term test 1 answers - university of...

Name: __________________________ Student Number: _____________________

University of Manitoba - Department of Chemistry

CHEM 2220 - Introductory Organic Chemistry II - Term Test 1

Thursday, February 17, 2011; 7-9 PM

This is a 2-hour test, marked out of 50 points total. Part marks are available on all questions. Put all answers in the spaces provided. If more space is required you may use the backs of the exam pages but be sure to indicate that you have done so. A spectroscopic data sheet is attached at the end of the exam.

QUESTION MARKS

1. Mechanism (10 Marks)



4. Reactions and Products (20 Marks)

5. Spectra and Structures (6 Marks)

TOTAL (50 Marks)

CHEM 2220 Test #1 of 8 Feb 17, 2011

1. (10 MARKS) Write detailed, stepwise mechanisms that account for the products of the reaction of pseudoionone with sulfuric acid in water. Highlight which of the two, -ionone or -ionone, is the predominant product of this reaction, and briefly explain why.

CHEM 2220 Test #1 of 8 Feb 17, 2011

2. (10 MARKS TOTAL) a. (6 MARKS) When the diol A was treated with phosphoric acid in boiling toluene,

bicyclic product B was formed. Write a detailed stepwise mechanism to explain the formation of B under these conditions.

b. (4 MARKS) If the chemists had continued heating for a much longer time, it is conceivable that B might have reacted further to give alkene C. Briefly explain why this process is so much slower than the observed reaction leading to B.

C

CHEM 2220 Test #1 of 8 Feb 17, 2011

3. (4 MARKS) Provide a stepwise mechanism for the following reaction.

CHEM 2220 Test #1 of 8 Feb 17, 2011

4. (20 MARKS) Provide the necessary reagents/solvents or starting materials or major products to correctly complete the following reactions. Mechanisms are NOT required. Give product stereochemistry where appropriate.

a. (2 Marks)

b. (2 Marks)

c. (2 Marks)

d. (4 Marks)

e. (2 Marks)

OH

OH

1. BH3 . THF2. H2O2, NaOH

CH3O

CHO

O

1. O3, MeOH2. Zn, AcOH

hultin

Sticky Note

The Zn/AcOH step is an alternative form of reduction that was discussed in Fox and Whitesell. In Klein, dimethylsufide is used instead.

hultin

Sticky Note

This reaction is not covered in Klein, and will not be part of the curriculum from 2012 onwards.

CHEM 2220 Test #1 of 8 Feb 17, 2011

f. (2 Marks)

g. (2 Marks)

h. (4 Marks)

CHEM 2220 Test #1 of 8 Feb 17, 2011

5. (6 MARKS) The spectra of an unknown organic compound A having the formula C5H9NO are shown below. What was the structure of compound A?

27.39 27.13

IR

13C NMR

1H NMR

Compound A

hultin

Sticky Note

Our bad! Correct formula is C5H9N.

CHEM 2220 Test #1 of 8 Feb 17, 2011

Spectroscopy “Crib Sheet” for 2.222 – 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


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, broadR–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

HH

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 RCNRCCR

ANSWER KEY

University of Manitoba - Department of Chemistry

CHEM 2220 - Introductory Organic Chemistry II - Term Test 1

Thursday, February 17, 2011; 7-9 PM

This is a 2-hour test, marked out of 50 points total. Part marks are available on all questions. Put all answers in the spaces provided. If more space is required you may use the backs of the exam pages but be sure to indicate that you have done so. A spectroscopic data sheet is attached at the end of the exam.

QUESTION MARKS




4. Reactions and Products (20 Marks)

5. Spectra and Structures (6 Marks)

TOTAL (50 Marks)

CHEM 2220 Test #1 ANSWERS Page 2 of 8 Feb 17, 2011

1. (10 MARKS) Write detailed, stepwise mechanisms that account for the products of the reaction of pseudoionone with sulfuric acid in water. Highlight which of the two, -ionone or -ionone, is the predominant product of this reaction, and briefly explain why.

O

H3O

OO

HAHB

HB

loss of HA

O

-ionone

loss of HB

O

-ionone

H

This reaction is occurring under reversible, thermodynamic conditions. The major product will be the more-stable isomer. The major product will be -ionone because its longer conjugated system will give it a lower energy.


2. (10 MARKS TOTAL) a. (6 MARKS) When the diol A was treated with phosphoric acid in boiling toluene,

bicyclic product B was formed. Write a detailed stepwise mechanism to explain the formation of B under these conditions.

b. (4 MARKS) If the chemists had continued heating for a much longer time, it is conceivable that B might have reacted further to give alkene C. Briefly explain why this process is so much slower than the observed reaction leading to B.

B

OH

H

OH2

HH

H H

elimination

C

1,2-shift of H

H

H HH

elimination

C

The elimination to give C requires formation of a primary carbocation intermediate, which is not energetically favorable. Note that it could perhaps rearrange to a more-stable benzylic secondary cation, which could then lose a proton, but either way it would initially have to start out as a high-energy cation.

hultin

Sticky Note

It was not necessary to show a detailed mechanism in your answer - I just drew it out to make things clearer. Note that my answer is implicitly assuming that the reaction is proceeding by an E1-like mechanism. This is reasonable given the fact that the only anion around (H2PO4 -) is a very weak base.


3. (4 MARKS) Provide a stepwise mechanism for the following reaction.


4. (20 MARKS) Provide the necessary reagents/solvents or starting materials or major products to correctly complete the following reactions. Mechanisms are NOT required. Give product stereochemistry where appropriate.

a. (2 Marks)

b. (2 Marks)

c. (2 Marks)

d. (4 Marks)

e. (2 Marks)

OH

OH

1. BH3 . THF2. H2O2, NaOH

CH3O

CHO

O

1. O3, MeOH2. Zn, AcOH

OsO4 (cat.), K3Fe(CN)6, quinuclidine, H2O/acetone


f. (2 Marks)

g. (2 Marks)

h. (4 Marks)


5. (6 MARKS) The spectra of an unknown organic compound A having the formula C5H9N are shown below. What was the structure of compound A?

27.39 27.13

IR

13C NMR

1H NMR

Compound A

Sorry about the typo in the formula on the original paper. C5H9N gives 2 degrees of unsaturation. Prominent IR band at about 2250 cm-1 indicates NITRILE. 13C Signal at about 122 ppm is consistent with nitrile carbon. The nitrile uses up our 2 degrees of unsaturation – no further rings or double bonds. 1H NMR shows four groups of signals in a 1:2:3:3 ratio. Note 1-proton multiplet at ~2.5, 2-proton multiplet at ~1.6, 3-proton doublet at ~1.3 and 3-proton triplet at ~1.1.

The 3-proton signals are obviously a CH3 with one neighbour (the doublet) and a CH3 with two neighbours (the triplet).


Spectroscopy “Crib Sheet” for 2.222 – 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


C H

1.5 – 2.5 C OH


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, broadR–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

HH

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 RCNRCCR

term test 1 answers - university of...

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