old mid-term 2007

8/3/2019 Old Mid-Term 2007

http://slidepdf.com/reader/full/old-mid-term-2007 1/6

CHEM 2210 Mid-Term Exam, October 31st, 2007Exam Part I

of 6

CHEM 2210 Introductory Organic Chemistry I: Structure and FunctionMid-Term Examination

Wednesday October 31, 2007Dr. N. R. Hunter and Dr. J. L. Sorensen

Total time = 120 minutes for both Part I and Part II

Do all questions. Each question is worth an equal number of points (1 each). Please use apencil to enter your answers onto the machine scoring sheet. Please ensure that your name is clearly printed on the scoring sheet and that your student number is properlyencoded in the top right hand area. You may keep this sheet and should record your answers on this sheet. Good practice is to choose the best response by circling theappropriate letter on this sheet. Only at the end of the exam, when you are certain of your choice should you fill in the appropriate circle on the machine sheet.

Good Luck!

1. The primary forces of attraction between two molecules of methane is:

A. London ForcesB. Hydrogen BondingC. Ionic BondingD. Dipole Dipole InteractionsE. Covalent Bonding

2. The H–C–H bond angle in methane is:

A. 105°B. 120°C. 180°D. 109.5°E. 107.5°

3. The correct name for the hydrocarbon displalyed below is:

A. 3,4,5,8-tetramethylnonaneB. 3,5,8,-trimethyldecaneC. 2,ethyl-5,7-dimethylnonaneD. 2,7,diethyl,4-methyloctaneE. 2,neo-pentyl-5,methylheptane

8/3/2019 Old Mid-Term 2007



of 6

4. Which statement best describes the π∗ antibonding molecular orbital.

A. It is the result of the in-phase overlap of two atomic p orbitals.B. It is the result of the out of phase overlap of two atomic s orbitals.C. It can hold only one electron.

D. It is the result of the out of phase overlap of two atomic p orbitals.E. It is the result of the in-phase overlap of two atomic s orbitals.

5. Which of the following molecules would have the lowest heat of hydrogenation?

A B C D E

6. Which of the following molecules are not aromatic?

S

A B C D E

7. Which of the following is a pair of resonance structures?

HO OHO

A

O HO

B

OH O

C

O OCH3

D

O O

E

8. Which is the best Lewis Dot Structure to describe methyl azide (CH3N3)?

CH3 N N N CH3 N N NCH3 N N NCH3 N N NCH3 N N N

A B C D E

8/3/2019 Old Mid-Term 2007



of 6

9. Rank the following radicals in order of increasing stability(i.e. least stable < most stable).

H

H H

A B C D A. A < B < C < DB. D < C < B < AC. C < B < A < DD. C < A < B < DE. D < A < B < C

10. What is the correct IUPAC name for the following molecule?

O

O

A. (E)-ethyl 3-hexeneoateB. (Z)-ethyl 3-hexeneoateC. (E)-ethyl 2-penteneoateD. (Z)-ethyl 2-penteneoateE. (E)-2-penteneoate ethyl ester

11 Which of the following molecules is the best match for the IR spectrum shownbelow?

O OH NH2

OHO NH2O

A B C D E

8/3/2019 Old Mid-Term 2007



of 6

12. Which of the following molecules is the best match for the IR spectrum shownbelow?

C

N

NH2

CH3O HO OCH3O

A B C D E

13. How many peaks (ignoring splitting) would you expect in the 1H NMR for thefollowing molecule?

O

A. 1; B. 2; C. 3; D. 4; E. 5

14. How many peaks would you expect in the 13C NMR for the followingmolecule?

CH3

NO2 A. 1; B. 2; C. 3; D. 4; E. 5

15. How many lines would the signal for the proton indicated by the arrow be splitinto?

Cl Cl

OHH

A. 2; B. 3; C. 4; D. 5; E. 6

8/3/2019 Old Mid-Term 2007



of 6

16. What is the relationship between the two molecules shown?

OH

OH A. Constitutional Isomers

B. Conformational IsomersC. Geometric IsomersD. EnantiomersE. Diasteroeomers

17. A tertiary carbocation is more stable than a methyl carbocation because:

A. The tertiary carbocation has an empty p orbital.B. The tertiary carbocation has a smaller HOMO–LUMO gap.C. The tertiary carbocation has a larger HOMO–LUMO gap.D. The tertiary carbocation has favorable molecular orbital interactions.

E. The tertiary carbocation is more planar than the methyl carbocation.

18. Place the following molecules in the order of increasing pKa(i.e. lowest pKa < highest pKa)

OH

C

OH

O

O

A B A. A < B < CB. A < C < BC. B < C < AD. C < B < A

E. C < A < B

19. Which of the following molecules is a tertiary (3°) amine?

NH2

NH2

NH2

NH

N

A B C D E Use the structures below to answer questions 20, 21 and 22.

NNH3 OHN

A B C D E

20. Which molecule can act as both electron pair donor and hydrogen atomdonor in hydrogen bonding?

21. Which molecule can act only as a hydrogen atom donor in hydrogenbonding?

22. Which molecule can act only as an electron pair donor in hydrogen bonding?

8/3/2019 Old Mid-Term 2007



of 6

Answer the questions that follow on the answer sheet provided. Be sure to include your nameand student number in the places indicated on the answer sheet. Be sure to write legibly!

1) Lovastatin is a widely prescribed cholesterol-lowering drug taken by millions of people worldwide.Lovastatin is produced by fermentation cultures of the fungus Aspergillus terreus and is amongthe top ten of all prescription drugs sales.

Carefully examine the chemical structure of lovastatin and answer the questions that follow.

O

O

HO O

O

H

a) What is the molecular formula of lovastatin? (1 mark)b) How many sp2 hybridized carbon atoms are there in lovastatin? (1 mark)

c) What is the multiplicity (i.e. splitting pattern) observed in the 1H NMR for the circled methygroup? (1 mark)d) Is the alcohol indicated by the arrow primary (1°); secondary (2°) or tertiary (3°)? (1 mark)e) How many sp3 hybridized carbon atoms are there in lovastatin? (1 mark)f) What is the multiplicity observed in the 1H NMR for the hydrogen atom in the box? (1 mark)

g) How many signals would you expect with a chemical shift greater than 100 ppm in the 13C NMR?(1 mark)

2) Attached (on opposite page) is the spectral data (UV, IR, MS, 1H NMR, and 13C NMR) for anunknown organic compound. Examine the data carefully and propose a structure for thisunknown molecule in the space provided. (Total 7 marks — part marks may be awarded)

3) Please provide the name of the compound below according to IUPAC rules (3 marks):

OH

Br

4) Please give the expected number of peaks (ignore splitting) in the 1H and 13C NMR spectra for

the following compounds (2 marks for each compound – 1 for 1H and 1 for 13C):

Cl

ClCl

OH

CH3O OCH3

a) b) c) d) 5) Assign a structure to the compound with molecular formula (C9H12), having no infrared absorption in

the region 3200 to 3400 cm –1 and the following spectral data (1H NMR data are given in the formatδ chemical shift (splitting pattern; integration) (5 marks):

1H NMR: δ 1.24 (d, 6H), 2.88 (septet, 1H), 7.25 multiplet, 5H)

13C NMR: δ 148.8, 128.3, 126.4, 125.8, 34.2, 24.0

old mid-term 2007

Documents