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(CHE 325) Organic Chemistry II Spring 2011

EXAM #1Name: KEYID#:

Check your exam to be sure it is complete. There are nine questions in this exam. Itis worth 100 points. To receive full credit for your submissions, make sure you doyour own work. Read each question carefully, and answer in clearly legible andcoherent English. Use the back of the page for additional space if needed. Partialcredit will be given, so show your work. Don't fall in love with individual problems. Dothe ones first which you can do easily. Pace yourself. Good luck!

N. Totah

1. (of 6)

2. (of 6)

3. (of 8)

4. (of 14)

5. (of 12)

6. (of 12)

7. (of 20)

8. (of 12)

9. (of 10)

Total: (100)

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1. (6 points) Give IUPAC names for the following compounds:

a.NO2

Br o-bromonitrobenzene

b.NH2

Cl Cl

CH2Cl3 2,6-dichloro-4-ethylaniline

2. (6 points) Write a complete, stepwise mechanism for the followingtransformation. Use curved arrows to denote electron flow.

+

H2SO4

H HH

OH2 or HSO4

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3a. (6 points) Consider the compounds below:

CH3O

OCH3

*

CH2 CH O CH3

CH3

CH3

A B

i. How many signals would you expect to see in the 1H NMR spectrum ofcompound A?

Answer: two

ii. What is the relationship between the two protons marked with an asterisk (*) incompound A? (circle one)

different homotopic enantiotopic diastereotopic

iii. The 13C NMR spectrum of compound B shows five signals. How many signals inthe corresponding DEPT-135 spectrum would point up?

Answer: four

3b. (2 points) What is the most likely structure of the molecule that gave thefollowing 1H NMR spectrum? Circle your answer.

OO

C D E F

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4a. (8 points) Label the following structures as aromatic, antiaromatic, ornonaromatic.

NN

aromaticnonaromatic

O

aromaticantiaromatic

H

4b. (6 points) Which of the following hydrocarbons (A or B) would you expect to bedeprotonated most readily in the presence of a base? Briefly explain. For full credit,you will need to discuss both compounds.

HH

HH

A B

Compound B is expected to be more acidic.In compound A, loss of a proton results in formation of an unstableantiaromatic compound (8 electrons; cycloheptatrienyl anion).However, deprotonation of compound B results in formation of anstable, aromatic system (6 electrons; cyclopentadienyl anion).Due to the stability associatied with aromaticity, compound B shouldthus be deprotonated more readily (e.g. is more acidic).

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5a. (4 points) Rank the following compounds in order of decreasing reactivity toElectrophilic Aromatic Substitution. Rank the most reactive compound 1, and the leastreactive compound 4. Place the number corresponding to the relative reactivity ofeach compound in the blank below it.

As(CH3)3OCH3

CH3

SO3HOCH3

NO2

2 4 1 3

Cl

5b. (4 points) Rank the following compounds in order of decreasing reactivity to

Nucleophilic Aromatic Substitution. Rank the most reactive compound 1, and the leastreactive compound 4. Place the number corresponding to the relative reactivity ofeach compound in the blank below it.

43 21

NO2

Cl

NO2

NO2NO2

ClCl

Cl

NO2

CH3

5c. (4 points) Trifluoromethylbenzene undergoes Electrophilic Aromatic Substitution.Do you expect this compound to react faster or more slowly than benzene underidentical reaction conditions (e.g. is the -CF3 group activating or deactivating)? Brieflyexplain your reasoning.

CF3

trifluoromethylbenzene Trifluoromethylbenzene will react more slowly than benzene (CF 3 isdeactivating. The CF3 group withdraws electron density from thering by induction making the ring a poorer nucleophile.

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6. (12 points) Match each of the compounds A, B, and C with the appropriate broadband decoupled 13C NMR spectra shown below. Briefly explain your reasoning. Note:for full credit, you rationale must unambiguousely exclude the other two compounds.Cite chemical shifts and identify relevant functionality in your discussion. Be specific.

CH3 O CH3 CH3 OH CH3 OH

O O OHA B C

13C NMR spectra:

i. Compound: B

TMS

ppm

Reasoning:

Has a C=O peak (ca. 180 ppm)Does not show peak at ca. 60ppm (C-O linkage)

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ii. Compound: C

TMS

ppmReasoning:

has no C=O at ca. 180ppm orshows C=C peaks ca. 100-150ppm

iii. Compound: A

TMS

ppm

Reasoning:

shows both C=O (180 ppm) and C-O (60ppm) peaks

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7. (20 points) Predict the major product(s) for the following reactions. Assumereagents are present in equimolar amounts unless otherwise indicated. Clearly depictany stereochemistry. "No Reaction" is a possible answer. CAUTION! Don't confusereactions that will occur slowly with those that will not occur!

a.CH3

O2N

HNO3

H2SO4

CH3

O2N NO2

b.

OCH3

CH3O

Cl

O

AlCl3

OCH3

CH3O

O

c.

Me3N

I2

CuCl2 Me3N I

d.

1. NaOH, 300C

2. H3O+

SO3H

CH3

CH3

OH

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e.OH

CH3 SO3

H2SO4

OH

CH3

OH

CH3+

HO3S

SO3H

f.

CH2

O2N

Br2

FeBr3

CH2

O2N

CH2

O2N

Br

Br

g.

KMnO4 (excess)

H2O

(step 1)

CO2H

CO2H

CH3Cl

AlCl3

(step 2)

product(s)of step 1

No Reaction

h.

NBS

CCl4, (PhCO)2O

Br

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8. (12 points) Propose a structure that is consistent with the spectral data provided.For full credit, show your work and briefly explain your reasoning. As part of youranswer, clearly correlate each peak in the 1H NMR spectrum to the structure of yourcompound (e.g. label the peaks).

F.W.: C10H14IR: 740 cm-11H NMR: integral values are listed above each peak

a

b

c

d

ea

a

a

4H

3H

2H 2H

3H

ppm

a

b

c

d

e

IR suggests this is a 1,2-disubstituted benzeneNMR supports structure as per labelled peaks

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9. (10 points) Show how you would synthesize the following compounds frombenzene or toluene. Use any other organic or inorganic reagents required. Avoidproduct mixtures if possible, but assume that ortho and para products can beseparated. Clearly indicate at what stage (if any) such separations are needed.

a.

Cl

CH3

NH2

HNO3

H2SO4

Cl2

FeCl3

CH3

NH2

1. Fe, H3O+

2. NaOH

or H2,Pd

CH3CH3(separate)

Cl

CH3

NH2

CH3

NO2

NO2

+

b.

SO3H

Cl

O

SO3

H2SO4

H2

Pd/C

O

SO3H

AlCl3

(+ ortho)