 4.0
2.2 – 2.7
Cl
C
H
3.0 – 4.0
4.5 – 6.5
Br
C
H
2.5 – 4.0
6.0 – 9.0
I
H
Aryl
H
Aromatic,
heteroaromatic
RCO2H
Y = O, NR, S
11
2.0 – 4.0
H
X–C–H
X = O, N, S, halide
Y
10
R3C–H
Aliphatic, alicyclic
Y
Y = O, NR, S
H
H
H
12
C
9
8
7
6
5
4
3
2
←δ
“Low Field”
1
0
“High Field”
13
C NMR – Typical Chemical Shift Ranges
CHx-Y
Y = O, N
Alkene
Ketone,
Aldehyde
Carbox. Acid
Aryl
CR3-CH2-CR3
CHx-C=O
CH3-CR3
Ester
Amide
220
200
180
160
140
120
100
80
60
40
20
←δ
IR – Typical Functional Group Absorption Bands
Group
C–H
C=C–H
C=C
C≡C–H
R–C≡C–R′
Aryl–H
Aryl C=C
Frequency
(cm-1)
2960 – 2850
3100 – 3020
1680 – 1620
3350 – 3300
2260 – 2100
3030 – 3000
1600, 1500
Intensity
Medium
Medium
Medium
Strong
Medium (R ≠ R′)
Medium
Strong
Group
RO–H
C–O
C=O
R2N–H
C–N
C≡N
RNO2
Frequency
(cm-1)
3650 – 3400
1150 – 1050
1780 – 1640
3500 – 3300
1230, 1030
2260 – 2210
1540
Intensity
Strong, broad
Strong
Strong
Medium, broad
Medium
Medium
Strong
0
ANSWER KEY
Page 1 of 15
Chemistry 2.222 Organic Chemistry II: Reactivity and Synthesis
FINAL EXAM – Winter 2003
Paper Number 546
Thursday April 24, 2003 6:00 – 9:00 pm
Frank Kennedy Brown Gym
Students are permitted to bring into the exam room ONE SHEET of 8½ x 11 paper, with any
HANDWRITTEN notes they wish (both sides). Molecular model kits are also permitted but no other
aids may be used.
The exam is in two parts. Answers to PART I are to be entered in the indicated spaces on the exam
paper itself.
An answer to ONE of the “Challenge Questions” in PART II may be written in the exam booklet for
EXTRA CREDIT.
PART I:
Question 1
(20 Marks)
Question 2
(10 Marks)
Question 3
(15 Marks)
Question 4
(15 Marks)
Question 5
(10 Marks)
SUB-TOTAL:
(70 Marks)
PART II (EXTRA CREDIT)
(7 Marks)
TOTAL:
In question 1, 3 questions were drawn from each of chapters 10-14. These
were divided into "predict the product", "give reagents" and "provide starting
compound" variations in an arbitrary manner.
Page 2 of 15
PART I: DO ALL QUESTIONS – There is choice in question 1 only.
1. (20 MARKS) Provide the missing product, starting compound or reagent/solvent/conditions
to correctly complete TEN of the following reactions. All reactions do in fact lead to
products. Clearly indicate which TEN responses you want marked!
Review Problem
11.8c, page 584.
You need a meta
director, but alkyl
groups are
o,p-directors.
O
1) Br2, AlBr3
CH2Cl2
2) Sn, HCl
Br
(a)
H
CH2I2, Zn, CH2Cl2
H
(I-CH2-Zn-I)
Simmons-Smith
reaction, see
Chapter 10,
page 515
H
H
(b)
O
Formation and
alkylation of
kinetic enolate,
see Chapter 13,
page 668.
O
LDA, THF, -78 oC
then
Br
(c)
O
Baeyer -Villiger
oxidation. This
one is shown in
Chapter 14, page
721.
O
O
(d)
Mercury II
catalyzed
hydration of a
terminal alkyne,
see Chapter 10,
page 503.
mCPBA, CH2Cl2
or any other specific peroxyacid reagent
Hg(OAc)2 (cat.)
H2SO4 (aq.)
(e)
O
Page 3 of 15
Electrophilic
bromination of the
more-activated
ring. This is
Supplementary
Problem 11.18e,
page 586.
O
Br2, AlBr3
O
(f)
Br
O
O
CH2Cl2
ortho isomer is also acceptable, although this would be the minor product
O
Amides are
reduced to amines
by LiAlH4. See
Chapter 12, page
601.
NH
LiAlH4, THF
NH
(H3O+ workup)
(g)
Dieckmann
condensation of
an unsymmetrical
diester. This
example is on
page 682 (Chapter
13).
O
NaOMe
MeOH
O
O
O
OMe
OMe
MeO
(h)
2 PhMgBr, Ether
w/u with aq. NH4Cl
Grignard reagents
add twice to esters
to make tertiary
alcohols. See
Chapter 12, page
637.
O
O
OMe
reagent, solvent, workup
O
O
OH
O
(i)
Br
An SN1 reaction
with rearrangement
of the intermediate
cation. See
Chapter 10, page
498.
HBr, CH2Cl2
MeO
(j)
OH
MeO
Page 4 of 15
(CH3CH2)2CuLi
THF, -78 oC
(acid w/u)
Conjugate addition
of a soft carbon
nucleophile. This is
essentially the same
as the example in
Chapter 13, page 678.
(k)
O
O
You could also use CH3CH2MgBr/CuI (cat.)/Ether/Acid workup
O
Amide formation.
See Chapter 12,
page 627 as well
as page 18-3.
Cl
or
O
O
C6H5NH2, Ether
N
O
H
O
(l)
This is the example
of Benzilic Acid
Rearrangement
from the text,
Chapter 14, page 707
NaOH
H2O
O
O
OH
COOH
(m)
Diazotization
followed by acid
hydrolysis forms
phenols. See
Chapter 11, page
556.
NH2
1) NaNO2,
HCl (aq.)
OH
2) H2SO4 (aq.)
heat
(n)
The Ester Enolate
Claisen
rearrangement.
This one differs by
one methyl group
from the example
(o)
in Chapter 14,
page 722, and
Exercise 14.17b.
LDA, THF
-78 oC
O
O
then heat
(H3O+ workup)
O
OH
Page 5 of 15
2. (10 MARKS) Propose a synthetic route to prepare 5-methylheptane-4-one (A) starting from
ethyl acetoacetate (B). You may use any additional organic starting materials having three
or fewer carbons, as well as any reagents or solvents you require. This transformation can
be accomplished in fewer than 5 steps. Remember that although a retrosynthetic
analysis may be useful to you in solving this problem, the answer I want is the
“forward synthesis” complete with reagents and products.
O
Make
starting from
OEt
B
A
O
O
O
O
NaOEt
OEt
O
EtOH
CH3CH2Br
O
NaOEt
OEt
O
OEt
EtOH
CH3Br
NaOEt
EtOH
CH3CH2Br
O
1) NaOH (aq)
2) H3O+
O
or
H3O+, heat
( and CO2 and EtOH)
( The Acetoacetic Ester synthesis of ketones)
This is Review Problem 13.6c.
O
or
LDA, THF -78 oC
CH3CH2Br
O
OEt
Page 6 of 15
3. (7 MARKS) Provide a detailed stepwise mechanism for the following transformation.
O
O
NaOCH3
CH3OH
O
O
O
+
O
O
OH
-O
O
O
- OMe
O
-O
O
O
O
OH
OH
O
O
O
O
-
+
OMe
OMe
O
O
-
O
O
OMe
O
OMe
O
O
OH
O
MeO H
OMe
OH
- OMe
Groutas problem #185, with the base explicitly listed as methoxide. The
original reagents were K2CO3/MeOH, which form methoxide in situ. The
mechanism is essentially base-promoted transesterification (Chapter 12,
page 624-625) followed by collapse of a hydrate anion to release
acetone (Chapter 12, pages 604-607), protonation of an enolate (Chapter
13, page 662), and another transesterification to make the lactone ring.
Page 7 of 15
(b) (8 MARKS) The Wieland-Miescher Ketone is an important starting material in steroid
synthesis. It can be prepared by the Robinson Annulation strategy shown here. Provide
a detailed stepwise mechanism for this reaction.
O
CH3
O
O
O
+
NaOH
CH3
H2O, 95 oC
O
O
H3C
O
-OH
H
O
-
O
CH3
O
O
H2O
O
O
-
O
O
O
HO
O
O
CH3
O
O
CH3
OH
H
HO -
O
-
H
O
CH3
O
H2O
O
O
O
-
The Robinson Annulation is discussed in Chapter 13, pages 679-680. This
problem is a slight modification of Supplementary Problem 13.13, as well
as being essentially the same as the example shown on page 679.
-
Page 8 of 15
4. (a) (5 MARKS) Compound A below is suggested to be a trace impurity formed when 4,4'di-(tert-butyl)biphenyl is prepared from biphenyl and t-butylchloride in CH2Cl2, using FeCl3
as a catalyst. Explain how A could be formed under these conditions.
+
major product
FeCl3
Cl
H
CH2Cl2
+
H
C
A - trace impurity
This is similar to
question 2 in the
Lab Manual
questions relating
to 4,4'-di-tertbutyl
biphenyl.
Compound A would
only be present in
trace amounts, if at
all, because FeCl3
is a relatively weak
Lewis Acid.
See Chapter 11,
pages 547-549.
The CH2 grouping comes from the solvent, CH2Cl2. Dichloromethane is not
very prone to reaction of this kind, but it may undergo reaction with Lewis
acids to a slight extent, especially if the reaction is left for an extended
period of time. This is probably what happened here.
H
H
C
Cl
FeCl3
H
δ
H
Cl
Cl
Cl
C
δ
FeCl3
Cl
CH2 FeCl4
Cl
CH2
Cl
H H2C
H2C
and then do the
Friedel-Crafts alkylation over
again with the other phenyl
group to form the product
Cl
(b) (5 MARKS) The aldehydes A and B below both undergo base-catalyzed condensation
reactions with acetone (C). The reaction of B under these conditions is much faster than
that of A. Briefly explain why this is so, taking into account electronic effects.
This is question #1
from the Lab Manual
problems related to the
Claisen-Schmidt
reaction.
The effect of
electron-withdrawing
and donating groups
on reactivity is first
demonstrated in
Section 11.4, with
regard to Electrophilic
Aromatic Substitution.
The concept is
extended to carbonyls
in Chapter 12, pages
602-603 and 620.
O
H3CO
O
O
O2N
H
A
H3C
H
B
CH3
C
These condensation reactions are Aldol processes, in which the enolate
of acetone attacks the aldehyde. The rates of the two processes will
depend on the degree of electrophilicity of the two aldehydes. Since B is
substituted with a strong electron-withdrawing group, whereas A carries
a strong electron-donating group, the carbonyl group of B will have a
greater degree of partial positive charge. Thus, reaction of B will be
faster.
O
O2N
H
These resonance forms
show that the nitro group
can directly withdraw
electron density from the
carbonyl.
O
O2N
H
O
O2N
O
O2N
H
H
O
O2N
H
Page 9 of 15
(c) (5 MARKS) Draw and label a diagram showing a reflux apparatus set-up. Give 2
advantages of using reflux conditions in a chemical reaction.
Top of condenser is OPEN!
Condenser
Water hose OUT
Water hose IN
Clamp attached to stand
Flask
Liquid
Boiling chips
Heat source
Advantages:
1) Reflux apparatus keeps solvent (and possibly also reagents and
products) from boiling away by recondensing it (them) and
returning the condensate to the flask.
2) Refluxing maintains a steady reaction temperature, which is the
boiling temperature of the solvent used.
Page 10 of 15
5. (10 MARKS) The IR, 13C NMR and 1H NMR spectra of an organic compound having the
molecular formula C7H15N are shown below. NB: In the 1H NMR spectrum, m=multiplet,
d=doublet, tr=triplet, q=quartet.
4000
3000
2000
m
1500
m
1000
d
q
500
tr
Draw the structure of the compound in the box below.
Formula shows 1 unsaturation.
IR shows NO NH! Bands at
2700-3100 are C-H stretches.
5 kinds of carbons, 2 of which are
clearly alkene.
5 kinds of H, ratio 1:2:2:4:6.
Quartet (4H) plus triplet (6H) can
ONLY mean two identical ethyl
groups!
H3C
N
H3C
CH2
Proton multiplets 5-6 ppm
can only be alkene. There
are three of them so it is a
monosubstituted alkene.
Remaining signal at ~3.1
ppm (2H, doublet) must be
CH2 with 1 neighbor, next
to alkene and N atom.
Page 11 of 15
PART II: CHALLENGE PROBLEM (EXTRA CREDIT).
DO ONE PROBLEM ONLY. Part II is worth up to 7 additional marks.
Write your answer in the exam booklet provided. Be sure your name and student number are on
the booklet. Insert the booklet into Part I when you hand it in.
A. SYNTHESIS
There is something seriously wrong with the following proposed synthesis of the drug
terfenadine. Identify the problem, and propose specific steps needed to make this route
workable.
H
N
Mg
Br
O
N
Ether
This is
Supplementary
Problem 15.12,
page 783.
OH
HO
HO
terfenadine
Answer: You cannot form a Grignard reagent from a halide that also contains an alcohol group.
The solution to the problem is to protect the alcohol with a group that is unaffected by
Grignard formation or reaction. This group must be removed from the final product.
Protection/deprotection is not necessarily straightforward, however. A benzyl ether could be
easily installed in the starting material, which would permit Grignard formation. This answer
(with appropriate reagents specified) would be acceptable for 6 out of 7 marks. (NB: making an
ester would NOT be suitable, because you can’t make a Grignard that contains a carbonyl group
either).
Br
N
N
Br
NaH
THF
HO
Br
O
Benzyl ether protects OH, will not
interfere with Grignard formation
OH group incompatible with
Grignard formation
However, in fact the removal of the benzyl group from the final product would likely damage
the product. Hydrogenation will cleave benzylic C-O bonds, but there are actually two such
groups in this molecule. We do not want to remove the existing hydroxyl group!
H2 (g)
Pd cat.
N
OH
N
MeOH
O
H H
HO
You can also cleave benzylic ethers using very strong acids, particularly HBr. However, this
would certainly lead to elimination reactions.
HBr
N
N
OH
O
Either or both of the hydroxyl
groups would eliminate
The 7th bonus mark would be given for commenting on the problem of deprotection. I don’t
require you to have a solution to this problem.
You could get around the difficulty by completely re-designing the synthesis. That is actually
not what the problem asked for, but I will evaluate any proposed alternates on their merits.
Page 12 of 15
B. MECHANISM
When carvone (A) is boiled in acidic water, it is converted to carvacrol (B). Give a mechanism
to explain this transformation.
O
OH
H3O+
100 oC
A
B
There are actually a couple of ways you could have written this process. It requires only
a series of protonation/deprotonation steps to move the double bonds around.
O
( + H+ )
O
H
H
+
+
H3O+
O
O ( - H+ )
(1,2-shift)
H3O+
H
( + H+ )
O
H
+
O
( - H+ )
H
H
H
H
The 1,2-shift step could also be written as an elimination/re-protonation sequence.
O
H
+
O
O
+
H
This is Groutas #126. This was definitely the easiest of the "challenge problems".
Page 13 of 15
C. SPECTROSCOPY
The unusual compound diketene has the molecular formula C4H4O2. Its 1H and 13C NMR
spectra are shown on the next page. Heating diketene in the presence of cyclopentadiene leads
to the formation of bicyclo[2.2.1]hept-5-ene-2-one. What is the structure of diketene?
Diketene
heat
+
O
O
H
Diketene has the structure:
O
H
H
H
The two vinylic hydrogens are double triplets. This is because each is
distinct, splitting the other into a doublet, which is then split by the two
other hydrogens (which are both equivalent). The CH2 group is a
double doublet because the signals are split by the two different vinylic
hydrogens.
The 13C NMR is harder to interpret. The signal at 165 ppm is the carbonyl
carbon, and the one at 42 ppm is the CH2 group. The alkene portion of the
structure is a bit strange. The carbon in the ring is at 147 ppm, but the terminal
carbon is upfield at 87 ppm. This spectrum really just told you that all 4 carbons
were different, and that one was probably a carbonyl
As its name implies, diketene is a dimer of the molecule ketene, and
they are interconverted by an electrocyclic reaction.
O
H
O
O
H
H
H
C
2
H
H
It is the monomer that undergoes a Diels-Alder reaction with cyclopentadiene.
You could see that if you worked backwards from the product in the reaction
shown.
This was the hardest of the three "challenge problems". The formula for
diketene indicates 3 degrees of unsaturation, which is a lot for such a small
structure. We have not said much about "double doublets" etc., but they
were discussed in Chapter 4, page 180 and they have occasionally popped
up in exercises.
Page 14 of 15
Diketene 13C NMR
Four signals, δ 165.18, 147.66,
87.06 and 42.40 ppm.
PPM
150.0
140.0
130.0
120.0
110.0
100.0
90.0
80.0
70.0
Diketene 1H NMR
Three signals, δ 4.877,
4.494 and 3.897 ppm.
Double triplet
PPM
PPM
4.900
4.896
4.892
4.90
4.888
4.884
4.880
4.876
4.872
4.80
4.868
4.864
4.860
4.856
4.70
50.0
Double doublet
Double triplet
PPM
60.0
4.504
4.502
4.500
4.498
4.496
4.494
4.492
4.490
4.488
PPM
4.486
3.906
3.904
3.902
3.900
3.898
3.896
3.894
3.892
3.890
3.888
4.484
4.852
4.60
4.50
4.40
4.30
4.20
4.10
4.00
3.90
3.80
3.886
Spectroscopy “Crib Sheet” for 2.222 – Introductory Organic Chemistry II
1
H NMR – Typical Chemical Shift Ranges
Type of Proton
C
Chemical Shift (δ)
CH3
Type of Proton
0.7 – 1.3
C
C
Chemical Shift (δ)
2.5 – 3.1
H
O
C
CH2
C
9.5 – 10.0
1.2 – 1.4
H
C
C
O
C
10.0 – 12.0
(solvent dependent)
1.4 – 1.7
H
OH
C
C
H
1.5 – 2.5
C
OH
1.0 – 6.0
(solvent dependent)
O
H
Aryl
C
H
2.1 – 2.6
O
C
H
3.3 – 4.0
2.2 – 2.7
Cl
C
H
3.0 – 4.0
4.5 – 6.5
Br
C
H
2.5 – 4.0
6.0 – 9.0
I
H
Aryl
H
Aromatic,
heteroaromatic
RCO2H
Y = O, NR, S
11
Y
10
Y = O, NR, S
H
H
9
8
7
6
5
4
3
2
←δ
1
0
“High Field”
C NMR – Typical Chemical Shift Ranges
CHx-Y
Y = O, N
Alkene
Ketone,
Aldehyde
Carbox. Acid
Aryl
CR3-CH2-CR3
CHx-C=O
CH3-CR3
Ester
Amide
220
R3C–H
Aliphatic, alicyclic
Y
“Low Field”
13
2.0 – 4.0
H
X–C–H
X = O, N, S, halide
H
12
C
200
180
160
140
120
100
80
60
40
20
←δ
IR – Typical Functional Group Absorption Bands
Group
C–H
C=C–H
C=C
C≡C–H
R–C≡C–R′
Aryl–H
Aryl C=C
Frequency
(cm-1)
2960 – 2850
3100 – 3020
1680 – 1620
3350 – 3300
2260 – 2100
3030 – 3000
1600, 1500
Intensity
Medium
Medium
Medium
Strong
Medium (R ≠ R′)
Medium
Strong
Group
RO–H
C–O
C=O
R2N–H
C–N
C≡N
RNO2
Frequency
(cm-1)
3650 – 3400
1150 – 1050
1780 – 1640
3500 – 3300
1230, 1030
2260 – 2210
1540
Intensity
Strong, broad
Strong
Strong
Medium, broad
Medium
Medium
Strong
0