Name: Student No: Page 1 of 15 - University of Manitoba · CHEM 2220 Final Exam 2014R Page 9 of 15 (e) During World War I, mustard gas was developed as a weapon. Mustard gas (bis(2-chloroethyl)sulfide)

Name: Student No: Page 1 of 15

CHEM 2220 Organic Chemistry II: Reactivity and Synthesis Prof. P.G. Hultin, Dr. H. Luong

FINAL EXAM – Winter Session 2014R

Tuesday April 22, 2013 6:00 pm – 9:00 pm Frank Kennedy Gold 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 and calculators (no text or graphics memory!) are

permitted but no other aids may be used.

Question 1 – Reactions and Products (30 Marks)

Question 2 – Synthesis (10 Marks)

Question 3 – Mechanism (12 Marks)

Question 4 – Mechanism grab-bag (32 Marks)

Question 5 – Laboratory (10 Marks)

Question 6 – Spectroscopy (6 Marks)

TOTAL: (100 Marks)

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CHEM 2220 Final Exam 2014R Page 2 of 15

1. (30 MARKS) Reactions and Products. Supply the missing molecular structure or reagent/solvent/reaction

conditions to correctly complete the following reactions. Show stereochemistry when necessary. Simple

aqueous acid or base workups can be assumed but any “special” workup conditions should be specified.

(a) (2 Marks)

(b) (2 Marks)

(c) (3 Marks)

(d) (3 Marks)

(e) (2 Marks)

(f) (2 Marks)

(g) (2 Marks)


(h) (2 Marks)

(i) (2 Marks)

(j) (2 Marks)

(k) (4 Marks)

(l) (4 Marks)


2. (10 MARKS) Propose a synthetic route (sequence of reactions) to prepare 3-phenylheptanoic acid

starting with diethyl malonate and benzaldehyde, plus any other reagents, solvents, or catalysts necessary.

This can be accomplished in 3 or 4 steps using reactions covered in class. Be sure to write out each reaction

in your route showing the starting compound, product and specific reaction conditions.


3. (12 MARKS TOTAL) In appropriate molecules, a series of Michael reactions can occur in sequence during a

single reaction to form quite complicated carbon skeletons. One such “cascade” reaction was reported by

Fukumoto and his co-workers in 1994.

(a) (2 Marks) Draw the structure of the enolate formed when the starting material was treated with LDA.

Show both resonance structures and include the positive counter ion.

(b) (3 Marks) In the structures below, certain carbons in the starting material have been labeled.

i) Indicate which carbons in the product correspond to these designated atoms, by applying the same

labels to the structure provided.

ii) Label the carbon in the product that corresponds to the nucleophilic site of the enolate you drew in part

(a).

iii) Label the bonds in the product that were formed during the reaction.


(c) (7 Marks) Based on your analysis in parts (a) and (b), write a detailed stepwise mechanism for this

reaction.


4. (32 MARKS TOTAL) Mechanism grab-bag!

(a) (4 Marks) Reaction of α-chloroketones with sodium borohydride produces epoxides. Write out a

mechanism for the following reaction. You do not have to show the fate of the boron during workup.

(b) (4 Marks) Draw a plausible mechanism for the following transformation.


(c) (4 Marks) Write a mechanism for the following variation on a reaction we discussed in class.

(d) (4 Marks) Alkaline hydrogen peroxide can promote a reaction similar to the Baeyer-Villiger oxidation.

Write a mechanism for the following reaction that accounts for the stereochemistry of the product.


(e) During World War I, mustard gas was developed as a weapon. Mustard gas (bis(2-chloroethyl)sulfide)

hydrolyzes to form HCl extremely rapidly in biological fluids, in contrast to ordinary alkyl chlorides that are

essentially inert under these conditions. Thus, it incapacitated soldiers who breathed it or who got it into

their eyes by causing acid burns.

i) (4 Marks) Write a mechanism and briefly explain why mustard gas undergoes hydrolysis so much

faster than does an analogous primary chloride.


ii) (6 Marks) Those soldiers who survived exposure to mustard gas were often horribly maimed, but they

also tended to have a high incidence of cancer later in life. It was discovered that this was because of

alkylation and cross-linking of DNA by the mustard agent. Draw a plausible structure for the mustard

gas cross-link between the DNA bases adenosine and thymidine, and based on your mechanism for

hydrolysis write a mechanism for the formation of this cross-link.

(f) (6 Marks) Like alkenes, simple cyclopropanes do not react with nucleophilic reagents. However,

certain types of substituents on the cyclopropyl ring permit nucleophilic ring opening to occur. Based on the

examples shown, draw a generalized mechanism and briefly explain this observation. (N.B. phenyl selenide

is a reasonably strong nucleophile.)


5. Lab Questions (10 MARKS Total)

After a long day of filling potholes in Winnipeg, Filmia Crater (a fictional city worker who lives next door to you)

came home with a terrible headache. As the kind neighbour that you are, you check your medicine cabinet and

there’s no acetaminophen (para-acetamidophenol, Compound C) to be found. However, in the organic

laboratory you have access to all the necessary materials to synthesize it. Assume you want to prepare about

15 g of Compound C in the most pure form possible (it’s a bad headache!).

To start off, you have a protected aminophenol (Compound A) to ensure that there’s reactivity only at the

amine. To remove the protecting group will only require heating with acid for a period of time.

In addition to Compound A, you are to use only the reagents and chemicals typically available in the CHEM

2220 laboratories.

Data Table: Most of the data are fictitious, and should NOT be used for real experimental purposes!

MW (g/mol) BP/MP (oC) Solubility (g/L)

Compound A 165 220/130 H2O: 1 Ethyl acetate: 50 Ethanol: 80

Compound B 207 150/20 H2O: 1 Ethyl acetate: 50 Ethanol: 80

Compound C 151 388/168 H2O: 0.1 Ethyl acetate: 20 Ethanol: 10

Acetic Anhydride 102 140/−73 H2O: 100 Ethyl acetate: 100 Ethanol: 100

In planning your synthesis you will need to make the following assumptions:

1. After the first step, there will be residual starting material (limiting reagent AND those in excess).

2. You need to isolate Compound B free of contaminants before proceeding to remove the protecting

group.

3. After the second step, there will be residual Compound B in the crude product mixture.

4. Since the reactions are incomplete, assume the isolated yield after each step will be 50%

There is some math involved but we have chosen numbers that can easily be approximated.


a) (7 Marks) Write out the procedure you will use to make acetaminophen.

b) (2 Marks) Product Analysis

You spotted a sample of the reaction

mixture in Step 1 on a TLC plate

towards the “end” of the reaction period,

using Compound A as a reference.

Draw what this TLC plate would look like

after you eluted it with ethyl acetate:

hexanes (1:1). Make sure to label

aspects of your plate and spots.

c) (1 Mark) In many of the laboratory experiments, you used brine to wash an organic layer. Why is

washing with brine necessary – would washing with plain distilled water have the same effect?


6. (6 MARKS Total) Spectroscopy. The molecular formula of an unknown organic compound is C7H14O2. The

IR, 13

C NMR and 1H NMR spectra of this compound are shown on the next page. Answer the following

questions about this compound.

(a) (0.5 MARK) What is the unsaturation number for this compound?

(b) (1.5 MARK) What functional group(s) does this compound contain? Indicate the specific evidence for

your conclusion.

(c) (1 MARK) What can you conclude from the number of 13

C NMR signals?

(d) (1 MARK) What can you conclude from the integral and splitting of the 1H NMR signal at 2.5 ppm?

(e) (2 MARKS) Draw the structure of this compound in the box below.

Structure for C7H14O2


Spectra for Question 6

IR

1H NMR

13C NMR

NB: all signals are single lines.

2H

tr

1H

heptet

2H

m

6H

d

3H

tr

Spectroscopy “Crib Sheet” for CHEM 2220 – Introductory Organic Chemistry II

1H NMR – Typical Chemical Shift Ranges

Type of Proton Chemical Shift (δ) Type of Proton Chemical Shift (δ)

C CH3 0.7 – 1.3 C C H 2.5 – 3.1

C CH2 C 1.2 – 1.4

O

H

9.5 – 10.0

C H

C

C

C

1.4 – 1.7

O

OH

10.0 – 12.0

(solvent dependent)

C H

1.5 – 2.5 C OH

1.0 – 6.0

(solvent dependent)

O

H

2.1 – 2.6 CO H

3.3 – 4.0

Aryl C H 2.2 – 2.7 CCl H

3.0 – 4.0

H

4.5 – 6.5 CBr H

2.5 – 4.0

Aryl H 6.0 – 9.0 CI H

2.0 – 4.0

13C NMR – Typical Chemical Shift Ranges

IR – Typical Functional Group Absorption Bands

Group Frequency

(cm-1

) Intensity Group

Frequency (cm

-1)

Intensity

C–H 2960 – 2850 Medium RO–H 3650 – 3400 Strong, broad

C=C–H 3100 – 3020 Medium C–O 1150 – 1050 Strong

C=C 1680 – 1620 Medium C=O 1780 – 1640 Strong

C≡C–H 3350 – 3300 Strong R2N–H 3500 – 3300 Medium, broad

R–C≡C–R′ 2260 – 2100 Medium (R ≠ R′) C–N 1230, 1030 Medium

Aryl–H 3030 – 3000 Medium C≡N 2260 – 2210 Medium

Aryl C=C 1600, 1500 Strong RNO2 1540 Strong

12 11 10 9 8 7 6 5 4 3 2 1 0

R3C–H

Aliphatic, alicyclic X–C–H

X = O, N, S, halide

Y

H H

Aromatic,

heteroaromatic

Y

H

RCO2H

Y = O, NR, S Y = O, NR, S

“Low Field” “High Field”

220 200 180 160 140 120 100 80 60 40 20 0

CH3-CR3

CHx-C=O

CR3-CH2-CR3

CHx-Y

Y = O, N Alkene

Aryl

Amide

Ester

Ketone, Aldehyde

Acid RCN

RCCR

ANSWER KEY Page 1 of 15

CHEM 2220 Organic Chemistry II: Reactivity and Synthesis Prof. P.G. Hultin, Dr. H. Luong

FINAL EXAM – Winter Session 2014R

Tuesday April 22, 2013 6:00 pm – 9:00 pm Frank Kennedy Gold 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 and calculators (no text or graphics memory!) are

permitted but no other aids may be used.

Question 1 – Reactions and Products (30 Marks)

Question 2 – Synthesis (10 Marks)

Question 3 – Mechanism (12 Marks)

Question 4 – Mechanism grab-bag (32 Marks)

Question 5 – Laboratory (10 Marks)

Question 6 – Spectroscopy (6 Marks)

TOTAL: (100 Marks)

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CHEM 2220 Final Exam 2014R ANSWERS Page 2 of 15

1. (30 MARKS) Reactions and Products. Supply the missing molecular structure or reagent/solvent/reaction

conditions to correctly complete the following reactions. Show stereochemistry when necessary. Simple

aqueous acid or base workups can be assumed but any “special” workup conditions should be specified.

(a) (2 Marks)

(b) (2 Marks)

(c) (3 Marks)

(d) (3 Marks)

(e) (2 Marks)

(f) (2 Marks)

See Klein section 9.11 and

Skill Builder 9.8.

See Klein pp 358-360.

See Klein section 23.11,

and SkillBuilder 23.6

The Strecker Synthesis was discussed in class, see also Klein pp 1194-1195.

Baeyer-Villiger oxidation,

Klein section 20.11.

Alkylation of a kinetic

enolate: Klein section 22.5.

Similar to Klein problem

22.86 d.


(g) (2 Marks)

(h) (2 Marks)

(i) (2 Marks)

(j) (2 Marks)

(k) (4 Marks)

(l) (4 Marks)

Crossed aldol addition under

irreversible conditions. Discussed in

class, also see Klein p 1052.

Wittig reaction. Klein pp 950-953.

Diels-Alder reaction. This

is Klein problem 17.44 a.

Acid-catalyzed hydrolysis of an acetal and

an ester forms a β-ketoacid, which

decarboxylates. See Klein sections 20.5

and 21.11, and also p 1063.

1) Hydroboration is the

only method we know

for anti-Markovnikov

hydration of alkenes.

2) Chromic acid oxidation

of primary alcohols

forms carboxylic acids.

3) Formation of acid

chloride using thionyl

chloride.

4) Amide formation: excess

amine plus acid

chloride.

Spectroscopic data indicate 3 kinds of aromatic hydrogens and 2 alkene hydrogens.

Also note 2 CH3 groups, and 2H doublet at 2.21 for CH2 next to alkene.


2. (10 MARKS) Propose a synthetic route (sequence of reactions) to prepare 3-phenylheptanoic acid

starting with diethyl malonate and benzaldehyde, plus any other reagents, solvents, or catalysts necessary.

This can be accomplished in 3 or 4 steps using reactions covered in class. Be sure to write out each reaction

in your route showing the starting compound, product and specific reaction conditions.

There are numerous ways in which the required synthesis could be accomplished. Any answer was judged on its

own merits.


3. (12 MARKS TOTAL) In appropriate molecules, a series of Michael reactions can occur in sequence during a

single reaction to form quite complicated carbon skeletons. One such “cascade” reaction was reported by

Fukumoto and his co-workers in 1994.

(a) (2 Marks) Draw the structure of the enolate formed when the starting material was treated with LDA.

Show both resonance structures and include the positive counter ion.

(b) (3 Marks) In the structures below, certain carbons in the starting material have been labeled.

i) Indicate which carbons in the product correspond to these designated atoms, by applying the same

labels to the structure provided.

ii) Label the carbon in the product that corresponds to the nucleophilic site of the enolate you drew in part

(a).

iii) Label the bonds in the product that were formed during the reaction.

These initial questions guide you towards the correct mechanism. There is only one possible

enolate that can be formed so this should have been straightforward. Identifying the labeled

atoms in the product is important because this locates where structural changes must occur. Look

at the functional groups and where the labels are relative to them in the starting material. Then,

spot the functional groups in the product and they will tell you where the labels probably are.


(c) (7 Marks) Based on your analysis in parts (a) and (b), write a detailed stepwise mechanism for this

reaction.

As you can see, the mechanism is actually extremely simple once you know WHAT structural

changes have to be accomplished. From part b-iii of the question we know we must make C-C

bonds from the enolate α-position to A, and from the ester α-position to C. Since we have a

nucleophile in the enolate ready to go, making this bond first is reasonable. We notice that the

result of doing this is to make the enolate of our ester. This permits us to make the second bond to

carbon C, leading to yet another enolate. There is no other structural change needed, so we can

assume that this enolate remains until the acid workup when it is protonated.


4. (32 MARKS TOTAL) Mechanism grab-bag!

(a) (4 Marks) Reaction of α-chloroketones with sodium borohydride produces epoxides. Write out a

mechanism for the following reaction. You do not have to show the fate of the boron during workup.

(b) (4 Marks) Draw a plausible mechanism for the following transformation.

This is Klein problem 21.75. The important features of a “correct” mechanism are seen in Klein

Mechanism 20.5 (page 925): 1) protonation on O followed by opening to give O-stabilized tertiary cation; 2)

hemiacetal intermediate; 3) protonation on O and collapse to release diol plus acetone. Then, the rest of

the mechanism is standard acid-catalyzed esterification (see Klein SkillBuilder 21.1 page 989).

This is Klein problem 14.57, and it is similar to Groutas #51. Basically, you know that borohydride

reduction of ketones forms alkoxide anion intermediates. You also know that epoxides can be

formed by treating vicinal halohydrins with base, and that this proceeds via the alkoxide. So, this

reaction is simply the combination of the two pathways.


(c) (4 Marks) Write a mechanism for the following variation on a reaction we discussed in class.

(d) (4 Marks) Alkaline hydrogen peroxide can promote a reaction similar to the Baeyer-Villiger oxidation.

Write a mechanism for the following reaction that accounts for the stereochemistry of the product.

This is Klein problem 17.68. It is simply a Diels-Alder reaction between

the diene and the alkyne, followed by the reverse of a Diels-Alder

reaction to release CO2.

The mechanism of a Baeyer-Villiger reaction involving a peroxyacid like mCPBA is shown in Klein

page 954, and we covered it in class. We discussed the behaviour of hydrogen peroxide in the

presence of NaOH when we talked about hydroboration reactions (see Klein Mechanism 9.3 page

415), and we saw that oxyanions can attack ketones (Mechanism 20.3 page 924). The mechanism

here is directly derived from consideration of the Baeyer-Villiger mechanism and the oxidative workup

in hydroboration/oxidation.


(e) During World War I, mustard gas was developed as a weapon. Mustard gas (bis(2-chloroethyl)sulfide)

hydrolyzes to form HCl extremely rapidly in biological fluids, in contrast to ordinary alkyl chlorides that are

essentially inert under these conditions. Thus, it incapacitated soldiers who breathed it or who got it into

their eyes by causing acid burns.

i) (4 Marks) Write a mechanism and briefly explain why mustard gas undergoes hydrolysis so much

faster than does an analogous primary chloride.

We saw something similar in the mechanism of forming an epoxide from a halohydrin (Klein

page 648), and we know that sulfides RSR are good nucleophiles (Klein Fig. 8.25 and page

655).

Part marks were given for suggesting that the electron-withdrawing nature of sulfur increased

the reactivity of the primary halide towards SN2 displacement.


ii) (6 Marks) Those soldiers who survived exposure to mustard gas were often horribly maimed, but they

also tended to have a high incidence of cancer later in life. It was discovered that this was because of

alkylation and cross-linking of DNA by the mustard agent. Draw a plausible structure for the mustard

gas cross-link between the DNA bases adenosine and thymidine, and based on your mechanism for

hydrolysis write a mechanism for the formation of this cross-link.

(f) (6 Marks) Like alkenes, simple cyclopropanes do not react with nucleophilic reagents. However,

certain types of substituents on the cyclopropyl ring permit nucleophilic ring opening to occur. Based on the

examples shown, draw a generalized mechanism and briefly explain this observation. (N.B. phenyl selenide

is a reasonably strong nucleophile.)

Since mustard reacts with water by an SN2 mechanism, it makes sense that it would react in a similar way

with other nucleophiles. Note that reaction with ring nitrogens in adenosine would either make cationic

products or disrupt the aromaticity of the ring so these are much less likely.

This is a slightly modified version of Groutas #172.


5. Lab Questions (10 MARKS Total)

After a long day of filling potholes in Winnipeg, Filmia Crater (a fictional city worker who lives next door to you)

came home with a terrible headache. As the kind neighbour that you are, you check your medicine cabinet and

there’s no acetaminophen (para-acetamidophenol, Compound C) to be found. However, in the organic

laboratory you have access to all the necessary materials to synthesize it. Assume you want to prepare about

15 g of Compound C in the most pure form possible (it’s a bad headache!).

To start off, you have a protected aminophenol (Compound A) to ensure that there’s reactivity only at the

amine. To remove the protecting group will only require heating with acid for a period of time.

In addition to Compound A, you are to use only the reagents and chemicals typically available in the CHEM

2220 laboratories.

Data Table: Most of the data are fictitious, and should NOT be used for real experimental purposes!

MW (g/mol) BP/MP (oC) Solubility (g/L)

Compound A 165 220/130 H2O: 1 Ethyl acetate: 50 Ethanol: 80

Compound B 207 150/20 H2O: 1 Ethyl acetate: 50 Ethanol: 80

Compound C 151 388/168 H2O: 0.1 Ethyl acetate: 20 Ethanol: 10

Acetic Anhydride 102 140/−73 H2O: 100 Ethyl acetate: 100 Ethanol: 100

In planning your synthesis you will need to make the following assumptions:

1. After the first step, there will be residual starting material (limiting reagent AND those in excess).

2. You need to isolate Compound B free of contaminants before proceeding to remove the protecting

group.

3. After the second step, there will be residual Compound B in the crude product mixture.

4. Since the reactions are incomplete, assume the isolated yield after each step will be 50%

There is some math involved but we have chosen numbers that can easily be approximated.


a) (7 Marks) Write out the procedure you will use to make acetaminophen.

b) (2 Marks) Product Analysis

You spotted a sample of the reaction

mixture in Step 1 on a TLC plate

towards the “end” of the reaction period,

using Compound A as a reference.

Draw what this TLC plate would look like

after you eluted it with ethyl acetate:

hexanes (1:1). Make sure to label

aspects of your plate and spots.

c) (1 Mark) In many of the laboratory experiments, you used brine to wash an organic layer. Why is

washing with brine necessary – would washing with plain distilled water have the same effect?

Brine is used after washing with another aqueous solution. Its purpose is to

crudely remove water before drying the organic layer over drying agent.

Replacing the brine with only water would not have the same effect.

Dissolve about 66 g of Compound A in acetic anhydride (50 mL minimum) to

acetylate the amine group. This reaction should not require heating or a catalyst.

The reaction can be monitored by TLC.

At the “end” of the reaction you have a mixture of Compound A and B and residual

acetic anhydride and acetic acid. Add water to quench the anhydride. Compound B

is a liquid and insoluble in water and therefore you can also wash with a basic

solution to remove the acetic acid. Compound A is removed only by an acidic wash.

Drying Compound B would be chemically wasteful since the next stage we’re heating

with acid (aqueous) anyway.

As Compound C is produced, we expect that a solid precipitates out (since it is

insoluble in water, and the phenol group can’t be deprotonated under acidic

conditions to make it more water soluble). Since there is Compound B present (as a

liquid) at the end of the reaction, just perform a suction filtration to isolate Compound

C and then wash the crystals with cold ethanol to remove Compound B contaminant.

Compound C can be further purified by recrystallization with ethanol.


6. (6 MARKS Total) Spectroscopy. The molecular formula of an unknown organic compound is C7H14O2. The

IR, 13

C NMR and 1H NMR spectra of this compound are shown on the next page. Answer the following

questions about this compound.

(a) (0.5 MARK) What is the unsaturation number for this compound?

(b) (1.5 MARK) What functional group(s) does this compound contain? Indicate the specific evidence for

your conclusion.

(c) (1 MARK) What can you conclude from the number of 13

C NMR signals?

(d) (1 MARK) What can you conclude from the integral and splitting of the 1H NMR signal at 2.5 ppm?

(e) (2 MARKS) Draw the structure of this compound in the box below.

Structure for C7H14O2

Unsaturation = 1

Contains an ester. IR ~1740 cm-1

, no OH band; 13

C NMR

~178 ppm; 1H NMR 2H triplet ~ 4 ppm suggests OCH2

6 13

C NMR signals but 7 carbons in the formula: one

signal represents 2 symmetrical carbons.

A 1-proton “heptet” (7 lines) means this proton has 6 identical nearest

neighbour protons. Since we also see a 6-proton doublet at ~1.2 ppm

this suggests an isopropyl group CH(CH3)2.

1. Chemical shift of the 1-H

heptet ~2.5 ppm suggests it

is next to the carbonyl of the

ester.

2. 2-H triplet at ~4 ppm means

OCH2CH2.

3. 3-H triplet at ~0.9 ppm

means methyl with 2

neighbours – only possibility

is OCH2CH2CH3 Note

multiplet at ~1.6 ppm is the

middle CH2, split by 2H of

one kind and 3H of another.

4. Only one way to put these

pieces together consistently.


Spectra for Question 6

IR

1H NMR

13C NMR

NB: all signals are single lines.

2H

tr

1H

heptet

2H

m

6H

d

3H

tr

Spectroscopy “Crib Sheet” for CHEM 2220 – Introductory Organic Chemistry II

1H NMR – Typical Chemical Shift Ranges

Type of Proton Chemical Shift (δ) Type of Proton Chemical Shift (δ)

C CH3 0.7 – 1.3 C C H 2.5 – 3.1

C CH2 C 1.2 – 1.4

O

H

9.5 – 10.0

C H

C

C

C

1.4 – 1.7

O

OH

10.0 – 12.0

(solvent dependent)

C H

1.5 – 2.5 C OH

1.0 – 6.0

(solvent dependent)

O

H

2.1 – 2.6 CO H

3.3 – 4.0

Aryl C H 2.2 – 2.7 CCl H

3.0 – 4.0

H

4.5 – 6.5 CBr H

2.5 – 4.0

Aryl H 6.0 – 9.0 CI H

2.0 – 4.0

13C NMR – Typical Chemical Shift Ranges

IR – Typical Functional Group Absorption Bands

Group Frequency

(cm-1

) Intensity Group

Frequency (cm

-1)

Intensity

C–H 2960 – 2850 Medium RO–H 3650 – 3400 Strong, broad

C=C–H 3100 – 3020 Medium C–O 1150 – 1050 Strong

C=C 1680 – 1620 Medium C=O 1780 – 1640 Strong

C≡C–H 3350 – 3300 Strong R2N–H 3500 – 3300 Medium, broad

R–C≡C–R′ 2260 – 2100 Medium (R ≠ R′) C–N 1230, 1030 Medium

Aryl–H 3030 – 3000 Medium C≡N 2260 – 2210 Medium

Aryl C=C 1600, 1500 Strong RNO2 1540 Strong

12 11 10 9 8 7 6 5 4 3 2 1 0

R3C–H

Aliphatic, alicyclic X–C–H

X = O, N, S, halide

Y

H H

Aromatic,

heteroaromatic

Y

H

RCO2H

Y = O, NR, S Y = O, NR, S

“Low Field” “High Field”

220 200 180 160 140 120 100 80 60 40 20 0

CH3-CR3

CHx-C=O

CR3-CH2-CR3

CHx-Y

Y = O, N Alkene

Aryl

Amide

Ester

Ketone, Aldehyde

Acid RCN

RCCR

Documents

Name: Student No: Page 1 of 15 - University of Manitoba · CHEM 2220 Final Exam 2014R Page 9 of 15 (e) During World War I, mustard gas was developed as a weapon. Mustard gas (bis(2-chloroethyl)sulfide)