Quiz 1-2 practice orgo chem PDF

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CHEMISTRY 343

February 28, 2007

Hour Exam 1 D. LEE

PRINT NAME and STUDENT ID:______________

Key________________________

There are 4 pages on this examination including this cover page; take a moment to verify this. Only answers written in permanent ink (not red) will be considered for regrading.

Problem 1

(10 points)

_________

Problem 2

(20 points)

_________

Problem 3

(10 points)

_________

Problem 4

(10 points)

_________

Problem 5

(10 points)

_________

Problem 6

(20 points)

_________

Problem 7

(20 points)

_________

Total

(100 points)

_________

1

1. (10 points) In the boxes provided, describe the indicated bonds in terms of their component molecular orbitals (appropriate resonance forms should be considered). Example H H

σC O

σC

sp2Osp2

N C N

π Cp Op

spNsp2

π

Cp Np

2. (a) (10 points) Ozone is found in the upper atmosphere where it absorbs UV radiation and thereby provides the surface of earth with a protective screen. Reaction of ozone with an alkene gives a molecule called carbonyl oxide as shown. Provide the best Lewis structure of carbonyl oxide by adding all appropriate multiple bonds (π-bonds if any), lone pairs of electrons, and formal charges at atoms (Zero formal charge doesn’t have to be indicated). Write additional resonance structures based on the structure you have completed. very minor H3C

H 3C

H 3C O O

O O

O O H 3C

H3C

H 3C A

B

H3C O O H3C

C

C'

Carbonyl oxide

(b) (5 points) Choose the most stable resonance structure from A–C that you have drawn in (a) and briefly explain why the resonance structure you have chosen is more stable than the other two? H3 C O O H3 C A

A is most stable resonance form because negative charge is on more electronegative oxygen compared to B where that is on the carbon and the lone pair electrons on the central oxygen in A are more delocalized compared to those of C (also the opposite charges are more close in A). C' is very minor contributor because the central oxygen has two positive charges.

(c) (5 points) From the structural information of carbonyl oxide in (a) and (b), redraw the most stable resonance structure with expected bond angle of C–O–O and predict the shape (linear, planar, tetrahedral, bent, etc.) of this molecule. H 3C

O CH3

bent or (trigonal) planar

O A

2

3. (10 points) Psymberin is a naturally occurring molecule that shows a potent and selective activity toward certain solid tumors. Identify five functional groups by circling around each functional group and name them. (*only five with no redundancy). ether Methyl OMe O

alkene

OH

Psymberin

OMe N H

(2°) amide

O

OH OH

O O

OH

OH phenol

ester

2° alcohol (hydroxyl group)

4. (a) (5 points) The following acid–base reaction should be carried out in an inert medium. Explain why you should avoid using H2O as the medium. inert solvent H3C

H

Na

H3C

NaNH2

+

+

NH3

In the presence of water the following acid-base reaction occurs, thus the strongest base available is OH-, which is not strong enough base to react with terminal acetylene (pKa = 25): Leveling Effect H2 O

HO-

NaNH2

+

+

NH3 pKa = 38

pKa = 15.7

(b) (5 points) Consider the equilibrium of the following acid–base reaction. Indicate the favorable direction by circling around either “A” or “B” A +

H3C

H 3C H

H

+

B

5. (a) (5 points) Draw the Newman projection formula for the lowest energy conformation (A) and highest energy conformation (B) of butane. CH3 H

H

H

H

H H

B

A

eclipsed conformation

anti conformation HH

CH3

CH3 CH 3

(b) (5 points) Briefly explain what makes the energy difference between conformations A and B. The eclipsing interactions between two pairs of C–H bond and between C–C bond (torsional strain) as well as the interactions between two CH3 groups (steric hindrance) in B increase its energy relative to A.

3

6. (a) (10 points) Draw the most stable conformation of the most stable 1,4-dimethyl cyclohexane (A) then, convert it to the corresponding ring-flipped conformation (B). CH3 CH3

H3C

ring-flip

A

B

CH3

(b) (5 points) What is the estimated energy difference between conformer A and B? (Use 3.8 kJ/mol for a gauche interaction in butane). Justify your estimations. gauche interactions

anti H

H

H

CH3 H

H

H

H CH3

H

H

H

H

H

H

H H

H

H H

H H

H

H

anti

Between an axial methyl and equatorial methyl on cyclohexane there are two gauche interaction difference. Thereofre, there are total four gauche interaction difference between A and B, which corresponds to 15.2 kj/mol energy difference.

(c) (5 points) Use the number you obtained in (b) and estimate the ratio of the more stable conformer to the less stable conformer at 25 °C (Use ΔG° = –5.7 logKeq).

-15.2 = -5.7 logKeq

Keq = 464

Thus, A : B = 464 : 1 or 99.8% : 0.2% 7. (a) (10 points) Draw the most stable conformation of 1,4-dichlorocyclohexane that has non-zero dipole moment. (b) (5 points) Identify each chloro group with “axial” and “equatorial”. (c) (5 points) Indicate that this compound is a “cis” or a “trans” isomer.

axial

Cl

equatorial

Cl

cis-1,4-dichlorocyclohexane

4...